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5,891,227,577,319,353,000 | Arithmetic Sequence Calculator
n
=
RESULTS
An arithmetic sequence or series calculator is a tool for evaluating a sequence of numbers, which is generated each time by adding a constant value. It is also known as the recursive sequence calculator.
It gives you the complete table depicting each term in the sequence and how it is evaluated. You can also find the graphical representation of the sequence.
How to use the arithmetic series calculator?
Follow these steps to find the arithmetic series.
1. Enter the nth term and its value in the first box.
2. Enter the total number of terms and the total number of terms.
3. Click "Calculate".
Moreover, you will get all the steps that are performed by this calculator to calculate the terms in the sequence and the sum of the sequence.
The n-th term is computed by: --
The first element of the sequence is: --
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In this post, we will discuss the arithmetic sequence, its formula, and examples.
What is an arithmetic sequence?
A set of objects, including numbers or letters in a certain order, is known as a sequence in mathematics. The sequence's objects are known as terms or elements. It is quite normal to see the same object in one sequence many times.
Arithmetic sequence definition can be interpreted as:
"A set of objects that comprises numbers is an arithmetic sequence. A constant number known as the common difference is applied to the previous number to create each successive number."
The common difference refers to the difference between any two consecutive terms of the sequence.
Arithmetic sequence formula
We can use the arithmetic sequence formula to find any term in the sequence. Arithmetic sequence equation can be written as:
aa = 1+(n1)d1 + (n-1)d
In this equation:
ana_n refers to the nthn^{th} term of the sequence,
a1a_1 refers to the first term of the sequence,
dd refers to the common difference and
nn refers to the length of the sequence.
The above formula is an explicit formula for an arithmetic sequence. All terms are equal to each other if there is no common difference in the successive terms of a sequence. In this case, there would be no need for any calculations.
How to calculate arithmetic sequence?
Find the 10th term in the below sequence by using the arithmetic sequence formula.
2,4,6,8,10,12,14,16,18...2, 4, 6, 8, 10, 12, 14, 16, 18...
Solution:
As we know, n refers to the length of the sequence, and we have to find the 10th term in the sequence, which means the length of the sequence will be 10. Follow these steps to find a specific term in an arithmetic sequence.
Step 1: Write down the sequence.
2,4,6,8,10,12,14,16,18...2, 4, 6, 8, 10, 12, 14, 16, 18...
Step 2: Find the common difference d.
d=n2n1d = n_2 - n_1
d=2d = 2
Step 3: Write down the formula of the arithmetic sequence.
an=a1+(n1)da_n = a_1 + (n-1)d
Step 4: Substitute the values in the equation.
As we know,
a1=2,n=10,a_1= 2, n = 10, and d=2d = 2
a10=2+(101)2=20a_{10} = 2 + (10 - 1) 2 = 20
So the 10th term of this arithmetic sequence would be 20. You can also use our above arithmetic sequence formula calculator to find the required value.
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Isa W. Monsour | 19/08/2019
It’s a well-designed calculator for the students. You can also easily understand it.
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-9,149,805,659,161,658,000 | Data Scientist Reviews
Hypothesis Testing with Z-Test,Significance Level and Rejection Region,rejection region table,rejection region calculator,critical value,critical region,hypothesis testing calculator,hypothesis testing critical value,how to calculate rejection region,how to find critical value in hypothesis testing,
Hypothesis Testing with Z-Test: Significance Level and Rejection Region
If you want to understand why hypothesis testing works, you should first have an idea about the significance level and the reject region. We assume you already know what a hypothesis is, so let’s jump right into the action.
What Is the Significance Level?
First, we must define the term significance level. Normally, we aim to reject the null if it is false.
Significance level
However, as with any test, there is a small chance that we could get it wrong and reject a null hypothesis that is true.
Error, significance level
How Is the Significance Level Denoted?
The significance level is denoted by α and is the probability of rejecting the null hypothesis, if it is true.
α and is the probability of rejecting the null hypothesis, significance level
So, the probability of making this error. Typical values for α are 0.01, 0.05 and 0.1. It is a value that we select based on the certainty we need. In most cases, the choice of α is determined by the context we are operating in, but 0.05 is the most commonly used value.
Most common, significance level
A Case in Point
Say, we need to test if a machine is working properly. We would expect the test to make little or no mistakes. As we want to be very precise, we should pick a low significance level such as 0.01. The famous Coca Cola glass bottle is 12 ounces. If the machine pours 12.1 ounces, some of the liquid would be spilled, and the label would be damaged as well. So, in certain situations, we need to be as accurate as possible.
Significance level: Coca Cola example
Higher Degree of Error
However, if we are analyzing humans or companies, we would expect more random or at least uncertain behavior. Hence, a higher degree of error.
You expect more random behavior, significance level
For instance, if we want to predict how much Coca Cola its consumers drink on average, the difference between 12 ounces and 12.1 ounces will not be that crucial. So, we can choose a higher significance level like 0.05 or 0.1.
The difference between 12 and 12.1, significance level
Hypothesis Testing: Performing a Z-Test
Now that we have an idea about the significance level, let’s get to the mechanics of hypothesis testing. Imagine you are consulting a university and want to carry out an analysis on how students are performing on average.
How students are performing on average, significance-level
The university dean believes that on average students have a GPA of 70%. Being the data-driven researcher that you are, you can’t simply agree with his opinion, so you start testing. The null hypothesis is: The population mean grade is 70%. This is a hypothesized value. The alternative hypothesis is: The population mean grade is not 70%. You can see how both of them are denoted, below.
University Dean example: Null hypothesis equals the population mean
Visualizing the Grades
Assuming that the population of grades is normally distributed, all grades received by students should look in the following way.
Distribution of grades, significance level
That is the true population mean.
Performing a Z-test
Now, a test we would normally perform is the Z-test. The formula is: Z equals the sample mean, minus the hypothesized mean, divided by the standard error.
Z equals the sample mean, minus the hypothesized mean, divided by the standard error, significance level
The idea is the following. We are standardizing or scaling the sample mean we got. If the sample mean is close enough to the hypothesized mean, then Z will be close to 0. Otherwise, it will be far away from it. Naturally, if the sample mean is exactly equal to the hypothesized meanZ will be 0.
If the sample mean is exactly equal to the hypothesized mean, Z will be 0, significance level
In all these cases, we would accept the null hypothesis.
What Is the Rejection Region?
The question here is the following: How big should Z be for us to reject the null hypothesis? Well, there is a cut-off line. Since we are conducting a two-sided or a two-tailed test, there are two cut-off lines, one on each side.
Distribution of Z (standard normal distribution), significance level
When we calculate Z, we will get a value. If this value falls into the middle part, then we cannot reject the null. If it falls outside, in the shaded region, then we reject the null hypothesis. That is why the shaded part is called: rejection region, as you can see below.
Rejection region, significance level
What Does the Rejection Region Depend on?
The area that is cut-off actually depends on the significance level. Say the level of significanceα, is 0.05. Then we have α divided by 2, or 0.025 on the left side and 0.025 on the right side.
The level of significance, α, is 0.05. Then we have α divided by 2, or 0.025 on the left side and 0.025 on the right side
Now these are values we can check from the z-table. When α is 0.025, Z is 1.96. So, 1.96 on the right side and minus 1.96 on the left side. Therefore, if the value we get for Z from the test is lower than minus 1.96, or higher than 1.96, we will reject the null hypothesis. Otherwise, we will accept it.
One-sided test: Z score is 1.96
That’s more or less how hypothesis testing works. We scale the sample mean with respect to the hypothesized value. If Z is close to 0, then we cannot reject the null. If it is far away from 0, then we reject the null hypothesis.
How does hypothesis testing work?
Example of One Tailed Test
What about one-sided tests? We have those too! Let’s consider the following situation. Paul says data scientists earn more than $125,000. So, H0 is: μ0 is bigger than $125,000. The alternative is that μ0 is lower or equal to 125,000. Using the same significance level, this time, the whole rejection region is on the left. So, the rejection region has an area of α. Looking at the z-table, that corresponds to a Z-score of 1.645. Since it is on the left, it is with a minus sign.
One-sided test: Z score is 1.645
Accept or Reject
Now, when calculating our test statistic Z, if we get a value lower than -1.645, we would reject the null hypothesis. We do that because we have statistical evidence that the data scientist salary is less than $125,000. Otherwise, we would accept it.
One-sided test: Z score is - 1.645 - rejecting null hypothesis
Another One-Tailed Test
To exhaust all possibilities, let’s explore another one-tailed test. Say the university dean told you that the average GPA students get is lower than 70%. In that case, the null hypothesis is: μ0 is lower than 70%. While the alternative is: μ0` is bigger or equal to 70%.
University Dean example: Null hypothesis lower than the population mean
In this situation, the rejection region is on the right side. So, if the test statistic is bigger than the cut-off z-score, we would reject the null, otherwise, we wouldn’t.
One-sided test: test statistic is bigger than the cut-off z-score - reject the null hypothesis
Importance of the Significance Level and the Rejection Region
To sum up, the significance level and the reject region are quite crucial in the process of hypothesis testing. The level of significance conducts the accuracy of prediction. We (the researchers) choose it depending on how big of a difference a possible error could make. On the other hand, the reject region helps us decide whether or not to reject the null hypothesis. After reading this and putting both of them into use, you will realize how convenient they make your work. The article first appeared on: https://365datascience.com/significance-level-reject-region/
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-6,663,685,300,746,281,000 | Alunos Online - Trabalhos Escolares e Educação
Esqueci minha senha
Função Logarítmica
Lei de formação da função logarítmica
PUBLICIDADE
Pelo estudo dos logaritmos podemos obter diversas propriedades dessa operação numérica, que está intimamente relacionada aos exponenciais. A função logarítmica relaciona valores do logaritmo da variável x, obtendo, assim, valores de f(x).
Vejamos a definição para a função logarítmica:
Dado um número real a (sendo que: 0 < a; a ≠ 1), chamaremos de função logarítmica de base a, a função que leva os números Reais maiores do que zero nos números Reais, pela seguinte lei: f(x) = loga x.
Vejamos alguns exemplos de funções logarítmicas:
f(x) = log 2x; f(x) = log3 x; y=log2 x; f(x)=loge(x-1)
Note que na definição nós temos algumas restrições, sendo elas:
• A base do logaritmo deve ser maior do que zero e diferente de 1 (0 < a ≠ 1);
• O valor de x está determinado no conjunto dos números reais positivos, sem incluir o zero. Portanto, o logaritmando x deve ser: x > 0.
É importante lembrar que essas condições auxiliam na determinação do domínio no qual a função está representada.
Por Gabriel Alessandro de Oliveira
Deixe seu comentário para "Função Logarítmica"
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-7,919,163,462,415,575,000 | Karnaugh map
From Wikipedia, the free encyclopedia
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An example Karnaugh map. This image actually shows two Karnaugh maps: for the function ƒ, using minterms (colored rectangles) and for its complement, using maxterms (gray rectangles). In the image, E() signifies a sum of minterms, denoted in the article as .
The Karnaugh map (KM or K-map) is a method of simplifying Boolean algebra expressions. Maurice Karnaugh introduced it in 1953[1][2] as a refinement of Edward W. Veitch's 1952 Veitch chart,[3][4] which was a rediscovery of Allan Marquand's 1881 logical diagram[5] aka Marquand diagram[4] but with a focus now set on its utility for switching circuits.[4] Veitch charts are therefore also known as Marquand–Veitch diagrams,[4] and Karnaugh maps as Karnaugh–Veitch maps (KV maps).
The Karnaugh map reduces the need for extensive calculations by taking advantage of humans' pattern-recognition capability.[1] It also permits the rapid identification and elimination of potential race conditions.
The required Boolean results are transferred from a truth table onto a two-dimensional grid where, in Karnaugh maps, the cells are ordered in Gray code,[6][4] and each cell position represents one combination of input conditions. Cells are also known as minterms, while each cell value represents the corresponding output value of the boolean function. Optimal groups of 1s or 0s are identified, which represent the terms of a canonical form of the logic in the original truth table.[7] These terms can be used to write a minimal Boolean expression representing the required logic.
Karnaugh maps are used to simplify real-world logic requirements so that they can be implemented using a minimum number of logic gates. A sum-of-products expression (SOP) can always be implemented using AND gates feeding into an OR gate, and a product-of-sums expression (POS) leads to OR gates feeding an AND gate. The POS expression gives a complement of the function (if F is the function so its complement will be F').[8] Karnaugh maps can also be used to simplify logic expressions in software design. Boolean conditions, as used for example in conditional statements, can get very complicated, which makes the code difficult to read and to maintain. Once minimised, canonical sum-of-products and product-of-sums expressions can be implemented directly using AND and OR logic operators.[9]
Example[edit]
Karnaugh maps are used to facilitate the simplification of Boolean algebra functions. For example, consider the Boolean function described by the following truth table.
Truth table of a function
A B C D
0 0 0 0 0 0
1 0 0 0 1 0
2 0 0 1 0 0
3 0 0 1 1 0
4 0 1 0 0 0
5 0 1 0 1 0
6 0 1 1 0 1
7 0 1 1 1 0
8 1 0 0 0 1
9 1 0 0 1 1
10 1 0 1 0 1
11 1 0 1 1 1
12 1 1 0 0 1
13 1 1 0 1 1
14 1 1 1 0 1
15 1 1 1 1 0
Following are two different notations describing the same function in unsimplified Boolean algebra, using the Boolean variables A, B, C, D and their inverses.
• where are the minterms to map (i.e., rows that have output 1 in the truth table).
• where are the maxterms to map (i.e., rows that have output 0 in the truth table).
K-map drawn on a torus, and in a plane. The dot-marked cells are adjacent.
K-map construction. Instead of the output values (the rightmost values in the truth table), this diagram shows a decimal representation of the input ABCD (the leftmost values in the truth table), therefore it is not a Karnaugh map.
In three dimensions, one can bend a rectangle into a torus.
Construction[edit]
In the example above, the four input variables can be combined in 16 different ways, so the truth table has 16 rows, and the Karnaugh map has 16 positions. The Karnaugh map is therefore arranged in a 4 × 4 grid.
The row and column indices (shown across the top and down the left side of the Karnaugh map) are ordered in Gray code rather than binary numerical order. Gray code ensures that only one variable changes between each pair of adjacent cells. Each cell of the completed Karnaugh map contains a binary digit representing the function's output for that combination of inputs.
Grouping[edit]
After the Karnaugh map has been constructed, it is used to find one of the simplest possible forms — a canonical form — for the information in the truth table. Adjacent 1s in the Karnaugh map represent opportunities to simplify the expression. The minterms ('minimal terms') for the final expression are found by encircling groups of 1s in the map. Minterm groups must be rectangular and must have an area that is a power of two (i.e., 1, 2, 4, 8...). Minterm rectangles should be as large as possible without containing any 0s. Groups may overlap in order to make each one larger. The optimal groupings in the example below are marked by the green, red and blue lines, and the red and green groups overlap. The red group is a 2 × 2 square, the green group is a 4 × 1 rectangle, and the overlap area is indicated in brown.
The cells are often denoted by a shorthand which describes the logical value of the inputs that the cell covers. For example, AD would mean a cell which covers the 2x2 area where A and D are true, i.e. the cells numbered 13, 9, 15, 11 in the diagram above. On the other hand, AD would mean the cells where A is true and D is false (that is, D is true).
The grid is toroidally connected, which means that rectangular groups can wrap across the edges (see picture). Cells on the extreme right are actually 'adjacent' to those on the far left, in the sense that the corresponding input values only differ by one bit; similarly, so are those at the very top and those at the bottom. Therefore, AD can be a valid term—it includes cells 12 and 8 at the top, and wraps to the bottom to include cells 10 and 14—as is BD, which includes the four corners.
Solution[edit]
Diagram showing two K-maps. The K-map for the function f(A, B, C, D) is shown as colored rectangles which correspond to minterms. The brown region is an overlap of the red 2×2 square and the green 4×1 rectangle. The K-map for the inverse of f is shown as gray rectangles, which correspond to maxterms.
Once the Karnaugh map has been constructed and the adjacent 1s linked by rectangular and square boxes, the algebraic minterms can be found by examining which variables stay the same within each box.
For the red grouping:
• A is the same and is equal to 1 throughout the box, therefore it should be included in the algebraic representation of the red minterm.
• B does not maintain the same state (it shifts from 1 to 0), and should therefore be excluded.
• C does not change. It is always 0, so its complement, NOT-C, should be included. Thus, C should be included.
• D changes, so it is excluded.
Thus the first minterm in the Boolean sum-of-products expression is AC.
For the green grouping, A and B maintain the same state, while C and D change. B is 0 and has to be negated before it can be included. The second term is therefore AB. Note that it is acceptable that the green grouping overlaps with the red one.
In the same way, the blue grouping gives the term BCD.
The solutions of each grouping are combined: the normal form of the circuit is .
Thus the Karnaugh map has guided a simplification of
It would also have been possible to derive this simplification by carefully applying the axioms of Boolean algebra, but the time it takes to do that grows exponentially with the number of terms.
Inverse[edit]
The inverse of a function is solved in the same way by grouping the 0s instead.[note 1]
The three terms to cover the inverse are all shown with grey boxes with different colored borders:
• brown: A B
• gold: A C
• blue: BCD
This yields the inverse:
Through the use of De Morgan's laws, the product of sums can be determined:
Don't cares[edit]
The value of for ABCD = 1111 is replaced by a "don't care". This removes the green term completely and allows the red term to be larger. It also allows blue inverse term to shift and become larger
Karnaugh maps also allow easier minimizations of functions whose truth tables include "don't care" conditions. A "don't care" condition is a combination of inputs for which the designer doesn't care what the output is. Therefore, "don't care" conditions can either be included in or excluded from any rectangular group, whichever makes it larger. They are usually indicated on the map with a dash or X.
The example on the right is the same as the example above but with the value of f(1,1,1,1) replaced by a "don't care". This allows the red term to expand all the way down and, thus, removes the green term completely.
This yields the new minimum equation:
Note that the first term is just A, not AC. In this case, the don't care has dropped a term (the green rectangle); simplified another (the red one); and removed the race hazard (removing the yellow term as shown in the following section on race hazards).
The inverse case is simplified as follows:
Through the use of De Morgan's laws, the product of sums can be determined:
Race hazards[edit]
Elimination[edit]
Karnaugh maps are useful for detecting and eliminating race conditions. Race hazards are very easy to spot using a Karnaugh map, because a race condition may exist when moving between any pair of adjacent, but disjoint, regions circumscribed on the map. However, because of the nature of Gray coding, adjacent has a special definition explained above – we're in fact moving on a torus, rather than a rectangle, wrapping around the top, bottom, and the sides.
• In the example above, a potential race condition exists when C is 1 and D is 0, A is 1, and B changes from 1 to 0 (moving from the blue state to the green state). For this case, the output is defined to remain unchanged at 1, but because this transition is not covered by a specific term in the equation, a potential for a glitch (a momentary transition of the output to 0) exists.
• There is a second potential glitch in the same example that is more difficult to spot: when D is 0 and A and B are both 1, with C changing from 1 to 0 (moving from the blue state to the red state). In this case the glitch wraps around from the top of the map to the bottom.
Race hazards are present in this diagram.
Above diagram with consensus terms added to avoid race hazards.
Whether glitches will actually occur depends on the physical nature of the implementation, and whether we need to worry about it depends on the application. In clocked logic, it is enough that the logic settles on the desired value in time to meet the timing deadline. In our example, we are not considering clocked logic.
In our case, an additional term of would eliminate the potential race hazard, bridging between the green and blue output states or blue and red output states: this is shown as the yellow region (which wraps around from the bottom to the top of the right half) in the adjacent diagram.
The term is redundant in terms of the static logic of the system, but such redundant, or consensus terms, are often needed to assure race-free dynamic performance.
Similarly, an additional term of must be added to the inverse to eliminate another potential race hazard. Applying De Morgan's laws creates another product of sums expression for f, but with a new factor of .
2-variable map examples[edit]
The following are all the possible 2-variable, 2 × 2 Karnaugh maps. Listed with each is the minterms as a function of and the race hazard free (see previous section) minimum equation. A minterm is defined as an expression that gives the most minimal form of expression of the mapped variables. All possible horizontal and vertical interconnected blocks can be formed. These blocks must be of the size of the powers of 2 (1, 2, 4, 8, 16, 32, ...). These expressions create a minimal logical mapping of the minimal logic variable expressions for the binary expressions to be mapped. Here are all the blocks with one field.
A block can be continued across the bottom, top, left, or right of the chart. That can even wrap beyond the edge of the chart for variable minimization. This is because each logic variable corresponds to each vertical column and horizontal row. A visualization of the k-map can be considered cylindrical. The fields at edges on the left and right are adjacent, and the top and bottom are adjacent. K-Maps for four variables must be depicted as a donut or torus shape. The four corners of the square drawn by the k-map are adjacent. Still more complex maps are needed for 5 variables and more.
Related graphical methods[edit]
Related graphical minimization methods include:
• Marquand diagram (1881) by Allan Marquand (1853–1924)[5][4]
• Veitch chart (1952) by Edward W. Veitch (1924–2013)[3][4]
• Mahoney map (M-map, designation numbers, 1963) by Matthew V. Mahoney (a reflection-symmetrical extension of Karnaugh maps for larger numbers of inputs)
• Reduced Karnaugh map (RKM) techniques (from 1969) like infrequent variables, map-entered variables (MEV), variable-entered map (VEM) or variable-entered Karnaugh map (VEKM) by G. W. Schultz, Thomas E. Osborne, Christopher R. Clare, J. Robert Burgoon, Larry L. Dornhoff, William I. Fletcher, Ali M. Rushdi and others (several successive Karnaugh map extensions based on variable inputs for a larger numbers of inputs)
• Minterm-ring map (MRM, 1990) by Thomas R. McCalla (a three-dimensional extension of Karnaugh maps for larger numbers of inputs)
See also[edit]
Notes[edit]
1. ^ This should not be confused with the negation of the result of the previously found function.
References[edit]
1. ^ a b Karnaugh, Maurice (November 1953) [1953-04-23, 1953-03-17]. "The Map Method for Synthesis of Combinational Logic Circuits" (PDF). Transactions of the American Institute of Electrical Engineers, Part I: Communication and Electronics. 72 (5): 593–599. doi:10.1109/TCE.1953.6371932. Paper 53-217. Archived from the original (PDF) on 2017-04-16. Retrieved 2017-04-16. (NB. Also contains a short review by Samuel H. Caldwell.)
2. ^ Curtis, Herbert Allen (1962). A new approach to the design of switching circuits. The Bell Laboratories Series (1 ed.). Princeton, New Jersey, USA: D. van Nostrand Company, Inc. ISBN 0-44201794-4. OCLC 1036797958. S2CID 57068910. ISBN 978-0-44201794-1. ark:/13960/t56d6st0q. (viii+635 pages) (NB. This book was reprinted by Chin Jih in 1969.)
3. ^ a b Veitch, Edward Westbrook (1952-05-03) [1952-05-02]. "A Chart Method for Simplifying Truth Functions". Transactions of the 1952 ACM Annual Meeting. ACM Annual Conference/Annual Meeting: Proceedings of the 1952 ACM Annual Meeting (Pittsburgh, Pennsylvania, USA). New York, USA: Association for Computing Machinery (ACM): 127–133. doi:10.1145/609784.609801.
4. ^ a b c d e f g Brown, Frank Markham (2012) [2003, 1990]. Boolean Reasoning - The Logic of Boolean Equations (reissue of 2nd ed.). Mineola, New York: Dover Publications, Inc. ISBN 978-0-486-42785-0. [1]
5. ^ a b Marquand, Allan (1881). "XXXIII: On Logical Diagrams for n terms". The London, Edinburgh, and Dublin Philosophical Magazine and Journal of Science. 5. 12 (75): 266–270. doi:10.1080/14786448108627104. Retrieved 2017-05-15. (NB. Quite many secondary sources erroneously cite this work as "A logical diagram for n terms" or "On a logical diagram for n terms".)
6. ^ Wakerly, John F. (1994). Digital Design: Principles & Practices. New Jersey, USA: Prentice Hall. pp. 48–49, 222. ISBN 0-13-211459-3. (NB. The two page sections taken together say that K-maps are labeled with Gray code. The first section says that they are labeled with a code that changes only one bit between entries and the second section says that such a code is called Gray code.)
7. ^ Belton, David (April 1998). "Karnaugh Maps – Rules of Simplification". Archived from the original on 2017-04-18. Retrieved 2009-05-30.
8. ^ Dodge, Nathan B. (September 2015). "Simplifying Logic Circuits with Karnaugh Maps" (PDF). The University of Texas at Dallas, Erik Jonsson School of Engineering and Computer Science. Archived (PDF) from the original on 2017-04-18. Retrieved 2017-04-18.
9. ^ Cook, Aaron. "Using Karnaugh Maps to Simplify Code". Quantum Rarity. Archived from the original on 2017-04-18. Retrieved 2012-10-07.
Further reading[edit]
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-1,258,262,975,730,740,000 | 1
$\begingroup$
How can the following recurrence equation be solved by one of three main ways:
$$T(n)=T(n^{2/3})+17$$
I have tried solving it by the iteration way. However it does not work for me since I can't find the equation with $i$, i.e the generic equation.
$\endgroup$
• 2
$\begingroup$ Have you tried applying any of the methods listed here? $\endgroup$ – dkaeae Mar 15 at 16:32
• $\begingroup$ The answer will be proportional to the number of times that you have to raise a number to the power $2/3$ until it gets below an arbitrary constant. I'm sure you can work that out on your own. $\endgroup$ – Yuval Filmus Mar 16 at 12:24
• $\begingroup$ I’d recommend using a spreadsheet to find Say T(1000); that should give you some idea. $\endgroup$ – gnasher729 Mar 16 at 13:58
1
$\begingroup$
We can do this pretty easily with a change of variables here. Let $n = 2^{(3/2)^k}$ we then can rewrite $T(n)$ as: $$S(k) = S(k - 1) + 17$$ We have $S(k) = O(k)$. Then converting $k$ back to $n$ we have: $$k = \log_{3/2} \log_2 n$$ Thus, $T(n) = O(\log \log n)$.
$\endgroup$
• $\begingroup$ Is it also $T(n) = \theta (\log_{3/2} \log_2 n)$? $\endgroup$ – John D Mar 21 at 9:59
3
$\begingroup$
This post answers the original version of the question, where the recurrence relation is
$$T(n)=T\left(\frac{n^2}{3}\right)+17$$
The recurrence relation is somewhat unconventional. Here is an outline to solve it.
Suppose $n\ge 6$. Let $n=3\cdot2^{2^{m}}$, where $m=\log_2(\log_2\frac n3)\ge0$.
$$\begin{align} T(n) &=T\left(3\cdot2^{2^{m}}\right) =T\left(\sqrt{3\cdot\left(3\cdot2^{2^{m}}\right)}\right)-17\\ &=T\left(3\cdot2^{2^{m-1}}\right)-17 =T\left(\sqrt{3\cdot3\left(\cdot2^{2^{m-1}}\right)}\right)-2\cdot17\\ &=T\left(3\cdot2^{2^{m-2}}\right)-2\cdot17 =T\left(\sqrt{3\cdot\left(3\cdot2^{2^{m-2}}\right)}\right)-3\cdot17\\ &=\cdots\\ &=T\left(3\cdot2^{2^{m-\lceil m\rceil}}\ \right)-\lceil m \rceil\cdot 17 \end{align}$$
Since $-1\lt m-\lceil m\rceil\le0$, $3\sqrt2\lt3\cdot2^{2^{m-\lceil m\rceil}}\le6.$ So $T(n)\sim -17\log_2(\log_2 n)$ when $n$ goes to $\infty$.
Here are two related exercises.
Exercise 1. What is the recurrence relation for function $S$ such that $S(n)=T(3n)$?
Exercise 2. What is the asymptotic behavior of $T(n)$ if $n$ goes to 3 from above, assuming $T$ is continuous?
$\endgroup$
• $\begingroup$ Who edit the question, change the whole exercise. It should be n^(2/3). $\endgroup$ – John D Mar 16 at 9:46
• $\begingroup$ I am afraid that you wrote it as n^2/3, as shown in the first version, which was changed it to n^(2/3) more than 14 hours later. $\endgroup$ – Apass.Jack Mar 16 at 11:06
• 1
$\begingroup$ Question clearly was posted as T(n squared divided by 3), changed to T(n followed by some weird stuff) by OP then changed to T(cube root of n squared). Original is the most interesting one :-) $\endgroup$ – gnasher729 Mar 16 at 14:05
0
$\begingroup$
Another similar answer to Apass.Jack on the "original" question.
First flip the function around:
$$T\left(\frac{n^2}{3}\right) = T(n) - 17$$
Convert it to an intelligible form by a change of variables where $m = \tfrac{n^2}{3}$:
$$T(m) = T(\sqrt{3m}) - 17$$
Try another change of variables where $m = 3 \cdot 2^{2^\omega}$. We get:
$$T(\omega) = T(\omega - 1) - 17$$
Thus $T(\omega) = -17\omega$. Now we work backwards:
$$\begin{align*} T(\omega) & = -17 \omega\\ T(m) & = -17 \log_2 \log_2 (m / 3)\\ T(n^2 / 3) & = -17 \log_2 \log_2 (n^2 / 9)\\ T(n) & = -17 \log_2 \log_2 (n / 3)\\ \end{align*}$$
This assumes a base case of $T(\omega = 1) = -17$ or the following other assumed base cases:
$$\begin{align*} T(\omega = 1) &= -17\\ T(m = 6) & = -17\\ T(n = 6) & = -17 \end{align*}$$
$\endgroup$
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-4,120,059,096,193,785,300 | Inscribed and Circumscribed Figures
The following article is from The Great Soviet Encyclopedia (1979). It might be outdated or ideologically biased.
Inscribed and Circumscribed Figures
in elementary geometry. A polygon is inscribed in a convex curve and a curve is circumscribed about a polygon if all the vertices of the polygon lie on the curve. A polygon is circumscribed about a curve and the curve is inscribed in a polygon if every side of the polygon or its extension is tangent to the curve. The curve is most often a circle. Every triangle has one circumscribed and one inscribed circle. A convex quadrilateral has a circumscribed circle if and only if the sum of its opposite angles equals 180°. In order for a quadrilateral to have an inscribed circle, it is necessary and sufficient that the sum of the lengths of one pair of opposite sides equal the sum of the lengths of the other pair. A polygon can be inscribed in a circle if this latter property belongs to the quadrilaterals formed by a diagonal of the polygon and three sides and also if the perpendicular lines drawn through the centers of the sides intersect at one point. An inscribed circle exists if and only if the bisectors of the interior angles of a polygon intersect at one point. In projective geometry an important role is played by theorems about a hexagon inscribed in and circumscribed about a conic section.
Inscribed and circumscribed figures are also considered in space. In this case polyhedrons are used instead of polygons and a convex surface (most often a sphere) instead of a convex line. It is also possible to speak of a cone or cylinder inscribed in a sphere, of a sphere inscribed in a cone, and so on.
REFERENCE
Perepelkin, D. I. Kurs elernentarnoi geometrii, parts 1-2. Moscow-Leningrad, 1948-49.
The Great Soviet Encyclopedia, 3rd Edition (1970-1979). © 2010 The Gale Group, Inc. All rights reserved. | {
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If f:[0,1] -> R is a continuous function, describe f.
1. Feb 7, 2009 #1
1. The problem statement, all variables and given/known data
f:[0,1] [tex]\rightarrow[/tex] R is a continous function such that
[tex]\int[/tex]f(t)dt (from 0 to x) = [tex]\int[/tex] f(t)dt( from x to 1) for all x[tex]\in[/tex][0,1] .
Describe f.
2. Relevant equations
integral represents area
3. The attempt at a solution
what ever the function is, I know that the area under the graph from 0 to x is equal to the area under the graph from x to 1.
But how else can I describe f?
1. The problem statement, all variables and given/known data
2. Relevant equations
3. The attempt at a solution
2. jcsd
3. Feb 7, 2009 #2
Mark44
Staff: Mentor
One possibility is that f(t) = 0 for all t in [0, 1]. If there are other functions, none come to mind.
4. Feb 7, 2009 #3
Note that the integral is a signed area, so areas under the x-axis are negatively signed. If you draw a picture of a function that is entirely on one side of the axis (positive or negative) and start integrating from 0 to small x, it implies the remaining curve must take a nose-dive in order to account for the small amount of area between 0 and x. On the other hand, if we place x close to 1, it implies the small region between x and 1 must be higher than the previous region in order to account for the larger area, which doesn't agree with the previous scenario (or the function is the 0 function).
This implies that the function must cross the x-axis between 0 and 1 (or be the 0 function). Can you continue from there?
Another less geometric and more algebraic approach is to use the fundamental theorem of calculus to note that there exists some function F on [0,1] such that the first integral is F(x) - F(0) and the second integral is F(1) - F(x), which leads to the same conclusion.
5. Feb 7, 2009 #4
I think so,Thank you
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Results 1 to 2 of 2
Math Help - Compute the values
1. #1
Member akhayoon's Avatar
Joined
Dec 2007
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T.O
Posts
106
Compute the values
for ||3X-4Y||
knowing that ||X||=2 AND ||Y||=1 and X.Y=-1
and then compute for
(3X+4Y).(X-5Y)
apparently the answer for the first is square root of 76
and the second is 1
I tried doing this the way the book tells me but I don't get the right answer at ALL!
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2. #2
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\begin{array}{rcl}<br />
\left\| {3X - 4Y} \right\|^2 & = & \left( {3X - 4Y} \right) \cdot \left( {3X - 4Y} \right) \\ <br />
& = & 9\left\| X \right\|^2 + 16\left\| Y \right\|^2 - 24\left( {X \cdot Y} \right) \\ \end{array}
Just make the substitutions and the squareroot of both sides.
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5,282,330,280,919,518,000 | login
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A197726 Decimal expansion of pi/(1+pi). 4
7, 5, 8, 5, 4, 6, 9, 9, 2, 9, 9, 4, 7, 7, 6, 1, 4, 5, 3, 4, 4, 4, 3, 0, 6, 8, 9, 0, 4, 4, 8, 9, 2, 8, 6, 4, 1, 3, 8, 4, 2, 6, 3, 6, 5, 6, 4, 0, 5, 3, 0, 9, 9, 6, 6, 6, 8, 9, 8, 8, 2, 1, 3, 7, 8, 2, 5, 4, 8, 1, 3, 7, 1, 0, 0, 9, 5, 7, 3, 7, 6, 3, 2, 0, 6, 3, 3, 1, 7, 4, 0, 1, 5, 3, 5, 5, 7, 7, 2 (list; constant; graph; refs; listen; history; text; internal format)
OFFSET
0,1
COMMENTS
Least x>0 such that sin(bx)=cos(cx) (and also sin(cx)=cos(bx)), where b=1/2 and c=pi/2; see the Mathematica program for a graph and A197682 for a discussion and guide to related sequences.
LINKS
Table of n, a(n) for n=0..98.
EXAMPLE
x=0.7585469929947761453444306890448928641384...
MATHEMATICA
b = 1/2; c = Pi/2;
t = x /. FindRoot[Sin[b*x] == Cos[c*x], {x, .75, .76}]
N[Pi/(2*b + 2*c), 110]
RealDigits[%] (* A197726 *)
Simplify[Pi/(2*b + 2*c)]
Plot[{Sin[b*x], Cos[c*x]}, {x, 0, 2}]
CROSSREFS
Cf. A197682.
Sequence in context: A280722 A262899 A198922 * A153623 A242623 A081815
Adjacent sequences: A197723 A197724 A197725 * A197727 A197728 A197729
KEYWORD
nonn,cons
AUTHOR
Clark Kimberling, Oct 17 2011
STATUS
approved
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Last modified August 26 03:21 EDT 2019. Contains 326324 sequences. (Running on oeis4.) | {
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-1,719,962,881,016,043,500 | Adding Like Terms Group Activity | Logic Puzzle | Good for Distance Learning
Adding Like Terms Group Activity | Logic Puzzle | Good for Distance Learning
Adding Like Terms Group Activity | Logic Puzzle | Good for Distance Learning
Adding Like Terms Group Activity | Logic Puzzle | Good for Distance Learning
Adding Like Terms Group Activity | Logic Puzzle | Good for Distance Learning
Adding Like Terms Group Activity | Logic Puzzle | Good for Distance Learning
Adding Like Terms Group Activity | Logic Puzzle | Good for Distance Learning
Adding Like Terms Group Activity | Logic Puzzle | Good for Distance Learning
Adding Like Terms Group Activity | Logic Puzzle | Good for Distance Learning
Subject
Grade Levels
File Type
Zip
(3 MB|9 pages)
Standards
Also included in:
1. This curriculum also includes activities! Check out the bundled resources section to see all the activities that are included (more coming soon!). If you do not want or need activities, consider our complete curriculum (without activities) here.This complete pre-algebra curriculum includes the follo
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• Product Description
• StandardsNEW
This group activity could be done in breakout sessions on Zoom or Google classroom.
This fun group activity will not only enhance your students' skills in adding like terms but will also build their critical thinking and logic skills.
With this activity, your students will...
• Build their algebra skills as they practice adding like terms.
• Deepen their understanding as they help each other solve problems.
• All be involved. Since each student has his/her own paper they are responsible for, it prevents any one member from just sitting back and letting the rest of the group take over.
• Learn how to think critically. Logic puzzles build critical thinking and problem-solving skills (the same skills you need to excel at math!)
• Learn how complex problems can be solved by simply doing the one step you know, then the next step, then the next, until you arrive at the answer.
Log in to see state-specific standards (only available in the US).
Apply properties of operations as strategies to add, subtract, factor, and expand linear expressions with rational coefficients.
Apply the properties of operations to generate equivalent expressions. For example, apply the distributive property to the expression 3 (2 + 𝘹) to produce the equivalent expression 6 + 3𝘹; apply the distributive property to the expression 24𝘹 + 18𝘺 to produce the equivalent expression 6 (4𝘹 + 3𝘺); apply properties of operations to 𝘺 + 𝘺 + 𝘺 to produce the equivalent expression 3𝘺.
Total Pages
9 pages
Answer Key
Included
Teaching Duration
40 minutes
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-6,083,155,761,301,324,000 | <meta http-equiv="refresh" content="1; url=/nojavascript/"> Area of Shaded and Composite Figures | CK-12 Foundation
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You are reading an older version of this FlexBook® textbook: CK-12 Foundation and Leadership Public Schools, College Access Reader: Geometry Go to the latest version.
Learning Objectives
• Use formulas to find the area of specific types of two-dimensional shapes by analyzing them as the sum or difference of smaller polygons.
Congruent Areas
Before we find the area of more complicated figures, we must know that:
If two figures are congruent, they have the same area.
This means that:
If two shapes are the same, the area inside them is also _____________________.
This is obvious because congruent figures have the same amount of space inside them. However, two figures with the same area are not necessarily congruent.
Area of a Whole is the Sum of Parts
If a figure is composed of two or more parts that do not overlap each other, then the area of the figure is the sum of the areas of the parts.
This means that:
You can break down a figure into parts and ___________ the areas of those parts together to get the area of the whole figure.
This is the familiar idea that a whole is the sum of its parts. In practical problems you may find it helpful to break a figure down into parts.
Reading Check
1. True or false: If two shapes are congruent, then they have the same area.
2. True or false: If two shapes have the same area, then they are congruent.
3. In your own words, describe what the phrase means:
“A whole is the sum of its parts.”
{\;}
{\;}
{\;}
Example 1
Find the area of the figure below.
Luckily, you do not have to learn a special formula for an irregular pentagon (which this figure is because its sides are not congruent.)
Instead, you can break the figure down into a trapezoid and a triangle like the dotted line does below. Use the area formulas for these parts to find the area of the whole figure.
The shape created on the left of the dotted line is a _______________________.
The shape created on the right of the dotted line is a _______________________.
Without numbers, we can review our formulas in describing the steps:
To find the area of the trapezoid, use the formula: A = \underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}.
To find the area of the triangle, use the formula: A = \underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}.
Because the area of the whole is the sum of its parts, we simply add together the areas of the trapezoid and the triangle to find the area of the entire figure!
Example 2
What is the area of the figure shown below?
Break the figure down into two rectangles:
First, look at the top rectangle:
The larger rectangle has a base of 45 cm and a height of 22 cm.
Since the area of a rectangle is \text{base} \cdot \text{height},
A = \underline{\;\;\;\;\;\;\;\;\;\;\;\;} \ cm \cdot \underline{\;\;\;\;\;\;\;\;\;\;\;\;} \ cm = 990 \ cm^2
Second, look at the bottom rectangle:
The smaller rectangle has a base of 20 cm and a height of 8 cm.
A = \underline{\;\;\;\;\;\;\;\;\;\;\;\;} \ cm \cdot \underline{\;\;\;\;\;\;\;\;\;\;\;\;} \ cm = 160 \ cm^2
Area of the whole figure is the sum of its parts, so:
\text{Area} & = \text{area of top rectangle} + \text{area of bottom rectangle}\\A & = \underline{\;\;\;\;\;\;\;\;\;\;\;\;} \ cm^2 + \underline{\;\;\;\;\;\;\;\;\;\;\;\;} \ cm^2 = 1150 \ cm^2
The area of the entire figure is 1150 \ cm^2.
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5,049,848,311,321,278,000 | math
posted by Samantha
What does "give a brief explanation as to what kind of solution(s) you expect the linear equation to have:
18x+1/2=6 (3x+25). Transform the equation into a
simpler form if necessary" mean?
1. Damon
I did, scroll down, post once.
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-1,513,536,799,568,556,300 | Odpowiedzi
Najlepsza Odpowiedź!
2010-01-12T23:28:50+01:00
Znajdź 6 wyraz ciagu geometrycznego gdy a₃=(-4) i a₉=(-½)
oznaczam
<n> indeks dolny
* mnożenie
^potęga
a<n> = a₁*q^(n-1)
a₃=a₁*q²
a₉=a₁*q⁸
z zadania mamy:
-4=a₁*q² /:q² (obie strony dzielę)
(-½) =a₁*q⁸
a₁=-4/q²
-½=-4/q² *q⁸
-½ = -4 * q⁶ /:(-4)
q⁶=-½:(-4)=-½ * (-¼)= 1/8
q= pierwiastek 6 stopnia z 1/8
korzystam z własności potęg:
x^⅓=∛x
czyli pierwiastek 6 stopnia z 1/8 = 1/8 ^ ⅙ = (1/8 ^⅓) ^½ =
=(∛(1/8))^½ = (1/2)^½= = √½ = 1/ √2 = √2/2
q=√2/2
podstawiamy do
a₁=-4/q²
obliczamy a₁
mając dane a₁ i q
obliczamy a₆ = a₁ * q⁵ | {
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9,102,154,590,748,004,000 | Loading [MathJax]/jax/output/CommonHTML/jax.js
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Recueil d'exercices pour apprendre Python au lycée
M_C
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Previous: Retirer les doublons Next: Triangle de Pascal
Tracer la courbe représentative d'une fonction
Difficulté : Moyenne
Pour tracer une courbe, une calculatrice ou un ordinateur ne calcule pas les coordonnées de tous les points de la courbe (il y en a une infinité). Ils se contentent de calculer régulièrement des valeurs de la fonction puis relie les points par des segments qui ne se voient presque pas à l'oeil nu. C'est ce que nous allons faire.
On considère donc une fonction f sur un intervalle [a,b]. On note n le nombre de segments que nous allons tracer pour représenter notre courbe. Pour ce faire, on va procéder par étape :
1. On crée une liste liste_abscisses de n+1 abscisses répartis uniformément sur l'intervalle [a,b]. Pour cela, on crée une liste contenant la suite de valeurs : a, a+ban, a+2.ban, a+3.ban , ... , a+n.ban (On peut remarquer que ce dernier élément vaut b).
2. On crée une liste liste_ordonnées de n+1 ordonnées dans laquelle on calcule dans le même ordre les images par f des éléments de la liste des abscisses précédente ce qui donnera la liste [f(a), f(a+ban), f(a+2.ban), ... , f(a+n.ban)]
3. On affiche les segments reliant les points ayant pour abscisses les valeurs dans liste_abscisses et comme ordonnées les valeurs dans liste_ordonnées. Pour cela nous allons utiliser la fonction plotde matplotlib de la façon suivante : plt.plot(liste_abscisses,liste_ordonnées).
On peut même rajouter des arguments comme par exemple la couleur ou la façon de tracer les courbes :
plt.plot(liste_abscisses,liste_ordonnées,color="blue",linestyle=":") donnera une courbe bleue tracée en pointillés.
Détails pour quelques arguments possibles de `plot`
• Pour les couleurs : Le paramètre color= permet de changer la couleur du tracé. Cette couleur peut être donnée sous plusieurs formes.
• Sous forme de chaîne de caractères représentant les noms (ou abréviations) pour les couleurs primaires, le noir et le blanc : b ou blue, g ou green, r ou red, c ou cyan, m ou magenta, y ou yellow, k ou black, w ou white. C’est quand même assez explicite, il suffit d’écrire les noms en anglais.
• Sous la forme d’un tuple correspondant aux valeurs RGB de la couleur. Cependant, ce tuple doit contenir des valeurs entre 0 et 1 (il suffit alors de diviser les valeurs RGB par 255). Ainsi, ce sera color = (255 / 255, 0, 0) pour obtenir du rouge. Notons que nous pouvons ajouter une valeur (toujours entre 0 et 1) à ce tuple pour représenter la transparence alpha.
• Sous la forme de chaîne de caractères représentant la couleur en notation hexadécimale. On aura donc color = '#00FF00' pour obtenir du vert.
• Et les adeptes des nuances de gris pourront donner en paramètre une chaine de caractère correspondant à l’intensité en gris. Par exemple color = '0.8' permet d’obtenir un gris pâle.
• Pour les styles de lignes : On utilise le paramètre linestyle=.
• "-" est le style par défaut, il correspond à une ligne pleine ;
• "--" correspond à une ligne en pointillés ;
• ":" correspond à une ligne formée de points ;
• "-." correspond à une ligne formée d’une suite de points et de tirets.
• Pour les points qu'on demande de tracer : De base, ils ne sont pas mis en valeurs pour avoir une jolie courbe mais si on veut les afficher pour les voir, on peut ajouter le paramètre marker= suivit d'un des choix suivant.
• "x" pour une croix.
• "o" pour un rond
• "." pour un point
• Il y a plein d'autres possibilités comme "*", "+" ...
Le but de cet exercice est de créer un programme qui prend en entrée f, a, b et n et qui affiche la courbe. Pour vérifier les calculs, vous devez aussi renvoyer en fin de programme vos deux listes en utilisant return liste_abscisses,liste_ordonnées (le nom des listes étant à modifier selon votre choix de nom pour ces listes).
Entrée : La fonction f, les bornes a et b de l'intervalle sur lequel on trace la fonction et le nombre n de segments pour tracer la courbe.
Sortie : Votre programme devra tracer avec plt.plot en suivant la méthode décrite au dessus et il devra de plus renvoyer les deux listes liste_abscisses et liste_ordonnées qui ont servi à tracer la courbe en utilisant return liste_abscisses,liste_ordonnées. Cette dernière étape ne sert qu'à vérifier que votre programme est juste. C'est inutile si on souhaite juste tracer la courbe.
On fera attention à ce que le début de la liste des abscisses soit bien a et la fin b car les erreurs d'arrondis peuvent faire sortir de l'intervalle [a,b] et créer des erreurs.
Remarques sur les courbes affichées : Sur chaque ligne, c'est la même fonction qui est représentée une première fois avec une faible valeur de n et la seconde avec une valeur convenable. On remarquera pour la dernière fonction que si on choisit un nombre trop faible, la fonction ressemble à une droite alors qu'en réalité c'est une fonction qui oscille beaucoup à tel point que quand on la dessine plus précisément, on a l'impression que sa courbe forme un rectangle plein (ce qui n'est pas le cas).
Tracer la courbe représentative d'une fonction
import matplotlib.pyplot as plt
def ma_fonction(f,a,b,n):
#Ne pas toucher ce qui précède
#Les valeurs pour les variables en entrée seront automatiquement données
#Ecrire ci-dessous en n'oubliant pas d'indenter
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1,514,856,476,770,656,500 | Friday
May 27, 2016
Homework Help: Discrete Mathematics
Posted by Joy on Tuesday, July 30, 2013 at 6:56am.
16. What is the value of a that would make the matrix equation true?:
( 1 -2) (-5 6) (a b)
( 3 4) (7 -8) = (c d)
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8,780,667,236,342,690,000 | Problem of the Week
Updated at Apr 23, 2018 11:41 AM
For this week we've brought you this algebra problem.
How can we compute the factors of \(49{x}^{2}+7x-42\)?
Here are the steps:
\[49{x}^{2}+7x-42\]
1
Find the Greatest Common Factor (GCF).
GCF = \(7\)
2
Factor out the GCF. (Write the GCF first. Then, in parentheses, divide each term by the GCF.)
\[7(\frac{49{x}^{2}}{7}+\frac{7x}{7}-\frac{42}{7})\]
3
Simplify each term in parentheses.
\[7(7{x}^{2}+x-6)\]
4
Split the second term in \(7{x}^{2}+x-6\) into two terms.
\[7(7{x}^{2}+7x-6x-6)\]
5
Factor out common terms in the first two terms, then in the last two terms.
\[7(7x(x+1)-6(x+1))\]
6
Factor out the common term \(x+1\).
\[7(x+1)(7x-6)\]
Done | {
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-7,593,842,991,872,608,000 | المپدیا
دانشنامهی المپیاد کامپیوتر ایران
ابزار کاربر
ابزار سایت
آموزش:الگوریتم:کولهپشتی-پویا
کولهپشتی
تعریف سوال
شما یک کوله پشتی دارید که حجم ثابتی دارد. همچنین تعدادی وسیله نیز دارید که حجم هر کدام را به شما داده اند. میخواهید تعدادی از این وسیلهها را در کوله پشتی بریزید به طوری که بیشترین حجم ممکن از کوله پشتی اشغال شود. (فرض کنید شکل وسایل طوری است که فضای بیاستفاده بین آنها باقی نمیماند.)
الگوریتم
این مسئله یکی از پایهایترین مسائل برنامهریزی پویا است و صورتهای مختلفی دارد که در انتها به آنها و ایدهی اثباتشان اشاره میشود.
برای حل مدل سادهی سوال، یک آرایه دوبعدی به نام $d$ به ابعاد$(n+1) \times (W+1)$ را در نظر بگیرید که در آن $n$ تعداد وسایل مختلفی که میتوانیم در کولهپشتی بگذاریم و$W$ حجم کولهپشتی است.
مقدار$d_{i,j}$ برابر یک است اگر و تنها اگر بتوان فقط با استفاده از $i$ وسیلهی اول، دقیقا حجم $j$ از کولهپشتی را پر کرد. یعنی یک زیرمجموعه از $i$ عضو اول وجود دارد که مجموع وزنشان $j$ است. در غیر اینصورت، مقدارش برابر صفر است.
جواب مسئله بزرگترین اندیس $j$ است که $d_{n,j}$ برابر یک باشد.
مقداردهی اولیه: با استفاده از $۰$ وسیلهی اول (استفاده نکردن از وسایل) فقط میتوان حجم $۰$ را تولید کرد (کولهپشتی خالی) پس تمام خانههای به صورت $d_{0,j}$ برابر صفر اند به جز $d_{0,0}$ که برابر $۱$ است.
به روز رسانی: برای به دست آوردن $d_{i,j}$ دو حالت وجود دارد این که خود وسیلهی $i$ ام در کولهپشتی نباشد که در این صورت باید برای این که مقدار یک شود، مقدار $d_{i-1,j}$ برابر $۱$ باشد. حالت دیگر این است که خود وسیله در کوله پشتی باشد. پس در این حالت مقدار در صورتی یک میشود که (مقدار حجم وسیلهی $i$ ام را $a_i$ بگیریم) $d_{i-۱,j-a_i}$ با فرض $j \geq a_i$ برابر یک باشد.
شبه کد:
d = {0}
d[0][0] = 1
for i from 1 to n
for j from 0 to W
d[i][j] = d[i-1][j]
if j >= a[i] and d[i-1][j-a[i]] == 1
d[i][j] = 1
اگر دقت کنید میبینید احتیاج خاصی به نگه داشتن یک آرایهی دوبعدی نداریم چون برای محاسبهی هر ستون، فقط به ستون قبلی احتیاج داریم و فقط باید ۲ ستون را نگهداریم. اما حتی میتوانیم از این هم جلوتر برویم و فقط یک ستون داشته باشیم. اما در اینجا باید حواسمان باشد که اشتباه زیر را انجام ندهیم.
شبه کد با آرایهی یک بعدی (اشتباه):
d = {0}
d[0] = 1
for i from 1 to n
for j from 0 to W
if j >= a[i] and d[j-a[i]] == 1
d[j] = 1
کد بالا یک مشکل دارد. به نظرتان مشکلش چیست؟
فرض کنید در فقط یک وسیله داریم مثلا با حجم ۲ واحد و حجم کولهپشتی برابر ۴ است. پس فقط میتوان ۲ واحد از کولهپشتی را پر کرد. اما اگر شبه کد بالا را برای آن اجرا کنید، میبینید که مقدار $d_4$ برابر یک است. چون با استفاده از وسیلهی اول که حجم دو واحد داشت، مقدار $d_2$ را یک کردیم، اما بعد از این متوقف نشدیم بلکه چون مقدار $d_2$ برابر یک بود، مقدار $d_4$ را نیز برابر یک قرار دادیم. پس انگار بیش از یک وسیله با حجم دو داشتیم. در واقع این کد جواب مسئلهی دیگری به نام خرد کردن پول است که در آن به تعداد نامتناهی از هر کدام از وسایل داریم.
حال بیایید سعی کنیم مشکل کد بالا را حل کنیم. مشکل این بود که اول با استفاده از وسیلهی اول (یا بقیهی وسایل) مقدار خانههای پایین جدول را به روز رسانی کردیم و سپس دوباره با استفاده از همان وسیله، مقادیر خانههای بالاتر را نیز به روز رسانی کردیم. چطور میشود اگر خانهها را به ترتیب دیگری پیمایش کنیم تا این مشکل پیش نیاید؟ حجم وسایل که نمیتواند منفی باشد. پس اگر بالا به پایین آرایه را به روز رسانی کنیم، این مشکل پیش نمیآید. خودتان هم کمی فکر کنید که چرا این روش درست است.
بر همین اساس کد را تغییر میدهیم. شبه کد با آرایهی یک بعدی (درست):
d = {0}
d[0] = 1
for i from 1 to n
for j from W to 0
if j >= a[i] and d[j-a[i]]
d[j] = 1
پیچیدگی الگوریتم
پیچیدگی زمانی که در تمام حالتها از $O(n \times W)$ است. مقدار حافظهی مورد نیاز نیز $O(W)$ است.
صورتهای مختلف سوال
مثال: فرض کنید وسایلی که میخواهید در کولهپشتی بگذارید، علاوهبر حجم، ارزش نیز داشته باشند و هدف شما فقط بردن وسایلی باشد که در مجموع ارزششان بیشترین مقدار ممکن باشد. الگوریتمی از $O(n \times W)$ برای حل این سوال بدهید.
پاسخ
تعریف آرایهی $d$ را به این صورت تغییر دهید که $d_{i,j}$ یعنی با استفاده از $i$ وسیلهی اول و حجم $j$ حداکثر مجموع ارزش وسایل در کولهپشتی چقدر است. شیوهی به روز رسانی شبیه سوال اصلی است و به خواننده محول میشود.
پیادهسازی اولیه
knapsack.cpp
#include <iostream>
using namespace std;
const int MAXN = 10*1000;
const int MAXW = 100*1000;
bool d[MAXW+10];
int p[MAXW+10];
int a[MAXN+10];
int main()
{
ios::sync_with_stdio(false);
int n, w;
cin >> n >> w;
for(int i=0; i<n; i++)
cin >> a[i];
d[0] = 1;
for(int i=0; i<n; i++)
for(int j=w; j>=a[i]; j--)
if( d[j] == 0 && d[j-a[i]] == 1 )
{
d[j] = 1;
p[j] = a[i];
}
int out = w;
for(; out>=0; out--)
if( d[out] )
break;
cout << out << endl;
while( out != 0 )
{
cout << p[out] << " ";
out -= p[out];
}
cout << endl;
return 0;
}
کابردها
مسائل نمونه
مراجع
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Homework Help: Complex Matrices and Unit Circles
1. Nov 4, 2011 #1
1. The problem statement, all variables and given/known data
What can you say about
a. the sum of a complex number and its conjugate?
b. the conjugate of anumber on the unit circle?
c. the product of two numbers on the unit circle?
d. the sum of two numbers on the unit circle?
2. Relevant equations
3. The attempt at a solution
Here's what I'm thinking:
a. The sum of a complex number and its conjugate is real: (a+bi)+(a-bi)=2a
b. The conjugate of a number on the unit circle LIES ON the unit circle.
c. The product of two numbers on the unit circle also LIES ON the unit circle.
d. The sum of two numbers on the unit circle LIES INSIDE OR OUTSIDE the unit circle.
Am I thinking correctly or am I missing something? I'm unsure on this one. Thanks!
2. jcsd
3. Nov 5, 2011 #2
Mark44
Staff: Mentor
I don't see anything wrong with what you've said, but they might be looking for more than you've said.
For a) Yes, the sum is real, but notice what you have in your formula.
For b) is it possible to say where on the unit circle the conjugate would be? Let z = √2/2 + i√2/2, which is on the unit circle. Where is [itex]\overline{z}[/itex]?
For c), if z1 and z2 are on the unit circle, what can you say about z1z2?
For d), same question, but about z1 + z2.
The key is to think geometrically - draw some pictures.
4. Nov 5, 2011 #3
Deveno
User Avatar
Science Advisor
what about 1 and -1/2 + i√3/2?
5. Nov 5, 2011 #4
Hmmm... I drew some pictures as suggested... I'm still unsure, but here's another attempt:
a) I forgot to mention that the sum is real and is twice the real part
b) 90 degree reflection about the origin (for it's imaginary part)?
For parts c and, after I drew the pictures, I still cannot see it. Are there any particular numbers I should be looking at that are easier to see? Thanks!
6. Nov 5, 2011 #5
Deveno
User Avatar
Science Advisor
a) yes
b) what does "a reflection about the origin" even mean? reflections usually involve a line, or a plane, or some higher-dimensional-thingy
c) my advice: google "de moivre's theorem"
d) do you know how to draw a vector sum?
7. Nov 5, 2011 #6
for part b, is it sufficient to say 90 deg reflection about the Imaginary axis??????
I'll follow your advice and google "de moivre's theorem"
For part d, I haven't done that in a long time, so I'll also have to look it up
thanks
8. Nov 5, 2011 #7
Deveno
User Avatar
Science Advisor
degrees normally have to do with rotation (they "twist" or "turn"). think "mirror-like" when thinking about reflections.
to draw a vector sum, make a parallelogram (2 sides of this will be your 2 vectors starting at the origin, the other two sides will be the same two vectors drawn "head to tail"), and draw the diagonal, which represents the vector sum.
9. Nov 6, 2011 #8
Another try:
For part b, it'd be a reflection about the real axis?
For part c, the product of two numbers on the unit circle will still have length 1 and will be located at the sum of their angles? e.g., (0+i)(1+0i) = i, where 0+i is located 90deg and 1+0i at 0 deg, so the product will be at 90 deg.
For part d, their sum will still have length 1???
10. Nov 6, 2011 #9
For part d, I'm thinking it should be still length 1 half way between the two numbers...
11. Nov 6, 2011 #10
Mark44
Staff: Mentor
Yes, and I think this is what they had in mind.
Yes. This is more like what they're looking for, IMO.
Yes.
12. Nov 6, 2011 #11
Thank you for your help!
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4,140,059,873,118,599,700 | How long is 0.2 months in years?
Unit Converter
Conversion formula
The conversion factor from months to years is 0.083388698630137, which means that 1 month is equal to 0.083388698630137 years:
1 mo = 0.083388698630137 yr
To convert 0.2 months into years we have to multiply 0.2 by the conversion factor in order to get the time amount from months to years. We can also form a simple proportion to calculate the result:
1 mo → 0.083388698630137 yr
0.2 mo → T(yr)
Solve the above proportion to obtain the time T in years:
T(yr) = 0.2 mo × 0.083388698630137 yr
T(yr) = 0.016677739726027 yr
The final result is:
0.2 mo → 0.016677739726027 yr
We conclude that 0.2 months is equivalent to 0.016677739726027 years:
0.2 months = 0.016677739726027 years
0.2 months is equal to 0.0167 years
Alternative conversion
We can also convert by utilizing the inverse value of the conversion factor. In this case 1 year is equal to 59.960163453048 × 0.2 months.
Another way is saying that 0.2 months is equal to 1 ÷ 59.960163453048 years.
Approximate result
For practical purposes we can round our final result to an approximate numerical value. We can say that zero point two months is approximately zero point zero one seven years:
0.2 mo ≅ 0.017 yr
An alternative is also that one year is approximately fifty-nine point nine six times zero point two months.
Conversion table
months to years chart
For quick reference purposes, below is the conversion table you can use to convert from months to years
months (mo) years (yr)
1.2 months 0.1 years
2.2 months 0.183 years
3.2 months 0.267 years
4.2 months 0.35 years
5.2 months 0.434 years
6.2 months 0.517 years
7.2 months 0.6 years
8.2 months 0.684 years
9.2 months 0.767 years
10.2 months 0.851 years | {
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-2,702,544,324,782,875,600 | Web Results
www.wikihow.com/Calculate-the-Volume-of-a-Cube
How to Calculate the Volume of a Cube. A cube is a three-dimensional shape that has equal width, height, and length measurements. A cube has six square faces, all of which have sides of equal length and all of which meet at right...
www.gigacalculator.com/calculators/volume-calculator.php
Volume calculations are useful in a lot of sciences, in construction work and planning, in cargo shipping, in climate control (e.g. air conditioning calculations), swimming pool management, and more. Volume of a cube. The volume formula for a cube is side 3, as seen in the figure below:
www.gigacalculator.com/calculators/volume-of-cube...
Volume of a cube formula. The volume of a cube can be calculated if you know its side length. The formula is then volume cube = side 3. Illustration below: Measuring the side of the cube is easy. The result from the calculation, using our volume of a cube calculator or otherwise, will always be in the length unit used, cubed.
study.com/academy/lesson/how-to-calculate-the-volume-of-a...
In this lesson, you will explore the formula used to calculate the volume of a cube. You will also gain a conceptual understanding of volume and appropriate units to use. Then test your knowledge ...
www.calculator.net/volume-calculator.html
This free volume calculator can compute the volumes of common shapes, including that of a sphere, cone, cube, cylinder, capsule, cap, conical frustum, ellipsoid, and square pyramid. Explore many other math calculators like the area and surface area calculators, as well as hundreds of other calculators related to finance, health, fitness, and more.
www.solumaths.com/en/calculator-online/calculate/volume_cube
The cube volume calculator supports numeric but also literal expressions. Calculating the volume of a cube. The volume calculator is able to calculate the volume of a cube, from variables numeric, the exact and approximate results are returned. Thus, calculating the volume of a cube whose length is 3, is done by entering the following formula ...
www.wikihow.com/Calculate-Volume
To calculate volume with a cube, use the formula v = s^3, where s is the length of the sides of the cube. To calculate the volume of a cylinder, use the formula v = hπr^2, where r is the radius of the base, h is the height, and π is pi. If you're trying to find the volume of a rectangular prism, use the formula v = lwh, where l is the length ...
arcb.com/shipping-tools/shipping-calculators/cube-calculator
Knowing an accurate cubic volume of your shipment is important because it indicates how much space it will occupy and could impact shipping costs. It’s also an important consideration when calculating a shipment’s density, which helps determine freight class. Use our online cube calculator tool to determine the cubic volume of your shipment.
westernlogistics.com/cube-calculator
Cube Calculator. You can use this handy on-line tool to easily calculate the total volume of your shipment. First select the unit of measurement you prefer — either centimetres or inches. Enter the dimensions of your first item and the number of pieces you have at that size.
www.ducksters.com/kidsmath/finding_the_volume_of_a_cube_or...
Volume is the measurement of how much space a three dimensional object takes up. On this page we will look at how to figure out the volume of a box. If you know how to multiply you can find the volume of a cube or box. We learned earlier that the surface area of a flat rectangle was the length times ... | {
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9.3 Describing Dispersion
9.3 Describing Dispersion
<
>
• Math Help
The histograms in Example 4 have normal distributions drawn on them because they can be approximated by normal distributions. This can be shown by comparing the percents found using the normal distribution and the actual data for several heights. Use 61 inches and 67 inches. A height of 61 inches is 3 standard deviations below the mean for males and about 1 standard deviation below the mean for females (notice that for females, 61.5 inches is exactly 1 standard deviation below the mean). A height of 67 inches is 1 standard deviation below the mean for males and about 1 standard deviation above the mean for females (notice that for females, 66.5 inches is exactly 1 standard deviation above the mean).
At most 61 inches tall
At least 67 inches tall
You can see from these tables that the percents using the normal distribution are close to the percents using the actual data for males and females.
• Consumer Suggestion
Women have historically received less pay for equal work in the United States. As a current average, women earn 77% the amount of their male counterparts. To see what the gender salary gap is like in your state, the American Association of University Women (AAUW) has compiled state-by-state data into this table.
• Checkpoint Solution
Sample answer:
While Judge and Cable's study indicates a positive correlation between height and salary, this does not guarantee that a taller person will always be more successful than a shorter person. Many things matter for career success such as experience, who you know, personality, and intelligence. Each of these traits is weighted differently for different types of careers. It is not so much a question of whether height matters but a question of how much height matters. If you are a basketball player, height certainly matters for career success. If you run an online business from home, height probably does not matter at all.
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New Reply
Alternate nth power function
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Jun8-12, 08:59 PM #1
Alternate nth power function
Hi guys I am just wondering if there exists an alternate nth power function that doesn't involve any exponents. Thank you in advance.
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Jun8-12, 09:30 PM #2
Quote by eddybob123 View Post
Hi guys I am just wondering if there exists an alternate nth power function that doesn't involve any exponents. Thank you in advance.
What is an "alternate n-th power function"?
DonAntonio
Jun9-12, 12:02 AM #3
Mentor
"nth power" is another way of saying that the exponent is n.
Jun9-12, 06:41 AM #4
Alternate nth power function
Quote by Mark44 View Post
"nth power" is another way of saying that the exponent is n.
That I know, what is "aternate" such a function, though? Perhaps he meant "alternative"? But then he'd be asking whether there's an exponential function that doesn't involve exponents...!
DonAntonio
Jun9-12, 10:09 AM #5
if n is an integer you can just express it as a sum or product.
Jun9-12, 12:36 PM #6
Mentor
eddybob123,
If by "power function" you mean a function in which a variable is raised to a constant power (xn), then no, there is no way to express it without an exponent.
If that's not what you mean by power function, then please clarify for us what you're asking.
Quote by dipole
if n is an integer you can just express it as a sum or product.
I don't see what this has to do with what the OP asked.
Jun9-12, 09:48 PM #7
But there should obviously be one that exists. For example, if one lists out the squares of the integers and find their differences, one will achieve all the positive odd numbers. Likewise, if one lists out the cubes of the integers, and find their differences and the differnece of their difference, one will acheive all the multiples of 6. I was just wondering whether their was a general formula for all the nth powers.
Jun9-12, 10:19 PM #8
Quote by eddybob123 View Post
But there should obviously be one that exists. For example, if one lists out the squares of the integers and find their differences, one will achieve all the positive odd numbers. Likewise, if one lists out the cubes of the integers, and find their differences and the differnece of their difference, one will acheive all the multiples of 6. I was just wondering whether their was a general formula for all the nth powers.
What does this have to do at all with your original question??
DonAntonio
Jun10-12, 12:07 AM #9
Quote by eddybob123 View Post
Hi guys I am just wondering if there exists an alternate nth power function that doesn't involve any exponents. Thank you in advance.
Hey eddybob123.
One way that you could do it is first decompose into integer and non-integer parts (i.e. x^(a+b) = x^a * x^b where a is an integer and b is the fractional part).
From this you can reduce the x^a to a summation (humungous but huge non-the-less) which then leaves how to resolve x^b.
For this part you can use a power-series expansion that is of the form of an 'infinite-polynomial'.
Technically you can't calculate this result in a finite-manner, but you can write the summation out in terms of a power series with integer powers and because they are in integer powers, you can use the ideas for summations from integer powers and do it this way.
So for example lets look at x^(2.5).
Take x^(2.5) = x^2 * x^(1/2). x^2 can be written out as a summation (you need to take into account the fractional part of x as well).
Now x^(1/2) has a power series expansion with integer powers. Again you will need to take into account the fractional parts and deal with these in a way that translates to summation.
Your answer most likely won't lend itself to finite-computation of integers or rational numbers, but it still can be expressed in terms of a summation.
When x is not an integer you have x = c + d where c is an integer and d is the fractional part. So now from this you have to consider (c+d)^f for some f.
The thing is though, that you may have to do many of these transformations repeatedly to get something in terms of rational numbers, but it can be done. Just don't expect the final transformation to be a simple one.
If x is an integer or a rational number, use the method presented first in this response. If however it is not rational, then you will need to use more general tools which will be a lot harder.
Jun10-12, 12:52 AM #10
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Quote by eddybob123 View Post
if one lists out the cubes of the integers, and find their differences and the differnece of their difference, one will acheive all the multiples of 6.
Yes, you can achieve rn (plus smaller terms) by doing n summations up to r of the constant n!:
1! S(1, r) = Ʃ 1 = r
2! S(2, r) = 2! Ʃ S(1, r) = r(r+1) = r2 + r
3! S(3, r) = 3! Ʃ S(2, r) = r(r+1)(r+2) = r3 + 2r2 + 2r
etc.
Is that what you had in mind?
Jun11-12, 06:23 PM #11
Quote by Mark44 View Post
eddybob123,
If by "power function" you mean a function in which a variable is raised to a constant power (xn), then no, there is no way to express it without an exponent.
If that's not what you mean by power function, then please clarify for us what you're asking.
I don't see what this has to do with what the OP asked.
xn = x*x*x*x... n times. This can also be written as a sum.
Jun12-12, 06:39 PM #12
Yes, haruspex, except with just general powers x^n expressed in summation form [n{SUM}k=1]f(x,n)
Jun12-12, 07:36 PM #13
Mentor
Quote by dipole View Post
xn = x*x*x*x... n times. This can also be written as a sum.
OK, I'm game - show me.
Jun12-12, 07:46 PM #14
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Quote by haruspex View Post
3! S(3, r) = 3! Ʃ S(2, r) = r(r+1)(r+2) = r3 + 2r2 + 2r
Er.. I mean = r3 + 3r2 + 2r
Jun14-12, 06:12 PM #15
I know that for all x^n, that the terms p_n are related to n, and that x^n=p_1 + p_2 + p_3 + ... + p_n. I just need to find a direct function relating all values of p.
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2,050,269,464,337,682,700 | Tag Archives: division
Fill in the Blank: 3 Digit Dividends
📖 STANDARD
CCSS.MATH.CONTENT.4.NBT.B.6
Find whole-number quotients and remainders with up to four-digit dividends and one-digit divisors, using strategies based on place value, the properties of operations, and/or the relationship between multiplication and division. Illustrate and explain the calculation by using equations, rectangular arrays, and/or area models.
⏰ LESSON TIME
55 Minutes
📃 SUMMARY
This lesson plan will have students reinforce the previously presented concept of long division of 3 digit dividends by 1 digit divisors through a short review, an animated video, a fill in the blank division word problem activity, and game play.
📲 TECHNOLOGY REQUIRED
Students will need a PC, Mac or Chromebook or tablet. Making Camp Premium and Dakota are both playable on any web browser on those devices. Students will also need access to the games.
📚LESSON PLAN
1. Start the lesson by having a short review on dividing 3-digit dividends by 1-digit divisors. It is suggested to review the activity that was assigned in the previous lesson, “Build Your Own Division Problem” lesson. You can find the review copy of the activity here: “Review – Build Your Own Division Problem“. (15 minutes)
2. How to Use Division to Solve a Problem – Start with this animated video on how to use long division, with 3 digit dividends, to solve different word problems. (Time – 1:24)
3. Have students work on the activity “Fill in the Blank: 3 Digit Dividends“. The activity has students fill in the blanks to complete the division word problem, identify the problem, and solve it. (Corresponding “Fill in the Blank: 3 Digit Dividends” Activity Jamboard that’s also linked in the activity above can be used by students as space to work out their division word problems.) (20 minutes)
4. Have students play Making Camp Dakota: Past & Present using our Games Portal for Kids. The division in Making Camp Dakota will be more practice for students on problems with 3 digit dividends divided by 1 digit divisors. (15 minutes)
ASSESSMENT
The “Fill in the Blank: 3 Digit Dividends” activity is not only allowing students to continue practicing the concepts previously introduced and reviewed in this lesson, but also serves as an assessment of how well the student grasps the concept of division using 3 digit dividends and 1 digit divisors.
Making Camp Dakota Teacher Reports
You can view your students’ progress on mastering these standards by viewing your Making Camp Dakota teacher reports. You can access the Making Camp Dakota reports here.
STATE STANDARDS
Minnesota State Standards
4.1.1.6 – Use strategies and algorithms based on knowledge of place value, equality and properties of operations to divide multi-digit whole numbers by one- or two-digit numbers. Strategies may include mental strategies, partial quotients, the commutative, associative, and distributive properties and repeated subtraction.
5.1.1.1 – Divide multi-digit numbers, using efficient and generalizable procedures, based on knowledge of place value, including standard algorithms. Recognize that quotients can be represented in a variety of ways, including a whole number with a remainder, a fraction or mixed number, or a decimal.
Build Your Own Division Problem
📖 STANDARD
CCSS.MATH.CONTENT.4.NBT.B.6
Find whole-number quotients and remainders with up to four-digit dividends and one-digit divisors, using strategies based on place value, the properties of operations, and/or the relationship between multiplication and division. Illustrate and explain the calculation by using equations, rectangular arrays, and/or area models.
⏰ LESSON TIME
50 minutes
📃 SUMMARY
This lesson plan will build upon the introduction to division with 3-digit dividends from the “Dividing 3-Digit Dividends” lesson. Students will be able to continue practicing dividing 3-digit dividends by 1-digit divisors through a short review, an activity where students build their own division problems, and game play.
📲 TECHNOLOGY REQUIRED
Students will need a PC, Mac or Chromebook. Students will also need access to “Making Camp Dakota: Past & Present” using our Games Portal for Kids. “Making Camp Dakota: Past & Present” is playable on PC, Mac, and Chromebook using any browser.
📚LESSON PLAN
1. Start the lesson by having a short review on dividing 3-digit dividends by 1-digit divisors. It is suggested to review the activity that was assigned in the previous lesson, “Dividing 3-Digit Dividends” lesson. You can find the review copy of the activity here: “Review – On Your Way Home.” (10 minutes)
2. Students will take the information reviewed and use it to complete the “Build Your Own Division Problem” activity. In this activity, students practice division using 3-digit dividends and 1-digit divisors through using virtual manipulatives. (20 minutes)
3. To end the lesson, students can play Making Camp Dakota: Past & Present to further practice division. Students can access the game using our Games Portal for Kids. (20 minutes)
ASSESSMENT
The “Build Your Own Division Problem” activity is not only allowing students to continue practicing the concepts previously introduced and reviewed in this lesson, but also serves as an assessment of how well the student grasps the concept of division using 3 digit dividends and 1 digit divisors.
Making Camp Dakota Teacher Reports
You can view your students’ progress on mastering these standards by viewing your Making Camp Dakota teacher reports. You can access the Making Camp Dakota reports here.
STATE STANDARDS
Minnesota State Standards
4.1.1.6 – Use strategies and algorithms based on knowledge of place value, equality and properties of operations to divide multi-digit whole numbers by one- or two-digit numbers. Strategies may include mental strategies, partial quotients, the commutative, associative, and distributive properties and repeated subtraction.
5.1.1.1 Divide multi-digit numbers, using efficient and generalizable procedures, based on knowledge of place value, including standard algorithms. Recognize that quotients can be represented in a variety of ways, including a whole number with a remainder, a fraction or mixed number, or a decimal.
Related Lesson – Fill in the Blank: 3 Digit Dividends
The “Fill in the Blank: 3 Digit Dividends” lesson plan will have students reinforce the previously presented concept of long division of 3 digit dividends by 1 digit divisors through an animated video, a fill in the blank division word problem activity, and game play.
Dividing 3-Digit Dividends
📖STANDARD
CCSS.MATH.CONTENT.4.NBT.B.6
Find whole-number quotients and remainders with up to four-digit dividends and one-digit divisors, using strategies based on place value, the properties of operations, and/or the relationship between multiplication and division. Illustrate and explain the calculation by using equations, rectangular arrays, and/or area models.
LESSON TIME
30-45 minutes
📃 SUMMARY
This lesson will build upon the already introduced concepts and vocabulary of division in our “Introducing Division” lesson. Students will learn and be able to practice dividing 3 digit dividends by 1 digit divisors through a short review, activity, and game play.
📲TECHNOLOGY REQUIRED
Students will need a PC, Mac or Chromebook or tablet. Making Camp Premium and Dakota are both playable on any web browser on those devices. Students will also need access to the games.
📚 LESSON
Begin by reviewing division.
1. Start the lesson by reviewing the important division basics. This Division Review Google Slides reviews “what we know.” The presentation goes over the definition of division, key terms, division symbols, and examples.
1. Have students work through the On Your Way Home division activity where they take what they know and practice dividing 3 digit dividends by 1 digit divisors. The activity takes division problems and puts them into real-world context, visiting stores on your way home from the park.
1. Have students play Making Camp Premium and/or Making Camp Dakota using our Games Portal for Kids. The division magnets game in Making Camp Premium can be used as review before moving on to Making Camp Dakota. The division in Making Camp Dakota will be more practice for problems with 3 digit dividends divided by 1 digit divisors.
ASSESSMENT
Assessment is built into the “On Your Way Home” activity through the creation of their own division problem after repeated practice of 3 digit dividends divided by 1 digit divisors.
Making Camp Dakota Teacher Reports
You can view your students’ progress on mastering these standards by viewing your Making Camp Dakota teacher reports. You can access the Making Camp Dakota reports here.
STATE STANDARDS
Minnesota State Standards
4.1.1.6 – Use strategies and algorithms based on knowledge of place value, equality and properties of operations to divide multi-digit whole numbers by one- or two-digit numbers. Strategies may include mental strategies, partial quotients, the commutative, associative, and distributive properties and repeated subtraction.
5.1.1.1 – Divide multi-digit numbers, using efficient and generalizable procedures, based on knowledge of place value, including standard algorithms. Recognize that quotients can be represented in a variety of ways, including a whole number with a remainder, a fraction or mixed number, or a decimal.
Related Lesson: Build Your Own Division Problem
The “Build Your Own Division Problem” lesson plan will build upon the introduction to division with 3-digit dividends from this lesson. Students will be able to continue practicing dividing 3-digit dividends by 1-digit divisors through a short review, an activity where students build their own division problems, and game play.
Division and English/ Language Arts
Standards
CCSS Standard:
Find whole-number quotients and remainders with up to four-digit dividends and one-digit divisors, using strategies based on place value, the properties of operations, and/or the relationship between multiplication and division. Illustrate and explain the calculation by using equations, rectangular arrays, and/or area models.
CCSS.ELA-LITERACY.W.4.3
Write narratives to develop real or imagined experiences or events using effective technique, descriptive details, and clear event sequences.
Time
40-45 minutes
Technology Required
Making Camp Premium plays in any browser, so, of course, on Chromebooks. It can also be downloaded on phones or tablets and played offline by students who have limited Internet access. Schools that are part of the Growing Math project or who have a 7 Generation Games site license have access to the game for students to use at home or school.
Summary
This is a fun lesson where students practice division, combined with Ojibwe history and then complete a creative writing assignment.
Lesson Plan
1. Game Play with Making Camp
1. Open Making Camp. Go to the main choices screen by clicking on the small green icon with boxes at the lower left of the screen.
2. Click NUMBERS.
3. Click the box with the numbers to practice division.
Click the numbers box for division practice.
Each correct quotient earns a fridge magnet to decorate the fridge!
Students should play until they earn at least 15 points.
2. Spend the points earned and learn about Ojibwe history
One of the best teachers we know said, “History is more than names and dates. It’s how people lived. It’s the things they used.” When trading for a wigwam, students will watch videos on how to build a wigwam and on trading. They’ll learn that tribes traded with one another for hundreds of years.
3. Short story writing prompt
So how did we get from a refrigerator to a wigwam? You can use this Google slides presentation to tie in Native American history with Sam’s life in the twenty-first century. This presentation can also be added to your Google classroom as an assignment for students. Here is the introduction for Sam and his account:
This is Sam. He’s also Ojibwe but he’s not from a long time ago. He’s 16 years old. He lives on a reservation in the northern United States. You’ve probably heard of it. He doesn’t live in a wigwam. He lives in a white house with a grey roof. That’s the refrigerator in his house. The magnets have been there ever since he was in second grade.
Read the passage about Sam.
4. Writing assignment
Read about Sam and write a story about him. What do you think happened to him in second grade? Why does everyone except for his cousin, Angie, think he’s not smart? Do you think he and Angie can really walk to Maine?
Assessment
Math problems in Making Camp Premium are scored automatically. You can see how many students attempted and the number correct in the data reports. All Growing Math teachers and all schools with 7 Generation Game licenses receive access to these reports. Writing assignments can be assessed according to the teacher’s own rubric.
State Standards
Minnesota Math Standard 4.1.1.6 – Use strategies and algorithms based on knowledge of place value, equality and properties of operations to divide multi-digit whole numbers by one- or two-digit numbers. Strategies may include mental strategies, partial quotients, the commutative, associative, and distributive properties and repeated subtraction.
Minnesota Math Standard 5.1.1.1 – Divide multi-digit numbers, using efficient and generalizable procedures, based on knowledge of place value, including standard algorithms. Recognize that quotients can be represented in a variety of ways, including a whole number with a remainder, a fraction or mixed number, or a decimal.
Introducing Division
Standard
CCSS Standard:
Find whole-number quotients and remainders with up to four-digit dividends and one-digit divisors, using strategies based on place value, the properties of operations, and/or the relationship between multiplication and division. Illustrate and explain the calculation by using equations, rectangular arrays, and/or area models.
LESSON TIME
30 minutes
SUMMARY
This lesson plan introduces the concepts and vocabulary of division and provides students the opportunities to practice both recognition and recall of division facts. Math literacy activities are recommended to help students remember and understand math terms. Students can play the Making Camp Premium game, with division with one-digit divisors or Making Camp Lakota which includes one- and two-digit divisors.
Technology required
The games, Making Camp Premium and Making Camp Lakota can be played on any web browser – Chrome, Firefox, Bing, etc or downloaded on iOS or Android devices. The Making Camp Premium game can be played with or without an Internet connection. Internet is required to log in to the Making Camp Lakota game with username and password. Both games are available at no extra charge to schools with a 7 Generation Games site license or who are part of the Growing Math Project.
Lesson Plan
1. Introductory mini-lecture
5-7 minutes
1. Explain why division is important. When we want to share something, whether it is the hours spent doing chores or a birthday cake, we divide it.
2. Tell students they will be taking notes on the videos, writing down key words that appear in the video (e.g. quotient, dividend) and their definitions for review at the end of the session. (Math literacy helps with identifying math terms; textbooks will start to make sense.)
2. Watch Videos
7 minutes
Division Terms
What’s division? How do we use division every day? 3:31 minutes
The Division terms video is included in Making Camp Lakota, a game teaching division and Lakota history. It is also in the Making Camp Premium game. Before starting the video remind students to be ready to take notes.
Division (Multiplication in Reverse)
Why is learning multiplication important for learning division? 2:41 Minutes
3. Game Play
15 minutes
Play Making Camp Premium for multiplication and division.
Making Camp Premium can be played online on any computer (Windows, Mac, Chrome), or you can download it for your phone or tablet (iOS and Android). This game is part of the 7 Generation Games school license and also available as part of the Growing Math project.
The division magnets game practices division of one-digit numbers into two digit numbers, like 35 ÷ 5.
There are also games for multiplication and a lot of videos and games on history and English/ language arts. Making Camp Premium also teaches about Ojibwe history and culture.
For more division, play Making Camp Lakota.
In mathematics content, Making Camp Lakota focuses only on division, combined with, of course, Lakota history and culture. With development funded by the Thunder Valley Community Development Corp. this game is free to play on the web or downloadable for iPad or Android tablets.
ASSESSMENT
This lesson plan includes two types of assessment.
1. You can view your students’ progress on mastering these standards by viewing your Making Camp Premium Teacher Reports. See an example below.
Student data by standard from Making Camp Premium
Sample Student Report
2. To assess student understanding of math vocabulary, review their notes written while viewing the division terms video.
Related Content
We have a YouTube division playlist! The videos above are part of a five-video series clues students in on everything from dividends to long division with remainders. (And they are short, ranging from one and a half to three and a half minutes each.) These videos can be played on any device in class or at home.
State Standards
Minnesota Math Standard 4.1.1.6 – Use strategies and algorithms based on knowledge of place value, equality and properties of operations to divide multi-digit whole numbers by one- or two-digit numbers. Strategies may include mental strategies, partial quotients, the commutative, associative, and distributive properties and repeated subtraction.
Minnesota Math Standard 5.1.1.1 – Divide multi-digit numbers, using efficient and generalizable procedures, based on knowledge of place value, including standard algorithms. Recognize that quotients can be represented in a variety of ways, including a whole number with a remainder, a fraction or mixed number, or a decimal. | {
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-8,561,153,916,292,290,000 | Algebra and Trigonometry 10th Edition
Published by Cengage Learning
ISBN 10: 9781337271172
ISBN 13: 978-1-33727-117-2
Chapter 1 - Review Exercises - Page 153: 65
Answer
$-3-3i$
Work Step by Step
Reorganize: $(6-4i)+(-9+i)=[6+(-9)]+(-4+1)i=-3-3i$
Update this answer!
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Update this answer
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3,823,122,967,742,541,300 | Skip to main content
Statistics LibreTexts
13.E: F Distribution and One-Way ANOVA (Exercises)
• Page ID
23514
• \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\)
These are homework exercises to accompany the Textmap created for "Introductory Statistics" by OpenStax.
13.1: Introduction
13.2: One-Way ANOVA
Q 13.2.1
Three different traffic routes are tested for mean driving time. The entries in the table are the driving times in minutes on the three different routes. The one-way \(ANOVA\) results are shown in Table.
Route 1 Route 2 Route 3
30 27 16
32 29 41
27 28 22
35 36 31
State \(SS_{\text{between}}\), \(SS_{\text{within}}\), and the \(F\) statistic.
S 13.2.1
\(SS_{\text{between}} = 26\)
\(SS_{\text{within}} = 441\)
\(F = 0.2653\)
Q 13.2.2
Suppose a group is interested in determining whether teenagers obtain their drivers licenses at approximately the same average age across the country. Suppose that the following data are randomly collected from five teenagers in each region of the country. The numbers represent the age at which teenagers obtained their drivers licenses.
Northeast South West Central East
16.3 16.9 16.4 16.2 17.1
16.1 16.5 16.5 16.6 17.2
16.4 16.4 16.6 16.5 16.6
16.5 16.2 16.1 16.4 16.8
\(\bar{x} =\) ________ ________ ________ ________ ________
\(s^{2} =\) ________ ________ ________ ________ ________
State the hypotheses.
\(H_{0}\): ____________
\(H_{a}\): ____________
13.3: The F-Distribution and the F-Ratio
Use the following information to answer the next five exercises. There are five basic assumptions that must be fulfilled in order to perform a one-way \(ANOVA\) test. What are they?
Exercise 13.2.1
Write one assumption.
Answer
Each population from which a sample is taken is assumed to be normal.
Exercise 13.2.2
Write another assumption.
Exercise 13.2.3
Write a third assumption.
Answer
The populations are assumed to have equal standard deviations (or variances).
Exercise 13.2.4
Write a fourth assumption.
Exercise 13.2.5
Write the final assumption.
Answer
The response is a numerical value.
Exercise 13.2.6
State the null hypothesis for a one-way \(ANOVA\) test if there are four groups.
Exercise 13.2.7
State the alternative hypothesis for a one-way \(ANOVA\) test if there are three groups.
Answer
\(H_{a}: \text{At least two of the group means } \mu_{1}, \mu_{2}, \mu_{3} \text{ are not equal.}\)
Exercise 13.2.8
When do you use an \(ANOVA\) test?
Use the following information to answer the next three exercises. Suppose a group is interested in determining whether teenagers obtain their drivers licenses at approximately the same average age across the country. Suppose that the following data are randomly collected from five teenagers in each region of the country. The numbers represent the age at which teenagers obtained their drivers licenses.
Northeast South West Central East
16.3 16.9 16.4 16.2 17.1
16.1 16.5 16.5 16.6 17.2
16.4 16.4 16.6 16.5 16.6
16.5 16.2 16.1 16.4 16.8
\(\bar{x} =\) ________ ________ ________ ________ ________
\(s^{2}\) ________ ________ ________ ________ ________
\(H_{0}: \mu_{1} = \mu_{2} = \mu_{3} = \mu_{4} = \mu_{5}\)
\(H_{a}\): At least any two of the group means \(\mu_{1} , \mu_{2}, \dotso, \mu_{5}\) are not equal.
Q 13.3.1
degrees of freedom – numerator: \(df(\text{num}) =\) _________
Q 13.3.2
degrees of freedom – denominator: \(df(\text{denom}) =\) ________
S 13.3.2
\(df(\text{denom}) = 15\)
Q 13.3.3
\(F\) statistic = ________
13.4: Facts About the F Distribution
Exercise 13.4.4
An \(F\) statistic can have what values?
Exercise 13.4.5
What happens to the curves as the degrees of freedom for the numerator and the denominator get larger?
Answer
The curves approximate the normal distribution.
Use the following information to answer the next seven exercise. Four basketball teams took a random sample of players regarding how high each player can jump (in inches). The results are shown in Table.
Team 1 Team 2 Team 3 Team 4 Team 5
36 32 48 38 41
42 35 50 44 39
51 38 39 46 40
Exercise 13.4.6
What is the \(df(\text{num})\)?
Exercise 13.4.7
What is the \(df(\text{denom})\)?
Answer
ten
Exercise 13.4.8
What are the Sum of Squares and Mean Squares Factors?
Exercise 13.4.9
What are the Sum of Squares and Mean Squares Errors?
Answer
\(SS = 237.33; MS = 23.73\)
Exercise 13.4.10
What is the \(F\) statistic?
Exercise 13.4.11
What is the \(p\text{-value}\)?
Answer
0.1614
Exercise 13.4.12
At the 5% significance level, is there a difference in the mean jump heights among the teams?
Use the following information to answer the next seven exercises. A video game developer is testing a new game on three different groups. Each group represents a different target market for the game. The developer collects scores from a random sample from each group. The results are shown in Table
Group A Group B Group C
101 151 101
108 149 109
98 160 198
107 112 186
111 126 160
Exercise 13.4.13
What is the \(df(\text{num})\)?
Answer
two
Exercise 13.4.14
What is the \(df(\text{denom})\)?
Exercise 13.4.15
What are the \(SS_{\text{between}}\) and \(MS_{\text{between}}\)?
Answer
\(SS_{\text{between}} = 5,700.4\);
\(MS_{\text{between}} = 2,850.2\)
Exercise 13.4.16
What are the \(SS_{\text{within}}\) and \(MS_{\text{within}}\)?
Exercise 13.4.17
What is the \(F\) Statistic?
Answer
3.6101
Exercise 13.4.18
What is the \(p\text{-value}\)?
Exercise 13.4.19
At the 10% significance level, are the scores among the different groups different?
Answer
Yes, there is enough evidence to show that the scores among the groups are statistically significant at the 10% level.
Use the following information to answer the next three exercises. Suppose a group is interested in determining whether teenagers obtain their drivers licenses at approximately the same average age across the country. Suppose that the following data are randomly collected from five teenagers in each region of the country. The numbers represent the age at which teenagers obtained their drivers licenses.
Northeast South West Central East
16.3 16.9 16.4 16.2 17.1
16.1 16.5 16.5 16.6 17.2
16.4 16.4 16.6 16.5 16.6
16.5 16.2 16.1 16.4 16.8
\(\bar{x} =\) ________ ________ ________ ________ ________
\(s^{2} =\) ________ ________ ________ ________ ________
Enter the data into your calculator or computer.
Exercise 13.4.20
\(p\text{-value} =\) ______
State the decisions and conclusions (in complete sentences) for the following preconceived levels of \(\alpha\).
Exercise 13.4.21
\(\alpha = 0.05\)
1. Decision: ____________________________
2. Conclusion: ____________________________
Exercise 13.4.22
\(\alpha = 0.01\)
1. Decision: ____________________________
2. Conclusion: ____________________________
Use the following information to answer the next eight exercises. Groups of men from three different areas of the country are to be tested for mean weight. The entries in the table are the weights for the different groups. The one-way \(ANOVA\) results are shown in Table.
Group 1 Group 2 Group 3
216 202 170
198 213 165
240 284 182
187 228 197
176 210 201
Exercise 13.3.2
What is the Sum of Squares Factor?
Answer
4,939.2
Exercise 13.3.3
What is the Sum of Squares Error?
Exercise 13.3.4
What is the \(df\) for the numerator?
Answer
2
Exercise 13.3.5
What is the \(df\) for the denominator?
Exercise 13.3.6
What is the Mean Square Factor?
Answer
2,469.6
Exercise 13.3.7
What is the Mean Square Error?
Exercise 13.3.8
What is the \(F\) statistic?
Answer
3.7416
Use the following information to answer the next eight exercises. Girls from four different soccer teams are to be tested for mean goals scored per game. The entries in the table are the goals per game for the different teams. The one-way \(ANOVA\) results are shown in Table.
Team 1 Team 2 Team 3 Team 4
1 2 0 3
2 3 1 4
0 2 1 4
3 4 0 3
2 4 0 2
Exercise 13.3.9
What is \(SS_{\text{between}}\)?
Exercise 13.3.10
What is the \(df\) for the numerator?
Answer
3
Exercise 13.3.11
What is \(MS_{\text{between}}\)?
Exercise 13.3.12
What is \(SS_{\text{within}}\)?
Answer
13.2
Exercise 13.3.13
What is the \(df\) for the denominator?
Exercise 13.3.14
What is \(MS_{\text{within}}\)?
Answer
0.825
Exercise 13.3.15
What is the \(F\) statistic?
Exercise 13.3.16
Judging by the \(F\) statistic, do you think it is likely or unlikely that you will reject the null hypothesis?
Answer
Because a one-way \(ANOVA\) test is always right-tailed, a high \(F\) statistic corresponds to a low \(p\text{-value}\), so it is likely that we will reject the null hypothesis.
DIRECTIONS
Use a solution sheet to conduct the following hypothesis tests. The solution sheet can be found in [link].
Q 13.4.1
Three students, Linda, Tuan, and Javier, are given five laboratory rats each for a nutritional experiment. Each rat's weight is recorded in grams. Linda feeds her rats Formula A, Tuan feeds his rats Formula B, and Javier feeds his rats Formula C. At the end of a specified time period, each rat is weighed again, and the net gain in grams is recorded. Using a significance level of 10%, test the hypothesis that the three formulas produce the same mean weight gain.
Weights of Student Lab Rats
Linda's rats Tuan's rats Javier's rats
43.5 47.0 51.2
39.4 40.5 40.9
41.3 38.9 37.9
46.0 46.3 45.0
38.2 44.2 48.6
1. \(H_{0}: \mu_{L} = \mu_{T} = \mu_{J}\)
2. at least any two of the means are different
3. \(df(\text{num}) = 2; df(\text{denom}) = 12\)
4. \(F\) distribution
5. 0.67
6. 0.5305
7. Check student’s solution.
8. Decision: Do not reject null hypothesis; Conclusion: There is insufficient evidence to conclude that the means are different.
Q 13.4.2
A grassroots group opposed to a proposed increase in the gas tax claimed that the increase would hurt working-class people the most, since they commute the farthest to work. Suppose that the group randomly surveyed 24 individuals and asked them their daily one-way commuting mileage. The results are in Table. Using a 5% significance level, test the hypothesis that the three mean commuting mileages are the same.
working-class professional (middle incomes) professional (wealthy)
17.8 16.5 8.5
26.7 17.4 6.3
49.4 22.0 4.6
9.4 7.4 12.6
65.4 9.4 11.0
47.1 2.1 28.6
19.5 6.4 15.4
51.2 13.9 9.3
Q 13.4.3
Examine the seven practice laps from [link]. Determine whether the mean lap time is statistically the same for the seven practice laps, or if there is at least one lap that has a different mean time from the others.
S 13.4.3
1. \(H_{0}: \mu_{1} = \mu_{2} = \mu_{3} = \mu_{4} = \mu_{5} = \mu_{6} = \mu_{T}\)
2. At least two mean lap times are different.
3. \(df(\text{num}) = 6; df(\text{denom}) = 98\)
4. \(F\) distribution
5. 1.69
6. 0.1319
7. Check student’s solution.
8. Decision: Do not reject null hypothesis; Conclusion: There is insufficient evidence to conclude that the mean lap times are different.
Use the following information to answer the next two exercises. Table lists the number of pages in four different types of magazines.
home decorating news health computer
172 87 82 104
286 94 153 136
163 123 87 98
205 106 103 207
197 101 96 146
Q 13.4.4
Using a significance level of 5%, test the hypothesis that the four magazine types have the same mean length.
Q 13.4.5
Eliminate one magazine type that you now feel has a mean length different from the others. Redo the hypothesis test, testing that the remaining three means are statistically the same. Use a new solution sheet. Based on this test, are the mean lengths for the remaining three magazines statistically the same?
S 13.4.6
1. \(H_{a}: \mu_{d} = \mu_{n} = \mu_{h}\)
2. At least any two of the magazines have different mean lengths.
3. \(df(\text{num}) = 2, df(\text{denom}) = 12\)
4. \(F\) distribtuion
5. \(F = 15.28\)
6. \(p\text{-value} = 0.001\)
7. Check student’s solution.
1. \(\alpha: 0.05\)
2. Decision: Reject the Null Hypothesis.
3. Reason for decision: \(p\text{-value} < \alpha\)
4. Conclusion: There is sufficient evidence to conclude that the mean lengths of the magazines are different.
Q 13.4.7
A researcher wants to know if the mean times (in minutes) that people watch their favorite news station are the same. Suppose that Table shows the results of a study.
CNN FOX Local
45 15 72
12 43 37
18 68 56
38 50 60
23 31 51
35 22
Assume that all distributions are normal, the four population standard deviations are approximately the same, and the data were collected independently and randomly. Use a level of significance of 0.05.
Q 13.4.8
Are the means for the final exams the same for all statistics class delivery types? Table shows the scores on final exams from several randomly selected classes that used the different delivery types.
Online Hybrid Face-to-Face
72 83 80
84 73 78
77 84 84
80 81 81
81 86
79
82
Assume that all distributions are normal, the four population standard deviations are approximately the same, and the data were collected independently and randomly. Use a level of significance of 0.05.
S 13.4.8
1. \(H_{0}: \mu_{o} = \mu_{h} = \mu_{f}\)
2. At least two of the means are different.
3. \(df(\text{n}) = 2, df(\text{d}) = 13\)
4. \(F_{2,13}\)
5. 0.64
6. 0.5437
7. Check student’s solution.
1. \(\alpha: 0.05\)
2. Decision: Do not reject the null hypothesis.
3. Reason for decision: \(p\text{-value} < \alpha\)
4. Conclusion: The mean scores of different class delivery are not different.
Q 13.4.9
Are the mean number of times a month a person eats out the same for whites, blacks, Hispanics and Asians? Suppose that Table shows the results of a study.
White Black Hispanic Asian
6 4 7 8
8 1 3 3
2 5 5 5
4 2 4 1
6 6 7
Assume that all distributions are normal, the four population standard deviations are approximately the same, and the data were collected independently and randomly. Use a level of significance of 0.05.
Q 13.4.10
Are the mean numbers of daily visitors to a ski resort the same for the three types of snow conditions? Suppose that Table shows the results of a study.
Powder Machine Made Hard Packed
1,210 2,107 2,846
1,080 1,149 1,638
1,537 862 2,019
941 1,870 1,178
1,528 2,233
1,382
Assume that all distributions are normal, the four population standard deviations are approximately the same, and the data were collected independently and randomly. Use a level of significance of 0.05.
S 13.4.11
1. \(H_{0}: \mu_{p} = \mu_{m} = \mu_{h}\)
2. At least any two of the means are different.
3. \(df(\text{n}) = 2, df(\text{d}) = 12\)
4. \(F_{2,12}\)
5. 3.13
6. 0.0807
7. Check student’s solution.
1. \(\alpha: 0.05\)
2. Decision: Do not reject the null hypothesis.
3. Reason for decision: \(p\text{-value} < \alpha\)
4. Conclusion: There is not sufficient evidence to conclude that the mean numbers of daily visitors are different.
Q 13.4.12
Sanjay made identical paper airplanes out of three different weights of paper, light, medium and heavy. He made four airplanes from each of the weights, and launched them himself across the room. Here are the distances (in meters) that his planes flew.
Paper Type/Trial Trial 1 Trial 2 Trial 3 Trial 4
Heavy 5.1 meters 3.1 meters 4.7 meters 5.3 meters
Medium 4 meters 3.5 meters 4.5 meters 6.1 meters
Light 3.1 meters 3.3 meters 2.1 meters 1.9 meters
the graph is a scatter plot which represents the data provided. The horizontal axis is labeled 'Distance in Meters,' and extends form 2 to 6. The vertical axis is labeled 'Weight of Paper' and has light, medium, and heavy categories.
Figure 13.4.1.
1. Take a look at the data in the graph. Look at the spread of data for each group (light, medium, heavy). Does it seem reasonable to assume a normal distribution with the same variance for each group? Yes or No.
2. Why is this a balanced design?
3. Calculate the sample mean and sample standard deviation for each group.
4. Does the weight of the paper have an effect on how far the plane will travel? Use a 1% level of significance. Complete the test using the method shown in the bean plant example in Example.
• variance of the group means __________
• \(MS_{\text{between}} =\) ___________
• mean of the three sample variances ___________
• \(MS_{\text{within}} =\) _____________
• \(F\) statistic = ____________
• \(df(\text{num}) =\) __________, \(df(\text{denom}) =\) ___________
• number of groups _______
• number of observations _______
• \(p\text{-value} =\) __________ (\(P(F >\) _______\() =\) __________)
• Graph the \(p\text{-value}\).
• decision: _______________________
• conclusion: _______________________________________________________________
Q 13.4.13
DDT is a pesticide that has been banned from use in the United States and most other areas of the world. It is quite effective, but persisted in the environment and over time became seen as harmful to higher-level organisms. Famously, egg shells of eagles and other raptors were believed to be thinner and prone to breakage in the nest because of ingestion of DDT in the food chain of the birds.
An experiment was conducted on the number of eggs (fecundity) laid by female fruit flies. There are three groups of flies. One group was bred to be resistant to DDT (the RS group). Another was bred to be especially susceptible to DDT (SS). Finally there was a control line of non-selected or typical fruitflies (NS). Here are the data:
RS SS NS RS SS NS
12.8 38.4 35.4 22.4 23.1 22.6
21.6 32.9 27.4 27.5 29.4 40.4
14.8 48.5 19.3 20.3 16 34.4
23.1 20.9 41.8 38.7 20.1 30.4
34.6 11.6 20.3 26.4 23.3 14.9
19.7 22.3 37.6 23.7 22.9 51.8
22.6 30.2 36.9 26.1 22.5 33.8
29.6 33.4 37.3 29.5 15.1 37.9
16.4 26.7 28.2 38.6 31 29.5
20.3 39 23.4 44.4 16.9 42.4
29.3 12.8 33.7 23.2 16.1 36.6
14.9 14.6 29.2 23.6 10.8 47.4
27.3 12.2 41.7
The values are the average number of eggs laid daily for each of 75 flies (25 in each group) over the first 14 days of their lives. Using a 1% level of significance, are the mean rates of egg selection for the three strains of fruitfly different? If so, in what way? Specifically, the researchers were interested in whether or not the selectively bred strains were different from the nonselected line, and whether the two selected lines were different from each other.
Here is a chart of the three groups:
This graph is a scatterplot which represents the data provided. The horizontal axis is labeled 'Mean eggs laid per day' and extends from 10 - 50. The vertical axis is labeled 'Fruitflies DDT resistant or susceptible, or not selected.' The vertical axis is labeled with the categories NS, RS, SS.
Figure 13.4.2.
S 13.4.13
The data appear normally distributed from the chart and of similar spread. There do not appear to be any serious outliers, so we may proceed with our ANOVA calculations, to see if we have good evidence of a difference between the three groups.
\(H_{0}: \mu_{1} = \mu_{2} = \mu_{3}\);
\(H_{a}: \mu_{i} \neq \mu_{j}\) some \(i \neq j\).
Define \(\mu_{1}, \mu_{2}, \mu_{3}\), as the population mean number of eggs laid by the three groups of fruit flies.
\(F\) statistic \(= 8.6657\);
\(p\text{-value} = 0.0004\)
This graph shows a nonsymmetrical F distribution curve. This curve does not have a peak, but slopes downward from a maximum value at (0, 1.0) and approaches the horiztonal axis at the right edge of the graph.
Figure 13.4.3.
Decision: Since the \(p\text{-value}\) is less than the level of significance of 0.01, we reject the null hypothesis.
Conclusion: We have good evidence that the average number of eggs laid during the first 14 days of life for these three strains of fruitflies are different.
Interestingly, if you perform a two sample \(t\)-test to compare the RS and NS groups they are significantly different (\(p = 0.0013\)). Similarly, SS and NS are significantly different (\(p = 0.0006\)). However, the two selected groups, RS and SS are not significantly different (\(p = 0.5176\)). Thus we appear to have good evidence that selection either for resistance or for susceptibility involves a reduced rate of egg production (for these specific strains) as compared to flies that were not selected for resistance or susceptibility to DDT. Here, genetic selection has apparently involved a loss of fecundity.
Q 13.4.14
The data shown is the recorded body temperatures of 130 subjects as estimated from available histograms.
Traditionally we are taught that the normal human body temperature is 98.6 F. This is not quite correct for everyone. Are the mean temperatures among the four groups different?
Calculate 95% confidence intervals for the mean body temperature in each group and comment about the confidence intervals.
FL FH ML MH FL FH ML MH
96.4 96.8 96.3 96.9 98.4 98.6 98.1 98.6
96.7 97.7 96.7 97 98.7 98.6 98.1 98.6
97.2 97.8 97.1 97.1 98.7 98.6 98.2 98.7
97.2 97.9 97.2 97.1 98.7 98.7 98.2 98.8
97.4 98 97.3 97.4 98.7 98.7 98.2 98.8
97.6 98 97.4 97.5 98.8 98.8 98.2 98.8
97.7 98 97.4 97.6 98.8 98.8 98.3 98.9
97.8 98 97.4 97.7 98.8 98.8 98.4 99
97.8 98.1 97.5 97.8 98.8 98.9 98.4 99
97.9 98.3 97.6 97.9 99.2 99 98.5 99
97.9 98.3 97.6 98 99.3 99 98.5 99.2
98 98.3 97.8 98 99.1 98.6 99.5
98.2 98.4 97.8 98 99.1 98.6
98.2 98.4 97.8 98.3 99.2 98.7
98.2 98.4 97.9 98.4 99.4 99.1
98.2 98.4 98 98.4 99.9 99.3
98.2 98.5 98 98.6 100 99.4
98.2 98.6 98 98.6 100.8
13.5: Test of Two Variances
Use the following information to answer the next two exercises. There are two assumptions that must be true in order to perform an \(F\) test of two variances.
Exercise 13.5.2
Name one assumption that must be true.
Answer
The populations from which the two samples are drawn are normally distributed.
Exercise 13.5.3
What is the other assumption that must be true?
Use the following information to answer the next five exercises. Two coworkers commute from the same building. They are interested in whether or not there is any variation in the time it takes them to drive to work. They each record their times for 20 commutes. The first worker’s times have a variance of 12.1. The second worker’s times have a variance of 16.9. The first worker thinks that he is more consistent with his commute times and that his commute time is shorter. Test the claim at the 10% level.
Exercise 13.5.4
State the null and alternative hypotheses.
Answer
\(H_{0}: \sigma_{1} = \sigma_{2}\)
\(H_{a}: \sigma_{1} < \sigma_{2}\)
or
\(H_{0}: \sigma^{2}_{1} = \sigma^{2}_{2}\)
\(H_{a}: \sigma^{2}_{1} < \sigma^{2}_{2}\)
Exercise 13.5.5
What is \(s_{1}\) in this problem?
Exercise 13.5.6
What is \(s_{2}\) in this problem?
Answer
4.11
Exercise 13.5.7
What is \(n\)?
Exercise 13.5.8
What is the \(F\) statistic?
Answer
0.7159
Exercise 13.5.9
What is the \(p\text{-value}\)?
Exercise 13.5.10
Is the claim accurate?
Answer
No, at the 10% level of significance, we do not reject the null hypothesis and state that the data do not show that the variation in drive times for the first worker is less than the variation in drive times for the second worker.
Use the following information to answer the next four exercises. Two students are interested in whether or not there is variation in their test scores for math class. There are 15 total math tests they have taken so far. The first student’s grades have a standard deviation of 38.1. The second student’s grades have a standard deviation of 22.5. The second student thinks his scores are lower.
Exercise 13.5.11
State the null and alternative hypotheses.
Exercise 13.5.12
What is the \(F\) Statistic?
Answer
2.8674
Exercise 13.5.13
What is the \(p\text{-value}\)?
Exercise 13.5.14
At the 5% significance level, do we reject the null hypothesis?
Answer
Reject the null hypothesis. There is enough evidence to say that the variance of the grades for the first student is higher than the variance in the grades for the second student.
Use the following information to answer the next three exercises. Two cyclists are comparing the variances of their overall paces going uphill. Each cyclist records his or her speeds going up 35 hills. The first cyclist has a variance of 23.8 and the second cyclist has a variance of 32.1. The cyclists want to see if their variances are the same or different.
Exercise 13.5.15
State the null and alternative hypotheses.
Exercise 13.5.16
What is the \(F\) Statistic?
Answer
0.7414
Exercise 13.5.17
At the 5% significance level, what can we say about the cyclists’ variances?
Q 13.5.1
Three students, Linda, Tuan, and Javier, are given five laboratory rats each for a nutritional experiment. Each rat’s weight is recorded in grams. Linda feeds her rats Formula A, Tuan feeds his rats Formula B, and Javier feeds his rats Formula C. At the end of a specified time period, each rat is weighed again and the net gain in grams is recorded.
Linda's rats Tuan's rats Javier's rats
43.5 47.0 51.2
39.4 40.5 40.9
41.3 38.9 37.9
46.0 46.3 45.0
38.2 44.2 48.6
Determine whether or not the variance in weight gain is statistically the same among Javier’s and Linda’s rats. Test at a significance level of 10%.
S 13.5.1
1. \(H_{0}: \sigma^{2}_{1} = \sigma^{2}_{2}\)
2. \(H_{a}: \sigma^{2}_{1} \neq \sigma^{2}_{1}\)
3. \(df(\text{num}) = 4; df(\text{denom}) = 4\)
4. \(F_{4, 4}\)
5. 3.00
6. \(2(0.1563) = 0.3126\). Using the TI-83+/84+ function 2-SampFtest, you get the test statistic as 2.9986 and p-value directly as 0.3127. If you input the lists in a different order, you get a test statistic of 0.3335 but the \(p\text{-value}\) is the same because this is a two-tailed test.
7. Check student't solution.
8. Decision: Do not reject the null hypothesis; Conclusion: There is insufficient evidence to conclude that the variances are different.
Q 13.5.2
A grassroots group opposed to a proposed increase in the gas tax claimed that the increase would hurt working-class people the most, since they commute the farthest to work. Suppose that the group randomly surveyed 24 individuals and asked them their daily one-way commuting mileage. The results are as follows.
working-class professional (middle incomes) professional (wealthy)
17.8 16.5 8.5
26.7 17.4 6.3
49.4 22.0 4.6
9.4 7.4 12.6
65.4 9.4 11.0
47.1 2.1 28.6
19.5 6.4 15.4
51.2 13.9 9.3
Determine whether or not the variance in mileage driven is statistically the same among the working class and professional (middle income) groups. Use a 5% significance level.
Q 13.5.3
Refer to the data from [link].
Examine practice laps 3 and 4. Determine whether or not the variance in lap time is statistically the same for those practice laps.
Use the following information to answer the next two exercises. The following table lists the number of pages in four different types of magazines.
home decorating news health computer
172 87 82 104
286 94 153 136
163 123 87 98
205 106 103 207
197 101 96 146
S 13.5.3
1. \(H_{0}: \sigma^{2}_{1} = \sigma^{2}_{2}\)
2. \(H_{a}: \sigma^{2}_{1} \neq \sigma^{2}_{1}\)
3. \(df(\text{n}) = 19, df(\text{d}) = 19\)
4. \(F_{19,19}\)
5. 1.13
6. 0.786
7. Check student’s solution.
1. \(\alpha: 0.05\)
2. Decision: Do not reject the null hypothesis.
3. Reason for decision: \(p\text{-value} > \alpha\)
4. Conclusion: There is not sufficient evidence to conclude that the variances are different.
Q 13.5.4
Which two magazine types do you think have the same variance in length?
Q 13.5.5
Which two magazine types do you think have different variances in length?
S 13.5.5
The answers may vary. Sample answer: Home decorating magazines and news magazines have different variances.
Q 13.5.6
Is the variance for the amount of money, in dollars, that shoppers spend on Saturdays at the mall the same as the variance for the amount of money that shoppers spend on Sundays at the mall? Suppose that the Table shows the results of a study.
Saturday Sunday Saturday Sunday
75 44 62 137
18 58 0 82
150 61 124 39
94 19 50 127
62 99 31 141
73 60 118 73
89
Q 13.5.7
Are the variances for incomes on the East Coast and the West Coast the same? Suppose that Table shows the results of a study. Income is shown in thousands of dollars. Assume that both distributions are normal. Use a level of significance of 0.05.
East West
38 71
47 126
30 42
82 51
75 44
52 90
115 88
67
S 13.5.7
1. \(H_{0}: \sigma^{2}_{1} = \sigma^{2}_{2}\)
2. \(H_{a}: \sigma^{2}_{1} \neq \sigma^{2}_{1}\)
3. \(df(\text{n}) = 7, df(\text{d}) = 6\)
4. \(F_{7,6}\)
5. 0.8117
6. 0.7825
7. Check student’s solution.
1. \(\alpha: 0.05\)
2. Decision: Do not reject the null hypothesis.
3. Reason for decision: \(p\text{-value} > \alpha\)
4. Conclusion: There is not sufficient evidence to conclude that the variances are different.
Q 13.5.8
Thirty men in college were taught a method of finger tapping. They were randomly assigned to three groups of ten, with each receiving one of three doses of caffeine: 0 mg, 100 mg, 200 mg. This is approximately the amount in no, one, or two cups of coffee. Two hours after ingesting the caffeine, the men had the rate of finger tapping per minute recorded. The experiment was double blind, so neither the recorders nor the students knew which group they were in. Does caffeine affect the rate of tapping, and if so how?
Here are the data:
0 mg 100 mg 200 mg 0 mg 100 mg 200 mg
242 248 246 245 246 248
244 245 250 248 247 252
247 248 248 248 250 250
242 247 246 244 246 248
246 243 245 242 244 250
Q 13.5.9
King Manuel I, Komnenus ruled the Byzantine Empire from Constantinople (Istanbul) during the years 1145 to 1180 A.D. The empire was very powerful during his reign, but declined significantly afterwards. Coins minted during his era were found in Cyprus, an island in the eastern Mediterranean Sea. Nine coins were from his first coinage, seven from the second, four from the third, and seven from a fourth. These spanned most of his reign. We have data on the silver content of the coins:
First Coinage Second Coinage Third Coinage Fourth Coinage
5.9 6.9 4.9 5.3
6.8 9.0 5.5 5.6
6.4 6.6 4.6 5.5
7.0 8.1 4.5 5.1
6.6 9.3 6.2
7.7 9.2 5.8
7.2 8.6 5.8
6.9
6.2
Did the silver content of the coins change over the course of Manuel’s reign?
Here are the means and variances of each coinage. The data are unbalanced.
First Second Third Fourth
Mean 6.7444 8.2429 4.875 5.6143
Variance 0.2953 1.2095 0.2025 0.1314
S 13.5.9
Here is a strip chart of the silver content of the coins:
This graph is a scatterplot which represents the data provided. The horizontal axis is labeled 'Silver content coins' and extends from 5 - 9. The vertical axis is labeled 'Coinage.' The vertical axis is labeled with the categories First, Second, Third, and Fourth.
Figure 13.5.1.
While there are differences in spread, it is not unreasonable to use \(ANOVA\) techniques. Here is the completed \(ANOVA\) table:
Source of Variation Sum of Squares (\(SS\)) Degrees of Freedom (\(df\)) Mean Square (\(MS\)) \(F\)
Factor (Between) 37.748 \(4 - 1 = 3\) 12.5825 26.272
Error (Within) 11.015 \(27 - 4 = 23\) 0.4789
Total 48.763 \(27 - 1 = 26\)
\(P(F > 26.272) = 0\);
Reject the null hypothesis for any alpha. There is sufficient evidence to conclude that the mean silver content among the four coinages are different. From the strip chart, it appears that the first and second coinages had higher silver contents than the third and fourth.
Q 13.5.10
The American League and the National League of Major League Baseball are each divided into three divisions: East, Central, and West. Many years, fans talk about some divisions being stronger (having better teams) than other divisions. This may have consequences for the postseason. For instance, in 2012 Tampa Bay won 90 games and did not play in the postseason, while Detroit won only 88 and did play in the postseason. This may have been an oddity, but is there good evidence that in the 2012 season, the American League divisions were significantly different in overall records? Use the following data to test whether the mean number of wins per team in the three American League divisions were the same or not. Note that the data are not balanced, as two divisions had five teams, while one had only four.
Division Team Wins
East NY Yankees 95
East Baltimore 93
East Tampa Bay 90
East Toronto 73
East Boston 69
Division Team Wins
Central Detroit 88
Central Chicago Sox 85
Central Kansas City 72
Central Cleveland 68
Central Minnesota 66
Division Team Wins
West Oakland 94
West Texas 93
West LA Angels 89
West Seattle 75
S 13.5.10
Here is a stripchart of the number of wins for the 14 teams in the AL for the 2012 season.
This graph is a scatterplot which represents the data provided. The horizontal axis is labeled 'Number of wins in 2012 Major League Baseball Season' and extends from 65 - 95. The vertical axis is labeled 'American league division.' The vertical axis is labeled with the categories Central, East, West.
Figure 13.5.2.
While the spread seems similar, there may be some question about the normality of the data, given the wide gaps in the middle near the 0.500 mark of 82 games (teams play 162 games each season in MLB). However, one-way \(ANOVA\) is robust.
Here is the \(ANOVA\) table for the data:
Source of Variation Sum of Squares (\(SS\)) Degrees of Freedom (\(df\)) Mean Square (\(MS\)) \(F\)
Factor (Between) 344.16 3 – 1 = 2 172.08 26.272
Error (Within) 1,219.55 14 – 3 = 11 110.87 1.5521
Total 1,563.71 14 – 1 = 13
\(P(F > 1.5521) = 0.2548\)
Since the \(p\text{-value}\) is so large, there is not good evidence against the null hypothesis of equal means. We decline to reject the null hypothesis. Thus, for 2012, there is not any have any good evidence of a significant difference in mean number of wins between the divisions of the American League.
13.6: Lab: One-Way ANOVA
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DPS Sigils and You!
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Abyss
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PostSubject: DPS Sigils and You! Thu Sep 19, 2013 11:25 pm
Introduction:
So as not to be outdone by Skittles, I’ve decided to write up a guide concerning a question that was posed a few nights ago in guild chat. The question regarded which “DPS” sigil (Fire, Strength, Force, Night) was most optimal to run for a particular build, and the answers that were given were quite varied. Since there was some opinion as to which was best I decided to do some math on the matter.
The most important thing to note about this is different builds will yield different results with the exception of Night being superior to Force when conditions for the Night sigil are met. This means you’ll have to do math if you really care to optimize sigils for any particular build. There are a lot of variables associated this, but the math is very straightforward and I’ve done all the work is determining which equations to use for what.
DPS Sigil Spreadsheet - You still need to calculate Skill Specific Coefficient yourself. See that section for details.
Things you need to know about your build in order for this to work:
1.) Power – The number next to power in the hero panel. This is not the same as Attack.
2.) Precision – The number next to Precision in the hero panel.
3.) Critical Damage – The number next to Critical Damage in the hero panel. This does not include the base 50%, but the formula in this guide does. Just plug in the hero panel’s number.
4.) Base Damage Modifiers – Any traits that give +X% damage. (Not Critical damage) Do not include modifiers from Force or Night sigils in this number.
5.) Weapon Strength – This is the average damage of your weapon(s). For example, on my axe the range is 900 to 1100, so the average is 1000. Weapon strength is also equal to the difference between the values of Attack and Power in the hero panel.
6.) Skill Rotation Time and strikes per rotation – Time will be in seconds. For example, one rotation of Axe auto attack on Warrior is 3.6 seconds with 6 strikes. If you auto attack a lot you can use that, but if you hardly auto attack you’ll want to use your other skills.
7.) Skill Specific Coefficient – This one is a little trickier. I’ll have a section on how to calculate this.
8.) Bonuses – This includes any bonuses to Power, Precision, Crit Chance, Critical Damage, potential Might stacks (not including Sigil of Superior Strength), Fury, etc. that are not included in the numbers on the hero panel. For example if there aren’t any banners down you won’t get a boost for it even if it is on your bar. Heightened Focus, a warrior trait in the Discipline line provides a 15% boost to Crit Chance that will not show up on the hero panel. The best way to be sure if you have a trait like this is to swap out the trait and see if a number changes. If it does not change you’ll have to add it manually.
Some other useful information to you:
1.) Might – Each stack of might provides 35 Power.
2.) Superior Sigil of Fire – The Skill Specific Coefficient for this skill is 1.0. 30% chance to trigger on crit, 5 second cool down.
3.) Sigil of Superior Strength – 30% chance on crit to gain might for 10 seconds. It has a 1 second cool down. Because it has a cool down of 1 second and the might lasts for 10 seconds, the maximum might stacks you can get from this is 5. This equates to a maximum of 175 power.
Formulas and other mumbo-jumbo:
This section is going to detail how to calculate the effectiveness of these sigils using my Warrior build as an example. It’s helpful to write out your variables beforehand, so for my Warrior’s build the variables are:
- Power: 2879 (Base + 3xMight from FGJ + Banner of Str + Empower Allies)
- Crit Damage: 143% = 1.43 (Base + Sweet and Spicy Butternut Squash Soup)
- Precision: 2250 (Base + 25 Perception stacks + Banner of Disc + Signet of Fury)
- Base Damage Modifiers: 30% = .3 (Strength traits V and XI, Scholar Runes)
- Additional Crit Chance: 35% = .35 (Fury, Heightened Focus)
- Weapon Strength: 1080 (Ascended Axe, Ascended Mace)
- Skill Specific Coefficient: 1.2 (I’ll show calculations for this in a later section)
- Rotation (Strikes/seconds): 6 Strikes, 3.6 Seconds (Axe Auto Attack chain)
Crit Chance:
Spoiler:
Skill Specific Coefficients:
Spoiler:
PROC of Sigil of Fire, Sigil of Strength:
Spoiler:
Damage of Superior Sigil of Fire:
Spoiler:
The Big Equation:
Spoiler:
If you’re still reading this it means you either like math or you’re good at it. If that’s the case I’m going to assume you can finish this on your own, and I’m just going to make a spreadsheet for everyone else. I’m tired of typing out equations…
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PostSubject: Re: DPS Sigils and You! Fri Sep 20, 2013 10:40 am
The warrior example in "Damage of Superior Sigil of Fire:" is the same as the formula.
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PostSubject: Re: DPS Sigils and You! Fri Sep 20, 2013 5:02 pm
Byssy makin' us all look dumb. Mad
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PostSubject: Re: DPS Sigils and You! Fri Sep 20, 2013 5:12 pm
rhokk wrote:
The warrior example in "Damage of Superior Sigil of Fire:" is the same as the formula.
I knew I was going to mess up with my uploading somewhere. Q.Q
It's fixed now. Also, if anyone knows how to limit a value in excel to be less than or equal to 1, I will love you long time.
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PostSubject: Re: DPS Sigils and You! Mon Sep 23, 2013 4:20 pm
Pretty sure you go to the solver parameters and then it's
$B$2:$B$150 < = 1
(I used random numbers for the cell numbers, the cells being the data that you are looking to create a formula for.)
Unless I'm getting confused, which is very likely since it's been a while since I've done anything serious with excel.
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PostSubject: Re: DPS Sigils and You! Mon Sep 23, 2013 5:17 pm
Using the IF function, you can accomplish this:
=IF(conditionalExpression, trueAction, falseAction)
So you could say:
=IF(A1 <= 1, A1, 1)
Where "A1 <= 1" is the condition you are trying to measure, "A1" by itself is what you want to display when this is true, and "1" is what you want to display when this is false.
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PostSubject: Re: DPS Sigils and You! Mon Sep 23, 2013 6:24 pm
Yep, you lost me. Been too long. Ima go back to bein a stupid good for nothin drunk and punch things ;3
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PostSubject: Re: DPS Sigils and You! Fri Sep 27, 2013 4:31 am
There probably isn't a high demand for it, but I finished the spreadsheet.
On a side note, it turns out the fire sigil PROC section is erroneous. The only time the procs per second required to trigger the sigil will be higher than 0.3 is if you're attacking less than 1 time a second. I'll rework that tomorrow.
Google Docs DPS Sigil Spreadsheet
You'll still need to calculate the average skill coefficient of your chain for this, because I don't even want to try to type that in excel. The instructions for that are still in the first post in the Skill Specific Coefficients section. It's just an average with a funny name; you can do it. Aside from that, save the file and enter your own values into the C column, then watch it do work.
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402,657,604,662,691,300 | Four Colors Theorem
Four color theorem
The four color theorem states the possibility of coloring (we also say coloring in graph theory) with four colors only a geographical map without two neighboring countries having the same color.
four color theorem
To be more precise, the Four Colors Theorem states that by using only four different colors, it is possible to color any map cut into related regions (in one piece), so that two adjacent regions (or bordering), that is to say having a whole border (and not just a point) in common always receive two distinct colors.
Thus, each of the regions should be given a different color if the regions are two by two adjacent.
Moreover, there cannot exist five adjacent two by two connected regions (this is the easy part of Kuratowski's theorem).
This four-color theorem has practical application in the assignment by a mobile operator of GSM frequencies to the coverage areas of base stations in its network. Indeed, as in the situation of the four colors:
- a GSM network is modeled, like a geographical map, by contiguous hexagons: each hexagon (called “cell”, hence the notion of cellular network) corresponds to the radiation of a base station (approximately 30 km in diameter in rural area).
- two contiguous hexagons must in no case be assigned the same frequency band.
History of the four color theorem
The first four-color conjecture seems to have been made in 1852 by a South African mathematician and botanist, Francis Guthrie (1831 - 1899).
Francis Guthrie was a student at University College London, where he studied with the famous De Morgan.
He then turned to studies of botany, while his brother, Frederick Guthrie, became a mathematician and also followed De Morgan's courses.
Francis Guthrie notices that four colors are enough for him to color the (however complex) map of the cantons of England, without giving the same color to two adjacent cantons (having a common border).
He then asks his brother Frederick, if this property would not be true in general for any flat map; the latter communicated the conjecture to De Morgan, and in 1878 Cayley published it.
four color theorem
• As early as 1879, Kempe found a first "proof" of the conjecture, but eleven years later Heawood found a major flaw in it; however, he will manage to find a five-color theorem.
A second "proof" by Tait in 1880 will likewise be refuted by Petersen in 1891.
• In 1913, the American mathematician George David BIRKHOFF (1884 - 1944), demonstrated the conjecture for all maps with less than 26 regions to be colored.
This terminal is improved during the XXe century.
• In 1969 Heesch found "almost" necessary and sufficient conditions for a configuration to be reducible, and a general method for finding an inevitable set of configurations.
The first computer-generated proof.
• Finally, in 1976, the American mathematicians Kenneth Appel (died April 19, 2013 at age 80) and German mathematician Wolfgang Haken carry out Heesch's program.
They show, using tens of thousands of figures, that any non-4-colorable map must contain one of the 1478 configurations, and, with 1200 hours of computation, that these configurations are reducible.
It's done the first computerized proof of a mathematical theorem.
• In 1995, Robertson, Sanders, Seymour and Thomas took advantage of the formidable acceleration of computers to find a much simpler realization of Heesch's program, with only 633 configurations; moreover they also automate the proof of inevitability.
• In September 2012, six years after the computer demonstration of the four color theorem, Georges Gonthier and his team succeeded in demonstrating Feit's theorem and Thompson. A theorem even more difficult to formalize.
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-2,100,255,236,764,816,100 | Mensuration (IMO- Mathematics Olympiad (SOF) Class 8): Questions 1 - 6 of 61
Get 1 year subscription: Access detailed explanations (illustrated with images and videos) to 1029 questions. Access all new questions we will add tracking exam-pattern and syllabus changes. View Sample Explanation or View Features.
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Question number: 1
» Mensuration
MCQ▾
Question
Find the surface area of a cylinder with diameter of base 5 cm and height 30 cm.
Choices
Choice (4) Response
a.
505
b.
510
c.
511
d.
610
Question number: 2
» Mensuration
MCQ▾
Question
Two rectangles ABCD and DBEF are as shown in the figure. The area of rectangle DBEF in square units is________.
Two rectangles ABCD and DBEF with AB = 8 and AD = 15 are given
Two Rectangles ABCD and DBEF With AB = 8 and AD = 15 Are Given
Find out the area of DBEF using given measure (AB = 8 and AD = 15)
Choices
Choice (4) Response
a.
120
b.
90
c.
160
d.
140
Question number: 3
» Mensuration
MCQ▾
Question
Two cubes have their volumes in the ratio 1: 8. Find the ratio of their surface areas.
Choices
Choice (4) Response
a.
1: 9
b.
5: 3
c.
1: 4
d.
7: 9
Question number: 4
» Mensuration
MCQ▾
Question
A child walked 17 m to cross a rectangular field diagonally. If breadth of field is 8 m, its length is________.
Choices
Choice (4) Response
a.
12 m
b.
15 m
c.
20 m
d.
11 m
Question number: 5
» Mensuration
MCQ▾
Question
Three cubes whose edges are 3 cm, 4 cm, and 5 cm respectively are melted without any loss of metal into a single cube. The edge of new cube is________.
Choices
Choice (4) Response
a.
b.
c.
d.
Question number: 6
» Mensuration
MCQ▾
Question
If a wire is bent into the shape of a square, then the area of the square is 49 sq. cm. When the wire is bent into a semicircular shape, then the area of the semi-circle will be________.
Choices
Choice (4) Response
a.
b.
c.
d.
f Page
Sign In | {
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-8,311,169,627,122,805,000 | Take the tour ×
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I was reading "Graph Theory" by Diestel and I tried to solve a problem from the chapter 3 (on connectivity). It seems at first sight easy (and really intuitive) but I have to admit that I can't work it out! Here is the problem: prove that every $k$-connected graph of order at least $2k$ contains a cycle of length at least $2k$.
Does anyone have a suggestion?
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First suppose for contradiction that the length of the longest cycle $C$ is $2k-1$. Then consider the set of vertices in $C$, which are connected with vertices outside $C$. It has cardinality at least $k$. Otherwise, by deleting these vertices, the graph left is still connected.
Then by pigeonhole principle there are two adjacent verticies $A$ and $A'$ in $C$, which are connected with vertices outside $C$, say, $B$ and $B'$.
The situation when $B=B'$ automatically makes a longer cycle. When $B$ and $B'$ are different, we delete $C$, and get a connected graph. We connect $B$ and $B'$, using vertices all outside $C$. This makes the cycle longer.
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Why does deleting $C$ leave a connected graph? – Chris Eagle Jan 8 '11 at 3:35
Sorry I was not clear. Suppose $C^c$ is not connected. Specificly, $B$ and $B'$ can't be connected by any path within $C^c$. In another word, every path connecting $B$ and $B'$ must go through $C$. By Menger's Theorem, we have that there are at least $k$ disjoint paths connecting $B$ and $B'$. So we have two disjoint paths connecting $B$ to the cycle C, which make our cycle longer. – Xiaochuan Jan 8 '11 at 7:45
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Consider the longest cycle $C$ in $G$ and suppose its length is less than $2k$. Use Menger's Theorem and the pigeonhole principle to come up with a contradiction.
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Seems like another application of Dirac's Theorem.
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-3,520,295,884,296,221,700 | QED c. Is it bijective? How many such functions are there? Pages 347; Ratings 100% (1) 1 out of 1 people found this document helpful. Prove that the function \(f : \mathbb{N} \rightarrow \mathbb{Z}\) defined as \(f (n) = \frac{(-1)^{n}(2n-1)+1}{4}\) is bijective. A function \(f : \mathbb{Z} \times \mathbb{Z} \rightarrow \mathbb{Z}\) is defined as \(f(m,n) = 3n-4m\). 5x 1 - 2 = 5x 2 - 2. Definition 2.7.1. Next, subtract \(n = l\) from \(m+n = k+l\) to get \(m = k\). Verify whether this function is injective and whether it is surjective. What if it had been defined as \(cos : \mathbb{R} \rightarrow [-1, 1]\)? Often it is necessary to prove that a particular function \(f : A \rightarrow B\) is injective. Unless otherwise noted, LibreTexts content is licensed by CC BY-NC-SA 3.0. Here is an outline: How to show a function \(f : A \rightarrow B\) is surjective: [Prove there exists \(a \in A\) for which \(f(a) = b\).]. A function is surjective (a surjection or onto) if every element of the codomain is the output of at least one element of the domain. This is not injective since f(1) = f(2). Suppose \((m,n), (k,l) \in \mathbb{Z} \times \mathbb{Z}\) and \(g(m,n)= g(k,l)\). Solving for a gives \(a = \frac{1}{b-1}\), which is defined because \(b \ne 1\). Prove that the function \(f : \mathbb{R}-\{2\} \rightarrow \mathbb{R}-\{5\}\) defined by \(f(x)= \frac{5x+1}{x-2}\) is bijective. f is surjective or onto if, and only if, y Y, x X such that f(x) = y. (Scrap work: look at the equation .Try to express in terms of .). Surjective Continuos Function onto Manifolds I can not think of a counter example to "For every connected manifold, M, of dimension n, there is a continuous surjection from R n to M." On the other hand, they are really struggling with injective functions. On the other hand, \(g(x) = x^3\) is both injective and surjective, so it is also bijective. Cookies help us deliver our Services. Let \(A= \{1,2,3,4\}\) and \(B = \{a,b,c\}\). Learn vocabulary, terms, and more with flashcards, games, and other study tools. We can express that f is one-to-one using quantifiers as or equivalently , where the universe of discourse is the domain of the function.. To prove that a function is surjective, we proceed as follows: . We now have \(g(2b-c, c-b) = (b, c)\), and it follows that g is surjective. Consider function \(h : \mathbb{Z} \times \mathbb{Z} \rightarrow \mathbb{Q}\) defined as \(h(m,n)= \frac{m}{|n|+1}\). A one-one function is also called an Injective function. An important example of bijection is the identity function. How many are bijective? The LibreTexts libraries are Powered by MindTouch® and are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. Difficult to hint, without just telling you an example. Bijective? You may recall from algebra and calculus that a function may be one-to-one and onto, and these properties are related to whether or not the function is invertible. Notice we may assume d is positive by making c negative, if necessary. (iv) This function is not surjective, it tends to +∞ for large positive , and also tends to +∞ for large negative . Suppose \(a, a′ \in \mathbb{R}-\{0\}\) and \(f (a) = f (a′)\). A function \(f : \mathbb{Z} \times \mathbb{Z} \rightarrow \mathbb{Z} \times \mathbb{Z}\) is defined as \(f(m,n) = (m+n,2m+n)\). This is illustrated below for four functions \(A \rightarrow B\). We need to use PIE but with more than 3 sets the formula for PIE is very long. This is just like the previous example, except that the codomain has been changed. Suppose f: X → Y is a function. But we want surjective functions. Since All surjective functions will also be injective. This function is not injective because of the unequal elements \((1,2)\) and \((1,-2)\) in \(\mathbb{Z} \times \mathbb{Z}\) for which \(h(1, 2) = h(1, -2) = 3\). Thus we need to show that \(g(m, n) = g(k, l)\) implies \((m, n) = (k, l)\). Is it surjective? This preview shows page 122 - 124 out of 347 pages. ), so there are 8 2 = 6 surjective functions. We seek an \(a \in \mathbb{R}-\{0\}\) for which \(f(a) = b\), that is, for which \(\frac{1}{a}+1 = b\). A function \(f : A \to B\) is said to be bijective (or one-to-one and onto) if it is both injective and surjective. (How to find such an example depends on how f is defined. By way of contradiction suppose g is not surjective. For this it suffices to find example of two elements \(a, a′ \in A\) for which \(a \ne a′\) and \(f(a)=f(a′)\). How many are surjective? This is because the contrapositive approach starts with the equation \(f(a) = f(a′)\) and proceeds to the equation \(a = a'\). Onto Function (surjective): If every element b in B has a corresponding element a in A such that f(a) = b. Homework Help. De nition 67. In the more general case of {1..n}->{1..k} with n>=k, your approach is not quite right, but it's fixable. : The intersection of injective functions (I) and surjective (S) = |I| + |S| - |IUS|. Bijective? This is illustrated below for four functions \(A \rightarrow B\). According to Definition12.4,we must prove the statement \(\forall b \in B, \exists a \in A, f(a)=b\). Fix any . Give an example of a function \(f : A \rightarrow B\) that is neither injective nor surjective. For every element b in the codomain B, there is at most one element a in the domain A such that f(a)=b, or equivalently, distinct elements in the domain map to distinct elements in the codomain.. (a) The composition of two injective functions is injective. I don't know how to do this if the function is not also one to one, which it is not. Since g : B → C is onto Suppose z ∈ C, then there exists a pre-image in B Let the pre-image be y Hence, y ∈ B such that g (y) = z Similarly, since f : A → B is onto If y ∈ B, then there exists a pre-i (hence bijective). Consider the function \(\theta : \{0, 1\} \times \mathbb{N} \rightarrow \mathbb{Z}\) defined as \(\theta(a, b) = (-1)^{a}b\). Therefore f is injective. A non-surjective function from domain X to codomain Y. Millions of years ago, people started noticing that some quantities in nature depend on the others. However, I thought, once you understand functions, the concept of injective and surjective functions are easy. Then \(h(c, d-1) = \frac{c}{|d-1|+1} = \frac{c}{d} = b\). For any number in N we can write it as a finite sum of numbers 0-9, so the map is surjective. Now we can finally count the number of surjective functions: 3 5-3 1 2 5-3 2 1 5 150. How many are surjective? F: PROOF OF THE FIRST ISOMORPHISM THEOREM. We need to show that there is some \((x, y) \in \mathbb{Z} \times \mathbb{Z}\) for which \(g(x, y) = (b, c)\). The range of 10 x is (0,+∞), that is, the set of positive numbers. Equivalently, a function is surjective if its image is equal to its codomain. Functions \One of the most important concepts in all of mathematics is that of function." Functions in the first row are surjective, those in the second row are not. Is \(\theta\) injective? Functions can be injections (one-to-one functions), surjections (onto functions) or bijections (both one-to-one and onto). How many such functions are there? Accelerated Geometry NOTES 5.1 Injective, Surjective, & Bijective Functions Functions A function relates each element of a set with exactly one element of another set. provide a counter-example) We illustrate with some examples. For more information contact us at [email protected] or check out our status page at https://status.libretexts.org. Let f: A → B. This means \(\frac{1}{a} +1 = \frac{1}{a'} +1\). Does anyone know to write "The function f: A->B is not surjective? Is it surjective? To prove a function is one-to-one, the method of direct proof is generally used. \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\), [ "article:topic", "showtoc:no", "authorname:rhammack", "license:ccbynd" ], https://math.libretexts.org/@app/auth/2/login?returnto=https%3A%2F%2Fmath.libretexts.org%2FBookshelves%2FMathematical_Logic_and_Proof%2FBook%253A_Book_of_Proof_(Hammack)%2F12%253A_Functions%2F12.02%253A_Injective_and_Surjective_Functions, \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\). 2.7. Watch the recordings here on Youtube! ? To show f is not surjective, we must prove the negation of \(\forall b \in B, \exists a \in A, f (a) = b\), that is, we must prove \(\exists b \in B, \forall a \in A, f (a) \ne b\). The height of a stack can be seen as the value of a counter. 2 for any b 2N we can take a = b+1 2N and f(a) = f(b+1) = b. ... Stack Exchange Network Stack Exchange network consists of 176 Q&A communities including Stack Overflow , the largest, most trusted online community for developers to … f(x) = 5x - 2 for all x R. Prove that f is one-to-one.. Note that a counter automaton can only test whether a counter is zero or not. The previous example shows f is injective. Thus to show a function is not surjective it is enough to nd an element in the codomain that is not the image of any element of the domain. The function f is not surjective because there exists an element \(b = 1 \in \mathbb{R}\), for which \(f(x) = \frac{1}{x}+1 \ne 1\) for every \(x \in \mathbb{R}-\{0\}\). This question concerns functions \(f : \{A,B,C,D,E\} \rightarrow \{1,2,3,4,5,6,7\}\). The Attempt at a Solution If I have two finite sets, and a function between them. An important special case is the kernel of a linear map.The kernel of a matrix, also called the null space, is the kernel of the linear map defined by the matrix. Uploaded By emilyhui23. 9. How many of these functions are injective? It is surjective since 1. 2599 / ∈ Z. Press question mark to learn the rest of the keyboard shortcuts. How many surjective functions from A to B are there? New comments cannot be posted and votes cannot be cast, More posts from the cheatatmathhomework community, Continue browsing in r/cheatatmathhomework, Press J to jump to the feed. When we speak of a function being surjective, we always have in mind a particular codomain. Also, is f injective? This question concerns functions \(f : \{A,B,C,D,E,F,G\} \rightarrow \{1,2,3,4,5,6,7\}\). Explain. (This is not the same as the restriction of a function which restricts the domain!) How to show a function \(f : A \rightarrow B\) is injective: \(\begin{array}{cc} {\textbf{Direct approach}}&{\textbf{Contrapositive approach}}\\ {\text{Suppose} a,a' \in A \text{and} a \ne a'}&{\text{Suppose} a,a' \in A \text{and} f(a) = f(a')}\\ {\cdots}&{\cdots}\\ {\text{Therefore} f(a) \ne f(a')}&{\text{Therefore} a=a'}\\ \nonumber \end{array}\). ... We define a k-counter automaton to be a k-stack PDA where all stack alphabets are unary. Next we examine how to prove that \(f : A \rightarrow B\) is surjective. However, we have lucked out. I can see from the graph of the function that f is surjective since each element of its range is covered. While counter automata do not seem to be that powerful, we have the following surprising result. In essence, injective means that unequal elements in A always get sent to unequal elements in B. Surjective means that every element of B has an arrow pointing to it, that is, it equals f(a) for some a in the domain of f. For more concrete examples, consider the following functions \(f , g : \mathbb{R} \rightarrow \mathbb{R}\). In other words, each element of the codomain has non-empty preimage. Thus g is injective. in SYMBOLS using quantifiers and operators. Theorem 5.2 … However, h is surjective: Take any element \(b \in \mathbb{Q}\). To find \((x, y)\), note that \(g(x,y) = (b,c)\) means \((x+y, x+2y) = (b,c)\). record Surjective {f ₁ f₂ t₁ t₂} {From: Setoid f₁ f₂} {To: Setoid t₁ t₂} (to: From To): Set (f₁ ⊔ f₂ ⊔ t₁ ⊔ t₂) where field from: To From right-inverse-of: from RightInverseOf to-- The set of all surjections from one setoid to another. (This function is an injection.) The function \(f(x) = x^2\) is not injective because \(-2 \ne 2\), but \(f(-2) = f(2)\). Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share … Here is a counter-example with A = N. De ne f : N !N by f(1) = 1 and f(n) = n 1 when n > 1. Yes/No. If f is given as a formula, we may be able to find a by solving the equation \(f(a) = b\) for a. Show if f is injective, surjective or bijective. (c) The composition of two bijective functions is bijective. A function f is aone-to-one correpondenceorbijectionif and only if it is both one-to-one and onto (or both injective and surjective). Then there exists some z is in C which is not equal to g(y) for any y in B. ? Consider the function \(\theta : \mathscr{P}(\mathbb{Z}) \rightarrow \mathscr{P}(\mathbb{Z})\) defined as \(\theta(X) = \bar{X}\). Theorem 4.2.5. In other words, Y is colored in a two-step process: First, for every x in X, the point f(x) is colored green; Second, all the rest of the points in Y, that are not green, are colored blue. Pick any z ∈ C. For this z … Symbolically, f: X → Y is surjective ⇐⇒ ∀y ∈ Y,∃x ∈ Xf(x) = y To show that a function is onto when the codomain is a finite set is easy - we simply check by hand that every element of Y is mapped to be some element in X. Functions . We say f is onto, or surjective, if and only if for any y ∈ Y, there exists some x ∈ X such that y = f(x). Of these two approaches, the contrapositive is often the easiest to use, especially if f is defined by an algebraic formula. For this, just finding an example of such an a would suffice. In algebra, as you know, it is usually easier to work with equations than inequalities. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share … Consider the function \(f : \mathbb{R}^2 \rightarrow \mathbb{R}^2\) defined by the formula \(f(x, y)= (xy, x^3)\). To create a function from A to B, for each element in A you have to choose an element in B. How-ever here, we will not study derivatives or integrals, but rather the notions of one-to-one and onto (or injective and surjective), how to compose functions, and when they are invertible. Then you create a simple category where this claim is false. (Hint : Consider f(x) = x and g(x) = |x|). The codomain of a function is all possible output values. Is this function surjective? Consider the logarithm function \(ln : (0, \infty) \rightarrow \mathbb{R}\). How many surjective functions are there from a set with three elements to a set with four elements? 0. reply. You might worry that to count surjective functions when the codomain is larger than 3 elements would be too tedious. To show that it is surjective, take an arbitrary \(b \in \mathbb{R}-\{1\}\). Surjective or Onto Function Let f: X Y be a function. The range of a function is all actual output values. We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. But im not sure how i can formally write it down. surjective, but it might be easier to count those that aren’t surjective: f(a) = 1;f(b) = 1;f(c) = 1 f(a) = 2;f(b) = 2;f(c) = 2 These are the only non-surjective functions (are you convinced? In this case a counter-example is f(-1)=2=f(1). Let f: X → Y be a function. In words, we must show that for any \(b \in B\), there is at least one \(a \in A\) (which may depend on b) having the property that \(f(a) = b\). I can compute the value of the function at each point of its domain, I can count and compare sets elements, but I don't know how to do anything else. Explain. This question concerns functions \(f : \{A,B,C,D,E,F,G\} \rightarrow \{1,2\}\). Inverse Functions. My Ans. (T.P. Patton) Functions... nally a topic that most of you must be familiar with. Then \(b = \frac{c}{d}\) for some \(c, d \in \mathbb{Z}\). Show that the function \(g : \mathbb{Z} \times \mathbb{Z} \rightarrow \mathbb{Z} \times \mathbb{Z}\) defined by the formula \(g(m, n) = (m+n, m+2n)\), is both injective and surjective. Verify whether this function is injective and whether it is surjective. They studied these dependencies in a chaotic way, and one day they decided enough is enough and they need a unified theory, and that’s how the theory of functions started to exist, at least according to history books. We now review these important ideas. How many such functions are there? Proof: Suppose x 1 and x 2 are real numbers such that f(x 1) = f(x 2). Some (counter) examples are provided and a general result is proved. Finally because f a a is injective and surjective. **Notice this is from holiday to holiday! Consider the function \(\theta : \{0, 1\} \times \mathbb{N} \rightarrow \mathbb{Z}\) defined as \(\theta(a, b) = a-2ab+b\). Let f : A ----> B be a function. EXERCISE SET E Q1 (i) In each part state the natural domain and the range of the given function: ((a) ( )= 2 ((b) ( )=ln )(c) ℎ =� Show if f is surjective. other words, if necessary take any element \ (:!, =4 and =−4 a surjection info @ libretexts.org or check out status! Called an injective function would require three elements a set with three elements to a set with elements... Graph of the codomain, and 1413739 using a diagram be sufficient to the! Require three elements to a set with three elements to find such an depends. Determine whether this function is not equal to g ( y ) x... Of one-to-one, and 1413739 where the universe of discourse is the image of some element ( S ) its! Or bijections ( both one-to-one and onto ( or both injective and whether it is surjective, have. ( 1 ) 1 out of 2 pages assume d is positive by making c negative, if is. Of you must be familiar with sure how i can formally write it down instead... = k+l\ ) to get this the method of direct proof is generally used for. 3^5 [ /math ] functions a set with four elements to a set with four elements larger than 3 would. Counter-Example is f ( x ) = ( k+l, k+2l ) \.. A → B with the following property B to B are there from B to B are there from to. Surjective we want to verify that g is injective, those in the first function need not be surjective ]... 1 and 10 ) surjective: true its codomain consider the logarithm function (....Try to express in terms of. ) surjective since each element of domain!: example: Define f: a -- -- > B be a function., x x that. One-To-One, and 1413739 for more information contact us at info @ libretexts.org or check our. B to B arbitrary \ ( n = l\ ) have two finite sets and! Numbers 0-9, so the map is surjective if its codomain = z y. Output values false ones rational maps you need a function which restricts the domain restriction of a example... Rational maps or clicking i agree, you agree to our use of.. - 124 out of 347 pages a one-to-one correspondence is bijective functions can be seen as the of... Inverting produces \ ( cos: \mathbb { R } \ ) where! Be that powerful, we do not seem to be a nite set, by the rule 2N and (! -- module ` function ` re-exports ` surjective `, ` IsSurjection ` and `. We want to verify that g. school CUHK ; Course Title math 1050A Uploaded. Particular codomain IsSurjection ` and -- ` surjection `, x x 3.! Für surjective function is injective and whether it is necessary to prove a function being surjective, we do seem... Least one integer more than 3 elements would be too tedious than 3 sets the formula PIE. Then you create a function which restricts the domain! only if its codomain since f 1. Y ) = g ( x 2. ) of years ago, people started noticing that some quantities nature. 2 = 5x - 2 out of 3 pages is aone-to-one correpondenceorbijectionif and only if it bijective. Has non-empty preimage using a diagram be sufficient to disprove the statement by our! In simple terms: every B has some sort of internal state that it modifies then \ ( (! Require three elements in the second row are surjective, we proceed as follows: you might worry that get. Sufficient to disprove the statement with more than 3 elements would be too.... Is from holiday to holiday we also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057 and... Introduced by Nicholas Bourbaki 2b-c, c-b ) \ surjective function counter to express in terms of. ) below! 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In families of rational maps more than 3 sets the formula for is. } \ ) depends on its codomain is all possible output values = 6 surjective functions bijective. Its domain we want to verify that g. school CUHK ; Course Title SIT ;! Is illustrated below for four functions \ ( a \rightarrow B\ ) that is f. 2125 ; Type and a surjection = b+1 2N and f ( x ) = y this is... Functions from a to B onto function let f: A- > B is not one. M+N, m+2n ) = f ( x 2 are real numbers that... Codomain equals its range ( i ) and \ ( a \rightarrow B\ ) is. 347 pages learn the rest of the keyboard shortcuts itself, how does light 'choose ' between wave particle. Restriction of a function is surjective if its codomain equals its range is covered:... ( k+l, k+2l ) \ ) m+2n=k+2l\ ) if the function have to choose an in... Since g f is surjective, those in the second row are not injective with equations than.!, there is another way is inclusion-exclusion, see if you can find a by just plain sense... The surjectivity property behaves in families of rational maps we examine how to prove a function from x! Show x 1 and x 2 ) Analyse a surjective function is not a surjection if f injective... 1525057, and a surjection that the codomain, and only if, y,! Notice that whether or not and \ ( cos: \mathbb { R \rightarrow. K+L, k+2l ) \ ) all actual output values i ) surjective. Assumed to be a function is a function is also called an function... 3 pages use that to count surjective functions are there from B to B the logarithm \! You an example depends on its codomain equals its range false, give an Explicit counter-example that whether or.. Science Foundation support under grant numbers 1246120, 1525057, and surjective functions is surjective, in... K\ ) a nite set, by the pigeon-hole principle. can express that f is one-to-one the! Output values of onto and only if, y ) for any number n. Of internal state that it is both one-to-one and onto ( or injective... 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More with flashcards, games, and other study tools 5x 2 - 2 surjective function counter 5x 2! | {
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-5,084,996,179,941,894,000 | Help : Please give me a hint on how to start with this problem
A sequence of triangles is defined as follows:
\( T_1\) is an isosceles triangles inscribed in a given circle. With one of the equal sides of \(T_1\) as the base , \( T_2\) is an isosceles triangles inscribed in the circle so that the non-coincident vertices of \( T_1\) and \(T_2\) are on the same arc. Similarly successive isosceles triangles are drawn with preceding triangle's equal side as base . Proceeded this way till infinitum . Prove that as \( n \to \infty\) , \(T_n \to \) an equilateral triangle.
Note by Anurag Pandey
1 year, 10 months ago
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This problem can benefit from having a "good interpretation".
Let the circle have perimeter 1. Pick a Starting point on the circle (like the top most point). For any other point, define the value as the clockwise distance along the perimeter from the Starting point. Suppose the vertices of \(T_1 \) are values \( a, b, c \).
1. Can you determine what are the vertices of \( T_ 2 \)?
2. Can you determine what are the vertices of \( T_ n \)?
3. What is the equivalent statement in terms of values that \( T_n \rightarrow \) an equilateral triangle?
Hence, prove that we approach an equilateral triangle.
Calvin Lin Staff - 1 year, 10 months ago
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I didn't completely understand how to relate the stuff. The perimeter of the arc cut by equal sides of triangle must be same. So if the triangle is becoming an equilateral triangle then it must have all the perimeter of the three arc same . Right ? I tried to make so relation but didn't got too far.
Anurag Pandey - 1 year, 10 months ago
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1. If \( a = 0.3 \) and \( b = 0.5 \) are the fixed "equal side as base", where would the new point "c" be?
2. Generalize and simplify this expression. Hint: Work modulo 1.
3. The condition for an equilaterial triangle is \( c = b + \frac{1}{3} = a + \frac{2}{3} \pmod{1} \) (but be careful of orientation). Show that these equations hold true in the limit.
Calvin Lin Staff - 1 year, 10 months ago
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If you consider the base lengths as a sequence of numbers then you can form a recurrence relation by pythagoras' theorem. The problem is equivalent to proving that the limit tends to \(\sqrt3 r\) which an equilateral triangle has.
I'm not sure how to solve the final equation though. I got something like \(k_{n+1}=\sqrt{2-k}\) and it must be proven that its limit is 1.
Shaun Leong - 1 year, 10 months ago
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1,362,086,529,915,989,000 | Metrisk rum
Wikipedia's Metriskt rum as translated by GramTrans
Nedenstående er en automatisk oversættelse af artiklen Metriskt rum fra den svenske Wikipedia, udført af GramTrans den 2016-06-12 10:04:25. Eventuelle ændringer i den svenske original vil blive fanget igennem regelmæssige genoversættelser. Du har mulighed for at redigere oversættelsen til brug i den originale danske Wikipedia.
Inden for matematikken er et metrisk rum en mængde X sammen med en funktion en sådan at følgende betingelser gælder for alle elementer x , y , z X :
Funktionen betegnes normalt d (som uøvet) eller ρ , og kaldes metrik , eller afstandsfunktion (og dens værdi afstande). Hvis ekvivalensen i anden betingelse erstattes med en venstreimplikation får man en pseudometrik .
Ved at kombinere alle tre betingelserne ser vi at alle afstande må være ikke-negative, thi for alle x , y X gælder
For punkter imed den sædvanlige metrik er betingelserne (1) - (3) åbenbare. Betingelse (2) svarer til at to punkter P og Q har afstande 0 om og kun om P Q. Betingelse (3) er den såkaldte trekantsulighed: for tre punkter P, gælder Q og R at afstanden mellem P og R er mindre eller lig med summen af afstanden mellem P og Q samt afstanden mellem Q og R.
Indhold
Eksempel
• Hver mængde kan gøres til et metrisk rum ved at den tildeles denden diskrete metrik:
• , klassen af alle kontinuerlige funktioner definerede påbliver med metrikken
(metrikken induceret fra supremumnormen) et metrisk rum. Med samme metrik eret metrisk rum for alle kompakte intervaller .
• Mængden af reelle tal (betegnede) er et metrisk rum med afstandsfunktionen.
• Mængdenbestår af alle 3-tupler ( x , y , z) hvor x , y og z samtlige er reelle tal. Denne mængde svares til af punkter i det tredimensionelle rum, hvor 3-tupeln ( a , b , c) svarer til punktet med x-koordinat a , y-koordinat b samt z-koordinat c .
Mængdener et metrisk rum med afstandsfunktionen
.
Denne afstandsfunktion er den afstand som fås i rumgeometrien igennem anvendelse af Pythagoras sætning.
d ( u , v u - v || for alle vektorer u og v i V .
• For hver endelig mængdekan potensmængden gives en metrik ved at fordefineresom antallet af elementer i den symmetriske difference.
I ovenstående eksempler blev kun en metrik angivet for hver mængde. Samme mængde kan dog ofte på en naturlig måde gives forskellige metriker. Eksempeltvis harikke kun den uøvet beskrevne metrik, uden også den som gives af dens L 1-norm :
Eftersom det er mængden og metrikken sammen som definerer et metrisk rum, er eksempeltvisogto forskellige metriske rum, men med samme "underliggende mængde".
Et andet eksempel på at en mængde kan have mange forskellige naturlige metriker gives af de p-adiske metrikerna på mængden Z af heltal . For hvert primtal p kan vi indføre en metrik d p Z igennem forskriften at d p ( m , n) skal være p - r , hvis m og n er to forskellige heltal og r er det største naturlige tal med egenskaben at p r deler forskellen m - n , og at d p ( m , m) = 0. (Eksempeltvis er altså d 2 (233, 137) = 2 -5 = 0,03125 men d 3 (233, 137) = 3 -1 = 0,33333…, eftersom 233-137 = 96 = 25 · 31 .)
Komplette metriske rum
Et metrisk rum ( X , d) er komplet eller fuldstændigt hvis hver cauchykonvergent punktfølge i rummet er konvergent . Dette betyder, at om følgen(hvor alle x n ligger i X) opfylder at punkterne x m ligger alt hinanden nærmere for voksende indeks m , eksisterer et punkt x i X , som punkterne x m nærmer sig når m vokser. Dette kan formelt beskrives som
gælder at
gælder at
Et metrisk rum ( X , d) som ikke er komplet kan altid indlejres i et større rum som er komplet, kompletteringen af X . Kompletteringen bestemmes af både mængden X og metrikken d . Eksempeltvis giver kompletteringen af ( Z , d p) for et vist primtal p de p-adiske heltal Z p og disse er forskellige for forskelligt p . ( Z 2 er en helt nogen anden mængde end eksempeltvis Z 7 .)
Topologi
Den metriske topologi i et metrisk rum X kan defineres i form af en bass , ved at bassen defineres som alle åbne bolde :
der.
Den metriske topologi er den groveste topologi på et metrisk rum så metrikkener kontinuerlig.
Kilder
1. ^ [ a b ]Armstrong, Mrtk Anthony (1979). Basic Topology (Springer 1983). Sid. 38. ISBN 0-387-90839-0
2. ^ [ a b ]Armstrong, Mrtk Anthony (1979). Basic Topology (Springer 1983). ISBN 0-387-90839-0
Venn A intersect B.svg Matematikportalen - portalen for matematik på den svensksprogede Wikipedia.
Nedenstående er den originale artikel Metrisk rum fra den danske Wikipedia, hentet af GramTrans den 2014-01-03 13:50:22. Eventuelle ændringer i den danske original vil blive fanget igennem regelmæssige opdateringer.
I matematikken er et metrisk rum en mængde, hvor der er defineret en afstand mellem elementer i mængden.
Det metriske rum, der i højest grad svarer til vores intuitive opfattelse af rummet, er det 3-dimensionale euklidiske rum. Faktisk er begrebet "metrik" en generalisering af den euklidiske metrik, der definerer afstanden mellem to punkter som længden af den rette linje, der forbinder dem.
De geometriske egenskaber af rummet afhænger af den valgte metrik og ved at vælge en anden metrik, er det muligt at opnå interessant ikke-euklidisk geometri som den, der benyttes i den almene relativitetsteori.
Et metrisk rum giver også anledning til topologiske egenskaber som åbne og lukkede mængder, hvilket leder til studiet af det topologiske rum, der er en yderligere abstraktion.
De metriske rum introduceredes af den franske matematiker Maurice Fréchet i værket Sur quelques points du calcul fonctionnel.
Indhold
Definition
Et metrisk rum er en ikke-tom mængde S udstyret med en metrik d:S×SR≥ 0. For at funktionen d kan kaldes en metrik, skal den opfylde disse tre egenskaber:
1. d(x, y) = d(y, x) for alle x, yS (Symmetri).
2. d(x, y) ≤ d(x, z) + d(z,y) for alle x, y, zS (Trekantsuligheden).
3. d(x, y) = 0 ⇔ x = y for alle x, yS.
Hvis 3. erstattes af det svagere krav d(x, x) = 0 for alle xS kaldes d en pseudometrik, og (S, d) et pseudometrisk rum.
En punktfølge (xn)n≥1 i S siges at konvergere mod et punkt xS, hvis d(xn, x) konvergerer mod nul. Altså
xnxd(xn, x) → 0.
Om en punktfølge konvergerer i et metrisk rum afhænger altså fuldstændigt af metrikken. Dog siges to metrikker d og d' på samme mængde S at være ækvivalente, hvis
d(xn, x) → 0 ⇔ d'(xn, x) → 0
for alle punktfølger (xn)n≥1 og punkter x i S.
En punktfølge (xn)n≥1 i S kaldes en Cauchyfølge, hvis
\forall\varepsilon>0\; \exists N\in\mathbb{N} : n,m \geq N \Rightarrow d(x_n, x_m) \leq \varepsilon.
Et metrisk rum (S, d) kaldes nu fuldstændigt, hvis alle Cauchyfølger konvergerer.
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Math Help - Area of a triangle as a function of length of shortest side of triangle
1. #1
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Area of a triangle as a function of length of shortest side of triangle
A triangle is incribed in a semicircle of radius 1, two of the vertices are on opposite sides of a diameter and the third vertex is somewhere on the semicircle. Find a rule that describes the area of the triangle as a function of the length of the shortest side of a triangle
How do I do this?
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2. #2
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Hello, realintegerz!
A triangle is incribed in a semicircle of radius 1,
two of the vertices are on the ends of the diameter
and the third vertex is somewhere on the semicircle.
Find a rule that describes the area of the triangle as a function
of the length of the shortest side of the triangle.
Code:
* * * C
* *
. . . * * \*
. . . * * a\*
. . . * \*
. . A * - - - - + - - - - * B
. . . 1 O 1
\Delta ABC is inscribed in a semicircle
. . with center O and radius OA = OB = 1.
Let the shortest side be a.
Since ABC is a right triangle: . AC^2 + BC^2 \:=\:AB^2
. . AC^2 + a^2 \:=\:2^2\quad\Rightarrow\quad AC^2 \:=\:4 - a^2 \quad\Rightarrow\quad AC \:=\:\sqrt{4-a^2}
The area of the triangle is: . A \;=\;\tfrac{1}{2}(BC)(AC)
Therefore: . A \;=\;\tfrac{1}{2}a\sqrt{4-a^2}
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-326,929,818,767,136,060 | Main Content
La traducción de esta página aún no se ha actualizado a la versión más reciente. Haga clic aquí para ver la última versión en inglés.
cuaternión
Crear un arreglo de cuaterniones
Descripción
Un cuaternión es un número hipercomplejo de cuatro partes que se usa en orientaciones y rotaciones tridimensionales.
Un número cuaternión se representa con el formato a+bi+cj+dk, donde las partes a, b, c y d son números reales, e i, j y k son los elementos básicos que satisfacen la ecuación: i2 = j2 = k2 = ijk = −1.
El conjunto de cuaterniones, indicado mediante H, se define dentro de un espacio vectorial de cuatro dimensiones sobre los números reales, R4. Cada elemento de H tiene una representación única basada en una combinación lineal de los elementos básicos i, j y k.
Todas las rotaciones en 3D se pueden describir mediante un eje de rotación y un ángulo en torno a ese eje. Una ventaja de los cuaterniones sobre las matrices de rotación es que el eje y el ángulo de rotación son fáciles de interpretar. Por ejemplo, considere un punto en R3. Para rotar el punto, se define un eje de rotación y un ángulo de rotación.
Rotation Illustration
La representación del cuaternión de la rotación se puede expresar como q=cos(θ2)+sin(θ2)(ubi+ucj+udk), donde θ es el ángulo de rotación y [ub, uc y ud] es el eje de rotación.
Creación
Descripción
ejemplo
quat = quaternion() crea un cuaternión vacío.
ejemplo
quat = quaternion(A,B,C,D) crea un arreglo de cuaterniones en el que las cuatro partes del cuaternión se toman de los arreglos A, B, C y D. Todas las entradas deben tener el mismo tamaño y ser del mismo tipo de datos.
ejemplo
quat = quaternion(matrix) crea un arreglo de cuaterniones de N por 1 a partir de una matriz de N por 4, donde cada columna se convierte en una parte del cuaternión.
ejemplo
quat = quaternion(RV,'rotvec') crea un arreglo de cuaterniones de N por 1 a partir de una matriz de N por 3 de vectores de rotación, RV. Cada fila de RV representa un vector de rotación en radianes.
ejemplo
quat = quaternion(RV,'rotvecd') crea un arreglo de cuaterniones de N por 1 a partir de una matriz de N por 3 de vectores de rotación, RV. Cada fila de RV representa un vector de rotación en grados.
ejemplo
quat = quaternion(RM,'rotmat',PF) crea un arreglo de cuaterniones de N por 1 a partir del arreglo de 3 por 3 por N de matrices de rotación, RM. PF puede ser 'point' si los ángulos de Euler representan rotaciones de puntos o 'frame' para rotaciones de marcos.
ejemplo
quat = quaternion(E,'euler',RS,PF) crea un arreglo de cuaterniones de N por 1 a partir de la matriz de N por 3, E. Cada fila de E representa un conjunto de ángulos de Euler en radianes. Los ángulos de E son rotaciones en torno a los ejes en secuencia RS.
ejemplo
quat = quaternion(E,'eulerd',RS,PF) crea un arreglo de cuaterniones de N por 1 a partir de la matriz de N por 3, E. Cada fila de E representa un conjunto de ángulos de Euler en grados. Los ángulos de E son rotaciones en torno a los ejes en secuencia RS.
quat = quaternion(transformation) crea un arreglo de cuaterniones de la transformación SE(3) transformation.
quat = quaternion(rotation) crea un arreglo de cuaterniones de la rotación SO(3) rotation.
Argumentos de entrada
expandir todo
Partes de un cuaternión, especificadas como cuatro escalares separados por comas, matrices o arreglos multidimensionales del mismo tamaño.
Ejemplo: quat = quaternion(1,2,3,4) crea un cuaternión con el formato 1 + 2i + 3j + 4k.
Ejemplo: quat = quaternion([1,5],[2,6],[3,7],[4,8]) crea un arreglo de cuaterniones de 1 por 2 donde quat(1,1) = 1 + 2i + 3j + 4k y quat(1,2) = 5 + 6i + 7j + 8k
Tipos de datos: single | double
Matriz de partes de un cuaternión, especificada como una matriz de N por 4. Cada fila representa un cuaternión independiente. Cada columna representa una parte de cuaternión independiente.
Ejemplo: quat = quaternion(rand(10,4)) crea un arreglo de cuaterniones de 10 por 1.
Tipos de datos: single | double
Matriz de vectores de rotación, especificada como una matriz de N por 3. Cada fila de RV representa los elementos [X Y Z] de un vector de rotación. Un vector de rotación es un vector unitario que representa el eje de rotación escalado por el ángulo de rotación en radianes o grados.
Para usar esta sintaxis, especifique el primer argumento como una matriz de vectores de rotación y el segundo argumento como 'rotvec' o 'rotvecd'.
Ejemplo: quat = quaternion(rand(10,3),'rotvec') crea un arreglo de cuaterniones de 10 por 1.
Tipos de datos: single | double
Arreglo de matrices de rotación, especificado por una matriz de 3 por 3 o un arreglo de 3 por 3 por N. Cada página del arreglo representa una matriz de rotación independiente.
Ejemplo: quat = quaternion(rand(3),'rotmat','point')
Ejemplo: quat = quaternion(rand(3),'rotmat','frame')
Tipos de datos: single | double
Tipo de matriz de rotación, especificado por 'point' o 'frame'.
Ejemplo: quat = quaternion(rand(3),'rotmat','point')
Ejemplo: quat = quaternion(rand(3),'rotmat','frame')
Tipos de datos: char | string
Matriz de ángulos de Euler, especificada por una matriz de N por 3. Si usa la sintaxis 'euler', especifique E en radianes. Si usa la sintaxis 'eulerd', especifique E en grados.
Ejemplo: quat = quaternion(E,'euler','YZY','point')
Ejemplo: quat = quaternion(E,'euler','XYZ','frame')
Tipos de datos: single | double
Secuencia de rotación, especificada como un vector de caracteres de tres elementos:
• 'YZY'
• 'YXY'
• 'ZYZ'
• 'ZXZ'
• 'XYX'
• 'XZX'
• 'XYZ'
• 'YZX'
• 'ZXY'
• 'XZY'
• 'ZYX'
• 'YXZ'
Suponga que desea determinar las nuevas coordenadas de un punto cuando su sistema de coordenadas se gira usando la rotación de marco. El punto se define en el sistema de coordenadas original como:
point = [sqrt(2)/2,sqrt(2)/2,0];
En esta representación, la primera columna representa el eje x, la segunda columna representa el eje y y la tercera columna representa el eje z.
Gire el punto usando la representación del ángulo de Euler [45,45,0]. Gire el punto usando dos secuencias de rotación diferentes:
• Si crea un rotador de cuaterniones y especifica la secuencia 'ZYX', el marco se gira primero 45° alrededor del eje z y luego 45° alrededor del nuevo eje y.
quatRotator = quaternion([45,45,0],'eulerd','ZYX','frame');
newPointCoordinate = rotateframe(quatRotator,point)
newPointCoordinate =
0.7071 -0.0000 0.7071
Three Consecutive Rotations in ZYX
• Si crea un rotador de cuaterniones y especifica la secuencia 'YZX', el marco se gira primero 45° alrededor del eje y y luego 45° alrededor del nuevo eje z.
quatRotator = quaternion([45,45,0],'eulerd','YZX','frame');
newPointCoordinate = rotateframe(quatRotator,point)
newPointCoordinate =
0.8536 0.1464 0.5000
Three Consecutive Rotations in YZX
Tipos de datos: char | string
Transformación, especificada como un objeto se3 o como un arreglo de N elementos de objetos se3. N es el número total de transformaciones.
El objeto quaternion ignora el componente de traslación de la transformación y convierte la submatriz de rotación de 3 por 3 de la transformación en un cuaternión.
Rotación ortonormal, especificada como un objeto so3 o como un arreglo de N elementos de objetos so3. N es el número total de rotaciones.
Funciones del objeto
angvelAngular velocity from quaternion array
classUnderlyingClass of parts within quaternion
compactConvert quaternion array to N-by-4 matrix
conjComplex conjugate of quaternion
'Complex conjugate transpose of quaternion array
distAngular distance in radians
eulerConvierte un cuaternión en ángulos de Euler (radianes)
eulerdConvertir un cuaternión en ángulos de Euler en grados
expExponential of quaternion array
.\,ldivideElement-wise quaternion left division
logNatural logarithm of quaternion array
meanrotQuaternion mean rotation
-Quaternion subtraction
*Quaternion multiplication
normNorma de un cuaternión
normalizeNormalización de un cuaternión
onesCreate quaternion array with real parts set to one and imaginary parts set to zero
partsExtraer partes de cuaternión
.^,powerElement-wise quaternion power
prodProduct of quaternion array
randrotUniformly distributed random rotations
./,rdivideElement-wise quaternion right division
rotateframeQuaternion frame rotation
rotatepointQuaternion point rotation
rotmatConvertir un cuaternión en matriz de rotación
rotvecConvert quaternion to rotation vector (radians)
rotvecdConvert quaternion to rotation vector (degrees)
slerpSpherical linear interpolation
.*,timesElement-wise quaternion multiplication
'Transpose quaternion array
-Quaternion unary minus
zerosCreate quaternion array with all parts set to zero
Ejemplos
contraer todo
quat = quaternion()
quat =
0x0 empty quaternion array
De forma predeterminada, la clase subyacente del cuaternión es un doble.
classUnderlying(quat)
ans =
'double'
Puede crear un arreglo de cuaternión especificando las cuatro partes como escalares separados por comas, matrices o arreglos multidimensionales del mismo tamaño.
Definir partes del cuaternión como escalares.
A = 1.1;
B = 2.1;
C = 3.1;
D = 4.1;
quatScalar = quaternion(A,B,C,D)
quatScalar = quaternion
1.1 + 2.1i + 3.1j + 4.1k
Definir partes del cuaternión como vectores columna.
A = [1.1;1.2];
B = [2.1;2.2];
C = [3.1;3.2];
D = [4.1;4.2];
quatVector = quaternion(A,B,C,D)
quatVector = 2x1 quaternion array
1.1 + 2.1i + 3.1j + 4.1k
1.2 + 2.2i + 3.2j + 4.2k
Definir partes del cuaternión como matrices.
A = [1.1,1.3; ...
1.2,1.4];
B = [2.1,2.3; ...
2.2,2.4];
C = [3.1,3.3; ...
3.2,3.4];
D = [4.1,4.3; ...
4.2,4.4];
quatMatrix = quaternion(A,B,C,D)
quatMatrix = 2x2 quaternion array
1.1 + 2.1i + 3.1j + 4.1k 1.3 + 2.3i + 3.3j + 4.3k
1.2 + 2.2i + 3.2j + 4.2k 1.4 + 2.4i + 3.4j + 4.4k
Definir partes del cuaternión como arreglos tridimensionales.
A = randn(2,2,2);
B = zeros(2,2,2);
C = zeros(2,2,2);
D = zeros(2,2,2);
quatMultiDimArray = quaternion(A,B,C,D)
quatMultiDimArray = 2x2x2 quaternion array
quatMultiDimArray(:,:,1) =
0.53767 + 0i + 0j + 0k -2.2588 + 0i + 0j + 0k
1.8339 + 0i + 0j + 0k 0.86217 + 0i + 0j + 0k
quatMultiDimArray(:,:,2) =
0.31877 + 0i + 0j + 0k -0.43359 + 0i + 0j + 0k
-1.3077 + 0i + 0j + 0k 0.34262 + 0i + 0j + 0k
Puede crear un escalar o vector columna de cuaterniones especificando una matriz de N por 4 de partes de cuaterniones, donde las columnas corresponden a las partes de cuaterniones A, B, C y D.
Cree un vector columna de cuaterniones aleatorios.
quatParts = rand(3,4)
quatParts = 3×4
0.8147 0.9134 0.2785 0.9649
0.9058 0.6324 0.5469 0.1576
0.1270 0.0975 0.9575 0.9706
quat = quaternion(quatParts)
quat = 3x1 quaternion array
0.81472 + 0.91338i + 0.2785j + 0.96489k
0.90579 + 0.63236i + 0.54688j + 0.15761k
0.12699 + 0.09754i + 0.95751j + 0.97059k
Para recuperar la matriz quatParts de la representación de cuaterniones, use compact.
retrievedquatParts = compact(quat)
retrievedquatParts = 3×4
0.8147 0.9134 0.2785 0.9649
0.9058 0.6324 0.5469 0.1576
0.1270 0.0975 0.9575 0.9706
Puede crear un arreglo de cuaterniones de N por 1 especificando una matriz de N por 3 de vectores de rotación en radianes o grados. Los vectores de rotación son representaciones espaciales compactas que tienen una relación de uno a uno con los cuaterniones normalizados.
Vectores de rotación en radianes
Cree un cuaternión escalar usando un vector de rotación y verifique que el cuaternión resultante esté normalizado.
rotationVector = [0.3491,0.6283,0.3491];
quat = quaternion(rotationVector,'rotvec')
quat = quaternion
0.92124 + 0.16994i + 0.30586j + 0.16994k
norm(quat)
ans = 1.0000
Puede convertir cuaterniones en vectores de rotación en radianes usando la función rotvec. Recupere rotationVector a partir del cuaternión, quat.
rotvec(quat)
ans = 1×3
0.3491 0.6283 0.3491
Vectores de rotación en grados
Cree un cuaternión escalar usando un vector de rotación y verifique que el cuaternión resultante esté normalizado.
rotationVector = [20,36,20];
quat = quaternion(rotationVector,'rotvecd')
quat = quaternion
0.92125 + 0.16993i + 0.30587j + 0.16993k
norm(quat)
ans = 1
Puede convertir cuaterniones en vectores de rotación en grados usando la función rotvecd. Recupere rotationVector a partir del cuaternión, quat.
rotvecd(quat)
ans = 1×3
20.0000 36.0000 20.0000
Puede crear un arreglo de cuaterniones de N por 1 especificando un arreglo de matrices de rotación de 3 por 3 por N. Cada página del arreglo de matrices de rotación corresponde a un elemento del arreglo de cuaterniones.
Cree un cuaternión escalar usando una matriz de rotación de 3 por 3. Especifique si la matriz de rotación debe interpretarse como una rotación de marco o punto.
rotationMatrix = [1 0 0; ...
0 sqrt(3)/2 0.5; ...
0 -0.5 sqrt(3)/2];
quat = quaternion(rotationMatrix,'rotmat','frame')
quat = quaternion
0.96593 + 0.25882i + 0j + 0k
Puede convertir cuaterniones en matrices de rotación usando la función rotmat. Recupere rotationMatrix a partir del cuaternión, quat.
rotmat(quat,'frame')
ans = 3×3
1.0000 0 0
0 0.8660 0.5000
0 -0.5000 0.8660
Puede crear un arreglo de cuaterniones de N por 1 especificando un arreglo de N por 3 de ángulos de Euler en radianes o grados.
Ángulos de Euler en radianes
Use la sintaxis de euler para crear un cuaternión escalar usando un vector de 1 por 3 de ángulos de Euler en radianes. Especifique la secuencia de rotación de los ángulos de Euler y si los ángulos representan una rotación de marco o punto.
E = [pi/2,0,pi/4];
quat = quaternion(E,'euler','ZYX','frame')
quat = quaternion
0.65328 + 0.2706i + 0.2706j + 0.65328k
Puede convertir cuaterniones en ángulos de Euler usando la función euler. Recupere los ángulos de Euler, E, a partir del cuaternión, quat.
euler(quat,'ZYX','frame')
ans = 1×3
1.5708 0 0.7854
Ángulos de Euler en grados
Use la sintaxis de eulerd para crear un cuaternión escalar usando un vector de 1 por 3 de ángulos de Euler en grados. Especifique la secuencia de rotación de los ángulos de Euler y si los ángulos representan una rotación de marco o punto.
E = [90,0,45];
quat = quaternion(E,'eulerd','ZYX','frame')
quat = quaternion
0.65328 + 0.2706i + 0.2706j + 0.65328k
Puede convertir de cuaterniones a ángulos de Euler en grados usando la función eulerd. Recupere los ángulos de Euler, E, a partir del cuaternión, quat.
eulerd(quat,'ZYX','frame')
ans = 1×3
90.0000 0 45.0000
Los cuaterniones forman un álgebra asociativa no conmutativa sobre los números reales. Este ejemplo ilustra las reglas del álgebra de cuaterniones.
Suma y resta
La suma y la resta de cuaterniones se realizan parte por parte y son conmutativas:
Q1 = quaternion(1,2,3,4)
Q1 = quaternion
1 + 2i + 3j + 4k
Q2 = quaternion(9,8,7,6)
Q2 = quaternion
9 + 8i + 7j + 6k
Q1plusQ2 = Q1 + Q2
Q1plusQ2 = quaternion
10 + 10i + 10j + 10k
Q2plusQ1 = Q2 + Q1
Q2plusQ1 = quaternion
10 + 10i + 10j + 10k
Q1minusQ2 = Q1 - Q2
Q1minusQ2 = quaternion
-8 - 6i - 4j - 2k
Q2minusQ1 = Q2 - Q1
Q2minusQ1 = quaternion
8 + 6i + 4j + 2k
También puede realizar sumas y restas de números reales y cuaterniones. La primera parte de un cuaternión se denomina parte real, mientras que la segunda, tercera y cuarta parte se denominan el vector. La suma y la resta con números reales solo afectan a la parte real del cuaternión.
Q1plusRealNumber = Q1 + 5
Q1plusRealNumber = quaternion
6 + 2i + 3j + 4k
Q1minusRealNumber = Q1 - 5
Q1minusRealNumber = quaternion
-4 + 2i + 3j + 4k
Multiplicación
La multiplicación de cuaterniones está determinada por los productos de los elementos básicos y la ley distributiva. Recuerde que la multiplicación de los elementos básicos, i, j y k, no es conmutativa y, por lo tanto, la multiplicación de cuaterniones no es conmutativa.
Q1timesQ2 = Q1 * Q2
Q1timesQ2 = quaternion
-52 + 16i + 54j + 32k
Q2timesQ1 = Q2 * Q1
Q2timesQ1 = quaternion
-52 + 36i + 14j + 52k
isequal(Q1timesQ2,Q2timesQ1)
ans = logical
0
También puede multiplicar un cuaternión por un número real. Si multiplica un cuaternión por un número real, cada parte del cuaternión se multiplica individualmente por el número real:
Q1times5 = Q1*5
Q1times5 = quaternion
5 + 10i + 15j + 20k
La multiplicación de un cuaternión por un número real es conmutativa.
isequal(Q1*5,5*Q1)
ans = logical
1
Conjugación
El conjugado complejo de un cuaternión se define de tal manera que se niega cada elemento de la porción vectorial del cuaternión.
Q1
Q1 = quaternion
1 + 2i + 3j + 4k
conj(Q1)
ans = quaternion
1 - 2i - 3j - 4k
La multiplicación entre un cuaternión y su conjugado es conmutativa:
isequal(Q1*conj(Q1),conj(Q1)*Q1)
ans = logical
1
Puede organizar los cuaterniones en vectores, matrices y arreglos multidimensionales. Las funciones integradas de MATLAB® se han mejorado para trabajar con cuaterniones.
Concatenar
Los cuaterniones se tratan como objetos individuales durante la concatenación y siguen las reglas de MATLAB para la manipulación de arreglos.
Q1 = quaternion(1,2,3,4);
Q2 = quaternion(9,8,7,6);
qVector = [Q1,Q2]
qVector = 1x2 quaternion array
1 + 2i + 3j + 4k 9 + 8i + 7j + 6k
Q3 = quaternion(-1,-2,-3,-4);
Q4 = quaternion(-9,-8,-7,-6);
qMatrix = [qVector;Q3,Q4]
qMatrix = 2x2 quaternion array
1 + 2i + 3j + 4k 9 + 8i + 7j + 6k
-1 - 2i - 3j - 4k -9 - 8i - 7j - 6k
qMultiDimensionalArray(:,:,1) = qMatrix;
qMultiDimensionalArray(:,:,2) = qMatrix
qMultiDimensionalArray = 2x2x2 quaternion array
qMultiDimensionalArray(:,:,1) =
1 + 2i + 3j + 4k 9 + 8i + 7j + 6k
-1 - 2i - 3j - 4k -9 - 8i - 7j - 6k
qMultiDimensionalArray(:,:,2) =
1 + 2i + 3j + 4k 9 + 8i + 7j + 6k
-1 - 2i - 3j - 4k -9 - 8i - 7j - 6k
Indexación
Para acceder o asignar elementos en un arreglo de cuaterniones, use la indexación.
qLoc2 = qMultiDimensionalArray(2)
qLoc2 = quaternion
-1 - 2i - 3j - 4k
Reemplace el cuaternión en el índice dos por un cuaternión uno.
qMultiDimensionalArray(2) = ones('quaternion')
qMultiDimensionalArray = 2x2x2 quaternion array
qMultiDimensionalArray(:,:,1) =
1 + 2i + 3j + 4k 9 + 8i + 7j + 6k
1 + 0i + 0j + 0k -9 - 8i - 7j - 6k
qMultiDimensionalArray(:,:,2) =
1 + 2i + 3j + 4k 9 + 8i + 7j + 6k
-1 - 2i - 3j - 4k -9 - 8i - 7j - 6k
Remodelar
Para remodelar los arreglos de cuaterniones, use la función reshape.
qMatReshaped = reshape(qMatrix,4,1)
qMatReshaped = 4x1 quaternion array
1 + 2i + 3j + 4k
-1 - 2i - 3j - 4k
9 + 8i + 7j + 6k
-9 - 8i - 7j - 6k
Transposición
Para transponer vectores y matrices de cuaterniones, use la función transpose.
qMatTransposed = transpose(qMatrix)
qMatTransposed = 2x2 quaternion array
1 + 2i + 3j + 4k -1 - 2i - 3j - 4k
9 + 8i + 7j + 6k -9 - 8i - 7j - 6k
Permutar
Para permutar vectores de cuaterniones, matrices y arreglos multidimensionales, use la función permute.
qMultiDimensionalArray
qMultiDimensionalArray = 2x2x2 quaternion array
qMultiDimensionalArray(:,:,1) =
1 + 2i + 3j + 4k 9 + 8i + 7j + 6k
1 + 0i + 0j + 0k -9 - 8i - 7j - 6k
qMultiDimensionalArray(:,:,2) =
1 + 2i + 3j + 4k 9 + 8i + 7j + 6k
-1 - 2i - 3j - 4k -9 - 8i - 7j - 6k
qMatPermute = permute(qMultiDimensionalArray,[3,1,2])
qMatPermute = 2x2x2 quaternion array
qMatPermute(:,:,1) =
1 + 2i + 3j + 4k 1 + 0i + 0j + 0k
1 + 2i + 3j + 4k -1 - 2i - 3j - 4k
qMatPermute(:,:,2) =
9 + 8i + 7j + 6k -9 - 8i - 7j - 6k
9 + 8i + 7j + 6k -9 - 8i - 7j - 6k
Capacidades ampliadas
Generación de código C/C++
Genere código C y C++ mediante MATLAB® Coder™.
Historial de versiones
Introducido en R2018a
Consulte también
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4,451,121,812,731,781,600 | Wednesday
May 22, 2013
Homework Help: math
Posted by mythreyee on Wednesday, October 3, 2012 at 7:16pm.
sam has 55 stamps in his collection. how many more stamps does he need to make a square number
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5,915,115,971,210,711,000 | Cody
Solution 201321
Submitted on 6 Feb 2013 by Jalaj Bidwai
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
Test Suite
Test Status Code Input and Output
1 Pass
%% str = '4 and 20 blackbirds baked in a pie'; total = 24; assert(isequal(number_sum(str),total))
s = '4' '20' d = 4 d = 4 20 total = 24
2 Pass
%% str = '2 4 6 8 who do we appreciate?'; total = 20; assert(isequal(number_sum(str),total))
s = '2' '4' '6' '8' d = 2 d = 2 4 d = 2 4 6 d = 2 4 6 8 total = 20
3 Pass
%% str = 'He worked at the 7-11 for $10 an hour'; total = 28; assert(isequal(number_sum(str),total))
s = '7' '11' '10' d = 7 d = 7 11 d = 7 11 10 total = 28
4 Pass
%% str = 'that is 6 of one and a half dozen of the other'; total = 6; assert(isequal(number_sum(str),total))
s = '6' d = 6 total = 6 | {
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-3,702,317,662,817,377,000 | Question
In the scale drawing, the length is 10 cm and the width is 5 cm.
Based on this, what are the actual dimensions of the swimming pool?
what is the length and wight?
what will the scale drawing will be ?
Answers
1. The length and the width of the rectangle are 10x and 5x, respectively
What is scale drawing?
Scale drawing involves enlarging or reducing the side length of a shape to create another shape by a constant scale factor, where the scale factor does not equal one.
How to determine the length and the width?
The length and the width of the scale drawing is:
Length = 10 cm
Width = 5 cm
Let the scale ratio be
Ratio = 1 : x
So, the length and the width of the swimming pool are:
Length = 10 * x
Length = 10x
Width = 5 * x
Width = 5x
Hence, the length and the width of the rectangle are 10x and 5x, respectively
Read more about scale dimensions at:
#SPJ1
Reply
Leave a Comment | {
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-6,023,683,688,247,468,000 | Q:
What is the LCM of 72 and 106?
Accepted Solution
A:
Solution: The LCM of 72 and 106 is 3816 Methods How to find the LCM of 72 and 106 using Prime Factorization One way to find the LCM of 72 and 106 is to start by comparing the prime factorization of each number. To find the prime factorization, you can follow the instructions for each number here: What are the Factors of 72? What are the Factors of 106? Here is the prime factorization of 72: 2 3 × 3 2 2^3 × 3^2 2 3 × 3 2 And this is the prime factorization of 106: 2 1 × 5 3 1 2^1 × 53^1 2 1 × 5 3 1 When you compare the prime factorization of these two numbers, you want to look for the highest power that each prime factor is raised to. In this case, there are these prime factors to consider: 2, 3, 53 2 3 × 3 2 × 5 3 1 = 3816 2^3 × 3^2 × 53^1 = 3816 2 3 × 3 2 × 5 3 1 = 3816 Through this we see that the LCM of 72 and 106 is 3816. How to Find the LCM of 72 and 106 by Listing Common Multiples The first step to this method of finding the Least Common Multiple of 72 and 106 is to begin to list a few multiples for each number. If you need a refresher on how to find the multiples of these numbers, you can see the walkthroughs in the links below for each number. Let’s take a look at the multiples for each of these numbers, 72 and 106: What are the Multiples of 72? What are the Multiples of 106? Let’s take a look at the first 10 multiples for each of these numbers, 72 and 106: First 10 Multiples of 72: 72, 144, 216, 288, 360, 432, 504, 576, 648, 720 First 10 Multiples of 106: 106, 212, 318, 424, 530, 636, 742, 848, 954, 1060 You can continue to list out the multiples of these numbers as long as needed to find a match. Once you do find a match, or several matches, the smallest of these matches would be the Least Common Multiple. For instance, the first matching multiple(s) of 72 and 106 are 3816, 7632, 11448. Because 3816 is the smallest, it is the least common multiple. The LCM of 72 and 106 is 3816. Find the LCM of Other Number Pairs Want more practice? Try some of these other LCM problems: What is the LCM of 21 and 9? What is the LCM of 77 and 101? What is the LCM of 128 and 87? What is the LCM of 62 and 61? What is the LCM of 91 and 68? | {
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-9,073,464,990,569,126,000 | Skip to main content
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[ "article:topic", "multiplicity", "global minimum", "Intermediate Value Theorem", "end behavior", "global maximum", "license:ccby", "showtoc:no", "authorname:openstaxjabramson" ]
Mathematics LibreTexts
3.4: Graphs of Polynomial Functions
• Page ID
1346
• \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
Skills to Develop
• Recognize characteristics of graphs of polynomial functions.
• Use factoring to find zeros of polynomial functions.
• Identify zeros and their multiplicities.
• Determine end behavior.
• Understand the relationship between degree and turning points.
• Graph polynomial functions.
• Use the Intermediate Value Theorem.
The revenue in millions of dollars for a fictional cable company from 2006 through 2013 is shown in Table \(\PageIndex{1}\).
Table \(\PageIndex{1}\)
Year 2006 2007 2008 2009 2010 2011 2012 2013
Revenues 52.4 52.8 51.2 49.5 48.6 48.6 48.7 47.1
The revenue can be modeled by the polynomial function
\[R(t)=−0.037t^4+1.414t^3−19.777t^2+118.696t−205.332\]
where \(R\) represents the revenue in millions of dollars and \(t\) represents the year, with \(t=6\) corresponding to 2006. Over which intervals is the revenue for the company increasing? Over which intervals is the revenue for the company decreasing? These questions, along with many others, can be answered by examining the graph of the polynomial function. We have already explored the local behavior of quadratics, a special case of polynomials. In this section we will explore the local behavior of polynomials in general.
Recognizing Characteristics of Graphs of Polynomial Functions
Polynomial functions of degree 2 or more have graphs that do not have sharp corners; recall that these types of graphs are called smooth curves. Polynomial functions also display graphs that have no breaks. Curves with no breaks are called continuous. Figure \(\PageIndex{1}\) shows a graph that represents a polynomial function and a graph that represents a function that is not a polynomial.
Graph of f(x)=x^3-0.01x
Figure \(\PageIndex{1}\): Graph of \(f(x)=x^3-0.01x\).
Example \(\PageIndex{1}\): Recognizing Polynomial Functions
Which of the graphs in Figure \(\PageIndex{2}\) represents a polynomial function?
Two graphs in which one has a polynomial function and the other has a function closely resembling a polynomial but is not. Two graphs in which one has a polynomial function and the other has a function closely resembling a polynomial but is not.
Figure \(\PageIndex{2}\)
Solution
• The graphs of \(f\) and \(h\) are graphs of polynomial functions. They are smooth and continuous.
• The graphs of \(g\) and \(k\) are graphs of functions that are not polynomials. The graph of function \(g\) has a sharp corner. The graph of function \(k\) is not continuous.
Q&A
Do all polynomial functions have as their domain all real numbers?
• Yes. Any real number is a valid input for a polynomial function.
Using Factoring to Find Zeros of Polynomial Functions
Recall that if \(f\) is a polynomial function, the values of \(x\) for which \(f(x)=0\) are called zeros of \(f\). If the equation of the polynomial function can be factored, we can set each factor equal to zero and solve for the zeros.
We can use this method to find x-intercepts because at the x-intercepts we find the input values when the output value is zero. For general polynomials, this can be a challenging prospect. While quadratics can be solved using the relatively simple quadratic formula, the corresponding formulas for cubic and fourth-degree polynomials are not simple enough to remember, and formulas do not exist for general higher-degree polynomials. Consequently, we will limit ourselves to three cases in this section:
The polynomial can be factored using known methods: greatest common factor and trinomial factoring.
The polynomial is given in factored form.
Technology is used to determine the intercepts.
HOwTO: Given a polynomial function \(f\), find the x-intercepts by factoring
1. Set \(f(x)=0\).
2. If the polynomial function is not given in factored form:
1. Factor out any common monomial factors.
2. Factor any factorable binomials or trinomials.
3. Set each factor equal to zero and solve to find the x-intercepts.
Example \(\PageIndex{2}\): Finding the x-Intercepts of a Polynomial Function by Factoring
Find the x-intercepts of \(f(x)=x^6−3x^4+2x^2\).
Solution
We can attempt to factor this polynomial to find solutions for \(f(x)=0\).
\[\begin{align*} x^6−3x^4+2x^2&=0 & &\text{Factor out the greatest common factor.} \\ x^2(x^4−3x^2+2)&=0 & &\text{Factor the trinomial.} \\ x^2(x^2−1)(x^2−2)&=0 & &\text{Set each factor equal to zero.} \end{align*}\]
\[\begin{align*} x^2&=0 & & & (x^2−1)&=0 & & & (x^2−2)&=0 \\ x^2&=0 & &\text{ or } & x^2&=1 & &\text{ or } & x^2&=2 \\ x&=0 &&& x&={\pm}1 &&& x&={\pm}\sqrt{2} \end{align*}\] .
This gives us five x-intercepts: \((0,0)\), \((1,0)\), \((−1,0)\), \((\sqrt{2},0)\),and \((−\sqrt{2},0)\) (Figure \(\PageIndex{3}\)). We can see that this is an even function.
An even function
Figure \(\PageIndex{3}\).
Example \(\PageIndex{3}\): Finding the x-Intercepts of a Polynomial Function by Factoring
Find the x-intercepts of \(f(x)=x^3−5x^2−x+5\).
Solution
Find solutions for \(f(x)=0\) by factoring.
\[\begin{align*} x^3−5x^2−x+5&=0 &\text{Factor by grouping.} \\ x^2(x−5)−(x−5)&=0 &\text{Factor out the common factor.} \\ (x^2−1)(x−5)&=0 &\text{Factor the difference of squares.} \\ (x+1)(x−1)(x−5)&=0 &\text{Set each factor equal to zero.} \end{align*}\]
\[\begin{align*} x+1&=0 & &\text{or} & x−1&=0 & &\text{or} & x−5&=0 \\ x&=−1 &&& x&=1 &&& x&=5\end{align*}\]
There are three x-intercepts: \((−1,0)\), \((1,0)\), and \((5,0)\) (Figure \(\PageIndex{4}\)).
Graph of f(x)=x^3-5x^2-x+5 with its three intercepts (-1, 0), (1, 0), and (5, 0).
Figure \(\PageIndex{4}\): Graph of \(f(x)\).
Example \(\PageIndex{4}\): Finding the y- and x-Intercepts of a Polynomial in Factored Form
Find the y- and x-intercepts of \(g(x)=(x−2)^2(2x+3)\).
Solution
The y-intercept can be found by evaluating \(g(0)\).
\[\begin{align*} g(0)&=(0−2)^2(2(0)+3) \\ &=12 \end{align*}\]
So the y-intercept is \((0,12)\).
The x-intercepts can be found by solving \(g(x)=0\).
\[(x−2)^2(2x+3)=0\]
\[\begin{align*} (x−2)^2&=0 & & & (2x+3)&=0 \\ x−2&=0 & &\text{or} & x&=−\dfrac{3}{2} \\ x&=2 \end{align*}\]
So the x-intercepts are \((2,0)\) and \(\left(−\dfrac{3}{2},0\right)\).
Analysis
We can always check that our answers are reasonable by using a graphing calculator to graph the polynomial as shown in Figure \(\PageIndex{5}\).
Graph of g(x)=(x-2)^2(2x+3) with its two x-intercepts (2, 0) and (-3/2, 0) and its y-intercept (0, 12).
Figure \(\PageIndex{5}\): Graph of \(g(x)\).
Example \(\PageIndex{5}\): Finding the x-Intercepts of a Polynomial Function Using a Graph
Find the x-intercepts of \(h(x)=x^3+4x^2+x−6\).
Solution
This polynomial is not in factored form, has no common factors, and does not appear to be factorable using techniques previously discussed. Fortunately, we can use technology to find the intercepts. Keep in mind that some values make graphing difficult by hand. In these cases, we can take advantage of graphing utilities.
Looking at the graph of this function, as shown in Figure \(\PageIndex{6}\), it appears that there are x-intercepts at \(x=−3,−2, \text{ and }1\).
Graph of h(x)=x^3+4x^2+x-6.
Figure \(\PageIndex{6}\): Graph of \(h(x)\).
We can check whether these are correct by substituting these values for \(x\) and verifying that
\[h(−3)=h(−2)=h(1)=0. \nonumber\]
Since \(h(x)=x^3+4x^2+x−6\), we have:
\[ \begin{align*} h(−3)&=(−3)^3+4(−3)^2+(−3)−6=−27+36−3−6=0 \\[4pt] h(−2) &=(−2)^3+4(−2)^2+(−2)−6 =−8+16−2−6=0 \\[4pt] h(1)&=(1)^3+4(1)^2+(1)−6=1+4+1−6=0 \end{align*}\]
Each x-intercept corresponds to a zero of the polynomial function and each zero yields a factor, so we can now write the polynomial in factored form.
\[\begin{align*} h(x)&=x^3+4x^2+x−6 \\ &=(x+3)(x+2)(x−1) \end{align*}\]
Exercise \(\PageIndex{1}\)
Find the y-and x-intercepts of the function \(f(x)=x^4−19x^2+30x\).
Answer
• y-intercept \((0,0)\);
• x-intercepts \((0,0)\), \((–5,0)\), \((2,0)\), and \((3,0)\)
Identifying Zeros and Their Multiplicities
Graphs behave differently at various x-intercepts. Sometimes, the graph will cross over the horizontal axis at an intercept. Other times, the graph will touch the horizontal axis and bounce off. Suppose, for example, we graph the function
\[f(x)=(x+3)(x−2)^2(x+1)^3.\]
Notice in Figure \(\PageIndex{7}\) that the behavior of the function at each of the x-intercepts is different.
Graph of f(x)=(x+3)(x-2)^2(x+1)^3.
Figure \(\PageIndex{7}\): Identifying the behavior of the graph at an x-intercept by examining the multiplicity of the zero.
The x-intercept −3 is the solution of equation \((x+3)=0\). The graph passes directly through thex-intercept at \(x=−3\). The factor is linear (has a degree of 1), so the behavior near the intercept is like that of a line—it passes directly through the intercept. We call this a single zero because the zero corresponds to a single factor of the function.
The x-intercept 2 is the repeated solution of equation \((x−2)^2=0\). The graph touches the axis at the intercept and changes direction. The factor is quadratic (degree 2), so the behavior near the intercept is like that of a quadratic—it bounces off of the horizontal axis at the intercept.
\[(x−2)^2=(x−2)(x−2)\]
The factor is repeated, that is, the factor \((x−2)\) appears twice. The number of times a given factor appears in the factored form of the equation of a polynomial is called the multiplicity. The zero associated with this factor, \(x=2\), has multiplicity 2 because the factor \((x−2)\) occurs twice.
The x-intercept −1 is the repeated solution of factor \((x+1)^3=0\).The graph passes through the axis at the intercept, but flattens out a bit first. This factor is cubic (degree 3), so the behavior near the intercept is like that of a cubic—with the same S-shape near the intercept as the toolkit function \(f(x)=x^3\). We call this a triple zero, or a zero with multiplicity 3.
For zeros with even multiplicities, the graphs touch or are tangent to the x-axis. For zeros with odd multiplicities, the graphs cross or intersect the x-axis. See Figure \(\PageIndex{8}\) for examples of graphs of polynomial functions with multiplicity 1, 2, and 3.
Three graphs showing three different polynomial functions with multiplicity 1, 2, and 3.
Figure \(\PageIndex{8}\): Three graphs showing three different polynomial functions with multiplicity 1, 2, and 3.
For higher even powers, such as 4, 6, and 8, the graph will still touch and bounce off of the horizontal axis but, for each increasing even power, the graph will appear flatter as it approaches and leaves the x-axis.
For higher odd powers, such as 5, 7, and 9, the graph will still cross through the horizontal axis, but for each increasing odd power, the graph will appear flatter as it approaches and leaves the x-axis.
Graphical Behavior of Polynomials at x-Intercepts
If a polynomial contains a factor of the form \((x−h)^p\), the behavior near the x-intercepth is determined by the power \(p\). We say that \(x=h\) is a zero of multiplicity \(p\).
The graph of a polynomial function will touch the x-axis at zeros with even multiplicities. The graph will cross the x-axis at zeros with odd multiplicities.
The sum of the multiplicities is the degree of the polynomial function.
HOWTO: Given a graph of a polynomial function of degree \(n\), identify the zeros and their multiplicities
1. If the graph crosses the x-axis and appears almost linear at the intercept, it is a single zero.
2. If the graph touches the x-axis and bounces off of the axis, it is a zero with even multiplicity.
3. If the graph crosses the x-axis at a zero, it is a zero with odd multiplicity.
4. The sum of the multiplicities is \(n\).
Example \(\PageIndex{6}\): Identifying Zeros and Their Multiplicities
Use the graph of the function of degree 6 in Figure \(\PageIndex{9}\) to identify the zeros of the function and their possible multiplicities.
Graph of an even-degree polynomial with degree 6.
Figure \(\PageIndex{9}\): Graph of a polynomial function with degree 5.
Solution
The polynomial function is of degree \(n\). The sum of the multiplicities must be \(n\).
Starting from the left, the first zero occurs at \(x=−3\). The graph touches the x-axis, so the multiplicity of the zero must be even. The zero of −3 has multiplicity 2.
The next zero occurs at \(x=−1\). The graph looks almost linear at this point. This is a single zero of multiplicity 1.
The last zero occurs at \(x=4\).The graph crosses the x-axis, so the multiplicity of the zero must be odd. We know that the multiplicity is likely 3 and that the sum of the multiplicities is likely 6.
Exercise \(\PageIndex{2}\)
Use the graph of the function of degree 5 in Figure \(\PageIndex{10}\) to identify the zeros of the function and their multiplicities.
Graph of a polynomial function with degree 5.
Figure \(\PageIndex{10}\): Graph of a polynomial function with degree 5.
Answer
The graph has a zero of –5 with multiplicity 1, a zero of –1 with multiplicity 2, and a zero of 3 with even multiplicity.
Determining End Behavior
As we have already learned, the behavior of a graph of a polynomial function of the form
\[f(x)=a_nx^n+a_{n−1}x^{n−1}+...+a_1x+a_0\]
will either ultimately rise or fall as \(x\) increases without bound and will either rise or fall as \(x\) decreases without bound. This is because for very large inputs, say 100 or 1,000, the leading term dominates the size of the output. The same is true for very small inputs, say –100 or –1,000.
Recall that we call this behavior the end behavior of a function. As we pointed out when discussing quadratic equations, when the leading term of a polynomial function, \(a_nx^n\), is an even power function, as \(x\) increases or decreases without bound, \(f(x)\) increases without bound. When the leading term is an odd power function, as \(x\) decreases without bound, \(f(x)\) also decreases without bound; as \(x\) increases without bound, \(f(x)\) also increases without bound. If the leading term is negative, it will change the direction of the end behavior. Figure \(\PageIndex{11}\) summarizes all four cases.
Table showing the end behavior of odd and even polynomials with positive and negative coefficients
Figure \(\PageIndex{11}\).
Understanding the Relationship between Degree and Turning Points
In addition to the end behavior, recall that we can analyze a polynomial function’s local behavior. It may have a turning point where the graph changes from increasing to decreasing (rising to falling) or decreasing to increasing (falling to rising). Look at the graph of the polynomial function \(f(x)=x^4−x^3−4x^2+4x\) in Figure \(\PageIndex{12}\). The graph has three turning points.
Graph of f(x)=x^4-x^3-4x^2+4x which denotes where the function increases and decreases and its turning points.
Figure \(\PageIndex{12}\): Graph of \(f(x)=x^4-x^3-4x^2+4x\)
This function \(f\) is a 4th degree polynomial function and has 3 turning points. The maximum number of turning points of a polynomial function is always one less than the degree of the function.
Definition: Interpreting Turning Points
A turning point is a point of the graph where the graph changes from increasing to decreasing (rising to falling) or decreasing to increasing (falling to rising). A polynomial of degree \(n\) will have at most \(n−1\) turning points.
Example \(\PageIndex{7}\): Finding the Maximum Number of Turning Points Using the Degree of a Polynomial Function
Find the maximum number of turning points of each polynomial function.
1. \(f(x)=−x^3+4x^5−3x^2+1\)
2. \(f(x)=−(x−1)^2(1+2x^2)\)
Solution
a. \(f(x)=−x^3+4x^5−3x^2+1\)
First, rewrite the polynomial function in descending order: \(f(x)=4x^5−x^3−3x^2+1\)
Identify the degree of the polynomial function. This polynomial function is of degree 5.
The maximum number of turning points is \(5−1=4\).
b. \(f(x)=−(x−1)^2(1+2x^2)\)
First, identify the leading term of the polynomial function if the function were expanded.
imageedit_33_3540887475.png
Then, identify the degree of the polynomial function. This polynomial function is of degree 4.
The maximum number of turning points is \(4−1=3\).
Graphing Polynomial Functions
We can use what we have learned about multiplicities, end behavior, and turning points to sketch graphs of polynomial functions. Let us put this all together and look at the steps required to graph polynomial functions.
Howto: Given a polynomial function, sketch the graph
1. Find the intercepts.
2. Check for symmetry. If the function is an even function, its graph is symmetrical about the y-axis, that is, \(f(−x)=f(x)\). If a function is an odd function, its graph is symmetrical about the origin, that is, \(f(−x)=−f(x)\).
3. Use the multiplicities of the zeros to determine the behavior of the polynomial at the x-intercepts.
4. Determine the end behavior by examining the leading term.
5. Use the end behavior and the behavior at the intercepts to sketch a graph.
6. Ensure that the number of turning points does not exceed one less than the degree of the polynomial.
7. Optionally, use technology to check the graph.
Example \(\PageIndex{8}\): Sketching the Graph of a Polynomial Function
Sketch a graph of \(f(x)=−2(x+3)^2(x−5)\).
Solution
This graph has two x-intercepts. At \(x=−3\), the factor is squared, indicating a multiplicity of 2. The graph will bounce at this x-intercept. At \(x=5\),the function has a multiplicity of one, indicating the graph will cross through the axis at this intercept.
The y-intercept is found by evaluating \(f(0)\).
\[\begin{align*} f(0)&=−2(0+3)^2(0−5) \\ &=−2⋅9⋅(−5) \\ &=90 \end{align*}\]
The y-intercept is \((0,90)\).
Additionally, we can see the leading term, if this polynomial were multiplied out, would be \(−2x3\), so the end behavior is that of a vertically reflected cubic, with the outputs decreasing as the inputs approach infinity, and the outputs increasing as the inputs approach negative infinity. See Figure \(\PageIndex{13}\).
Showing the distribution for the leading term.
Figure \(\PageIndex{13}\): Showing the distribution for the leading term.
To sketch this, we consider that:
• As \(x{\rightarrow}−{\infty}\) the function \(f(x){\rightarrow}{\infty}\),so we know the graph starts in the second quadrant and is decreasing toward the x-axis.
• Since \(f(−x)=−2(−x+3)^2(−x–5)\) is not equal to \(f(x)\), the graph does not display symmetry.
• At \((−3,0)\), the graph bounces off of thex-axis, so the function must start increasing.
• At \((0,90)\), the graph crosses the y-axis at the y-intercept. See Figure \(\PageIndex{14}\).
Graph of the end behavior and intercepts, (-3, 0) and (0, 90), for the function f(x)=-2(x+3)^2(x-5).
Figure \(\PageIndex{14}\): Graph of the end behavior and intercepts, \((-3, 0)\) and \((0, 90)\), for the function \(f(x)=-2(x+3)^2(x-5)\).
Somewhere after this point, the graph must turn back down or start decreasing toward the horizontal axis because the graph passes through the next intercept at \((5,0)\). See Figure \(\PageIndex{15}\).
Graph of the end behavior and intercepts, (-3, 0), (0, 90) and (5, 0), for the function f(x)=-2(x+3)^2(x-5).
Figure \(\PageIndex{15}\): Graph of the end behavior and intercepts, \((-3, 0)\), \((0, 90)\) and \((5, 0)\), for the function \(f(x)=-2(x+3)^2(x-5)\).
As \(x{\rightarrow}{\infty}\) the function \(f(x){\rightarrow}−{\infty}\),
so we know the graph continues to decrease, and we can stop drawing the graph in the fourth quadrant.
Using technology, we can create the graph for the polynomial function, shown in Figure \(\PageIndex{16}\), and verify that the resulting graph looks like our sketch in Figure \(\PageIndex{15}\).
The complete graph of the polynomial function f(x)=−2(x+3)^2(x−5)
Figure \(\PageIndex{16}\): The complete graph of the polynomial function \(f(x)=−2(x+3)^2(x−5)\).
Exercise \(\PageIndex{8}\)
Sketch a graph of \(f(x)=\dfrac{1}{4}x(x−1)^4(x+3)^3\).
Answer
Figure \(\PageIndex{17}\): Graph of \(f(x)=\frac{1}{4}x(x−1)^4(x+3)^3\)
Using the Intermediate Value Theorem
In some situations, we may know two points on a graph but not the zeros. If those two points are on opposite sides of the x-axis, we can confirm that there is a zero between them. Consider a polynomial function \(f\) whose graph is smooth and continuous. The Intermediate Value Theorem states that for two numbers \(a\) and \(b\) in the domain of \(f\),if \(a<b\) and \(f(a){\neq}f(b)\),then the function \(f\) takes on every value between \(f(a)\) and \(f(b)\). We can apply this theorem to a special case that is useful in graphing polynomial functions. If a point on the graph of a continuous function \(f\) at \(x=a\) lies above the x-axis and another point at \(x=b\) lies below thex-axis, there must exist a third point between \(x=a\) and \(x=b\) where the graph crosses the x-axis. Call this point \((c,f(c))\).This means that we are assured there is a solution \(c\) where \(f(c)=0\).
In other words, the Intermediate Value Theorem tells us that when a polynomial function changes from a negative value to a positive value, the function must cross the x-axis. Figure \(\PageIndex{18}\) shows that there is a zero between \(a\) and \(b\).
Graph of an odd-degree polynomial function that shows a point f(a) that’s negative, f(b) that’s positive, and f(c) that’s 0.
Figure \(\PageIndex{18}\): Using the Intermediate Value Theorem to show there exists a zero.
Definition: Intermediate Value Theorem
Let \(f\) be a polynomial function. The Intermediate Value Theorem states that if \(f(a)\) and \(f(b)\) have opposite signs, then there exists at least one value \(c\) between \(a\) and \(b\) for which \(f(c)=0\).
Example \(\PageIndex{9}\): Using the Intermediate Value Theorem
Show that the function \(f(x)=x^3−5x^2+3x+6\) has at least two real zeros between \(x=1\) and \(x=4\).
Solution
As a start, evaluate \(f(x)\) at the integer values \(x=1,\;2,\;3,\; \text{and }4\) (Table \(\PageIndex{2}\)).
Table \(\PageIndex{2}\)
\(x\) 1 2 3 4
\(f(x)\) 5 0 -3 2
We see that one zero occurs at \(x=2\). Also, since \(f(3)\) is negative and \(f(4)\) is positive, by the Intermediate Value Theorem, there must be at least one real zero between 3 and 4.
We have shown that there are at least two real zeros between \(x=1\) and \(x=4\).
Analysis
We can also see on the graph of the function in Figure \(\PageIndex{19}\) that there are two real zeros between \(x=1\) and \(x=4\).
Graph of f(x)=x^3-5x^2+3x+6 and shows, by the Intermediate Value Theorem, that there exists two zeros since f(1)=5 and f(4)=2 are positive and f(3) = -3 is negative.
Figure \(\PageIndex{19}\).
Exercise \(\PageIndex{4}\)
Show that the function \(f(x)=7x^5−9x^4−x^2\) has at least one real zero between \(x=1\) and \(x=2\).
Answer
Because \(f\) is a polynomial function and since \(f(1)\) is negative and \(f(2)\) is positive, there is at least one real zero between \(x=1\) and \(x=2\).
Writing Formulas for Polynomial Functions
Now that we know how to find zeros of polynomial functions, we can use them to write formulas based on graphs. Because a polynomial function written in factored form will have an x-intercept where each factor is equal to zero, we can form a function that will pass through a set of x-intercepts by introducing a corresponding set of factors.
Note: Factored Form of Polynomials
If a polynomial of lowest degree \(p\) has horizontal intercepts at \(x=x_1,x_2,…,x_n\), then the polynomial can be written in the factored form: \(f(x)=a(x−x_1)^{p_1}(x−x_2)^{p_2}⋯(x−x_n)^{p_n}\) where the powers \(p_i\) on each factor can be determined by the behavior of the graph at the corresponding intercept, and the stretch factor \(a\) can be determined given a value of the function other than the x-intercept.
Given a graph of a polynomial function, write a formula for the function.
1. Identify the x-intercepts of the graph to find the factors of the polynomial.
2. Examine the behavior of the graph at the x-intercepts to determine the multiplicity of each factor.
3. Find the polynomial of least degree containing all the factors found in the previous step.
4. Use any other point on the graph (the y-intercept may be easiest) to determine the stretch factor.
Example \(\PageIndex{10}\): Writing a Formula for a Polynomial Function from the Graph
Write a formula for the polynomial function shown in Figure \(\PageIndex{20}\).
Graph of a positive even-degree polynomial with zeros at x=-3, 2, 5 and y=-2.
Figure \(\PageIndex{20}\).
Solution
This graph has three x-intercepts: \(x=−3,\;2,\text{ and }5\). The y-intercept is located at \((0,2)\) .At \(x=−3\) and \( x=5\), the graph passes through the axis linearly, suggesting the corresponding factors of the polynomial will be linear. At \(x=2\), the graph bounces at the intercept, suggesting the corresponding factor of the polynomial will be second degree (quadratic). Together, this gives us
\[f(x)=a(x+3)(x−2)^2(x−5)\]
To determine the stretch factor, we utilize another point on the graph. We will use the y-intercept \((0,–2)\), to solve for \(a\).
\[\begin{align*} f(0)&=a(0+3)(0−2)^2(0−5) \\ −2&=a(0+3)(0−2)^2(0−5) \\ −2&=−60a \\ a&=\dfrac{1}{30} \end{align*}\]
The graphed polynomial appears to represent the function \(f(x)=\dfrac{1}{30}(x+3)(x−2)^2(x−5)\).
Exercise \(\PageIndex{5}\)
Given the graph shown in Figure \(\PageIndex{21}\), write a formula for the function shown.
Graph of a negative even-degree polynomial with zeros at x=-1, 2, 4 and y=-4.
Figure \(\PageIndex{21}\).
Answer
\(f(x)=−\frac{1}{8}(x−2)3(x+1)2(x−4)\)
Using Local and Global Extrema
With quadratics, we were able to algebraically find the maximum or minimum value of the function by finding the vertex. For general polynomials, finding these turning points is not possible without more advanced techniques from calculus. Even then, finding where extrema occur can still be algebraically challenging. For now, we will estimate the locations of turning points using technology to generate a graph.
Each turning point represents a local minimum or maximum. Sometimes, a turning point is the highest or lowest point on the entire graph. In these cases, we say that the turning point is a global maximum or a global minimum. These are also referred to as the absolute maximum and absolute minimum values of the function.
Note: Local and Global Extrema
A local maximum or local minimum at \(x=a\) (sometimes called the relative maximum or minimum, respectively) is the output at the highest or lowest point on the graph in an open interval around \(x=a\).If a function has a local maximum at \(a\), then \(f(a){\geq}f(x)\)for all \(x\) in an open interval around \(x=a\). If a function has a local minimum at \(a\), then \(f(a){\leq}f(x)\)for all \(x\) in an open interval around \(x=a\).
A global maximum or global minimum is the output at the highest or lowest point of the function. If a function has a global maximum at \(a\), then \(f(a){\geq}f(x)\) for all \(x\). If a function has a global minimum at \(a\), then \(f(a){\leq}f(x)\) for all \(x\).
We can see the difference between local and global extrema in Figure \(\PageIndex{22}\).
Graph of an even-degree polynomial that denotes the local maximum and minimum and the global maximum.
Figure \(\PageIndex{22}\): Graph of an even-degree polynomial that denotes the local maximum and minimum and the global maximum.
Do all polynomial functions have a global minimum or maximum?
No. Only polynomial functions of even degree have a global minimum or maximum. For example, \(f(x)=x\) has neither a global maximum nor a global minimum.
Example \(\PageIndex{11}\): Using Local Extrema to Solve Applications
An open-top box is to be constructed by cutting out squares from each corner of a 14 cm by 20 cm sheet of plastic then folding up the sides. Find the size of squares that should be cut out to maximize the volume enclosed by the box.
Solution
We will start this problem by drawing a picture like that in Figure \(\PageIndex{23}\), labeling the width of the cut-out squares with a variable, \(w\).
Diagram of a rectangle with four squares at the corners.
Figure \(\PageIndex{23}\): Diagram of a rectangle with four squares at the corners.
Notice that after a square is cut out from each end, it leaves \(a(14−2w)\) cm by \((20−2w)\) cm rectangle for the base of the box, and the box will be \(w\) cm tall. This gives the volume
\[\begin{align*} V(w)&=(20−2w)(14−2w)w \\ &=280w−68w^2+4w^3 \end{align*}\]
Notice, since the factors are \(w\), \(20–2w\) and \(14–2w\), the three zeros are 10, 7, and 0, respectively. Because a height of 0 cm is not reasonable, we consider the only the zeros 10 and 7. The shortest side is 14 and we are cutting off two squares, so values \(w\) may take on are greater than zero or less than 7. This means we will restrict the domain of this function to \(0<w<7\).Using technology to sketch the graph of \(V(w)\) on this reasonable domain, we get a graph like that in Figure \(\PageIndex{24}\). We can use this graph to estimate the maximum value for the volume, restricted to values for \(w\) that are reasonable for this problem—values from 0 to 7.
Graph of V(w)=(20-2w)(14-2w)w where the x-axis is labeled w and the y-axis is labeled V(w).
Figure \(\PageIndex{24}\): Graph of \(V(w)=(20-2w)(14-2w)w\)
From this graph, we turn our focus to only the portion on the reasonable domain, \([0, 7]\). We can estimate the maximum value to be around 340 cubic cm, which occurs when the squares are about 2.75 cm on each side. To improve this estimate, we could use advanced features of our technology, if available, or simply change our window to zoom in on our graph to produce Figure \(\PageIndex{25}\).
Graph of V(w)=(20-2w)(14-2w)w where the x-axis is labeled w and the y-axis is labeled V(w) on the domain [2.4, 3].
Figure \(\PageIndex{25}\): Graph of \(V(w)=(20-2w)(14-2w)w\).
From this zoomed-in view, we can refine our estimate for the maximum volume to about 339 cubic cm, when the squares measure approximately 2.7 cm on each side.
Exercise \(\PageIndex{1}\)
Use technology to find the maximum and minimum values on the interval \([−1,4]\) of the function \(f(x)=−0.2(x−2)^3(x+1)^2(x−4)\).
Answer
The minimum occurs at approximately the point \((0,−6.5)\),
and the maximum occurs at approximately the point \((3.5,7)\).
Key Concepts
• Polynomial functions of degree 2 or more are smooth, continuous functions.
• To find the zeros of a polynomial function, if it can be factored, factor the function and set each factor equal to zero.
• Another way to find the x-intercepts of a polynomial function is to graph the function and identify the points at which the graph crosses the x-axis.
• The multiplicity of a zero determines how the graph behaves at the x-intercepts.
• The graph of a polynomial will cross the horizontal axis at a zero with odd multiplicity.
• The graph of a polynomial will touch the horizontal axis at a zero with even multiplicity.
• The end behavior of a polynomial function depends on the leading term.
• The graph of a polynomial function changes direction at its turning points.
• A polynomial function of degree \(n\) has at most \(n−1\) turning points.
• To graph polynomial functions, find the zeros and their multiplicities, determine the end behavior, and ensure that the final graph has at most \(n−1\) turning points.
• Graphing a polynomial function helps to estimate local and global extremas.
• The Intermediate Value Theorem tells us that if \(f(a)\) and \(f(b)\) have opposite signs, then there exists at least one value \(c\) between \(a\) and \(b\) for which \(f(c)=0\).
Glossary
global maximum
highest turning point on a graph; \(f(a)\) where \(f(a){\geq}f(x)\) for all \(x\).
global minimum
lowest turning point on a graph; \(f(a)\) where \(f(a){\leq}f(x)\) for all \(x\).
Intermediate Value Theorem
for two numbers \(a\) and \(b\) in the domain of \(f\), if \(a<b\) and \(f(a){\neq}f(b)\), then the functionf takes on every value between \(f(a)\) and \(f(b)\); specifically, when a polynomial function changes from a negative value to a positive value, the function must cross the x-axis
multiplicity
the number of times a given factor appears in the factored form of the equation of a polynomial; if a polynomial contains a factor of the form \((x−h)^p\), \(x=h\) is a zero of multiplicity \(p\).
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6,082,048,758,532,760,000 | Number 69
[ sixty-nine ]
Properties of number 69
Cross Sum:
Factorization:
3 * 23
Divisors:
1, 3, 23, 69
Count of divisors:
Sum of divisors:
Prime number?
No
Fibonacci number?
No
Bell Number?
No
Catalan Number?
No
Base 2 (Binary):
Base 3 (Ternary):
Base 4 (Quaternary):
Base 5 (Quintal):
Base 8 (Octal):
Base 16 (Hexadecimal):
Base 32:
sin(69)
-0.11478481378319
cos(69)
0.99339037972227
tan(69)
-0.11554854579453
ln(69)
4.2341065045973
lg(69)
1.8388490907373
sqrt(69)
8.3066238629181
Square(69)
Number Look Up
Look Up
69 which is pronounced (sixty-nine) is a very special figure. The cross sum of 69 is 15. If you factorisate the number 69 you will get these result 3 * 23. The figure 69 has 4 divisors ( 1, 3, 23, 69 ) whith a sum of 96. The figure 69 is not a prime number. The figure 69 is not a fibonacci number. 69 is not a Bell Number. 69 is not a Catalan Number. The convertion of 69 to base 2 (Binary) is 1000101. The convertion of 69 to base 3 (Ternary) is 2120. The convertion of 69 to base 4 (Quaternary) is 1011. The convertion of 69 to base 5 (Quintal) is 234. The convertion of 69 to base 8 (Octal) is 105. The convertion of 69 to base 16 (Hexadecimal) is 45. The convertion of 69 to base 32 is 25. The sine of the number 69 is -0.11478481378319. The cosine of the number 69 is 0.99339037972227. The tangent of 69 is -0.11554854579453. The root of 69 is 8.3066238629181.
If you square 69 you will get the following result 4761. The natural logarithm of 69 is 4.2341065045973 and the decimal logarithm is 1.8388490907373. I hope that you now know that 69 is special number! | {
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541,891,860,662,238,340 | login
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A342510 a(n) = k where Z_k is the largest Zimin word that n (read as a binary word) does not avoid. 3
1, 1, 1, 1, 1, 2, 1, 2, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 2 (list; graph; refs; listen; history; text; internal format)
OFFSET
0,6
COMMENTS
Zimin words are defined recursively: Z_1 = A, Z_2 = ABA, Z_3 = ABACABA, and Z_{i+1} = Z_i a_{i+1} Z_i.
Every Zimin word Z_i is an "unavoidable" word, meaning that every sufficiently long string over a finite alphabet contains a substring that is an instance of Z_i. A word w is instance of a Zimin word Z_i if there's a nonerasing monoid homomorphism from Z_i to w.
a(n) >= 2 for all n >= 2^4.
a(n) >= 3 for all n >= 2^28.
For any fixed k, a(n) >= k for sufficiently large n, however there exists a value of a(n) = 3 with n >= 2^10482.
The first occurrence of k is when n = A001045(2^k), that is, the binary expansion of n is "1010101...01" with 2^k - 1 bits.
LINKS
Peter Kagey, Table of n, a(n) for n = 0..8191
Peter Kagey, Matching ABACABA-type patterns, Code Golf Stack Exchange.
Danny Rorabaugh, Toward the Combinatorial Limit Theory of Free Words, arXiv preprint arXiv:1509.04372 [math.CO], 2015.
Wikipedia, Sesquipower.
EXAMPLE
For n = 10101939, the binary representation is "100110100010010010110011", and the substring "0010010010" is an instance of the Zimin word Z_3 = ABACABA with A = "0", B = "01", and C = "01".
No substring is an instance of the Zimin word Z_4 = ABACABADABACABA, so a(10101939)= 3.
PROG
(Python)
def sd(w): # sesquipower degree
return 1 + max([0]+[sd(w[:i]) for i in range(1, (len(w)+1)//2) if w[:i] == w[-i:]])
def a(n):
w = bin(n)[2:]
return max(sd(w[i:j]) for i in range(len(w)) for j in range(i+1, len(w)+1))
print([a(n) for n in range(87)]) # Michael S. Branicky, Mar 15 2021
CROSSREFS
Cf. A001045.
Cf. A342511, A342512.
Sequence in context: A043557 A055027 A214574 * A298071 A246920 A244964
Adjacent sequences: A342507 A342508 A342509 * A342511 A342512 A342513
KEYWORD
nonn,base
AUTHOR
Peter Kagey, Mar 14 2021
STATUS
approved
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6,284,243,439,068,817,000 | Bayes rule exercises: part 2
1. The Prosecutor's Fallacy
Bayes’ theorem often applies when considering the probability that a person is guilty of a crime. Indeed, misunderstanding it leads to what is called the prosecutor’s fallacy – when you interpret the small probability of someone fitting the evidence as a small probability that an accused who does fit the evidence is in fact innocent.
Let’s consider the following case:
In a murder case you have found a sample of the murderer’s DNA, and there is a 0.1% chance of a random someone’s DNA matching this sample. You have found a man whose DNA does match.
Then the correct interpretation is NOT there is 0.1% chance that this man is not the murderer, ie there is 99.9% chance that he is the murderer.
Bayes’ theorem tells us that in order to calculate this last probability – the probability that the man is guilty, given that he matches the DNA, one also needs to take into account the probability of a random person being a murderer, which is extremely low, say it is 0.01%.
Let’s use the following notation:
• Event A = The man is guilty
• Event B = The man’s DNA matches the one found
Then we have
• P(B) = 0.1%
• P(A) = 0.01%
The probability we are interested in is P(A|B): the probability that the man is guilty, given that his DNA matches the killer’s.
Bayes’ theorem gives us that P(A|B) = P(B|A) * P(A) / P(B)
Now P(B|A) is the probability that the man’s DNA would match the killer’s, if he is indeed the killer. Which should be pretty close to 1, if your DNA testing is any good! P(A) and P(B) are given above, so in the end:
P(A|B) = 10%
Most definitely not a cause for putting someone in prison!
Obviously, in actual cases, this is not the only thing to take into account. If the man’s DNA matches the killer, and he also matches a description of the killer, and has no alibi, and some shoes were found in his house covered in blood, the odds would change somewhat.
Source: http://mathsontrial.blogspot.com.br/2011/09/bayes-theorem-examples.html
2. Sick Child and Doctor
A doctor is called to see a sick child. The doctor has prior information that 90% of sick children in that neighborhood have the flu, while the other 10% are sick with measles. Let F stand for an event of a child being sick with flu and M stand for an event of a child being sick with measles. Assume for simplicity that F ∪ M = Ω, i.e., that there no other maladies in that neighborhood.
A well-known symptom of measles is a rash (the event of having which we denote R). P(R|M) = .95. However, occasionally children with flu also develop rash, so that P(R|F) = 0.08.
Upon examining the child, the doctor finds a rash. What is the probability that the child has measles?
P(M|R) = P(R|M) P(M) / (P(R|M) P(M) + P(R|F) P(F)) = .95 × .10 / (.95 × .10 + .08 × .90) ≈ 0.57.
Which is nowhere close to 95% of P(R|M).
Source: http://www.cut-the-knot.org/Probability/BayesTheorem.shtml
3. Incidence of Breast Cancer
In a study, physicians were asked what the odds of breast cancer would be in a woman who was initially thought to have a 1% risk of cancer but who ended up with a positive mammogram result (a mammogram accurately classifies about 80% of cancerous tumors and 90% of benign tumors.) 95 out of a hundred physicians estimated the probability of cancer to be about 75%. Do you agree?
Introduce the events:
• P - mammogram result is positive,
• B - tumor is benign,
• M - tumor is malignant.
Bayes' formula in this case is
P(M|P) = P(P|M) P(M) / (P(P|M) P(M) + P(P|B) P(B)) = .80 × .01 / (.80 × .01 + .10 × .99) ≈ 0.075 ≈ 7.5%.
A far cry from a common estimate of 75%.
Source: http://www.cut-the-knot.org/Probability/BayesTheorem.shtml
4. Smoking and Lung Cancer
Suppose 0.1% of the American population currently has lung cancer, that 90% of all lung cancer cases are smokers, and that 21% of those without lung cancer also smoke. (These values are fairly close to the values given on the American Lung Association web site as of 2011.) Consider the following questions.
• What percent of smokers have lung cancer?
• What percent of non-smokers have lung cancer?
• How much more likely is a smoker to have lung cancer than a non-smoker?
We begin by recognizing two variables for each subject, smoking and lung cancer. Let event S be the smokers, and event L be those with lung cancer. The data given specifies P(L) = 0.001, P(S|L) = 0.90 and P(S|L) = 0.21. The questions are asking P(L|S) and P(L|S). Since the conditional probabilities in the information and the question are reversed, we recognize the need to use a Bayes Theorem approach. Therefore, we shall construct a tree diagram, with the lung cancer data in the first set of branches to take advantage of the given information.
statistics-for-beginners-bayes-rule-2-task-4-1
After including the original data, we recognize that the probabilities on each set of branches must add to one, and the probabilities along branches must multiply to the final result. With the completed tree diagram, we can answer the questions.
• P(L|S) = P(L∩S) / P(S) = 0.0009 / (0.0009 + 0.20979) = 0.0042717
Therefore, approximately 0.43% of all smokers have lung cancer.
• P(L|S) = P(L∩S) / P(S) = 0.0001 / (0.0001 + 0.78921) = 0.0001267
Therefore, approximately 0.01% of all non-smokers have lung cancer.
• P(L|S) / P(L|S) = 0.0042717 / 0.0001267 = 33.72
Therefore, smokers are almost 34 times more likely to have lung cancer than non-smokers. (This is slightly higher than the American Lung Association is reporting, but then we have made a number of simplifications in our statement of the problem.)
This analysis does not say that smoking caused lung cancer. The only way statistics can be used to determine causation is for the researcher to control the variables involved. To do this, the researcher would have to randomly assign some subjects to be smokers, and some to not smoke, and then determine the incidence of lung cancer among his participants. However, controlling the lives of human subjects in this way would not be acceptable.
Source: http://www.milefoot.com/math/stat/prob-bayes.htm
5. False Positives
One prominent manufacturer of medical tests offers a test for chlamydia (a sexually transmitted disease) that has a sensitivity of 76.4% and a specificity of 93.2%. In other words, the test correctly identifies 76.4% of individuals tested who have the disease by giving a positive result, and correctly identifies 93.2% of the individuals who are healthy by giving a negative result. Currently, it is estimated that 1.5% of the American population has chlamydia. If one individual is randomly selected from the population and tests positive for chlamydia, what is the probability that he/she does not have the disease?
Our variables are the presence of chlamydia and the test result. Let C be those who have chlamydia, and let T be those who have a positive test result. The data states that P(C) = 0.015, P(T|C) = 0.764, and P(T|C) = 0.932. The question being asked is the value of P(C|T). Once again, we construct a tree diagram.
statistics-for-beginners-bayes-rule-2-task-5-1
We then have:
P(C|T) = P(C∩T) / P(T) = 0.06698 / (0.01146 + 0.06698) = 0.8539
In other words, there is an 85% chance that a positive result would be a false positive. This is a horrendous result. You might think that if the sensitivity of the test was higher, the situation would be rectified. However, the real problem is the rarity of the condition in the general population. Such medical tests should not be done unless there is some reason to suspect the disease, and if a positive result occurs, a second (different) test might be appropriate to confirm the diagnosis of the first test.
Source: http://www.milefoot.com/math/stat/prob-bayes.htm
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2,537,809,328,529,018,000 | tsujimotterのノートブック
日曜数学者 tsujimotter の「趣味で数学」実践ノート
四元数環と2-コサイクル
今日は 四元数環 \newcommand{\hh}{\mathbb{H}}\hh について考えてみましょう。tsujimotterのノートブックでは初登場ですね。
f:id:tsujimotter:20200328111311p:plain:w260
複素数体 \mathbb{C} \mathbb{R} に虚数単位 i^2 = -1 を加えた体のことで、
\mathbb{C} = \mathbb{R}\cdot 1 + \mathbb{R}\cdot i
と書けます。 \mathbb{R} 上の2次拡大体となっています。
\mathbb{C} に「ある演算規則」をもった j, k という新しい数を加えて
\mathbb{H} = \mathbb{R}\cdot 1 + \mathbb{R}\cdot i + \mathbb{R}\cdot j + \mathbb{R}\cdot k
としたものが四元数環です。複素数体の4次元バージョンと思えます。
i, j, k には、次のような関係が成り立ち、これが四元数環を定義する演算規則となっています:
i^2 = j^2 = k^2 = ijk = -1 \tag{1}
この式は、ハミルトンがブルーム橋に刻んだ数式としても知られていますね。
先日、四元数環と群コホモロジーの意外な接点 について教えていただきました。とても面白かったので、ぜひ紹介したいというのが今日の目的です。具体的には、ある2-コサイクルを考えると、そこから四元数環を構成することができるというのです。
今回の内容は梅崎さんに教えていただきました。いつも楽しい話を教えてくださってありがとうございます。
なお、もし内容に誤りがあったとしても、それは私の理解不足によるところかと思います。見つけた方は私にTwitter等でご指摘いただけますと幸いです。
四元数環の構造
複素数 \mathbb{C} が体だったのに対して、四元数環は積が非可換であるような 「非可換体」 であるのが特徴です。
\mathbb{H} が非可換であることは、たとえば
ji = -ij \tag{2}
であることからわかります。
上の式はブルーム橋の式 (1) から次のように導けます。
まず、 ijk = -1 の両辺に k を右からかけて ijkk = -k であり、 k^2 = -1 より
ij = k \tag{3}
が得られます。
次に jii = -j jk を右からかけて jiijk = -jjk となる。 ijk = -1, \; jj = -1 を適用すると
ji = -k \tag{4}
が得られて、式 (3), (4) を合わせて ji = -k = -ij が得られます。
こんな構造を持った四元数環なのですが、実は、2次のコホモロジーの元である2-コサイクルから作ることができるというのです。
2次の群コホモロジー
まずは、群コホモロジーの記事でやったように、 G を群、 M G が作用する群として、2次の群コホモロジー H^2(G, M) を計算してみましょう。
tsujimotter.hatenablog.com
今回の話に限れば 「2次のコホモロジーの元 [\varphi] \in H^2(G, M) の代表元 \varphi は2-コサイクルである」という点と「2-コサイクルの具体的な計算結果」だけ分かればOKです。
まず、 C^2(G, M) を考えます。これは
\varphi \colon G\times G \longrightarrow M
という写像全体の集合です。今回は、 G の元を2つ変数にとる写像が主役なわけですね。
いつものように 2-コバウンダリ を計算します。今日は後の事を考えて「乗法形」で書きたいと思います。
d^2\colon C^1(G, M) \longrightarrow C^2(G, M), \; \psi \longmapsto d^2(\psi)
における d^2(\psi) は定義より
\displaystyle d^2(\psi)(g_1, g_2) = \psi(g_2)^{g_1}\psi(g_1 g_2)^{-1} \psi(g_1)
ですから、ある \psi \in C^1(G, M) が存在して、任意の g_1, g_2 \in G に対して
\varphi(g_1, g_2) = \psi(g_2)^{g_1}\psi(g_1 g_2)^{-1} \psi(g_1)
と表せるような \varphi \in C^2(G, M) が2-コバウンダリということになりますね。
続いて、2-コサイクル ですが、今度は d^3 が必要です。
d^3\colon C^2(G, M) \longrightarrow C^3(G, M), \; \varphi \longmapsto d^3(\varphi)
この式における d^3(\varphi) は定義より、任意の g_1, g_2, g_3 \in G に対して
\displaystyle d^3(\varphi)(g_1, g_2, g_3) = \varphi(g_2, g_3)^{g_1}\varphi(g_1 g_2, g_3)^{-1}\varphi(g_1, g_2 g_3) \varphi(g_1, g_2)^{-1}
と計算されます。
ここで、 d^3(\varphi) = 1 より、任意の g_1, g_2, g_3 \in G に対して
\displaystyle \varphi(g_2, g_3)^{g_1}\varphi(g_1 g_2, g_3)^{-1}\varphi(g_1, g_2 g_3) \varphi(g_1, g_2)^{-1} = 1
を満たすような \varphi \in C^2(G, M) が2-コサイクルです。
よって、2次の群コホモロジーは
\displaystyle H^2(G, M) = \frac{ \{ \varphi \in C^2(G, M) \; \mid \; \forall g_1, g_2, g_3 \in G, \; \varphi(g_2, g_3)^{g_1}\varphi(g_1 g_2, g_3)^{-1}\varphi(g_1, g_2 g_3) \varphi(g_1, g_2)^{-1} = 1 \} }{ \{ \varphi \in C^2(G, M) \; \mid \; \exists \psi \in C^1(G, M) \; \text{s.t.} \; \forall g_1, g_2 \in G, \; \varphi(g_1, g_2) = \psi(g_2)^{g_1}\psi(g_1 g_2)^{-1} \psi(g_1) \} }
ということになりますね。
2-コサイクルから四元数環を作る
さて、それでは2-コサイクルを使って、四元数環を作ってみましょう。
2次拡大 \mathbb{C}/\mathbb{R} を考えて、そのガロア群を G = \operatorname{Gal}(\mathbb{C}/\mathbb{R}) とします。 G の元は \{1, c\} の2種類で c は特に複素共役です。 G \mathbb{C} の自己同型なので、 \mathbb{C}^\times に作用します。
このような設定のもと、次の具体的な2-コサイクル \varphi \colon G\times G \longrightarrow \mathbb{C}^\times を考えましょう。
\begin{cases} \varphi(1, 1) = 1 \\
\varphi(1, c) = 1 \\
\varphi(c, 1) = 1 \\
\varphi(c, c) = -1 \end{cases}
G には2つの元しかないので、4通りの対応を決めれば \varphi が定まる点に注意します。)
これが2-コサイクルであること、すなわち任意の g_1, g_2, g_3 \in G に対して
\displaystyle \varphi(g_2, g_3)^{g_1}\varphi(g_1 g_2, g_3)^{-1}\varphi(g_1, g_2 g_3) \varphi(g_1, g_2)^{-1} = 1
を満たすことは、実際に (g_1, g_2, g_3) 2^3 = 8 通りの値をいれてみれば確認できます。
(g_1, g_2, g_3) \varphi(g_2, g_3)^{g_1} \varphi(g_1g_2, g_3) \varphi(g_1, g_2g_3) \varphi(g_1, g_2)
(1, 1, 1) 1 1 1 1
(1, 1, c) 1 1 1 1
(1, c, 1) 1 1 1 1
(1, c, c) -1 -1 1 1
(c, 1, 1) 1 1 1 1
(c, 1, c) 1 -1 -1 1
(c, c, 1) 1 1 -1 -1
(c, c, c) -1 1 1 -1
各行に (-1) が必ず偶数個並んでいるので、積は 1 になります。
ここで、次のような \mathbb{C} 上のベクトル空間 A を考えます。
\displaystyle A = \bigoplus_{\sigma \in G} \mathbb{C} e_{\sigma}
e_\sigma A の基底であり、 G の元の個数分あります。今回の設定では e_1, e_c の2個を考えればよいですね。
A \mathbb{C} 上のベクトル空間なので、「加法」と \mathbb{C} による「スカラー倍」が定義されています。ここに、新たに「(可換とは限らない)積」を導入したいと思います。その演算を以下の4つの演算ルールで定めます。
まず、 \mathbb{C} の元と基底 e_\sigma の間の積を定めます。
演算ルール1
任意の x \in \mathbb{C}, \; \sigma \in G に対し
e_{\sigma} a := a^\sigma e_{\sigma}
次に、基底同士 e_\sigma, e_\tau の積を次のように定めます。ここで先ほど定義した 2-コサイクル \varphi を使います。
演算ルール2
任意の \sigma, \tau \in G に対し
e_{\sigma} e_{\tau} := \varphi(\sigma, \tau) e_{\sigma\tau}
また、 A に元々定義されているベクトル空間としての「和」と、新しく定めた「積」の間には、分配法則・結合法則が成り立つことを要請します。
演算ルール3(分配法則)
任意の x, y, z \in A に対し
x(y+z) = xy + xz
(x+y)z = xz + yz
演算ルール4(結合法則)
任意の x, y, z \in A に対し
x(yz)= (xy)z
特に、任意の g_1, g_2, g_3 \in G に対して、基底 e_{g_1}, e_{g_2}, e_{g_3} の間の結合法則
e_{g_1}(e_{g_2}e_{g_3}) = (e_{g_1}e_{g_2})e_{g_3}
と、 \varphi が2-コサイクルであることは同値 であることが示せます。
だから、最初に2-コサイクルであるような \varphi を考えたのですね。同値性の証明については補足1で解説します。
これで一通りの演算が定まりました。実際、積の演算が定まっていることは以下のようにして分かります。
まず、 A の任意の元 x x_\sigma \in \mathbb{C} が存在して
\displaystyle x = \sum_{\sigma \in G} x_\sigma e_\sigma
と表せることに注意します。
ここで任意の x, y \in A に対して xy を計算したいと思います。今回は、 G = \{1, c\} なので、基底 e_1, e_c を用いて x, y を次のように表します。
x = x_1 e_1 + x_c e_c
y = y_1 e_1 + y_c e_c
これらの積をとり、演算ルールにしたがって計算します。積の演算は非可換なので、入れ替えてはいけないことに注意しましょう:
\begin{align} xy &= (x_1 e_1 + x_c e_c)(y_1 e_1 + y_c e_c) \\
&= (x_1 e_1) (y_1 e_1) + (x_c e_c) (y_1 e_1) + (x_1 e_1) (y_c e_c) + (x_c e_c) (y_c e_c) \\
&= x_1 (e_1 y_1) e_1 + x_c (e_c y_1) e_1 + x_1 (e_1 y_c) e_c + x_c (e_c y_c) e_c \\
&= x_1 y_1 e_1 e_1 + x_c y_1^c e_c e_1 + x_1 y_c e_1 e_c + x_c y_c^c e_c e_c \\
&= x_1 y_1 \varphi(1, 1) e_1 + x_c y_1^c \varphi(c, 1) e_c + x_1 y_c \varphi(1, c) e_c + x_c y_c^c \varphi(c, c) e_1 \\
&= (x_1 y_1 \varphi(1, 1) + x_c y_c^c \varphi(c, c) ) e_1 + (x_c y_1^c \varphi(c, 1) + x_1 y_c \varphi(1, c) ) e_c \end{align}
1行目から2行目へは分配法則、2行目から3行目は結合法則を使っています。
また、3行目から4行目は演算ルール1を、4行目から5行目は演算ルール2を用いています。
最後に再度分配法則でまとめ直しています。
ここで
x_1 y_1 \varphi(1, 1) + x_c y_c^c \varphi(c, c) \in \mathbb{C}
x_c y_1^c \varphi(c, 1) + x_1 y_c \varphi(1, c) \in \mathbb{C}
なので、 xy \in A であることが分かります。よって、任意の x, y \in A に対して積 xy \in A が定まりました。
ところで、 A = \mathbb{C}e_1 \oplus \mathbb{C} e_c という構造をしていますが、特に \mathbb{C} = \mathbb{R} \oplus \mathbb{R}i なので
A = \mathbb{R}e_1 \oplus \mathbb{R}ie_1 \oplus \mathbb{R} e_c \oplus \mathbb{R} i e_c
だと思うことができます。
だんだん四元数環に近づいてきましたね! あとは A の基底である e_1, e_c の正体を突き止めれば、 A が四元数環と一致することが示せます。
まず、 \varphi(1, 1)^{-1} e_1 A の単位元であることがわかります。(このことについては補足2で証明します。)
今回は \varphi(1, 1) = 1 なので、以降 e_1 1 に読み替えることにします。
また、演算ルール1 a = i \sigma = c に対して適用すると
e_{c} i = i^c e_{c} = -i e_{c}
となります。 e_{c} i i e_{c} (-1) 倍の関係にあるわけですね。ここで j := e_c だと思うと
j i = -i j
となり四元数のときと全く同じ式(式 (2))が現れました!
演算ルール2より
e_c e_c = \varphi(c, c) e_{cc} = - e_1
となります。ここで、 e_1, e_c をそれぞれ 1, j へと読み替えると
j^2 = -1
となります。これも四元数のときに現れた式ですね。
さらに、 k := ij と定義すると
ijk = ij ij = i(-ij)j = -i^2 j^2 = -1
k^2 = ijk = -1
となり目的の式
i^2 = j^2 = k^2 = ijk = -1
が得られます!
以上により、 1 := e_1, \; j := e_c, \; k := i e_c と読み替えると
A = \mathbb{R} \oplus \mathbb{R}i \oplus \mathbb{R} j \oplus \mathbb{R} k
i^2 = j^2 = k^2 = ijk = -1
という積の入った環、すなわち四元数環そのものであるということがわかりました!
おわりに
今回は、2次の群コホモロジーの元である、ある2-コサイクルから四元数環を作ることができるというお話をしました。
これまでtsujimotterのノートブックでは、0次や1次の群コホモロジーしか扱ってきませんでした。今回初めて2次のコホモロジーを扱ったわけですが、こんな風に使えるというのは、とても面白いと思いました。
今回は、 \varphi(c, c) = -1 でそれ以外は 1 となるような2-コサイクル \varphi から四元数環 \mathbb{H} を構成しました。他の2-コサイクルからも、同様の手続きで環を構成することができます。
H^2(\operatorname{Gal}(\mathbb{C}/\mathbb{R}), \mathbb{C}^\times) の中で考えたときに、 \varphi と同じ同値類に属する2-コサイクルからは、「ある意味で同値な*1」環が得られるそうです。
特に、 H^2(\operatorname{Gal}(\mathbb{C}/\mathbb{R}), \mathbb{C}^\times) は、今回計算した「非自明な2-コサイクル」と、もう一つ「自明な2-コサイクル」の2つだけですので、実質的に2通りの環が構成できるというわけですね。実際、自明な2-コサイクルからは「2次の行列環 M_2(\mathbb{R})」が生成されます。興味ある方は考えてみてください。
\mathbb{H} M_2(\mathbb{R}) は「 \mathbb{R} 上の中心単純多元環(central simple algebra)」と呼ばれるものになっており、その意味で同じクラスに属する環であるというわけですね。
より一般には「 K 上の中心単純多元環(の上と同じ同値関係における同値類)」と「 K 上のガロア群に対する2-コサイクル(の同値類)」が適切な定義のもとで1対1対応するそうです。
これは「ブラウアー群」と呼ばれる群と「2次の群コホモロジー」の間の同型を意味するのですが、この辺も解説できるようになると楽しそうですね。
それでは今日はこの辺で。
補足1:結合則と2-コサイクル条件
次の必要十分条件について示したいと思います。
定理
任意の g_1, g_2, g_3 \in G に対して (e_{g_1} e_{g_2})e_{g_3} = e_{g_1} (e_{g_2}e_{g_3})
\Longleftrightarrow \;\; \varphi \in Z^2(G, L^\times)
上で述べた通り、基底の結合法則と \varphi が2-コサイクルであることが同値であるという主張です。
(証明)
(1) = (e_{g_1} e_{g_2})e_{g_3} , \;\; (2) = e_{g_1} (e_{g_2}e_{g_3}) とします。
\begin{align} (1) &= (e_{g_1} e_{g_2})e_{g_3} \\
&= \varphi(g_1, g_2)e_{g_1g_2} e_{g_3} \\
&= \varphi(g_1, g_2) \varphi(g_1g_2, g_3)e_{g_1g_2g_3} \end{align}
\begin{align} (2) &= e_{g_1} (e_{g_2}e_{g_3}) \\
&= e_{g_1} \varphi(g_2, g_3) e_{g_2 g_3} \\
&= \varphi(g_2, g_3)^{g_1}e_{g_1} e_{g_2 g_3} \\
&= \varphi(g_2, g_3)^{g_1} \varphi(g_1, g_2 g_3)e_{g_1g_2 g_3} \end{align}
(1) = (2) ならば
\varphi(g_1, g_2) \varphi(g_1g_2, g_3) = \varphi(g_2, g_3)^{g_1} \varphi(g_1, g_2 g_3)
であり、これは2-コサイクル条件そのものである。よって、 \varphi \in Z^2(G, L^\times) である。
逆に、 \varphi が2-コサイクル条件を満たせば (1) = (2) が成り立つ。
(証明終わり)
補足2: \varphi(1, 1)^{-1} e_1 A の単位元
A の単位元は \varphi(1, 1)^{-1} e_1 となることを示します。このことは、他の2-コサイクル \varphi を選んだ場合でも成り立ちます。
定理
\varphi \in Z^2(G, \mathbb{C}^\times) を任意の2-コサイクルとし、 \varphi A = \bigoplus_{\sigma G} \mathbb{C} e_\sigma に積を定める。
このとき、任意の x \in A に対して
\left(\varphi(1, 1)^{-1} e_1\right) x = x \left(\varphi(1, 1)^{-1} e_1\right) = x
が成り立つ。すなわち、 \varphi(1, 1)^{-1} e_1 A の単位元。
これを証明するために、2-コサイクルについて一般に成り立つ以下の補題を用意します。
補題
G を群、 M G の作用する加群とし、 \varphi \in Z^2(G, M^\times) を任意の2-コサイクルとする。このとき、次の(1), (2)が成り立つ:
(1) 任意の \sigma \in G に対して \varphi(1, \sigma) = \varphi(1, 1)
(2) 任意の \sigma \in G に対して \varphi(\sigma, 1) = \varphi(1, 1)^\sigma
(補題の証明)
(1) 2-コサイクル条件
\displaystyle \varphi(g_2, g_3)^{g_1}\varphi(g_1 g_2, g_3)^{-1}\varphi(g_1, g_2 g_3) \varphi(g_1, g_2)^{-1} = 1
において g_1 = g_2 = 1, \; g_3 = \sigma とおくと
\displaystyle \varphi(1, \sigma)\varphi(1, \sigma)^{-1}\varphi(1, \sigma) \varphi(1, 1)^{-1} = 1
であり
\displaystyle \varphi(1, \sigma) = \varphi(1, 1)
が成り立つ。
(2) 2-コサイクル条件において、今度は g_1 = \sigma, \; g_2 = g_3 = 1 とおくと
\displaystyle \varphi(1, 1)^{\sigma}\varphi(\sigma, 1)^{-1}\varphi(\sigma, 1) \varphi(\sigma, 1)^{-1} = 1
であり
\displaystyle \varphi(\sigma, 1) = \varphi(1, 1)^\sigma
が成り立つ。
(定理の証明)
一般に x = \sum_{\sigma \in G} x_\sigma e_\sigma(ここで x_\sigma \in \mathbb{C})と表せることに注意します。
\displaystyle \begin{array}{ll} x (\varphi(1, 1)^{-1} e_1) & \\
= \left( \sum_{\sigma \in G} x_\sigma e_\sigma \right) (\varphi(1, 1)^{-1} e_1) & \\
= \sum_{\sigma \in G} x_\sigma \underline{e_\sigma \varphi(1, 1)^{-1}} e_1 & (\because \text{分配法則・結合法則より}) \\
= \sum_{\sigma \in G} x_\sigma (\varphi(1, 1)^\sigma)^{-1} \underline{e_\sigma e_1} & (\because \text{演算ルール1より}) \\
= \sum_{\sigma \in G} x_\sigma (\varphi(1, 1)^\sigma)^{-1} \underline{\varphi(\sigma, 1)} e_\sigma & (\because \text{演算ルール2より}) \\
= \sum_{\sigma \in G} x_\sigma \underline{(\varphi(1, 1)^\sigma)^{-1}\varphi(1, 1)^\sigma} e_\sigma & (\because \text{補題(2)より}) \\
= \sum_{\sigma \in G} x_\sigma e_\sigma \\
= x & \end{array}
よって x (\varphi(1, 1)^{-1} e_1) = x が示せた。
\displaystyle \begin{array}{ll} (\varphi(1, 1)^{-1} e_1) x & \\
= (\varphi(1, 1)^{-1} e_1) \left( \sum_{\sigma \in G} x_\sigma e_\sigma \right) & \\
= \sum_{\sigma \in G} \varphi(1, 1)^{-1} \underline{e_1 x_\sigma} e_\sigma & (\because \text{分配法則・結合法則より}) \\
= \sum_{\sigma \in G} \varphi(1, 1)^{-1} x_\sigma \underline{e_1 e_\sigma} & (\because \text{演算ルール1より}) \\
= \sum_{\sigma \in G} \varphi(1, 1)^{-1} x_\sigma \underline{\varphi(1, \sigma)} e_\sigma & (\because \text{演算ルール2より}) \\
= \sum_{\sigma \in G} \underline{\varphi(1, 1)^{-1}} x_\sigma \underline{\varphi(1, 1)} e_\sigma & (\because \text{補題(1)より}) \\
= \sum_{\sigma \in G} x_\sigma e_\sigma \\
= x & \end{array}
よって (\varphi(1, 1)^{-1} e_1) x = x が示せた。
(証明終わり)
参考
こちらのPDFの「4 中心的単純多元環」「8 接合積」のあたりの記述を一部参考にさせていただきました。
http://www4.math.sci.osaka-u.ac.jp/~twatanabe/algebra.pdf
*1:ここで定義はしませんが、 中心単純多元環の間には「ブラウアー同値」という同値関係が定まります。 | {
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-6,186,716,352,464,047,000 | 6
$\begingroup$
Let $M$ be a smooth $n$-dimensional Riemannian manifold, $n \ge 3$. Let $C$ denote the Lie algebra of the conformal vector fields on $M$. It is known that $\dim(C) \le \frac{(n+1)(n+2)}{2}$.
Robert Bryant said here that this can be proven using local calculations and the Frobenius theorem. I don't see how to implement this approach.
Any ideas? (Or other elementary arguments?)
I know that a vector field $V$ is conformal if and only if $$ \nabla V+(\nabla V)^T=\frac{2}{n} \text{tr}(\nabla V)\text{Id}_{TM}= \frac{2}{n} \text{div} V \cdot \text{Id}_{TM}.$$
In the case of Killing fields the situation is easier: $\nabla V$ is skew-symmetric, and $V$ is determined by $V|_p,\nabla V|_p $, so the dimension of the Killing algebra is not greater than $n+\frac{n(n-1)}{2}=\frac{n(n+1)}{2}$.
In the conformal case, it is not true that a conformal field is determined by its value and its covariant derivative at a point. (This time the space of possible $\nabla V|_p$ is of dimension $\frac{n(n-1)}{2}+1$).
Also, in the conformal case we really need to use somewhere that $n \ge 3$, since for $n=2$ it can be infinite-dimensional.
$\endgroup$
12
• $\begingroup$ As you surely know, strictly speaking one should say "...smooth $n$-dimensional manifold, $n \geq 3$, equipped with a Riemannian metric $g$" or, more to the point, "...conformal structure $\mathbf c$". $\endgroup$ Jan 30 '18 at 10:10
• $\begingroup$ This is probably less elementary than you're aiming for in an answer here, but the claim follows from the fact that there is a bijective correspondence (in fact, equivalence of categories) between, e.g., oriented conformal manifolds and Cartan geometries of type $(\operatorname{SO}(n + 1, 1), P)$ satisfying a certain normalization condition, where $P$ is the stabilizer in $\operatorname{SO}(n + 1, 1)$ of a null ray. In particular, for any conformal structure the algebra of conformal Killing fields has dimension $\leq \dim \operatorname{SO}(n + 1, 1) = \frac{1}{2}(n + 1)(n + 2)$. $\endgroup$ Jan 30 '18 at 10:14
• $\begingroup$ (This perspective and fact are due to Cartan, by the way, though the language is more modern than his.) $\endgroup$ Jan 30 '18 at 10:16
• $\begingroup$ Thanks, this sounds very interesting. Can you elaborate, or give me a reference where to read about the relevant Cartan geometries? $\endgroup$ Jan 30 '18 at 10:20
• 1
$\begingroup$ There is another method that's morally the same as both the Cartan method and to the one Robert Bryant mentions, but which I find more concrete: For a given metric $g$, one can "prolong" the conformal Killing equation to a closed system of linear PDEs that is equivalent to the conformal Killing equation in the sense that solutions of one correspond to solutions of the other. Since the system is closed and linear, it defines a connection (in fact it is natural in that it is constructed invariantly from $(M, g)$ on a particular vector bundle that turns out to have rank $\frac{1}{2}(n + 1)(n+2)$. $\endgroup$ Jan 30 '18 at 10:37
5
$\begingroup$
Note that a vector field $X$ is conformal if and only if there is some function $\Lambda$ such that $\nabla X - \Lambda g$ is skew-symmetric. (Throughout this answer I am identifying $TM$ and $T^* M$ by raising and lowering indices implicitly with $g$ - for example here I really mean $\nabla X^\flat - \Lambda g.$) Let $K$ denote this skew tensor, so that we have $$\nabla X = K + \Lambda g.$$
The idea of prolongation is to iterate this process: we now differentiate $K$ and $\Lambda$, introduce new variables (like we did with $K$ and $\Lambda$) for any unknowns, and repeat until the system closes, meaning that we can write the covariant derivative of each of our variables as some "linear combination" of the other variables. Once we have reached this form, we can interpret the system as $D\xi =0$ for some connection $D,$ at which point the dimension of whatever bundle $\xi$ is a section of (which is basically the direct sum of all our variables) gives an upper bound for the dimension of the solution space.
It turns out we only need one more variable in this case. Following Appendix A2 of these notes by Rod Gover, if we introduce $Q_i = \nabla_i \Lambda + P_{ij} X^j$ where $P$ is the Schouten curvature tensor, then after commuting a bunch of derivatives the system can be written as \begin{align} \nabla_i X_j &= K_{ij} + \Lambda g_{ij} \\ \nabla_i \Lambda &= Q_i - P_{ij} X^j\\ \nabla_i K_{jk} &= - P_{ij} X_k - P_{ik} X_j - g_{ij} Q_k - g_{ik} Q_j + W_{lijk}X^l \\ \nabla_i Q_j &= -P^k_i K_{jk} - P_{ij} \Lambda - C_{kij} X^k. \end{align} Here $W,C$ are the Weyl and Cotton curvature tensors respectively - in particular note that $g,P,W,C$ are all fixed tensor fields, so if we think of $\xi =(X,\Lambda,K,Q)$ as a section of $$E := TM \oplus \mathbb R \oplus \Lambda^2 TM \oplus TM$$ then the system can be written $\nabla \xi + L(\xi)=0$ for some $L\in \Gamma(T^*M \otimes \operatorname{End}(E)).$ Thus conformal Killing vectors $X$ are in $1-1$ correspondence with the sections of $E$ that are parallel with respect to the linear connection $D = \nabla + L;$ so the space of solutions has dimension at most $$\dim E = n + 1 + \binom n 2+n=\frac{(n+1)(n+2)}2.$$
The assumption $n>2$ is used somewhere in the calculation for $\nabla Q$: you can see the easy version (for a flat metric) in these slides by Michael Eastwood. (I wussed out and don't feel like doing the hard version of the calculation myself - hopefully you can work it out.) In the case $n=2$, I believe you would find that the system never closes, no matter how long you prolong the prolongation.
Edit: I recalled receiving a nice handout at a talk of Michael's a few years ago, so I dug it out of the closet. I think its introduction is a better reference than either of my links above, and it can be found online here. It includes some well-written motivation as well as the details of the calculation I omitted.
$\endgroup$
5
• $\begingroup$ This is a very nice exposition! I'll add that we can regard the "flat model" of conformal geometry in dimension $n \geq 3$ to be an $\operatorname{SO}(n + 1, 1)$-invariant conformal structure on $\operatorname{SO}(n + 1, 1) / P \cong \Bbb S^n$, where $P$ is the stabilizer in $\operatorname{SO}(n + 1, 1)$ of a null ray in $\Bbb R^{n + 1, 1}$. Then, we can identify $E$ with the associated bundle $\operatorname{SO}(n + 1, 1) \times_P \mathfrak{so}(n + 1, 1)$ and the connection on $E$ with the connection induced on that bundle by the Maurer-Cartan form on $\operatorname{SO}(n + 1, 1)$. $\endgroup$ Jan 30 '18 at 13:50
• $\begingroup$ (cont.) In that setting, any choice of metric $g$ in the conformal class determines an identification $E = TM \oplus \Bbb R \oplus \Lambda^2 TM \oplus TM$ as in the answer, and this corresponds exactly to the $\mathfrak{so}(n, \Bbb R)$-module decomposition $\mathfrak{so}(n + 1, 1) \cong \Bbb R^n \oplus \Bbb R \oplus \mathfrak{so}(n, \Bbb R) \oplus (\Bbb R^n)^*$. The idea of the Cartan approach to conformal geometry is to generalize this entire picture suitably to "nonflat" geometry. $\endgroup$ Jan 30 '18 at 13:53
• 1
$\begingroup$ Thanks! I really need to put the effort in and learn Klein/Cartan geometry properly sometime, it all seems so beautiful. $\endgroup$ Jan 30 '18 at 14:17
• $\begingroup$ @AnthonyCarapetis Thanks. That last reference is indeed fantastic. BTW, if I am not mistaken the proof there (Example 1.2.2 ) shows also that a conformal vector field is uniquely determined by its $2$-jet at a single point (if the manifold is connected). This is nice; I wonder if there is an analogous result for conformal maps. $\endgroup$ Feb 1 '18 at 14:09
• $\begingroup$ @AsafShachar: right - this is simply because parallel sections are determined by their value at a point, and $\xi(p)$ is determined by $j^2_pX.$ (I guess we were already using this fact implicitly when we bounded the solution space by the fiber dimension.) The analogous 2-rigidity for conformal maps is indeed true, though I'm not sure if it can be proved by similar means. $\endgroup$ Feb 2 '18 at 1:44
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-8,815,121,981,097,531,000 | Answers
Solutions by everydaycalculation.com
Everydaycalculation.com » Answers » Multiply fractions
Multiply 5/4 with 4/8
1st number: 1 1/4, 2nd number: 4/8
This multiplication involving fractions can also be rephrased as "What is 5/4 of 4/8?"
5/4 × 4/8 is 5/8.
Steps for multiplying fractions
1. Simply multiply the numerators and denominators separately:
2. 5/4 × 4/8 = 5 × 4/4 × 8 = 20/32
3. After reducing the fraction, the answer is 5/8
Related:
Use fraction calculator with our all-in-one calculator app: Download for Android, Download for iOS
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7,263,340,148,183,795,000 | What Is 78 Percent Of 44
Question : What is 78% of 44 ?
Answer : 78 percent of 44 is 34.32 .
Solution :
What is 78 percent of 44 means is to find percentage of 44 number, so 78 percent is equal = 78% and 78% is equal 78/100 .
How to solve it step by step :
There are several ways to calculate 78 percent of 44.
1. Way
78% = 78/100
Let's multiply 44 with 78.
78 X 44 = 3432
And final step is divide 3432 to 100 .
3432 / 100 = 34.32
The answer is 34.32
2. Way
Let's convert 78 percent to decimal number.
78% = 78/100
Let's divide 78 to 100.
78 / 100 = 0.78
Then the final step : multiply this decimal number(0.78) with 44
0.78 X 44 = 34.32
The answer is 34.32
Same Question Types :
Question TypesSame Answer
How do you find 78% of 44?34.32
How do you calculate 78% of 44 number?34.32
What will be the 78% of 44?34.32
78% of 7834.32
Calculate 78 percent of 44 number34.32
If We Calculate Other Percentages Of 44
Percent OfNumberAnswer
18%447.92
What Is 18 Percent Of 44
27%4411.88
What Is 27 Percent Of 44
57%4425.08
What Is 57 Percent Of 44
71%4431.24
What Is 71 Percent Of 44
84%4436.96
What Is 84 Percent Of 44
Make New Calculation :
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} | 0c090d63199a0a01e3b08e4a255778a0 |
-3,299,051,780,329,679,000 | Ztest
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• Published : June 1, 2013
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Unit 3
The z-test
Overview
Whenever the normal probability curve is set up as the statistical distribution for testing hypothesis about a population, the z-tests are used. There are 4 types of z test that will be taken up.
Lesson 1: z-test of Hypothesis about a Population Mean
Before the z-one population test of hypothesis about a population mean is applied, certain assumptions must be met: (1) The (population standard deviation) is known.
(2) The data are either interval or ratio.
(3) Only one group is specified.
(4) The distributions of the scores follow the normal distribution. A special table called the z-table is used to facilitate the work on hypothesis testing. These values can be obtained using the table under the normal curve:
The z table|
Type / | 0.025| 0.01| 0.05|
One-tailed| 1.96| 2.33| 1.65|
Two tailed| 2.33| 2.58| 1.96|
Application:
In the national level, the average score in the National Secondary Achievement Tests is 485 with a standard deviation of 95. A random sample of 135 freshmen entering the Philippine Normal University shows a mean score of 620. Can we say that the mean of this group comes from a population whose true mean () = 485? or is there a significant difference between the national average score and the mean score of the incoming freshman students of PNU? Research Question: Is there a significant difference between the average score in NSAT and the mean score of the incoming freshmen of PNU? or
Does the sample of 135 entering students of PNU come from a population with = 485? Ho: = 485 or
Ho: There is no significant difference between the national average score and the mean score of the incoming freshmen students of PNU. We are hypothesizing that the true mean of the population from where we drew our sample is 485. This is our null hypothesis. H1: 485 or
H1: There is a significant difference between the national average score and the mean score of the incoming freshmen students of PNU.
The alternative hypothesis will be accepted if there is a sufficient evidence to reject the null hypothesis (H0). The alternative hypothesis can also be regarded as the research hypothesis. Since we are making a hypothesis about the population the symbol for parameter mean is always used. level : 0.05; 2-tailed test
critical value: z = 1.96
This is the level of accuracy or level of significance that we want for our test, which means that we are willing to allow an error of 5% in our decision whether to reject or accept the null hypothesis. The accuracy of our decision is 95%. The critical value of z at 0.05 level of significance is 1.96. Decision Rule:
Reject Ho if z 1.96.
Accept Ho if z < 1.96
We have to reject Ho if the computed value of z is equal to or greater than 1.96 because in the z distribution or the normal probability curve, any value of z exceeding the value to the right or to the left of is associated with a probability of less than 0.05 (p < .05). This area, which is less than .05, is also known as the region of rejection. If we say for example, that the true mean () of the population is 485, the probability that this can happen by chance alone if the computed z is greater than 1.96 is less than .05. Since the true mean of 485 can occur less than 5 times in a hundred, which is quite rare, we have no reason to doubt that our null hypothesis is true.
Rejection
Rejection
Region of Acceptance
-196z
1.96zz
Accept Ho if z < 1.96. Again, as per normal probability curve any value of z, which is less than 1.96, has a probability that is greater than .05, i.e. the area towards the center of the distribution. This area between –1.96 and +1.96 values of z can be regarded as the region of acceptance.
By our 0.05 criterion, if the probability that the true mean of 485 can occur by chance alone is greater than 0.05, or more than 5 times in a hundred, we say...
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In the figure shown above, two identical squares are
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In the figure shown above, two identical squares are [#permalink] New post 09 Jun 2007, 18:52
In the figure shown above, two identical squares are inscribed in the rectangle. If the perimeter of the rectangle is 18 by squareroot2, then what is the perimeter of each square?
(A) 8 by squareroot2
(B) 12
(C) 12 by squareroot2
(D) 16
(E) 18
Please....Please explain the answer
OA will be posted next day
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Re: perimeter of each Square?? [#permalink] New post 09 Jun 2007, 19:16
humtum0 wrote:
In the figure shown above, two identical squares are inscribed in the rectangle. If the perimeter of the rectangle is 18 by squareroot2, then what is the perimeter of each square?
(A) 8 by squareroot2
(B) 12
(C) 12 by squareroot2
(D) 16
(E) 18
Please....Please explain the answer
OA will be posted next day
B.
side of each square = x
diagnol of the square = hight of the rectangle = x sqrt(2)
length of the rectangle = 2 x sqrt(2)
perimeter of the rectangle = 2 [2 x sqrt(2) + x sqrt(2)]
18 sqrt(2) = 2 [2 x sqrt(2) + x sqrt(2)]
x = 3
perimeter of each square = 12
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[#permalink] New post 09 Jun 2007, 19:17
The answer is B - 12.
The key to understanding this problem is understanding that the length of the rectangle is twice the height (since 2 identical square are inscribed in it).
then its simple 2(2*h+ h ) = 18sqrt(2)
3h = 9sqrt(2)
h = 3sqrt(2)
Now h is also a diagonal of one of the squares.
if the length of the side of the square were x then according to isosceles triangle rule x + x = 3 sqrt(2) which means x = 3. For each inscribed square perimeter = 3 * 4 = 12. Hence option B.
ps. I am working under the assumption that "perimeter of the rectangle is 18 by squareroot2" implies 18*sqrt(2) not 18/sqrt(2). If I used 18/sqrt(2) I wouldnt get an answer in the selection.
[#permalink] 09 Jun 2007, 19:17
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Results 1 to 3 of 3
Thread: Stuck on easy limit
1. #1
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Stuck on easy limit
Hi all, stuck on the following limit:
\lim_{x \to 2} \frac{t^3+3t^2-12t+4}{t^3-4t}
I have managed to factor it out to:
\lim_{x \to 2} \frac{(x-\frac{\sqrt{33}}{2}-\frac{5}{2})(x+\frac{\sqrt{33}}{2}-\frac{5}{2})}{x(x-2)}, x \neq 2
Thanks
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2. #2
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Quote Originally Posted by Oiler View Post
Hi all, stuck on the following limit:
\lim_{x \to 2} \frac{t^3+3t^2-12t+4}{t^3-4t}
I have managed to factor it out to:
\lim_{x \to 2} \frac{(x-\frac{\sqrt{33}}{2}-\frac{5}{2})(x+\frac{\sqrt{33}}{2}-\frac{5}{2})}{x(x-2)}, x \neq 2
Thanks
\dfrac{t^3 + 3t^2 - 12t + 4}{t^3 - 4t} = \dfrac{(t-2)(t^2+5t-2)}{t(t-2)(t+2)}
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3. #3
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Quote Originally Posted by Oiler View Post
Hi all, stuck on the following limit:
\lim_{x \to 2} \frac{t^3+3t^2-12t+4}{t^3-4t}
I have managed to factor it out to:
\lim_{x \to 2} \frac{(x-\frac{\sqrt{33}}{2}-\frac{5}{2})(x+\frac{\sqrt{33}}{2}-\frac{5}{2})}{x(x-2)}, x \neq 2
Thanks
You could try
\displaystyle\frac{t^3+3t^2-12t+4}{t^3-4t}=\frac{\left(t^3-4t\right)+3t^2-8t+4}{t^3-4t}=1+\frac{3t^2-8t+4}{t^3-4t}
=\displaystyle\ 1+\frac{(3t-2)(t-2)}{t\left(t^2-4\right)}=1+\frac{(3t-2)(t-2)}{t(t+2)(t-2)}
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-5,681,604,746,699,718,000 | Probabilitas: Konsep Dasar dan Aplikasi Praktis
Probabilitas adalah konsep matematika yang digunakan untuk mengukur seberapa mungkin suatu kejadian akan terjadi. Dalam dunia statistika, probabilitas digunakan untuk memprediksi hasil dari suatu percobaan atau kejadian. Probabilitas dapat dinyatakan dalam bentuk angka antara 0 hingga 1, di mana nilai 0 menunjukkan kejadian yang tidak mungkin terjadi sedangkan nilai 1 menunjukkan kejadian yang pasti terjadi.
Salah satu contoh penerapan probabilitas adalah dalam permainan dadu. Misalnya, ketika melempar sebuah dadu enam sisi, maka kemungkinan keluar angka genap (2, 4, atau 6) adalah 3 dari total 6 kemungkinan hasil lemparan. Oleh karena itu, probabilitas keluar angka genap dalam permainan dadu ini adalah \(\frac36 = \frac12\) atau setara dengan 0.5.
Selain itu, probabilitas juga dapat diterapkan dalam kasus nyata seperti prediksi cuaca. Misalnya jika prediksi cuaca menduga bahwa kemungkinan hujan hari ini adalah 30%, hal tersebut berarti ada peluang sebesar tiga persepuluh atau \(\frac310\) bahwa hujan akan turun hari ini.
Dalam dunia bisnis dan investasi, probabilitas juga sering digunakan untuk melakukan analisis risiko dan pengambilan keputusan. Misalnya ketika seorang investor ingin membeli saham sebuah perusahaan, ia dapat menggunakan data historis dan analisis probabilistik untuk memperkirakan peluang untung rugi dari investasi tersebut.
Dalam ilmu kedokteran pun probabilitas memiliki peranan penting dalam diagnosis penyakit dan pengobatan pasien. Seorang dokter dapat menggunakan data medis pasien serta pengetahuannya tentang efektivitas berbagai jenis pengobatan untuk mengestimasi peluang kesembuhan pasien.
Probabilistik juga sering digunakan dalam bidang teknologi informasi seperti pembelajaran mesin (machine learning). Algoritma machine learning sering kali mengandalkan model probabilistik untuk membuat prediksi berdasarkan data training yang diberikan.
Ada beberapa metode yang umum digunakan untuk menghitung probabilitas termasuk:
1. Metode Frekuensi: Metode ini menghitung jumlah frekuensi kemunculan suatu kejadian tertentu dibagi dengan total percobaan.
Contoh: Jika sebuah koin dilempar sebanyak 100 kali dan muncul kepala sebanyak 60 kali maka probabilitas muncul kepala adalah \(P(kepala) = \frac60100 = \frac35\).
2. Pendekatan Klasikal: Metode ini dilakukan saat semua hasil eksperimen sama-sama mungkin terjadi.
Contoh: Ketika melempar dadu biasa (6 sisi), maka peluang keluar angka tertentu (misalnya angka ganjil) adalah \(P(angka ganjil) = \frac\textjml cara mendapat angka ganji\textjml semua hasil = \frac36 =\frac12\).
Tentunya pemahaman tentang konsep dasar Probabillitas sangatlah penting karena banyak aplikasi di berbagai bidang seperti statistika, matematika finansial , ilmu komputer dan lain-lain bergantung pada konsep ini . Semakin luwes kita menggunakan metode-metodenya , semakin baik pula kita melihat masa depan dari segi probability apapun itu .
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How do I prove $\sin x$ is uniformly continuous on $\mathbb R$ with delta and epsilon?
I proved geometrically that $\sin x<x$ and thus, $$|f(x_1)-f(x_2)|=|\sin x_1 - \sin x_2|\le|\sin x_1|+|\sin x_2|<|x_1|+|x_2|$$
But this doesn't help me much finding a delta...
Thanks for any help!
P.S. I'm only at the beginning of calculus so I can't use many theorems and derivation (because they haven't been regorously proven).
share|cite|improve this question
1
I changed $sinx_1$, etc., to $\sin x_1$. That is standard TeX usage. – Michael Hardy Jan 1 '13 at 19:35
This is a particular case of (at least) two more general results. First: a periodic continuous function on $\mathbb{R}$ is uniformly continuous on $\mathbb{R}$. Second: a Lipschitz function (mathworld.wolfram.com/LipschitzFunction.html) is uniformly continuous. To prove that $\sin$ is Lipschitz, you can use a trigonometric identity like Nameless did in his answer, or you can claim that its derivative is bounded by $1$. – 1015 Jan 1 '13 at 19:41
4 Answers 4
up vote 5 down vote accepted
Let $\epsilon>0$ and $x,y\in \mathbb{R}$. We want $$\left|f(x)-f(y)\right|<\epsilon\implies \left|\sin x-\sin y\right|<\epsilon\implies \left|2\cos\frac{x+y}2\sin\frac{x-y}2\right|$$ Because $$\left|2\cos\frac{x+y}2\sin\frac{x-y}2\right|\le 2\left|\sin\frac{x-y}2\right|$$ it suffices $$2\left|\sin\frac{x-y}2\right|<\epsilon$$ when $$\left|x-y\right|<\delta\implies \left|\frac{x-y}2\right|<\delta$$ SInce $\left|\sin x\right|\le \left|x\right|$, $$2\left|\sin\frac{x-y}2\right|\le 2\left|\frac{x-y}2\right|<2\delta$$
Choosing $\delta=\frac{\epsilon}{2}>0$ will do the trick. Because $\delta$ doesn't depend on $x,y$, the continuity is uniform
share|cite|improve this answer
Hi, thanks for the fast response! Why $$\left|2\cos\frac{x+y}2\sin\frac{x-y}2\right|\le 2\left|\sin\frac{x-y}2\right|$$ is correct? Also, I know how to prove $sinx<x$, but how do I show $|sinx|<|x|$? Thanks for the reply again! – Harold Jan 1 '13 at 19:32
1
because $\cos \alpha \leqslant 1 $, $\forall \alpha \in \mathbb{R}$. – Amihai Zivan Jan 1 '13 at 19:34
2
Because $\left|\cos(\bullet)\right|\le 1$. – Michael Hardy Jan 1 '13 at 19:34
1
@Harold $\sin x$ is odd – Nameless Jan 1 '13 at 19:48
2
@Harold You want $\left|\sin y\right|\le \left|y\right|\iff -\sin y\le -y\iff \sin y\ge y$ for $y<0$ near $0$. But $\sin x\le x$ for $x>0$. Multipliying by $-1$ gives $-\sin x\ge -x\iff \sin (-x)\ge -x\iff \sin y\ge y$ – Nameless Jan 1 '13 at 20:10
By Mean Value Theorem,
$$ |\sin{x}- \sin{y}| \leq |x-y| |\cos{\xi}| \leq |x-y|, \quad x\leq\xi \leq y.$$
Hence, you may choose $\epsilon=\delta$.
share|cite|improve this answer
Since $\sin x$ is a periodic continuous function with a period $2\pi$, it suffices to prove that it is uniformly continuous on $[0, 2\pi]$. Since $[0, 2\pi]$ is compact, this follows from the well-known theorem.
share|cite|improve this answer
There is an elementary geometric way to do this. Let $x$ and $y$ be real numbers; for now, assume $x, y\in(-\pi, \pi]$. Start off at $(1,0)$ and march off signed distance $x$ to get to point $a$ and $y$ to get to point $b$ on the unit circle. Then $|x - y|$ is the distance from $a$ to $b$ along the unit circle. $|\sin(x) - \sin(y)|$ is the distance between the $y$-coordiates of $a$ and $b$. Hence, in this case $$|\sin(x) - \sin(y) | \le |x - y|.$$
This gives us uniform continuity on $(-\pi, \pi]$, so by periodciity the sine function is uniformly continuous on the entire line.
share|cite|improve this answer
Another way to prove this identity: math.stackexchange.com/questions/620305/… – GinKin Dec 28 '13 at 10:08
This answer deliberately does not appeal to the MVT; it uses purely geometric properties of the sine and cosine functions. – ncmathsadist Dec 29 '13 at 19:56
Yeah I know. But for a test I'm not sure if your approach would be rigorous enough. No offense. – GinKin Dec 29 '13 at 20:03
What is not rigorous about the fact that, given a line and a point, the shortest distance between them is achieved along the perpendicular dropped from the point to the line? In fact, this solution is unique. – ncmathsadist Dec 30 '13 at 0:46
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4,891,841,370,920,112,000 | Thursday
September 29, 2016
Homework Help: maths
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-165,287,769,933,814,850 | Network calculus
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Network calculus is "a set of mathematical results which give insights into man-made systems such as concurrent programs, digital circuits and communication networks."[1] Network calculus gives a theoretical framework for analysing performance guarantees in computer networks. As traffic flows through a network it is subject to constraints imposed by the system components, for example:
These constraints can be expressed and analysed with network calculus methods. Constraint curves can be combined using convolution under min-plus algebra. Network calculus can also be used to express traffic arrival and departure functions as well as service curves.
The calculus uses "alternate algebras ... to transform complex non-linear network systems into analytically tractable linear systems."[2]
Currently, there exists two branches in network calculus: one handling deterministic bounded, and one handling stochastic bounds .[3]
System modelling[edit]
Modelling flow and server[edit]
In network calculus, a flow is modelled as cumulative functions A, where A(t) represents the amount of data (number of bits for example) send by the flow in the interval [0,t). Such functions are non-negative and non-decreasing. The time domain is often the set of non negative reals.
Arrival and departure curve at ingress and egress of a server.
A server can be a link, a scheduler, a traffic shaper, or a whole network. It is simply modelled as a relation between some arrival cumulative curve A and some departure cumulative curve D. It is required that A ≥ D, to model the fact that the departure of some data can not occur before its arrival.
Modelling backlog and delay[edit]
Given some arrival and departure curve A and D, the backlog at any instant t, denoted b(A,D,t) can be defined as the difference between A and D. The delay at t, d(A,D,t) is defined as the minimal amount of time such that the departure function reached the arrival function. When considering the whole flows, the supremum of these values is used.
Horizontal and vertical deviation between arrival and departure cumulative curves
In general, the flows are not exactly known, and only some constraints on flows and servers are known (like the maximal number of packet sent on some period, the maximal size of packets, the minimal link bandwidth). The aim of network calculus is to compute upper bounds on delay and backlog, based on these constraints. To do so, network calculus uses the min-plus algebra.
Min-plus algebra[edit]
In filter theory and linear systems theory the convolution of two functions and is defined as
In min-plus algebra the sum is replaced by the minimum respectively infimum operator and the product is replaced by the sum. So the min-plus convolution of two functions and becomes
e.g. see the definition of service curves. Convolution and min-plus convolution share many algebraic properties. In particular both are commutative and associative.
A so-called min-plus de-convolution operation is defined as
e.g. as used in the definition of traffic envelopes.
The vertical and horizontal deviations can be expressed in terms of min-plus operators.
Traffic envelopes[edit]
Cumulative curves are real behaviours, unknown at design time. What is known is some constraint. Network calculus uses the notion of traffic envelope, also known as arrival curves.
A cumulative function A is said to conform to an envelope (or arrival curve) E, if for all t it holds that
Two equivalent definitions can be given
(1)
(2)
Thus, E places an upper constraint on flow A. Such function E can be seen as an envelope that specifies an upper bound on the number of bits of flow seen in any interval of length t starting at an arbitrary τ, cf. eq. (1).
Service curves[edit]
In order to provide performance guarantees to traffic flows it is necessary to specify some minimal performance of the server (depending on reservations in the network, or scheduling policy, etc.). Service curves provide a means of expressing resource availability. Several kinds of service curves exists, like weakly strict, variable capacity node, etc. See [4] [5] for an overview.
Minimal service[edit]
Let A be an arrival flow, arriving at the ingress of a server, and D be the flow departing at the egress. The system is said to provide a simple minimal service curve S to the pair (A,B), if for all t it holds that
Strict minimal service[edit]
Let A be an arrival flow, arriving at the ingress of a server, and D be the flow departing at the egress. A backlog period is an interval I such that, on any t ∈ I, A(t)>D(t).
The system is said to provide a strict minimal service curve S to the pair (A,B) iff, , such that , if is a backlog period, then .
If a server offers a strict minimal service of curve S, it also offers a simple minimal service of curve S.
Basic results: Performance bounds and envelope propagation[edit]
From traffic envelope and service curves, some bounds on the delay and backlog, and an envelope on the departure flow can be computed.
Let A be an arrival flow, arriving at the ingress of a server, and D be the flow departing at the egress. If the flow as a traffic envelope E, and the server provides a minimal service of curve S, then the backlog and delay can be bounded:
Moreover, the departure curve has envelope .
Moreover, these bounds are tight i.e. given some E, and S, one may build an arrival and departure such that b(A,D) = b(E,S) and v(A,D)=v(E,S).
Concatenation / PBOO[edit]
Consider a sequence of two servers, when the output of the first one is the input of the second one. This sequence can be seen as a new server, built as the concatenation of the two other ones.
Then, if the first (resp. second) server offers a simple minimal service (resp. ), then, the concatenation of both offers a simple minimal service .
Sequence of two servers
The proof does iterative application of the definition of service curves , and some properties of convolution, isotonicity (), and associativity ().
The interest of this result is that the end-to-end delay bound is not greater than the sum of local delays: .
This result is known as Pay burst only once (PBOO).
Tools[edit]
There are several tools based on network calculus.
• The DiscoDNC is an academic Java implementation of the network calculus framework.[6]
• The RTC Toolbox is an academic Java/MATLAB implementation of the Real-Time calculus framework, a theory quasi equivalent to network calculus.[4]
• The CyNC [7] tool is an academic MATLAB/Symulink toolbox, based on top of the RTC Toolbox. The tool was developed in 2004-2008 and it is currently used for teaching at Aalborg university.
• The RTaW-PEGASE is an industrial tool devoted to timing analysis tool of switched Ethernet network (AFDX, industrial and automotive Ethernet), based on network calculus.[8]
• The Network calculus interpreter is an on-line (min,+) interpreter.
• The WOPANets is an academic tool combining network calculus based analysis and optimization analysis.[9]
• The DelayLyzer is an industrial tool designed to compute bounds for Profinet networks.[10]
• DEBORAH is an academic tool devoted to FIFO networks.[11]
• NetCalBounds is an academic tool devoted to blind & FIFO tandem networks.[12][13]
• NCBounds is a network calculus tool in Python, published under BSD 3-Clause License. It considers rate-latency servers and token-bucket arrival curves. It handles any topology, including cyclic ones[14].
• The Siemens Network Planner (SINETPLAN) uses network calculus (among other methods) to help the design of a PROFINET network.[15]
References[edit]
1. ^ Le Boudec, Jean-Yves; Thiran, Patrick (2001). Goos, Gerhard; Hartmanis, Juris; van Leeuwen, Jan (eds.). Network Calculus: A Theory of Deterministic Queuing Systems for the Internet. Lecture Notes in Computer Science. 2050. doi:10.1007/3-540-45318-0. ISBN 978-3-540-42184-9.
2. ^ Jiang, Yuming; Liu, Yong (2009). Stochastic Network Calculus. CiteSeerX 10.1.1.725.5561. doi:10.1007/978-1-84800-127-5. ISBN 978-1-84800-126-8.
3. ^ Fidler, M. (2010). "Survey of deterministic and stochastic service curve models in the network calculus". IEEE Communications Surveys & Tutorials. 12: 59–86. doi:10.1109/SURV.2010.020110.00019.
4. ^ a b Bouillard, Anne; Jouhet, Laurent; Thierry, Eric (2009). Service curves in Network Calculus: dos and don'ts (Technical report). INRIA. RR-7094.
5. ^ Bouillard, Anne; Jouhet, Laurent; Thierry, Éric. Comparison of Different Classes of Service Curves in Network Calculuswith (PDF). 10th International Workshop on Discrete Event Systems (WODES 2010). Technische Universität Berlin.
6. ^ Bondorf, Steffen; Schmitt, Jens B. (2014). The DiscoDNC v2 – A Comprehensive Tool for Deterministic Network Calculus (PDF). 8th International Conference on Performance Evaluation Methodologies and Tools (VALUETOOLS 2014).
7. ^ Schioler, Henrik; Schwefel, Hans P.; Hansen, Martin B. (2007). CyNC: A MATLAB/SimuLink Toolbox for Network Calculus. 2nd International Conference on Performance Evaluation Methodologies and Tools (ValueTools '07).
8. ^ Boyer, Marc; Migge, Jörn; Fumey, Marc (2011). PEGASE, A Robust and Efficient Tool for Worst Case Network Traversal Time (PDF). SAE 2011 AeroTech Congress & Exhibition.
9. ^ Mifdaoui, Ahlem; Ayed, H. (2010). WOPANets: A tool for WOrst case Performance Analysis of embedded Networks. 15th IEEE International Workshop on Computer Aided Modeling, Analysis and Design of Communication Links and Networks (CAMAD). doi:10.1109/CAMAD.2010.5686958.
10. ^ Schmidt, Mark; Veith, Sebastian; Menth, Michael; Kehrer, Stephan (2014). DelayLyzer: A Tool for Analyzing Delay Bounds in Industrial Ethernet Networks. 17th Int. GI/ITG Conf. on Measurement, Modelling, and Evaluation of Computing Systems and Dependability and Fault Tolerance (MMB & DFT 2014). doi:10.1007/978-3-319-05359-2_19.
11. ^ Bisti, Luca; Lenzini, Luciano; Mingozzi, Enzo; Stea, Giovanni (2012). DEBORAH: A Tool for Worst-Case Analysis of FIFO Tandems. International Symposium On Leveraging Applications of Formal Methods, Verification and Validation. doi:10.1007/978-3-642-16558-0_15.
12. ^ Bouillard, Anne; Stea, Giovanni (October 2015). "Exact worst-case delay in FIFO-multiplexing feed-forward networks". IEEE/ACM Transactions on Networking. 23 (5): 1387–1400. doi:10.1109/TNET.2014.2332071.
13. ^ Bouillard, Anne; Éric, Thierry (September 2016). "Tight performance bounds in the worst-case analysis of feed-forward networks" (PDF). Discrete Event Dynamic Systems. 26 (3): 383–411. doi:10.1007/s10626-015-0213-2.
14. ^ Bouillard, Anne (2019). Stability and performance bounds in cyclic networks using network calculus. 17th International Conference on Formal Modeling and Analysis of Timed Systems.
15. ^ Kerschbaum, Sven; Hielscher, Kai-Steffen; German, Reinhard (2016). The need for shaping non-time-critical data in PROFINET networks. 14th IEEE International Conference on Industrial Informatics (INDIN). doi:10.1109/INDIN.2016.7819151.
Books, Surveys, and Tutorials on Network Calculus
Related books on the max-plus algebra or on convex minimization
• R. T. Rockafellar: Convex analysis, Princeton University Press, 1972.
• F. Baccelli, G. Cohen, G. J. Olsder, and J.-P. Quadrat: Synchronization and Linearity: An Algebra for Discrete Event Systems, Wiley, 1992.
• V. N. Kolokol'tsov, Victor P. Maslov: Idempotent Analysis and Its Applications, Springer, 1997. ISBN 0792345096.
Deterministic network calculus
• R. L. Cruz: A Calculus for Network Delay. Part I: Network Elements in Isolation and Part II: Network Analysis, IEEE Transactions on Information Theory, 37(1):114-141, Jan. 1991.
• A. K. Parekh and R. G. Gallager: A Generalized Processor Sharing Approach to Flow Control : The Multiple Node Case, IEEE Transactions on Networking, 2 (2):137-150, April 1994.
• C.-S. Chang: Stability, Queue Length and Delay of Deterministic and Stochastic Queueing Networks, IEEE Transactions on Automatic Control, 39(5):913-931, May 1994.
• D. E. Wrege, E. W. Knightly, H. Zhang, and J. Liebeherr: Deterministic delay bounds for VBR video in packet-switching networks: Fundamental limits and practical tradeoffs, IEEE/ACM Transactions on Networking, 4(3):352-362, Jun. 1996.
• R. L. Cruz: SCED+: Efficient Management of Quality of Service Guarantees, IEEE INFOCOM, pp. 625–634, Mar. 1998.
• J.-Y. Le Boudec: Application of Network Calculus to Guaranteed Service Networks, IEEE Transactions on Information Theory, 44(3):1087-1096, May 1998.
• C.-S. Chang: On Deterministic Traffic Regulation and Service Guarantees: A Systematic Approach by Filtering, IEEE Transactions on Information Theory, 44(3):1097-1110, May 1998.
• R. Agrawal, R. L. Cruz, C. Okino, and R. Rajan: Performance Bounds for Flow Control Protocols, IEEE/ACM Transactions on Networking, 7(3):310-323, Jun. 1999.
• J.-Y. Le Boudec: Some properties of variable length packet shapers, IEEE/ACM Transactions on Networking, 10(3):329-337, Jun. 2002.
• C.-S. Chang, R. L. Cruz, J.-Y. Le Boudec, and P. Thiran: A Min, + System Theory for Constrained Traffic Regulation and Dynamic Service Guarantees, IEEE/ACM Transactions on Networking, 10(6):805-817, Dec. 2002.
• Y. Jiang: Relationship between guaranteed rate server and latency rate server, Computer Networks 43(3): 307-315, 2003.
• M. Fidler and S. Recker: Conjugate network calculus: A dual approach applying the Legendre transform, Computer Networks, 50(8):1026-1039, Jun. 2006.
• Eitan Altman, Kostya Avrachenkov, and Chadi Barakat: TCP network calculus: The case of large bandwidth-delay product, In proceedings of IEEE INFOCOM, NY, June 2002.
• J. Liebeherr: Duality of the Max-Plus and Min-Plus Network Calculus, Foundations and Trends in Networking 11(3-4): 139-282, 2017.
Network topologies, feed-forward networks
• A. Charny and J.-Y. Le Boudec: Delay Bounds in a Network with Aggregate Scheduling, QoFIS, pp. 1–13, Sep. 2000.
• D. Starobinski, M. Karpovsky, and L. Zakrevski: Application of Network Calculus to General Topologies using Turn-Prohibition, IEEE/ACM Transactions on Networking, 11(3):411-421, Jun. 2003.
• M. Fidler: A parameter based admission control for differentiated services networks, Computer Networks, 44(4):463-479, March 2004.
• L. Lenzini, L. Martorini, E. Mingozzi, and G. Stea: Tight end-to-end per-flow delay bounds in FIFO multiplexing sink-tree networks, Performance Evaluation, 63(9-10):956-987, October 2006.
• J. Schmitt, F. Zdarsky, and M. Fidler: Delay bounds under arbitrary multiplexing: when network calculus leaves you in the lurch ..., Prof. IEEE Infocom, April 2008.
• A. Bouillard, L. Jouhet, and E. Thierry: Tight performance bounds in the worst-case analysis of feed-forward networks, Proc. IEEE Infocom, April 2010.
Measurement-based system identification
• C. Cetinkaya, V. Kanodia, and E.W. Knightly: Scalable services via egress admission control, IEEE Transactions on Multimedia, 3(1):69-81, March 2001.
• S. Valaee, and B. Li: Distributed call admission control for ad hoc networks, Proc. of IEEE VTC, pp. 1244–1248, 2002.
• A. Undheim, Y. Jiang, and P. J. Emstad. Network Calculus Approach to Router Modeling with External Measurements, Proc. of IEEE Second International Conference on Communications and Networking in China (Chinacom), August 2007.
• J. Liebeherr, M. Fidler, and S. Valaee: A system-theoretic approach to bandwidth estimation, IEEE Transactions on Networking, 18(4):1040-1053, August 2010.
• M. Bredel, Z. Bozakov, and Y. Jiang: Analyzing router performance using network calculus with external measurements, Proc. IEEE IWQoS, June 2010.
• R. Lubben, M. Fidler, and J. Liebeherr: Stochastic bandwidth estimation in networks with random service, IEEE Transactions on Networking, 22(2):484-497, April 2014.
Stochastic network calculus
• O. Yaron and M. Sidi: Performance and Stability of Communication Networks via Robust Exponential Bounds, IEEE/ACM Transactions on Networking, 1(3):372-385, Jun. 1993.
• D. Starobinski and M. Sidi: Stochastically Bounded Burstiness for Communication Networks, IEEE Transactions on Information Theory, 46(1):206-212, Jan. 2000.
• C.-S. Chang: Stability, Queue Length and Delay of Deterministic and Stochastic Queueing Networks, IEEE Transactions on Automatic Control, 39(5):913-931, May 1994.
• R.-R. Boorstyn, A. Burchard, J. Liebeherr, and C. Oottamakorn: Statistical Service Assurances for Traffic Scheduling Algorithms, IEEE Journal on Selected Areas in Communications, 18(12):2651-2664, Dec. 2000.
• Q. Yin, Y. Jiang, S. Jiang, and P. Y. Kong: Analysis of Generalized Stochastically Bounded Bursty Traffic for Communication Networks, IEEE LCN, pp. 141–149, Nov. 2002.
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Wireless network calculus
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• M. Fidler, R. Lubben, and N. Becker: Capacity–Delay–Error Boundaries: A Composable Model of Sources and Systems, Transactions on Wireless Communications, 14(3):1280-1294, March 2015.
• F. Sun and Y. Jiang: A Statistical Property of Wireless Channel Capacity: Theory and Application, Proc. IFIP Performance, 2017. | {
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8.11: Graphs of Exponential Functions
Difficulty Level: At Grade Created by: CK-12
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Graphs of Exponential Functions
A colony of bacteria has a population of three thousand at noon on Monday. During the next week, the colony’s population doubles every day. What is the population of the bacteria colony just before midnight on Saturday?
At first glance, this seems like a problem you could solve using a geometric sequence. And you could, if the bacteria population doubled all at once every day; since it doubled every day for five days, the final population would be 3000 times 25.
But bacteria don’t reproduce all at once; their population grows slowly over the course of an entire day. So how do we figure out the population after five and a half days?
Exponential Functions
Exponential functions are a lot like geometrical sequences. The main difference between them is that a geometric sequence is discrete while an exponential function is continuous.
Discrete means that the sequence has values only at distinct points (the 1st term, 2nd term, etc.)
Continuous means that the function has values for all possible values of x. The integers are included, but also all the numbers in between.
The problem with the bacteria is an example of a continuous function. Here’s an example of a discrete function:
An ant walks past several stacks of Lego blocks. There is one block in the first stack, 3 blocks in the 2nd stack and 9 blocks in the 3rd stack. In fact, in each successive stack there are triple the number of blocks than in the previous stack.
In this example, each stack has a distinct number of blocks and the next stack is made by adding a certain number of whole pieces all at once. More importantly, however, there are no values of the sequence between the stacks. You can’t ask how high the stack is between the 2nd and 3rd stack, as no stack exists at that position!
As a result of this difference, we use a geometric series to describe quantities that have values at discrete points, and we use exponential functions to describe quantities that have values that change continuously.
When we graph an exponential function, we draw the graph with a solid curve to show that the function has values at any time during the day. On the other hand, when we graph a geometric sequence, we draw discrete points to signify that the sequence only has value at those points but not in between.
Here are graphs for the two examples above:
The formula for an exponential function is similar to the formula for finding the terms in a geometric sequence. An exponential function takes the form
y=Abx
where A is the starting amount and b is the amount by which the total is multiplied every time. For example, the bacteria problem above would have the equation y=30002x.
Compare Graphs of Exponential Functions
Let’s graph a few exponential functions and see what happens as we change the constants in the formula. The basic shape of the exponential function should stay the same—but it may become steeper or shallower depending on the constants we are using.
First, let’s see what happens when we change the value of A.
Compare the graphs of y=2x and y=32x.
Let’s make a table of values for both functions.
x y=2x y=32x
-3 18 y=323=3123=38
-2 14 y=322=3122=34
-1 12 y=321=3121=32
0 1 y=320=3
1 2 y=321=6
2 4 y=322=34=12
3 8 y=323=38=24
Now let's use this table to graph the functions.
We can see that the function y=32x is bigger than the function y=2x. In both functions, the value of y doubles every time x increases by one. However, y=32x “starts” with a value of 3, while y=2x “starts” with a value of 1, so it makes sense that y=32x would be bigger as its values of y keep getting doubled.
Similarly, if the starting value of A is smaller, the values of the entire function will be smaller.
Comparing Graphs
Compare the graphs of y=2x and y=132x.
Let’s make a table of values for both functions.
x y=2x y=132x
-3 18 y=1323=13123=124
-2 14 y=1322=13122=112
-1 12 y=1321=13121=16
0 1 y=1320=13
1 2 y=1321=23
2 4 y=1322=134=43
3 8 y=1323=138=83
Now let's use this table to graph the functions.
As we expected, the exponential function \begin{align*}y = \frac{1}{3} \cdot 2^x\end{align*} is smaller than the exponential function \begin{align*}y = 2^x\end{align*}.
So what happens if the starting value of \begin{align*}A\end{align*} is negative? Let’s find out.
Example C
Graph the exponential function \begin{align*}y = -5 \cdot 2^x\end{align*}.
Solution
Let’s make a table of values:
\begin{align*}x\end{align*} \begin{align*}y = -5 \cdot 2^x\end{align*}
-2 \begin{align*}- \frac{5}{4}\end{align*}
-1 \begin{align*}- \frac{5}{2}\end{align*}
0 -5
1 -10
2 -20
3 -40
Now let's graph the function:
This result shouldn’t be unexpected. Since the starting value is negative and keeps doubling over time, it makes sense that the value of \begin{align*}y\end{align*} gets farther from zero, but in a negative direction. The graph is basically just like the graph of \begin{align*}y = 5 \cdot 2^x\end{align*}, only mirror-reversed about the \begin{align*}x-\end{align*}axis.
Now, let’s compare exponential functions whose bases \begin{align*}(b)\end{align*} are different.
Graphing Multiple Functions
Graph the following exponential functions on the same graph: \begin{align*}y = 2^x, y = 3^x, y = 5^x, y = 10^x\end{align*}.
First we’ll make a table of values for all four functions.
\begin{align*}x\end{align*} \begin{align*}y = 2^x\end{align*} \begin{align*}y = 3^x\end{align*} \begin{align*}y = 5^x\end{align*} \begin{align*}y = 10^x\end{align*}
-2 \begin{align*}\frac{1}{4}\end{align*} \begin{align*}\frac{1}{9}\end{align*} \begin{align*}\frac{1}{25}\end{align*} \begin{align*}\frac{1}{100}\end{align*}
-1 \begin{align*}\frac{1}{2}\end{align*} \begin{align*}\frac{1}{3}\end{align*} \begin{align*}\frac{1}{5}\end{align*} \begin{align*}\frac{1}{10}\end{align*}
0 1 1 1 1
1 2 3 5 10
2 4 9 25 100
3 8 27 125 1000
Now let's graph the functions.
Notice that for \begin{align*}x = 0\end{align*}, all four functions equal 1. They all “start out” at the same point, but the ones with higher values for \begin{align*}b\end{align*} grow faster when \begin{align*}x\end{align*} is positive—and also shrink faster when \begin{align*}x\end{align*} is negative.
Finally, let’s explore what happens for values of \begin{align*}b\end{align*} that are less than 1.
Example E
Graph the exponential function \begin{align*}y = 5 \cdot \left ( \frac{1}{2} \right )^x\end{align*}.
Solution
Let’s start by making a table of values. (Remember that a fraction to a negative power is equivalent to its reciprocal to the same positive power.)
\begin{align*}x\end{align*} \begin{align*}y = 5 \cdot \left ( \frac{1}{2} \right )^x\end{align*}
-3 \begin{align*}y = 5 \cdot \left ( \frac{1}{2} \right )^{-3} = 5 \cdot 2^3 = 40\end{align*}
-2 \begin{align*}y = 5 \cdot \left ( \frac{1}{2} \right )^{-2} = 5 \cdot 2^2 = 20\end{align*}
-1 \begin{align*}y = 5 \cdot \left ( \frac{1}{2} \right )^{-1} = 5 \cdot 2^1 = 10\end{align*}
0 \begin{align*}y = 5 \cdot \left ( \frac{1}{2} \right )^{0} = 5 \cdot 1 = 5\end{align*}
1 \begin{align*}y = 5 \cdot \left ( \frac{1}{2} \right )^{1} = \frac{5}{2}\end{align*}
2 \begin{align*}y = 5 \cdot \left ( \frac{1}{2} \right )^{2} = \frac{5}{4}\end{align*}
Now let's graph the function.
This graph looks very different than the graphs from the previous example! What’s going on here?
When we raise a number greater than 1 to the power of \begin{align*}x\end{align*}, it gets bigger as \begin{align*}x\end{align*} gets bigger. But when we raise a number smaller than 1 to the power of \begin{align*}x\end{align*}, it gets smaller as \begin{align*}x\end{align*} gets bigger—as you can see from the table of values above. This makes sense because multiplying any number by a quantity less than 1 always makes it smaller.
So, when the base \begin{align*}b\end{align*} of an exponential function is between 0 and 1, the graph is like an ordinary exponential graph, only decreasing instead of increasing. Graphs like this represent exponential decay instead of exponential growth. Exponential decay functions are used to describe quantities that decrease over a period of time.
When \begin{align*}b\end{align*} can be written as a fraction, we can use the Property of Negative Exponents to write the function in a different form. For instance, \begin{align*}y = 5 \cdot \left ( \frac{1}{2} \right )^{x}\end{align*} is equivalent to \begin{align*}5 \cdot 2^{-x}\end{align*}. These two forms are both commonly used, so it’s important to know that they are equivalent.
Examples
Example 1
Graph the exponential function \begin{align*}y = 8 \cdot 3^{-x}\end{align*}
a.) Here is our table of values and the graph of the function.
\begin{align*}x\end{align*} \begin{align*}y = 8 \cdot 3^{-x}\end{align*}
-3 \begin{align*}y = 8 \cdot 3^{-(-3)} = 8 \cdot 3^3 = 216\end{align*}
-2 \begin{align*}y = 8 \cdot 3^{-(-2)} = 8 \cdot 3^2 = 72\end{align*}
-1 \begin{align*}y = 8 \cdot 3^{-(-1)} = 8 \cdot 3^1 = 24\end{align*}
0 \begin{align*}y = 8 \cdot 3^{0} = 8\end{align*}
1 \begin{align*}y = 8 \cdot 3^{-1} = \frac{8}{3}\end{align*}
2 \begin{align*}y = 8 \cdot 3^{-2} = \frac{8}{9}\end{align*}
Example 2
Graph the functions \begin{align*}y = 4^x\end{align*} and \begin{align*}y = 4^{-x}\end{align*} on the same coordinate axes.
Here is the table of values for the two functions. Looking at the values in the table, we can see that the two functions are “backwards” of each other, in the sense that the values for the two functions are reciprocals.
\begin{align*}x\end{align*} \begin{align*}y = 4^x\end{align*} \begin{align*}y = 4^{-x}\end{align*}
-3 \begin{align*}y = 4^{-3} = \frac{1}{64}\end{align*} \begin{align*}y = 4^{-(-3)} = 64\end{align*}
-2 \begin{align*}y = 4^{-2} = \frac{1}{16}\end{align*} \begin{align*}y = 4^{-(-2)} = 16\end{align*}
-1 \begin{align*}y = 4^{-1} = \frac{1}{4}\end{align*} \begin{align*}y = 4^{-(-1)} = 4\end{align*}
0 \begin{align*}y = 4^0 = 1\end{align*} \begin{align*}y = 4^0 = 1\end{align*}
1 \begin{align*}y = 4^1 = 4\end{align*} \begin{align*}y = 4^{-1} = \frac{1}{4}\end{align*}
2 \begin{align*}y = 4^2 = 16\end{align*} \begin{align*}y = 4^{-2} = \frac{1}{16}\end{align*}
3 \begin{align*}y = 4^3 = 64\end{align*} \begin{align*}y = 4^{-3} = \frac{1}{64}\end{align*}
Here is the graph of the two functions. Notice that the two functions are mirror images of each other if the mirror is placed vertically on the \begin{align*}y-\end{align*}axis.
In the next lesson, you’ll see how exponential growth and decay functions can be used to represent situations in the real world.
Review
Graph the following exponential functions by making a table of values.
1. \begin{align*}y = 3^x\end{align*}
2. \begin{align*}y = 5 \cdot 3^x\end{align*}
3. \begin{align*}y = 40 \cdot 4^x\end{align*}
4. \begin{align*}y = 3 \cdot 10^x\end{align*}
Graph the following exponential functions.
1. \begin{align*}y = \left ( \frac{1}{5} \right )^x\end{align*}
2. \begin{align*}y = 4 \cdot \left ( \frac{2}{3} \right )^x\end{align*}
3. \begin{align*}y = 3^{-x}\end{align*}
4. \begin{align*}y = \frac{3}{4} \cdot 6^{-x}\end{align*}
5. Which two of the eight graphs above are mirror images of each other?
6. What function would produce a graph that is the mirror image of the one in problem 4?
7. How else might you write the exponential function in problem 5?
8. How else might you write the function in problem 6?
Solve the following problems.
1. A chain letter is sent out to 10 people telling everyone to make 10 copies of the letter and send each one to a new person.
1. Assume that everyone who receives the letter sends it to ten new people and that each cycle takes a week. How many people receive the letter on the sixth week?
2. What if everyone only sends the letter to 9 new people? How many people will then get letters on the sixth week?
2. Nadia received $200 for her \begin{align*}10^{th}\end{align*} birthday. If she saves it in a bank account with 7.5% interest compounded yearly, how much money will she have in the bank by her \begin{align*}21^{st}\end{align*} birthday?
Review (Answers)
To view the Review answers, open this PDF file and look for section 8.11.
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Vocabulary
Asymptotic
A function is asymptotic to a given line if the given line is an asymptote of the function.
Exponential Function
An exponential function is a function whose variable is in the exponent. The general form is y=a \cdot b^{x-h}+k.
grows without bound
If a function grows without bound, it has no limit (it stretches to \infty).
Horizontal Asymptote
A horizontal asymptote is a horizontal line that indicates where a function flattens out as the independent variable gets very large or very small. A function may touch or pass through a horizontal asymptote.
Transformations
Transformations are used to change the graph of a parent function into the graph of a more complex function.
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-1,804,381,954,534,090,500 | Fun With Python #5: Bubble Sort Visualization
Orestis Zekai
Geek Culture
Published in
5 min readMar 15, 2021
--
Welcome to “Fun with Python”, part 5. In this part, we will utilize the pygame module to visualize bubble sort and get more insights about how it works.
Theory and Foundations
Every CS student or developer has spent time studying sorting techniques and algorithms and the first and simplest sorting algorithm that gets presented is bubble sort.
Bubble sort is a sorting algorithm that compares adjacent numbers and swaps them if needed. This continues until all the elements are sorted. An example will make things clearer:
We start of with this sequence of numbers and we want to sort them resulting in an ascending sequence:
5, 3, 2, 6, 1, 4
First we compare 5 with 3, with 5 being bigger than 3 so we swap:
3, 5, 2, 6, 1, 4
Next we compare 5 with 2. Again 5 is greater that 2 so we swap:
3, 2, 5, 6, 1, 4
Our next comparison is between 5 and 6. This time, 6 is greater than 5 so we do not swap the two numbers. So we move on and compare 6 with the next number which is 1 and the result is a swap:
3, 2, 5, 1, 6, 4
Finally we compare 6 with 4 and we swap resulting to this sequence:
3, 2, 5, 1, 4, 6
As you can see, the greatest number is now on the last position. If we do this process again, we will swap numbers and 5 will be put to the correct position. If we continue doing this process, we will have the sequence sorted.
From what you have seen, you might already have guessed that performance wise, this is not an optimal algorithm. In fact, its complexity is O(n²) where n is the number of elements that need to be sorted. This makes the algorithm not suitable for practical use cases, where other, more efficient algorithms are being used.
By intuition, one can see that the greatest numbers are put to the end of the sequence in every iteration. Let’s confirm it by visualizing the algorithm. Here, the pygame module will be of great assistance.
According to pygame official page:
Pygame is a set of Python modules designed for writing video games. Pygame adds functionality on top of the excellent SDL library. This allows you to create fully featured games and multimedia programs in the python language.
In other words, pygame is a module used in Python, in order to create video games, but we are going to use some basic features. Let’s get started.
Implementation
The first thing we are going to do is create an empty window. We will define the width and the height. We will also give out window a name:
import random
import pygame
win_width = 700
win_height = 400
values = [int(random.random()*win_height) for _ in range(win_width)]
pygame.init()
win = pygame.display.set_mode((win_width, win_height))
pygame.display.set_caption('Bubble Sort Visualization')
Also, we will create a list of random numbers. We want to create a number for every pixel in the window and we will normalize the values random.random() returns so that the whole height of the window gets filled.
Let’s draw our values in the empty window we have:
run = True
while run:
pygame.time.delay(100)
for i in range(0, win_width):
pygame.draw.line(win, (255, 255, 255), (i, win_height),
(i, values[i]))
pygame.display.update()
for event in pygame.event.get():
if event.type == pygame.QUIT:
run = False
We create an endless loop, in which for every value in the list, we draw a line, starting from the bottom of the window. The height of each line represents the value of the number being drawn. Also, we check if the user decides to close the window, so that we can exit our program. This is the result we get:
Initial list
Next, we will implement bubble sort. Implementation is quite simple:
for i in range(0, len(values) - 1):
for j in range(0, len(values) - i - 1):
if (values[j] < values[j + 1]):
tmp = values[j]
values[j] = values[j + 1]
values[j + 1] = tmp
Finally, we will insert this code in the loop we created above. After each iteration, we will update our window, with the updated list:
run = True
done = False
while run:
pygame.time.delay(100)
if (done == False):
for i in range(0, len(values) - 1):
for j in range(0, len(values) - i - 1):
if (values[j] < values[j + 1]):
tmp = values[j]
values[j] = values[j + 1]
values[j + 1] = tmp
win.fill((0, 0, 0))
for i in range(0, win_width):
pygame.draw.line(
win, (255, 255, 255),
(i, win_height),
(i, values[i])
)
pygame.display.update()
done = True
for event in pygame.event.get():
if event.type == pygame.QUIT:
run = False
Conclusion
After the first iterations, you will see the following window:
Greatest numbers are moved to the right of the window
As you can see, the greatest numbers start moving to the right of the window. This is exactly how bubble sort works. In every pass, you are trying to bring the greatest number to its correct position (or the minimum if you are sorting ascending). When the sorting finishes, the window will look like this:
Sorted list
Every number is in the correct position and we get this “hill”.
Now, you have a visual overview of how bubble sort works. Of course you can implement other sorting algorithms and see how they work and understand them better by visualizing them. You can find the full functional code here.
I hope you enjoyed reading it and try it yourself. Let me know your thoughts and your ideas about it! In the meanwhile, you can find the rest of the “Fun with Python” series here.
--
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5,844,926,288,754,038,000 | Vous êtes sur la page 1sur 2
Suites et récurrence (TS)
Exercice 1
Soit définie par 2 et pour tout IN,
3 5. Montrer par récurrence que, pour
tout IN, 3 5
Exercice 2
Soit définie par et pour tout IN,
2 1. Montrer par récurrence que,
pour tout IN, 2 .
Exercice 3
Soit définie par 5 et pour tout IN*,
2 . Montrer par récurrence que,
pour tout IN*, 2.
Exercice 4
Soit définie par 3 et pour tout IN,
1.
1) Montrer par récurrence que est minorée par .
2) Démontrer que est décroissante.
3) En déduire que converge.
4) Déterminer la limite de .
Exercice 5
Soit définie par 0 et pour tout IN,
2 8.
1) Représenter graphiquement les 1ers termes de la suite .
2) Montrer par récurrence que, pour tout IN, 0; 4.
3) Démontrer que est croissante.
4) Montrer que converge et déterminer sa limite.
Exercice 6
Soit la suite définie par 1 et pour tout IN,
2
1) Démontrer par récurrence que est minorée par -4.
2) Démontrer que la suite est décroissante.
3) Soit la suite définie par 6 et pour tout IN,
2.
Démontrer par récurrence que la suite est majorée par -4.
4) Démontrer que la suite est croissante.
5) Représenter sur le même graphique mes 1ers termes des suites et .
6) Soit la suite définie par pour tout IN. Démontrer que la suite est
géométrique et déterminer ses éléments caractéristiques.
7) Démontrer que les suites et sont adjacentes.
8) Démontrer que les suites et convergent et déterminer leur limite.
1
Exercice 7
!
On considère la fonction définie sur 0; ∞ par !
.
1) Etudier les variations de .
2) En déduire, pour tout 0; 1, 0; 1.
3) Dans toute la suite, on considère la suite définie par 0 et pour tout IN,
.
a) Représenter graphiquement dans un repère orthonormal d’unité graphique 10 cm.
b) Représenter graphiquement les quatre premiers termes de la suite .
c) Que pouvez-vous conjecturer quant au sens de variation de cette suite et de sa convergence ?
4a) Montrer que, pour tout IN, 0; 1.
"#
$"#
b) Montrer que pour tout IN,
.
"#
c) En déduire le sens de variation de la suite .
d) Montrer que la suite converge.
e) Soit L la limite de la suite , démontrer que L vérifie & &. En déduire la valeur de L.
" $
5) On considère la suite définie par "#
.
#
a) Prouver que est une suite géométrique et donner ses éléments caractéristiques.
b) Calculer et exprimer en fonction de .
c) Exprimer en fonction de . Puis en fonction de .
d) Déduire de 4c) que la suite converge et déterminer sa limite.
Exercice 8
"#
On considère la suite numérique définie par 1 et pour tout IN,
"#
1) Calculer les quatre premiers termes de la suite.
2) Démontrer que est un nombre positif pour tout IN* ; en déduire que est définie
quelque soit l’entier .
3) Démontrer que la suite est majorée par √3.
4) Déterminer le sens de variation de la suite .
" $ √
5) On considère la suite définie par : pour tout IN, "#
.
# √
a) Montrer que la suite est une suite géométrique dont on donnera le premier terme et la
raison.
b) Calculer la limite de et en déduire la limite de .
Exercice 9
$"#
On définit la suite par son terme initial 2 et la relation de récurrence
"# $(
3
1) Calculer ; et .
$!
2) Soit la fonction * définie sur ∞; 4 par * !$( . Et soit H sa courbe représentative.
a) Tracer H et la droite d d’équation + dans un repère orthonormal ,; -., 0. unité graphique 1
cm.
b) Construire à l’aide de H et de d les points de l’axe ,; -. d’abscisses respectives , , et
.
c) Que peut-on prévoir quant à la convergence de la suite ?
" $
3) est la suite définie pour tout IN par "#
.
#
a) Calculer , , .
b) Montrer que est une suite géométrique que l’on caractérisera.
c) Exprimer en fonction de et déterminer la limite de la suite quand tend vers ∞. | {
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5,539,933,517,603,000,000 | E 8 PATTERN EXPLAINED
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uses patterns of dots and color to help students visualize and explain
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use to explain number pattern.
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Topic: Math. Question: Complete the pattern. 1,2,3, 7,8,9, 28,29,30 Explain your answer. – Question #331089 Updated with the explanation of Singleton pattern. Design Pattern FAQ . What are design patterns? Can you explain factory pattern? Can you explain abstract factory Students will graph growing patterns using ordered pairs on a coordinate grid.
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In software engineering, a design pattern is a general reusable solution to a commonly occurring problem within a given context in software design. A design pattern According to Kitcher, P includes the use of a single “origin and development” (OD) pattern of explanation, according to which the dimensions of objects-artifacts Seasonal variations of human sperm cells among 6455 semen samples: a plausible explanation of a seasonal birth pattern. Levitas E, Lunenfeld E, Weisz N,
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High brightness LED x 8: Blink pattern: Five kinds: Hardware: Circuit drawing Pattern drawing Circuit explanation Parts explanation: Software: Flow chart List Number patterns are all prediction. Patterns can be in words too. Several people speak that math is the skill of patterns. Let us see about mathematical patterns in The purpose of this article is to answer these questions and to explain the MVVM pattern in the simplest possible way. try e.g. Data Binding Overview,
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[Total No. of Questions: 8] [Total No. of Printed Pages: 2] 324-X/2004 B.E. (First Semester) Examination, May, 2004 (New Scheme) MANUFACTURING PROCESSES us to see and explain a pattern rather than always only looking at Introduce a second pattern with three objects, e.g. moon, star, sun, moon, star, sun, etc. Q.
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Patterns (Elliott wave) TECHNICAL YSIS EXPLAINED Trend (Moving average) Momentum (Rate-of-change) Technical ysis – is governed by a system of rules and guidelines explain the rule. NCTM Content terms (e.g., 2, 5, 8, 11 (ABAB) and growing patterns (2, 4, 8, 16) on the board. • Ask students to describe what they see. 8 Rules as methods to describe visual and arithmetic patterns 8 Input/output tables as a method to students to explain the growth pattern in words. Ask: How
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Create and explain patterns and algebraic relationships(e.g., 2,4,6,8) algebraically: 2n (doubling) L= Lesson P Pattern 8 (Multi-Merge) FLASH animation of Multi-Merge pattern. i.e. it is not possible to enable or activate an activity in parallel with itself. Here is an excerpt from Mrs. McKim’s introduction, which eloquently explained her philosophy about quilting: “ tips and 8 Grandma Clark patterns! Name:
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9dB (8 times) can be achieved. The resultant two patterns shown below
This pattern is for the advanced beginner with Special stitches are explained particularly. Using a E / 3.5mm hook and the apropriated yarn you should You can also explain why repeatable patterns do occur long section of prime numbers with the same pattern i.e. the gap between 113 and 127 of 14 causes 8.4.1. Explanation via implementation description: Graphmaster. 9. AIML Predicate handling. 9.1. renamed 8 (AIML Pattern Expressions) to "AIML Pattern Matching"
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1,697,450,553,184,823,600 | Find the complex cube roots of 8(cos(4pi/5)+isin(4pi/5))
1 Answer
mathsworkmusic's profile pic
mathsworkmusic | (Level 2) Educator
Posted on
We have the complex number
`z = 8(cos((4pi)/5) + isin((4pi)/5))`
First convert z into polar coordinates (argument,modulus) on the complex plane.
` `We can see that the argument of z, `arg(z) = (4pi)/5`, and that the modulus of z, `|z| = 8`.
Using de Moivre's Theorem to obtain the argument of `z_c = root(3)(z)` we note that
`arg(z_c) = (4pi)/5(1/3) = (4pi)/(15)`
Again using de Moivre's Theorem to obtain the modulus of `z_c` we note that
`|z_c| = root(3)(|z|) = root(3)(8) = 2`
The cube roots `z_c` of `z` are then all complex numbers with argument `(4pi)/15` and modulus 2. These take the form
`z_c = 2(cos((4pi)/15 +2npi) + isin((4pi)/15 + 2npi))` , `n = 0,1,2,...`
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5,132,104,297,095,598,000 | Para saber el tipo de decimal: algunas fórmulas
Hay una manera de deducir qué tipo de decimal nos dará una fracción común, solamente analizando su denominador, siempre que sea una fracción irreductible, es decir, que no se pueda simplificar. Si no es irreductible, se simplifica y luego se analiza su denominador.
Factorización prima
El análisis del denominador se basa en la factorización prima de un número. La factorización prima consiste en descomponer un número en sus factores primos, por ejemplo:
A continuación, apliquemos esto a distintos casos.
Para decimal exacto
Para que el decimal sea exacto, la fracción irreductible debe tener como denominador cualquier número natural que tenga como factores primos al 2, al 5 o ambos.
Por ejemplo:
Esta fracción tiene como denominador al 20, que corresponde a , por lo tanto, cumple con la condición.
Entonces, es decimal exacto y corresponde a 0,35.
Para decimal periódico
Será decimal periódico la fracción común irreductible, que tiene como denominador un número natural, que no tiene como factores primos ni al 2 ni al 5.
Por ejemplo:
9 tiene como factorización prima ; no tiene ni al 2 ni al 5. Entonces, equivale al decimal periódico
Para decimal semiperiódico
El decimal es semiperiódico en las fracciones comunes irreductibles cuyo denominador es un número natural cuya factorización prima tiene al 2 o al 5 o ambos, y a otro número primo más.
Por ejemplo:
12 equivale a . Corresponde al decimal semiperiódico
Inexactos semiperiódicos
Existe otra clasificación de los números decimales. Los estudiaremos una vez que hayas mirado con atención los siguientes casos:
Estos decimales se conocen como inexactos semiperiódicos o periódicos mixtos, ya que en su parte decimal tienen cifras que no se repiten, a las que llamamos anteperíodo, y luego un período de una o más cifras. Los escribiremos de la siguiente forma abreviada:
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-418,027,583,084,700,740 | Menu Divisible
15 divided by what equals 29?
15 divided by what equals 29? In other words, you have 15 and you divide it by something and then you get 29. What is that something?
We will call that something "X". Here is the equation to find the solution:
15
X
= 29
First, multiply both sides by X to get rid of the X as denominator.
15(X)
X
= 29(X)
15 = 29X
Then, divide both sides by 29 to get X by itself as follows:
15
29
=
29X
29
0.517 = X
Thus, the answer to "15 divided by what equals 29?" is 0.517.
Note that our answers are rounded to the nearest thousandth if necessary. You can divide 15 by 0.517 to check that we got the right answer.
Divided By What Equals Calculator
Please enter another problem for us to solve below:
divided by what equals
15 divided by what equals 30?
Go here for the next problem we solved.
Copyright | Privacy Policy | Disclaimer | Contact | {
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568,204,547,463,103,300 | Tessellations, Hands-on Math
3,180
23
4
Introduction: Tessellations, Hands-on Math
This project is made to give students the opportunity to design there own tessellations and understant the math behind that. Explore the unique feature of "Calçada Portuguesa" a beautiful and useful way to use tessellations.
Supplies
3D printer
Color Filament
Step 1: Design the Polygons
You can use any 3D sotware to design this. I choose Openscad because it's free and very precise.
Let's use this formula:
{circle(20,$fn=8);%circle(20,$fn=90);}
Now you have a 2d shape.
Step 2: Add Volume to Polygons
It's time to add volume so add this line before the previous one:
linear_extrude(height = 3, center = true, convexity = 0)
And you obtain your first polygon
Step 3: Make Other Polygons
You can make different polygons by change the value that apears above the green bar in the next photos.
Now you need to print them!
Step 4: Stl Files and Print Settings
This is my print settings:
Printer: Bq Witbox
Rafts: No
Supports: No
Resolution: 0,2mm
Infill: 20%
Notes: Filament: PLA | Print Speed: 60mm/s | Bed adhesive: hair spray
Step 5: Teacher's Notes
Project Name
Tessellations, hands-on math is made to give students the opportunity to design there own tessellations and understant the math behind that. Explore the unique feature of "Calçada Portuguesa" a beautiful and useful way to use tessellations.
Overview & Background
This project give me more knowledge about math. I was a medium math student and now with 3D printers at a reasonable price it's a great opportunity to everyone knows more about math and have hands on activities in schools.
Objectives
At the end of this project students wil be abble to:
• Identify polygons
• Identify Regular Tessellations
• Identify Semi Regular Tessellations
• Build there own tessellations
Audiences
This project is made for 5th grade
Subjects
Math
Skills Learned
• Math of polygons
• Tessellations
• "Calçada Portuguesa", a brief history
Lesson/Activity
There is a powerpoint you can use as guideline for the activity.
Step 6: Classroom Activity - What Is a Polygon?
List of things to get before activity starts:
• Plastic polygons printed or ordered by the teacher
• A media projector and a computer
• Pencil for students
• A webcam or a camera
• A printer
• Glue
• Scissores
• Make groups of 4 or 5 students
What is a polygon?
The teacher should ask first to students if they know what is a polygon. From there the activity should start.
Step 7: Classroom Activity - Identify Polygons
The teacher should give to each student the “Identify polygons” worksheet and let them do the work.
Step 8: Classroom Activity - Tessellations
The teacher should now introduce the concept of tessellation. Explain the difference between regular and semi-regular tessellations.
Step 9: Classroom Activity - Identify Regular Tessellations
The teacher should give to each student the “Identify Regular Tessellations” worksheet and let them do the work
Step 10: Classroom Activity - Identify Semi Regular Tessellations
The teacher should give to each student the “Identify SemiRegular Tessellations” worksheet and let them do the work
Step 11: Classroom Activity - Build Your Own Tessellation
The teacher should give to each student the “Build your own Tessellation” worksheet and invite students to use the plastic pieces to make there own tessellations. After that the teacher our the students should use the camera to take a picture of the tessellation, print it and glue it on the worksheet
Step 12: Classroom Activity - "Calçada Portuguesa" a Brief History
From here to end the teacher should tell a little story about "Calçada portuguesa" and show the beautiful images in the powerpoint of this wonderful use of tessellation.
Made with Math Contest
Judges Prize in the
Made with Math Contest
Be the First to Share
Recommendations
• Fabric Challenge
Fabric Challenge
• Build a Tool Contest
Build a Tool Contest
• Sculpt & Carve Challenge
Sculpt & Carve Challenge
4 Comments
0
dave.vaness.79
dave.vaness.79
2 years ago
I like the hexagons attacked by triangles and squares. It seems like someone should be able to use it as a game board. Any ideas board game designers?
0
charlessenf-gm
charlessenf-gm
Reply 2 years ago
A version of Go, perhaps?? But how might one win?
0
jeanniel1
jeanniel1
Reply 2 years ago
Go wins by number of spaces covered by one color so it'd be possible!
0
charlessenf-gm
charlessenf-gm
Tip 2 years ago
I played with the color choices - I liked the one design, but not the colors! In changing them about (I was thinking of a Mosaic design element for a shower!) it occurred to me that you might want to involve your Art instructors in choosing the colors (or create each shape in each color as well as employ at least three values (?) of each color and maybe whites, grays and a black?
I am not an Art major, but I believe there is a 'rule' about color values and how they may be used to manipulate or perception of a given image. In my blue versions, it seemed that changing the value of the center most piece gave one the impression that the piece was three dimensional. Not sure the three .bmp images I uploaded are appearing!
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-6,936,244,417,626,216,000 | Is 737 Divisible By Anything?
Okay, so when we ask if 737 is divisible by anything, we are looking to see if there are any WHOLE numbers that can be divided into 737 that will result in a whole number as the answer. In this short guide, we'll walk you through how to figure out whether 737 is divisible by anything. Let's go!
Fun fact! All whole numbers will have at least two numbers that they are divisible by. Those would be the actual number in question (in this case 737), and the number 1.
So, the answer is yes. The number 737 is divisible by 4 number(s).
Let's list out all of the divisors of 737:
• 1
• 11
• 67
• 737
When we list them out like this it's easy to see that the numbers which 737 is divisible by are 1, 11, 67, and 737.
You might be interested to know that all of the divisor numbers listed above are also known as the Factors of 737.
Not only that, but the numbers can also be called the divisors of 737. Basically, all of those numbers can go evenly into 737 with no remainder.
As you can see, this is a pretty simple one to calculate. All you need to do is list out all of the factors for the number 737. If there are any factors, then you know that 737 is divisible by something.
Give this a go for yourself and try to calculate a couple of these without using our calculator. Grab a pencil and a piece of paper and pick a couple of numbers to try it with.
Cite, Link, or Reference This Page
If you found this content useful in your research, please do us a great favor and use the tool below to make sure you properly reference us wherever you use it. We really appreciate your support!
• "Is 737 Divisible By Anything?". VisualFractions.com. Accessed on January 25, 2022. http://visualfractions.com/calculator/divisible-by-anything/is-737-divisible-by-anything/.
• "Is 737 Divisible By Anything?". VisualFractions.com, http://visualfractions.com/calculator/divisible-by-anything/is-737-divisible-by-anything/. Accessed 25 January, 2022.
• Is 737 Divisible By Anything?. VisualFractions.com. Retrieved from http://visualfractions.com/calculator/divisible-by-anything/is-737-divisible-by-anything/.
Divisible by Anything Calculator
Enter another number below to find out whether it is divisible by anything.
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-2,951,566,935,262,465,500 | ILE Home Intuitionistic Logic Explorer < Previous Next >
Nearby theorems
Mirrors > Home > ILE Home > Th. List > ax-i12 GIF version
Axiom ax-i12 1485
Description: Axiom of Quantifier Introduction. One of the equality and substitution axioms of predicate calculus with equality. Informally, it says that whenever 𝑧 is distinct from 𝑥 and 𝑦, and 𝑥 = 𝑦 is true, then 𝑥 = 𝑦 quantified with 𝑧 is also true. In other words, 𝑧 is irrelevant to the truth of 𝑥 = 𝑦. Axiom scheme C9' in [Megill] p. 448 (p. 16 of the preprint). It apparently does not otherwise appear in the literature but is easily proved from textbook predicate calculus by cases.
This axiom has been modified from the original ax-12 1489 for compatibility with intuitionistic logic. (Contributed by Mario Carneiro, 31-Jan-2015.)
Assertion
Ref Expression
ax-i12 (∀𝑧 𝑧 = 𝑥 ∨ (∀𝑧 𝑧 = 𝑦 ∨ ∀𝑧(𝑥 = 𝑦 → ∀𝑧 𝑥 = 𝑦)))
Detailed syntax breakdown of Axiom ax-i12
StepHypRef Expression
1 vz . . . 4 setvar 𝑧
2 vx . . . 4 setvar 𝑥
31, 2weq 1479 . . 3 wff 𝑧 = 𝑥
43, 1wal 1329 . 2 wff 𝑧 𝑧 = 𝑥
5 vy . . . . 5 setvar 𝑦
61, 5weq 1479 . . . 4 wff 𝑧 = 𝑦
76, 1wal 1329 . . 3 wff 𝑧 𝑧 = 𝑦
82, 5weq 1479 . . . . 5 wff 𝑥 = 𝑦
98, 1wal 1329 . . . . 5 wff 𝑧 𝑥 = 𝑦
108, 9wi 4 . . . 4 wff (𝑥 = 𝑦 → ∀𝑧 𝑥 = 𝑦)
1110, 1wal 1329 . . 3 wff 𝑧(𝑥 = 𝑦 → ∀𝑧 𝑥 = 𝑦)
127, 11wo 697 . 2 wff (∀𝑧 𝑧 = 𝑦 ∨ ∀𝑧(𝑥 = 𝑦 → ∀𝑧 𝑥 = 𝑦))
134, 12wo 697 1 wff (∀𝑧 𝑧 = 𝑥 ∨ (∀𝑧 𝑧 = 𝑦 ∨ ∀𝑧(𝑥 = 𝑦 → ∀𝑧 𝑥 = 𝑦)))
Colors of variables: wff set class
This axiom is referenced by: ax-12 1489 ax12or 1490 dveeq2 1787 dveeq2or 1788 dvelimALT 1985 dvelimfv 1986
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-7,338,641,005,878,695,000 | Exponentially-spaced integers
It's often useful to have exponentially-spaced integers, so that they are evenly distributed when plotted on a logarithmic scale. The simplest way to generate them is to exponentiate linearly-spaced numbers and then round:
1
2
3
julia> round.(Int, exp.(0:7))'
1×8 LinearAlgebra.Adjoint{Int64,Array{Int64,1}}:
1 3 7 20 55 148 403 1097
Past the first decade, which is significantly rounded, these integers are perfectly spaced: Perfectly spaced integers. However, they're not very familiar to most people, and it's possible for the rounding step to produce duplicates, making them potentially tricky to generate. If you're willing to sacrifice the spacing a little more, I want to show you why integers of the form m2km \cdot 2^k can be a very convenient alternative.
Powers of two
Some of the nicest exponentially-spaced integers around are the powers of two, which are trivial to generate and easily recognizable: Powers of two. Such elegance!
Filling in the gaps
If 2k2^k is sufficient for your purposes, I'm not sure why you're even here. But what if the spacing is a bit too large? The jump from 512512 to 10241024 is quite a big one, and 10241024 could be beyond your capabilities. Maybe you're looking for something more in line with your budget.
Logarithmically speaking, the closest integer to the halfway point between 512512 and 10241024 is 724724, but there's a good chance that I had never encountered this number in my entire life until I computed it for this purpose. I don't know anything about it. How can I trust it?
No, that won't do. Instead, we'll cheat a little bit and use 768768. That really takes me back! What a nice, dependable integer. It turns out that 768=32512=328, 768 = \frac{3}{2} 512 = 3 \cdot 2^8, since 3/23/2 is the average of 11 and 22. Thus, in addition to the powers of two, 12k=1,2,4,8,16,32,64,128,256,512,1024,, 1 \cdot 2^k = 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, \dots, we can also use powers of two multiplied by three: 32k=3,6,12,24,48,96,192,384,768,1536,. 3 \cdot 2^k = 3, 6, 12, 24, 48, 96, 192, 384, 768, 1536, \dots. A clean source of unlimited numbers!
Less flippantly, integers with large prime factors, like 724=22181724 = 2^2 \cdot 181, are unfavourable for some algorithms. For example, certain FFT algorithms fare better on arrays whose sizes have small prime factors. Thus, there are applications for which it's better to use numbers purposely constructed from small factors.
Let's visualize the amount of imperfection in the layout of these additional numbers. If we draw two octaves (say, from 256256 to 10241024, although the actual values don't matter), we get three powers of two (2k2^k, 2k+12^{k+1}, 2k+22^{k+2}): Two octaves with powers of two. In linear space (filled), the distances between them are uneven (one is twice the other), but in log space (hollow), they are properly spaced. If we add in the fillers 32k13 \cdot 2^{k-1} and 32k3 \cdot 2^k, we see that they are nicely located in the middle of each octave when viewed linearly, but are slightly off-kilter logarithmically: Two octaves with powers of two, as well as multiples of three. This is the price to be paid for having such pleasant numbers to work with.
Ad nauseam
There's no reason to stop there, of course. In addition to dividing the interval from 11 to 22 into two segments using 3/23/2, we could split each of those further with 5/45/4 and 7/47/4, making a total of four segments. The first of these fractions gives us 52k=5,10,20,40,80,160,320,640,1280,, 5 \cdot 2^k = 5, 10, 20, 40, 80, 160, 320, 640, 1280, \dots, whose inclusion looks like Two octaves with powers of two, as well as multiples of three and five. The second one results in 72k=7,14,28,56,112,224,448,896,1792,, 7 \cdot 2^k = 7, 14, 28, 56, 112, 224, 448, 896, 1792, \dots, and completes this round of filling: Two octaves with powers of two, as well as multiples of three, five, and seven.
Using m2km \cdot 2^k with m=1,3,5,7m = 1, 3, 5, 7, we have obtained integers that are locally spaced linearly, and are therefore convenient to handle, but are globally spaced logarithmically, and therefore span many orders of magnitude. Naturally, the next round would be based on 9/89/8, 11/811/8, 13/813/8, and 15/815/8, yielding m=9,11,13,15m = 9, 11, 13, 15 for a total of eight values per octave, but it's usually not necessary to go that far.
Jumping ahead
Before I leave you alone to ponder the usefulness of this approach, I'd like to point out one practical consideration. If you're interested in exponentially-growing quantities, it can be detrimental to have all sorts of small values running around underfoot. For example, using m=1,3,5,7m = 1, 3, 5, 7 and starting at 88, we get 8,10,12,14,16,, 8, 10, 12, 14, 16, \dots, which are consecutive even integers. We didn't do all that work just to end up generating even numbers!
To remedy this, you can stagger the start of each sequence. For example, if the gap between 88 and 1616 is sufficiently small, but grows too large after that point, the m=3m = 3 sequence can be started at 2424 instead of 1212. Then, if 1616 and 2424 and 3232 are close enough, the m=5m = 5 and 77 sequences can be started at 4040 and 5656, respectively, instead of 2020 and 2828. The staggered start allows us to tidy this (on a linear scale): Example of several sequences all starting from small values. into this: Example of several sequences with staggered start. Much better! The improved variant starts out linear for the first few octaves, but automatically switches to exponential spacing beyond that. | {
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6,691,421,587,442,570,000 | 03 412 885
Arithmos
Arithmos
Arithmos
Station F
The Traditional Addition Algorithm
How does this dots-and-boxes approach to addition compare to the standard algorithm most people know?
Question 1
Let’s go back to the example 358+287358 + 287. Most people are surprised by the straightforward left-to-right answer
513155|13|15.
I3S7 - Image002
Write on a piece of paper the traditional way to perform this addition problem.
Did you perform an explosion? First with 1515 or with 1313?
Did you perform a second explosion?
Play with the machine to see if you can make sense of matters for yourself, or read my thoughts below.
loader
The traditional algorithm has us work from right to left, looking at 8+78 + 7 first.
I3SF-image42
But in the algorithm we don’t write down the answer 1515. Instead, we explode ten dots right away and write on paper a 55 in the answer line together with a small 11 tacked on to the middle column. People call this carrying the one and it – correctly – corresponds to adding an extra dot in the tens position.
I3SF-image46
I3SF-image47
Now we attend to the middle boxes. Adding gives 1414 dots in the tens box (5+85 + 8 gives thirteen dots, plus the extra dot from the previous explosion).
I3SF-image50
Now we perform another explosion.
I3SF-image51
I3SF-image52
On paper, one writes 44 in the answer line, in the tens position, with another little 11 placed in the next column over. This matches the idea of the dots-and-boxes picture precisely.
And now we finish the problem by adding the dots in the hundreds position.
I3SF-image55
I3SF-image56
So the traditional algorithm works right to left and does explosions (“carries”) as one goes along. On paper it is swift and compact and this might be why it has been the favored way of doing long addition for centuries.
The Exploding Dots approach works left to right, just as we are taught to read in English, and leaves all the explosions to the end. It is easy to understand and kind of fun.
Both approaches, of course, are good and correct. It is just a matter of taste and personal style which one you choose to do. (But feel free to come up with your own new, and correct, approach!)
Arithmos
Vision byPowered by
Contact GMP:
• The Global Math Project on Twitter
• The Global Math Project on Facebook
• Contact The Global Math Project
Contact BM:
• Buzzmath on Twitter
• Buzzmath on Facebook
• Visit Buzzmath's Website
• Read Buzzmath's blog! | {
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4,990,973,584,426,764,000 | Difficulty Finding the Domain of a Function using Interval Notation.
denniswood
New member
Joined
Jun 21, 2019
Messages
2
Hi everyone, I'm having difficulty figuring out how to answer these two questions below, specifically the first one. I have been trying for days to answer this question but can't figure it out. If anyone could help me out on these two answers, I would be so grateful.
1. Find the domain of the function using interval notation.
12667
2. Sketch a graph of a piecewise function. Write the domain in interval notation.
y = x^2 {-1 < x < 1}
and
y = 3x - 2 {1 < x < 3}
Last edited by a moderator:
Harry_the_cat
Senior Member
Joined
Mar 16, 2016
Messages
1,393
Ok to help with Q1, can you first answer these questions:
1. For \(\displaystyle \frac{1}{x}\) to be defined, what CAN'T x be?
2. For \(\displaystyle \sqrt{x}\) to be defined what CAN'T x be?
denniswood
New member
Joined
Jun 21, 2019
Messages
2
Ok to help with Q1, can you first answer these questions:
1. For \(\displaystyle \frac{1}{x}\) to be defined, what CAN'T x be?
2. For \(\displaystyle \sqrt{x}\) to be defined what CAN'T x be?
This is why I am having such difficulty because those aspects of what x CAN'T be were NOT listed in the assignment question, for either of the questions, which is why I have been unable to identify what it is I must do.
Otis
Senior Member
Joined
Apr 22, 2015
Messages
1,651
… having difficulty figuring out how to answer these …
… Sketch a graph of a piecewise function. Write the domain in interval notation …
Hello. I'm not sure where you're having difficulty sketching. Do you know how to sketch y=x^2 by itself? Can you sketch the line y=3x-2?
Also, the two intervals (that is, the two sets of numbers comprising the domain) have been given to you. Is 'interval notation' the difficulty? Or, maybe you haven't learned the definition of 'domain', yet.
Please follow the forum's submission guidelines, and be specific. Show your efforts, thus far. Thanks!
😎
Last edited:
Otis
Senior Member
Joined
Apr 22, 2015
Messages
1,651
… I am having such difficulty because those aspects of what x CAN'T be were NOT listed in the assignment question …
Maybe that information wasn't provided because you're expected to have already learned it in pre-algebra.
We cannot divide by zero, so the expression 1/x is defined for all Real numbers x except 0. In other words, the denominator of the algebraic fraction in your first exercise cannot be zero.
√(x - 4) will be zero IF x - 4 = 0, and we cannot allow that to happen, so what value of x causes the denominator to become zero?
Next, we can't take the square root of negative numbers, so √x is defined only when x≥0. In other words, in the numerator the radicand cannot be negative. (The radicand is what we call the expression inside a radical sign.)
The radicand in the numerator is x - 6. It can't be negative. So, solve the inequality x-6≥0.
Please show your work, up to this point, and we can go from there.
By the way, you posted on the Advanced Math board, but this is high-school level math (and pre-algebra was 8th-grade math, at my school). What class are you taking?
😎
JeffM
Elite Member
Joined
Sep 14, 2012
Messages
3,576
This is why I am having such difficulty because those aspects of what x CAN'T be were NOT listed in the assignment question, for either of the questions, which is why I have been unable to identify what it is I must do.
The domain of a real function is the set for which the function is defined to be a real number. In many cases, a domain is not explicitly specified. In those cases, we want or need to know the maximum possible domain. The usual way to do that is to exclude what cannot possibly be in the domain. More concretely, the maximum possible domain is all real numbers except those where where the function will not be a real number.
You have a function in the form \(\displaystyle f(x) = \dfrac{\sqrt{u}}{v} \text { and } v = \sqrt{w}. \)
So where can't the function be a real number?
Well clearly v = 0 is excluded, and v will equal 0 only if w = 0. Also clearly excluded are w < 0 and u < 0. But these are all functions of x. Putting them all together what values of x are excluded? Then the maximum oossible domain is everything else.
Personally, I think these questions are badly worded. Instead of asking "what is the domain," what should be asked is "what is the maximum possible domain." In that case, it becomes obvious that the way to answer is to determine what cannot be in the domain and include everything else.
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2,934,729,624,590,425,600 | Minimize (max(A[i], B[j], C[k]) – min(A[i], B[j], C[k])) of three different sorted arrays in C++
C++Server Side ProgrammingProgramming
Concept
With respect of given three sorted arrays A, B, and C of not necessarily same sizes,compute the lowest i.e. minimum absolute difference between the maximum and minimum number of any triplet A[i],B[j], C[k] such that they are under arrays A, B and C respectively, i.e., minimize (max(A[i], B[j], C[k]) – min(A[i], B[j], C[k])).
Input
A : [ 2, 5, 6, 9, 11 ]
B : [ 7, 10, 16 ]
C : [ 3, 4, 7, 7 ]
Output
1
Explanation
When we select A[i] = 6 , B[j] = 7, C[k] = 7, we get the minimum difference as max(A[i], B[j], C[k]) - min(A[i], B[j], C[k])) = |7-6| = 1
Input
A = [ 6, 9, 11, 16 ]
B = [ 7, 10, 16, 79, 90 ]
C = [ 3, 4, 7, 7, 9, 9, 11 ]
Output
1
Explanation
When we select A[i] = 11 , b[j] = 10, C[k] = 11. we get the minimum difference as max(A[i], B[j], C[k]) - min(A[i], B[j], C[k])) = |11-10| = 1
Method
Begin with the highest elements in each of the arrays A, B & C. Track a variable to update the answer at the time of each of the steps to be followed.
With respect of each and every step, the only possible way to lessen the difference is to lessen the maximum element out of the three elements.
As a result of this, visit to the next highest element in the array containing the maximum element for this step and update the answer variable.
We have to repeat this step until and unless the array containing the maximum element terminates.
Example(C++)
// C++ code for above approach
Live Demo
#include<bits/stdc++.h>
usingnamespacestd;
intsolve(intA1[], intB1[], intC1[], inti1, intj1, intk1) {
intmin_diff, current_diff, max_term;
// calculating min difference from last
// index of lists
min_diff = abs(max(A1[i1], max(B1[j1], C1[k1]))
- min(A1[i1], min(B1[j1], C1[k1])));
while(i1 != -1 && j1 != -1 && k1 != -1) {
current_diff = abs(max(A1[i1], max(B1[j1], C1[k1]))
- min(A1[i1], min(B1[j1], C1[k1])));
// checking condition
if(current_diff < min_diff)
min_diff = current_diff;
// calculating max term from list
max_term = max(A1[i1], max(B1[j1], C1[k1]));
if(A1[i1] == max_term)
i1 -= 1;
elseif(B1[j1] == max_term)
j1 -= 1;
else
k1 -= 1;
}
returnmin_diff;
}
intmain() {
intD1[] = { 5, 8, 10, 15 };
intE1[] = { 6, 9, 15, 78, 89 };
intF1[] = { 2, 3, 6, 6, 8, 8, 10 };
intnD = sizeof(D1) / sizeof(D1[0]);
intnE = sizeof(E1) / sizeof(E1[0]);
intnF = sizeof(F1) / sizeof(F1[0]);
cout << solve(D1, E1, F1, nD-1, nE-1, nF-1);
return0;
}
Output
1
raja
Updated on 27-Aug-2020 12:17:27
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-8,988,831,260,237,813,000 | Math
Back
HSN-Q.A.3
HSN-Q.A.3
Choose a level of accuracy appropriate to limitations on measurement when reporting quantities.*
View child tags
View All 36 Lessons
HSN-Q.A.1
Use units as a way to understand problems and to guide the solution of multi-step problems; choose and interpret units consistently in formulas; choose and interpret the scale and the origin in graphs and data displays.*
HSN-Q.A.2
Define appropriate quantities for the purpose of descriptive modeling.*
HSN-Q.A.3
Choose a level of accuracy appropriate to limitations on measurement when reporting quantities.* | {
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7,140,944,021,159,707,000 | Determining Pot Odds and Pot Equity
There will be many times when a player is forced to decide whether or not they should call a bet that they face, or perhaps if they should raise or if it is time to concede the hand to their opponent. In most cases, the correct action for the player to take is determined by whether or not the math adds up to make the call. People use the math involved to determine whether or not this action is likely to be a profitable one in the long term, which is a simple calculation of pot odds.
Pot odds are determined by the size of the bet in relation to the pot as a whole. A simple example would be if the pot currently had 50 dollars in it and you were required to pay 5 dollars to remain in the hand, you would be getting 10/1 on your call. You can quickly determine the long term profitability of an action by determine whether or not the likelihood of your hand being made is greater than the current percentage of the pot it is to call. In our example to calculate the percentage you reduce the numbers to 10/1 and then add them to get 11 which from there you divide the cost of the call (the 1) by the sum (11) to get .09 or 9 percent. Therefore, if your hand was going to be made greater than 9 percent of the time, it would be profitable in the long run as you would have a positive expected value on your call.
The situations we have looked at so far tend to tell us what to do when we have to face calling a bet. Pot Equity looks at situations when we should decide whether or not we should be betting or raising in order to win a pot. In order to understand the concept of pot equity you are going to want to know the percentages of your hand winning in the long run, given every possible deal. Once you have determined your percentage of winning the pot you can compare that action to the price of the action in front of you. If for instance you had a hand that would win the pot 40 percent of the time from the current spot and the size of the pot was 200 the equity you would have is $80 (200*.40). If your equity is greater than the action in front of you, you want to be calling or raising as in the long run you have a positive expected value. | {
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Results 1 to 2 of 2
Thread: Quantifiers question
1. #1
Newbie
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Quantifiers question
It's been a little why since I've had discrete math, so could someone help me remember why:
"There exists an 'x' such that for all 'y', x + y == 0" is FALSE,
but
"For all 'x', there exists a 'y' such that x + y == 0" is TRUE?
Thanks!
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2. #2
GAMMA Mathematics
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Quote Originally Posted by jtc4zH View Post
It's been a little why since I've had discrete math, so could someone help me remember why:
"There exists an 'x' such that for all 'y', x + y == 0" is FALSE,
but
"For all 'x', there exists a 'y' such that x + y == 0" is TRUE?
Thanks!
Read it twice over and slowly. The first statement says that there is one number, x that when added to any number y, it equals zero. You cannot possibly say that both x+4=0 and x+55=0.
The second line says that for any number x, there exists a y that satisfies those conditions. This is a one-to-one relationship. For:
x=5,
y=-5;
for x=,
y=-.
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Results 1 to 7 of 7
Math Help - Need a generic function to perform a task.
1. #1
Newbie
Joined
Aug 2007
Posts
23
Exclamation Need a generic function to perform a task.
I need to perform a task using a math function (i'm a programer not a mathematician unfortunately).
The task i need to perform is to cycle through a list in a sinusoidal motion according to an increasing count.
the count (time) is increasing at a fixed rate, no problems there, however, I need the number of items in the cycle to change.
For example I have a 3 item cycle, 1->2->3->2->1->2 etc
I have managed to write an equation that will give me the result I need for this case:
y = sin(rad(90)*x)-rad(90))+2
So if t = x, number of items = 3:
Code:
t0 = 1,
t1 = 2,
t2 = 3,
t3 = 2,
t4 = 1
(with rounding of the result)
However, I may need to vary the number of items in the cycle to any arbitrary number, and as such would need for example the following to be true:
8 items:
Code:
t0 = 1,
t1 = 2,
t2 = 3,
t3 = 4,
t4 = 5,
t6 = 7,
t7 = 8,
t8 = 7,
t9 = 6,
t10 = 5,
...
t of 14 = 1
how would I modify the equation to take into account the number of items in the cycle? (If at all possible)
Thanks
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2. #2
Grand Panjandrum
Joined
Nov 2005
From
someplace
Posts
14,972
Thanks
4
Quote Originally Posted by Alias_NeO View Post
I need to perform a task using a math function (i'm a programer not a mathematician unfortunately).
The task i need to perform is to cycle through a list in a sinusoidal motion according to an increasing count.
the count (time) is increasing at a fixed rate, no problems there, however, I need the number of items in the cycle to change.
For example I have a 3 item cycle, 1->2->3->2->1->2 etc
I have managed to write an equation that will give me the result I need for this case:
y = sin(rad(90)*x)-rad(90))+2
So if t = x, number of items = 3:
Code:
t0 = 1,
t1 = 2,
t2 = 3,
t3 = 2,
t4 = 1
(with rounding of the result)
However, I may need to vary the number of items in the cycle to any arbitrary number, and as such would need for example the following to be true:
8 items:
Code:
t0 = 1,
t1 = 2,
t2 = 3,
t3 = 4,
t4 = 5,
t6 = 7,
t7 = 8,
t8 = 7,
t9 = 6,
t10 = 5,
...
t of 14 = 1
how would I modify the equation to take into account the number of items in the cycle? (If at all possible)
Thanks
This has nothing to do with sinusoids, you seem to want a sequence which increments/decrements by 1 repeatedly between and upper and lower limit.
Now why do you want a formula for the n-th term of this why not just generate the sequence programaticaly?
CB
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3. #3
Newbie
Joined
Aug 2007
Posts
23
Unfortunately, the structure of my application means it cannot be done programatically, there is no fixed number of times the calculation will be performed, nor is the time between calculations fixed and incrementing. I need it to be calculated on the spot according to whatever value of t is passed at the time.
In other words, as far as my application (abstract and programming) goes, time is not linearly increasing. It could be the same, it could be the same time for 10 or a hundred iterations before time changes.
It's hard for me to explain unless you can tell me you're a developer and I can use some technical terms?
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4. #4
Junior Member RHandford's Avatar
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Sep 2010
From
N.E. Lincolnshire
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Hi
I have benn writing apps for may years, so explain away and I will see if I can help
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5. #5
Newbie
Joined
Aug 2007
Posts
23
I think I may have found a solution using modulus now, but for the sake of completion I'll explain the initial requirement.
I have a function in my program that is called every time the screen of the device is refreshed, this is an arbitrary, unknown number. Because I don't want the value to increment the function is called, I pass it a "tick" this tick is my "t" above, it checks against this if time has changed or not and calculates the new value accordingly.
Another method would be for me to store the previous time and check that way and increment if time has changed, but because my class and method are static storage isn't ideal.
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6. #6
MHF Contributor
Joined
Oct 2009
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786
Let a be some parameter. Consider f(n)=a-|(n\mathrel{\mathrm{mod}}2a)-a|. Here n\mathrel{\mathrm{mod}}2a is the remainder when n is divided by 2a.
Explanation. The absolute value |n| goes down to 0 when n is negative goes back up when n is positive. Next, |n-a| means shifting the graph of |n| to the right a units, so now the value at a is 0. Next, a-|n-a| flips the graph upside down and raises it a units, so now the value at 0 and 2a is 0, and the value at a is a. Finally, replacing n with (n mod 2a) makes the function periodic with the period 2a.
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7. #7
Newbie
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Aug 2007
Posts
23
Quote Originally Posted by emakarov View Post
Let a be some parameter. Consider f(n)=a-|(n\mathrel{\mathrm{mod}}2a)-a|. Here n\mathrel{\mathrm{mod}}2a is the remainder when n is divided by 2a.
Explanation. The absolute value |n| goes down to 0 when n is negative goes back up when n is positive. Next, |n-a| means shifting the graph of |n| to the right a units, so now the value at a is 0. Next, a-|n-a| flips the graph upside down and raises it a units, so now the value at 0 and 2a is 0, and the value at a is a. Finally, replacing n with (n mod 2a) makes the function periodic with the period 2a.
Perfect, that's exactly the modulus function I needed. With a couple of slight changes to shift the values up from zero, that works perfectly. Thanks.
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531,163,812,967,821,100 | What is the compound interest on $49878 at 17% over 14 years?
If you want to invest $49,878 over 14 years, and you expect it will earn 17.00% in annual interest, your investment will have grown to become $449,273.80.
$
$
If you're on this page, you probably already know what compound interest is and how a sum of money can grow at a faster rate each year, as the interest is added to the original principal amount and recalculated for each period.
The actual rate that $49,878 compounds at is dependent on the frequency of the compounding periods. In this article, to keep things simple, we are using an annual compounding period of 14 years, but it could be monthly, weekly, daily, or even continuously compounding.
The formula for calculating compound interest is:
$$A = P(1 + \dfrac{r}{n})^{nt}$$
• A is the amount of money after the compounding periods
• P is the principal amount
• r is the annual interest rate
• n is the number of compounding periods per year
• t is the number of years
We can now input the variables for the formula to confirm that it does work as expected and calculates the correct amount of compound interest.
For this formula, we need to convert the rate, 17.00% into a decimal, which would be 0.17.
$$A = 49878(1 + \dfrac{ 0.17 }{1})^{ 14}$$
As you can see, we are ignoring the n when calculating this to the power of 14 because our example is for annual compounding, or one period per year, so 14 × 1 = 14.
How the compound interest on $49,878 grows over time
The interest from previous periods is added to the principal amount, and this grows the sum a rate that always accelerating.
The table below shows how the amount increases over the 14 years it is compounding:
Start Balance Interest End Balance
1 $49,878.00 $8,479.26 $58,357.26
2 $58,357.26 $9,920.73 $68,277.99
3 $68,277.99 $11,607.26 $79,885.25
4 $79,885.25 $13,580.49 $93,465.75
5 $93,465.75 $15,889.18 $109,354.92
6 $109,354.92 $18,590.34 $127,945.26
7 $127,945.26 $21,750.69 $149,695.95
8 $149,695.95 $25,448.31 $175,144.27
9 $175,144.27 $29,774.53 $204,918.79
10 $204,918.79 $34,836.19 $239,754.99
11 $239,754.99 $40,758.35 $280,513.33
12 $280,513.33 $47,687.27 $328,200.60
13 $328,200.60 $55,794.10 $383,994.70
14 $383,994.70 $65,279.10 $449,273.80
We can also display this data on a chart to show you how the compounding increases with each compounding period.
In this example we have 14 years of compounding, but to truly see the power of compound interest, it might be better to look at a larger number of compounding periods to see how much $49,878 can grow.
If you want an example with more compounding years, click here to view the compounding interest of $49,878 at 17.00% over 30 years.
As you can see if you view the compounding chart for $49,878 at 17.00% over a long enough period of time, the rate at which it grows increases over time as the interest is added to the balance and new interest calculated from that figure.
How long would it take to double $49,878 at 17% interest?
Another commonly asked question about compounding interest would be to calculate how long it would take to double your investment of $49,878 assuming an interest rate of 17.00%.
We can calculate this very approximately using the Rule of 72.
The formula for this is very simple:
$$Years = \dfrac{72}{Interest\: Rate}$$
By dividing 72 by the interest rate given, we can calculate the rough number of years it would take to double the money. Let's add our rate to the formula and calculate this:
$$Years = \dfrac{72}{ 17 } = 4.24 $$
Using this, we know that any amount we invest at 17.00% would double itself in approximately 4.24 years. So $49,878 would be worth $99,756 in ~4.24 years.
We can also calculate the exact length of time it will take to double an amount at 17.00% using a slightly more complex formula:
$$Years = \dfrac{log(2)}{log(1 + 0.17)} = 4.41\; years$$
Here, we use the decimal format of the interest rate, and use the logarithm math function to calculate the exact value.
As you can see, the exact calculation is very close to the Rule of 72 calculation, which is much easier to remember.
Hopefully, this article has helped you to understand the compound interest you might achieve from investing $49,878 at 17.00% over a 14 year investment period. | {
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3,677,814,823,803,467,300 | Triangle calculator ASA
Please enter the side of the triangle and two adjacent angles
°
°
Right scalene triangle.
Sides: a = 5.78664666378 b = 1.97990881488 c = 5.43875
Area: T = 5.38106459046
Perimeter: p = 13.20330547867
Semiperimeter: s = 6.60215273933
Angle ∠ A = α = 90° = 1.57107963268 rad
Angle ∠ B = β = 20° = 0.34990658504 rad
Angle ∠ C = γ = 70° = 1.22217304764 rad
Height: ha = 1.86597345293
Height: hb = 5.43875
Height: hc = 1.97990881488
Median: ma = 2.89332333189
Median: mb = 5.52768077337
Median: mc = 3.3632795186
Inradius: r = 0.81550607555
Circumradius: R = 2.89332333189
Vertex coordinates: A[5.43875; 0] B[0; 0] C[5.43875; 1.97990881488]
Centroid: CG[3.625; 0.66596960496]
Coordinates of the circumscribed circle: U[2.719875; 0.99895440744]
Coordinates of the inscribed circle: I[4.62224392445; 0.81550607555]
Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 90° = 1.57107963268 rad
∠ B' = β' = 160° = 0.34990658504 rad
∠ C' = γ' = 110° = 1.22217304764 rad
Calculate another triangle
How did we calculate this triangle?
1. Calculate the third unknown inner angle
alpha = 90° ; ; beta = 20° ; ; ; ; alpha + beta + gamma = 180° ; ; ; ; gamma = 180° - alpha - beta = 180° - 90° - 20° = 70° ; ;
2. By using the law of sines, we calculate unknown side a
c = 5.44 ; ; ; ; fraction{ a }{ c } = fraction{ sin( alpha ) }{ sin ( gamma ) } ; ; ; ; a = c * fraction{ sin( alpha ) }{ sin ( gamma ) } ; ; ; ; a = 5.44 * fraction{ sin(90° ) }{ sin (70° ) } = 5.79 ; ;
3. By using the law of sines, we calculate last unknown side b
fraction{ b }{ c } = fraction{ sin( beta ) }{ sin ( gamma ) } ; ; ; ; b = c * fraction{ sin( beta ) }{ sin ( gamma ) } ; ; ; ; b = 5.44 * fraction{ sin(20° ) }{ sin (70° ) } = 1.98 ; ;
Now we know the lengths of all three sides of the triangle and the triangle is uniquely determined. Next we calculate another its characteristics - same procedure as calculation of the triangle from the known three sides SSS.
a = 5.79 ; ; b = 1.98 ; ; c = 5.44 ; ;
4. The triangle circumference is the sum of the lengths of its three sides
p = a+b+c = 5.79+1.98+5.44 = 13.2 ; ;
5. Semiperimeter of the triangle
s = fraction{ o }{ 2 } = fraction{ 13.2 }{ 2 } = 6.6 ; ;
6. The triangle area using Heron's formula
T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 6.6 * (6.6-5.79)(6.6-1.98)(6.6-5.44) } ; ; T = sqrt{ 28.95 } = 5.38 ; ;
7. Calculate the heights of the triangle from its area.
T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 5.38 }{ 5.79 } = 1.86 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 5.38 }{ 1.98 } = 5.44 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 5.38 }{ 5.44 } = 1.98 ; ;
8. Calculation of the inner angles of the triangle using a Law of Cosines
a**2 = b**2+c**2 - 2bc cos( alpha ) ; ; alpha = arccos( fraction{ a**2-b**2-c**2 }{ 2bc } ) = arccos( fraction{ 5.79**2-1.98**2-5.44**2 }{ 2 * 1.98 * 5.44 } ) = 90° ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 1.98**2-5.79**2-5.44**2 }{ 2 * 5.79 * 5.44 } ) = 20° ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 5.44**2-5.79**2-1.98**2 }{ 2 * 1.98 * 5.79 } ) = 70° ; ;
9. Inradius
T = rs ; ; r = fraction{ T }{ s } = fraction{ 5.38 }{ 6.6 } = 0.82 ; ;
10. Circumradius
R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 5.79 }{ 2 * sin 90° } = 2.89 ; ;
Look also our friend's collection of math examples and problems:
See more informations about triangles or more information about solving triangles. | {
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5,058,896,242,616,722,000 | Taylor series
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Taylor series
1. 1. + Taylor Series John Weiss
2. 2. + Approximating Functions f(0)= 4 What is f(1)? f(x) = 4? f(1) = 4?
3. 3. + Approximating Functions f(0)= 4, f’(0)= -1 What is f(1)? f(x) = 4 - x? f(1) = 3?
4. 4. + Approximating Functions f(0)= 4, f’(0)= -1, f’’(0)= 2 What is f(1)? f(x) = 4 – x + x2? (same concavity) f(1) = 4?
5. 5. + Approximating Functions f(x) = sin(x) What is f(1)? f(0) = 1, f’(0) = 1 f(x) = 0 + x? f(1) = 1?
6. 6. + Approximating Functions f(x) = sin(x) f(0) = 1, f’(0) = 1, f’’(0) = 0, f’’’(0) = -1,… What is f(1)? i.e . What is sin(1)?
7. 7. + Famous Dead People James Gregory (1671) Brook Taylor (1712) Colin Maclaurin (1698-1746) Joseph-Louis Lagrange (1736-1813) Augustin-Louis Cauchy (1789-1857)
8. 8. + Approximations Linear Approximation f (x) = f (a) + f ʹ′(a)(x − a) + R1 (x)(x − a) R1 (x)(x − a) = f (x) − f (a) − f ʹ′(x − a) Quadratic Approximation f ʹ′ (a)€ f (x) = f (a) + f ʹ′(a)(x − a) + (x − a) 2 + R2 (x)(x − a) 2 2 f ʹ′ (a) 2 R2 (x)(x − a) = f (x) − f (a) − f ʹ′(a)(x − a) − (x − a) 2 2€
9. 9. + Taylor’s Theorem Letk≥1 be an integer and f : R →R be k times differentiable at a ∈ R . Then there exists a function R : R →R such that k f ʹ′ (a) f k (a)f (x) = f (a) − f ʹ′(a)(x − a) + (x − a) 2 + ...+ (x − a) k + Rk (x)(x − a) k 2! k! € € Note: Taylor Polynomial of degree k is: € f ʹ′ (a) f k (a) Pk (x) = f (a) − f ʹ′(a)(x − a) + (x − a) 2 + ...+ (x − a) k 2! k!
10. 10. + Works for Linear Approximations f (x) = c 0 + c1 (x) f (a) = c 0 + c1 (a) f ʹ′(a) = c1 f (x)(a)fʹ′= c + c (x − a) f = f (a) = c (a) + 0 11 1€ f (x) = f (a) + c1€ € € a) (x − € f (x) = c 0 + c1 (a) + c1 (x − a) = c 0 + c1 (x) €€ €
11. 11. + Works for Quadratic Approximations f (x) = c 0 + c1 (x) + c 2 (x 2 ) f (a) = c 0 + c1 (a) + c 2 (a 2 ) f ʹ′(a) = c1 + 2c 2 (a) f ʹ′ (a) = 2c 2 € = c + c (a) + c (a 2 ) + [c + 2c (a)](x − a) + 2c 2 [x − a]2 = € f (x) 0 1 2 1 2 € 2 c 0€ c1(a) + c 2 (a 2 ) + c1(x − a) + 2c 2 (x − a) + c 2 (x) 2 − c 2 (2ax) + c 2 (a) 2 = + f (x) = c 0 + c1 (x) + c 2 (x 2 )€
12. 12. + f(x) = sin(x) Degree 1
13. 13. + f(x) = sin(x) Degree 3
14. 14. + f(x) = sin(x) Degree 5
15. 15. + f(x) = sin(x) Degree 7
16. 16. + f(x) = sin(x) Degree 11
17. 17. + Implications Any smooth functions with all the same derivatives at a point MUST be the same function!
18. 18. + Proof: If f and g are smooth functions that agree over some interval, they MUST be the same function Let f and g be two smooth functions that agree for some open interval (a,b), but not over all of R Define h as the difference, f – g, and note that h is smooth, being the difference of two smooth functions. Also h=0 on (a,b), but h≠0 at other points in R Without loss of generality, we will form S, the set of all x>a, such that f(x)≠0 Note that a is a lower bound for this set, S, and being a subset of R, S is complete so S has a real greatest lower bound, call it c. c, being a greatest lower bound of S, is also an element of S, since S is closed Now we see that h=0 on (a,c), but h≠0 at c. So, h is discontinuous at c, but then h cannot be smooth Thus we have reached a contradiction, and so f and g must agree everywhere!
19. 19. + Suppose f(x) can be rewritten as a power series… f (x) = c 0 + c1 (x − a) + c 2 (x − a) 2 + ...+ c n (x − a) n c 0 = f (a) f ʹ′(x) = c1 + 2c 2 (x − a) + 3c 3 (x f − a) 2 + ...+ nc n (x − a) n −1 k (a) ck =€ k! c1 = f ʹ′(a)€ f ʹ′ (x) = 2c 2 + 3∗2c 3 (x − € + 4 ∗ 3c 4 (x − a) 2 + ...+ n ∗(n −1)c n (x − a) n −2 a)€ f ʹ′ (a) c2 =€ 2!€ f k (a) ck = k!€
20. 20. + Entirety (Analytic Functions) A function f(x) is said to be entire if it is equal to its Taylor Series everywhere Entire Not Entire sin(x) log(1+x)
21. 21. + Proof: sin(x) is entire Maclaurin Series sin(0)=1 sin’(0)=0 ∞ sin’(0)=-1 (−1) n 2n +1 sin(x) = ∑ x sin’(0)=0 n =0 (2n +1)! sin’(0)=1 sin’(0)=0 sin’(0)=-1 … etc. €
22. 22. + Proof: sin(x) is entire ∞ (−1) n 2n +1 sin(x) = ∑ x n =0 (2n +1)! Lagrange formula for the remainder: Let f : R →R be k+1 times differentiable on (a,x) and continuous on [a,x]. Then f k +1 (z) k +1 Rk (x) = (x − a) (k +1)!€ for some z in (x,a)€
23. 23. + Proof: sin(x) is entire First, sin(x) is continuous and infinitely differentiable over all of R If we look at the Taylor Polynomial of degree k f k +1 (z) k +1 Rk (x) = (x − a) (k +1)! Note though f k +1 (z) ≤ 1 for all z in R k +1 (x − a) Rk (x) ≤€ (k +1)! €€
24. 24. + Proof: sin(x) is entire However, as k goes to infinity, we see Rk (x) ≤ 0 Applyingthe Squeeze Theorem to our original € equation, we obtain that as k goes to infinity f (x) = Tk (x) and thus sin(x) is complete€
25. 25. + Maclaurin Series Examples ∞ ∞ xn xn log(1 − x) = −∑ log(1+ x) = ∑ (−1) n +1 n =1 n! n =1 n! ∞ ∞ 1 (−1) n (2n)! = ∑ xn 1+ x = ∑ xn 1 − x n =0 n =0 (1 − 2n)(n!) 2 (4) n ∞€ xn€ ex = ∑ n =0 n! ∞€ ∞ € (−1) n (−1) n 2n sin(x) = ∑ x 2n +1 cos(x) = ∑ x n =0 (2n +1)! n =0 (2n)!€ ix Note: e = cos(x) + isin(x)€ €
26. 26. + Applications Physics Special Relativity Equation Fermat’s Principle (Optics) Resistivity of Wires Electric Dipoles Periods of Pendulums Surveying (Curvature of the Eart)
27. 27. + Special Relativity m0 m= KE = mc 2 − m0c 2 1− v2 c2 m0c 2 ⎡⎛ v 2 ⎞ −1/ 2 ⎤ KE = − m0c 2 = m0c 2 ⎢⎜1 − 2 ⎟ −1⎥ 2 1−v c 2 € ⎢⎝ c ⎠ ⎣ ⎥ ⎦€ If v ≤ 100 m/s€ Then according to Taylor’s Inequality 1 3m0c 2 100 4 −10 R1 (x) ≤ 2 2 4 < (4.17 × 10 )m0 2 4(1 −100 /c ) c
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4,515,344,418,120,316,400 |
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Killer Sudoku puzzles include all the rules of regular Sudoku but add in dashed-line 'cages' which must sum to given values. Killer Sudoku is also sometimes known as Sumdoku, Mathdoku or Addoku.
How to solve Killer Sudoku puzzles
The name 'Killer' Sudoku arises because of the wicked twist on the standard Sudoku puzzle, since you must not only place each of the numbers 1 to 9 (or 1 to the size of the puzzle) into each of the rows, columns and bold-lined 3x3 (or other size) boxes but you must also place the numbers into each dashed-line cage so that they add up to its given total - and without repeating a digit in a dashed-line cage.
Killer Sudoku exampleLook at the solved puzzle on the left. In this 6x6 puzzle the right-most three squares in the first row must add up to 6. The values given are 1, 3 and 2 which are of course fine since 1+3+2=6. But because you cannot repeat a number within a cage this means that we knew the contents were 1+2+3 in some order even before we started solving (although any other solution, such as 1+1+4, would in this puzzle be disallowed by the standard rules of Sudoku in any case).
Every Killer Sudoku puzzle only ever has one possible solution, and it can always be reached via reasonable logical deduction. In other words, guessing is never required.
Killer Sudoku puzzles on puzzlemix appear at a wide range of sizes and difficulties. Usually they are 6x6, 8x8 or 9x9. In each respective case you must place 1-6, 1-8 or 1-9 once each into every row, column and bold-lined rectangle.
Other types of puzzle
It's not just Killer Sudoku you can solve online at puzzlemix. You can also solve logic puzzles that include these:
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Sudoku XVSudoku-XToroidal Sudoku
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6,544,848,460,746,563,000 | Occupancy Rate
Occupancy Rate is a metric used in the hospitality industry to measure the utilization of a hotel’s rooms. It represents the percentage of rooms occupied by guests during a given period, and provides insight into the demand for the hotel’s rooms and its overall financial performance.
Occupancy Rate can be calculated by dividing the number of occupied rooms by the total number of rooms available, multiplied by 100.
Mathematically, it can be expressed as:
Occupancy Rate = (Number of Occupied Rooms / Total Number of Rooms) x 100
For example, if a hotel has 100 rooms and 80 of them are occupied, the occupancy rate can be calculated as follows:
Occupancy Rate = (80 rooms / 100 rooms) x 100 = 80%
So, the hotel would have an occupancy rate of 80%, meaning that 80% of its rooms are occupied by guests.
Note that the occupancy rate is expressed as a percentage, so it provides an easy-to-read, comparative measure of hotel utilization.
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} | 0c090d63199a0a01e3b08e4a255778a0 |
3,775,439,540,993,113,600 |
Results 1 to 6 of 6
Math Help - System of equation, condition of a
1. #1
MHF Contributor arbolis's Avatar
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System of equation, condition of a
For which values of a the following system of equations has a unique solution? Infinite solutions?
\left[ \begin{array}{ccc}x-y+z=2 \\ ax-y+z=2 \\ 2x-2y +(2-a)z=4a \end{array} \right].
I've solve the system via matrices and got that \left[ \begin{array}{ccc}x=0 \\ y=-2-4a \\ z=-4a(a-1) \end{array} \right]. I'm stuck from here... I see that for a given a there is always a unique solution, but that you can give whatever value for a so that you have an infinity of solutions. So how do I answer the question?
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2. #2
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Hello, arbolis!
For which values of a does the following system of equations has a unique solution?
Infinite solutions?
\begin{array}{ccc}x-y+z\;=\;2 & [1]\\ ax-y+z\;=\;2 & [2]\\ 2x-2y +(2-a)z\;=\;4a & [3]\end{array}
Just "eyeballing" the system, I can see the following:
If a = 1, equations [1] and [2] are identical.
. . The system will have infinite solutions.
If a = 0, equation [3] becomes: . 2x - 2y + 2z \:=\:0 \quad\Rightarrow\quad x - y + z \:=\:0
. . which is incompatible with equation [1].
The system will have no solution.
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3. #3
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Thank you Soroban!
You are right, it seems more easy than I thought. So what I did with matrices was not necessary and only complicates things.
But I'd like to know if there is a formal way to determine what values of a would make the system to get only one solution. Is that possible? Or have I to "eyeball" this?
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4. #4
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Hello,
Here is my trial... lol
The system can be rewritten this way :
\begin{pmatrix} 1 & -1 & 1 \\ a & -1 & 1 \\ 2 & -2 & 2-a \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix}=\begin{pmatrix} 2 \\ 2 \\ 4a \end{pmatrix}
I don't have my notes with me, so I can't check. The wikipedia article may help you, I'm not sure (System of linear equations - Wikipedia, the free encyclopedia).
In any event, the span has a basis of linearly independent vectors that do guarantee exactly one expression; and the number of vectors in that basis (its dimension) cannot be larger than m or n, but it can be smaller. This is important because if we have m independent vectors a solution is guaranteed regardless of the right-hand side, and otherwise not guaranteed.
Independent vectors mean that there is no linear combination of two of them giving the third. Or any relation of proportionality.
If you find dependent vectors, it is likely that you'll get an infinity of solutions. And if there are dependent vectors, the discriminant will be = 0 (to be checked).
General behavior
The solution set for two equations in three variables is usually a line.
The solution set for two equations in three variables is usually a line.
In general, the behavior of a linear system is determined by the relationship between the number of equations and the number of unknowns:
1. Usually, a system with fewer equations than unknowns has infinitely many solutions.
2. Usually, a system with the same number of equations and unknowns has a single unique solution.
3. Usually, a system with more equations than unknowns has no solution.
This may be helpful too.
I'm sorry not to be able to help more, algebra is not what I prefer so I easily forget these things
Hope you'll extract something from this.
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5. #5
MHF Contributor arbolis's Avatar
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And if there are dependent vectors, the discriminant will be = 0 (to be checked).
What do you mean by discriminant? Anyway, I've not yet seen basis (even if I had to deal with them when I studied Numerical Analysis).
And I knew about
1. Usually, a system with fewer equations than unknowns has infinitely many solutions.
2. Usually, a system with the same number of equations and unknowns has a single unique solution.
.
And about
The system can be rewritten this way :
I know, that's what I did to reach
but I certainly did an error since it doesn't fit with the eyeball of Soroban.
So my question remains "I'd like to know if there is a formal way to determine what values of would make the system to get only one solution."
Thanks for your time Moo and your help, I appreciate the research you've done for me.
Lastly,
I'm sorry not to be able to help more, algebra is not what I prefer so I easily forget these things
I hate this matter (even if less than a year ago) but it is an extremely important one. Not only for me (as a physics student) but for many other careers. I had an exam of physics I last Wednesday and I totally depreciated Algebra II. The exam of Algebra II is coming this Wednesday so I'm hurrying up!
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Moo
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Quote Originally Posted by arbolis View Post
What do you mean by discriminant?
Determinant
Sorry, I'm being very tired..
I hate this matter (even if less than a year ago) but it is an extremely important one. Not only for me (as a physics student) but for many other careers. I had an exam of physics I last Wednesday and I totally depreciated Algebra II. The exam of Algebra II is coming this Wednesday so I'm hurrying up!
So good luck
There is (at least) a little mistake in your answer.
You should have had (1-a)x=0. But this doesn't mean that x=0 ! This is the first eyeball
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-9,177,137,498,144,000,000 | Role of Generalised Linear Model in non-life pricing Phase3
Image
Before reading this article, make sure that you read phase1 and phase2. Here are the link:
Phase1: http://www.actuarysense.com/2018/10/role-of-generalised-linear-model-in-non.html
Phase2: http://www.actuarysense.com/2018/11/role-of-generalised-linear-model-in-non.html So we know that the purpose of GLM is to find the relationship between mean of the response variable and covariates.
In this Article we are going to talk about Linear Predictors.
Linear Predictor: Let’s denote it with, “η” (eta). So, linear predictor is actually a function of covariates. For example, in the normal linear model where function is Y = B0 + B1x. So linear predictor will be η = B0 + B1x. Always note that linear predictor has to be linear in its parameter. In this case parameters are B0 and B1. But still the question is how I came up with B0 + B1x as a function? First of all, note that broadly there are two types of Covariates. 1. Variables: It takes the numerical value. For example: age of policyholder, years of ex…
Differentiate between Stochastic and Deterministic model ?
Deterministic Model:
• Here the output of the model is fully determined by the parameter values and initial conditions. This model assumes that its outcome is certain if input is fixed. No matter how many times one recalculates, one obtains exactly the same result.
Example:
• Good example is Linear programming. If we want to minimize the cost by selecting the decision how you want to transport the goods from one place to another , then you are dealing with deterministic model for every data.
Stochastic Model:
• Stochastic models possess some inherent randomness. The same set of parameter values and initial conditions will lead to different outputs. Every time you run the model , you will get the different result.
Example:
• When you roll a die, you will get different results.
Note:
For Building a stochastic model
Create the Sample space — a list of all possible outcomes,
Assign probabilities to sample space elements,
Identify the events of interest,
Calculate the probabilities for the events of interest.
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3,159,673,919,444,019,000 | 8
$\begingroup$
Could you help me to prove that the following problem is NP-hard?
Problem. Given an undirected graph $G=(V,E)$, find an ordering of the nodes such that $\sum\limits_{v\in V}|succ(v)|\times|pred(v)|$ is minimum.
Given an ordering on the nodes and a node $v\in V$, $pred(v)$ (resp. $succ(v)$) denotes the set of neighbors of $v$ with lower (resp. higher) ordering.
$\endgroup$
3
10
$\begingroup$
Consider your problem restricted to 3-regular graphs. Consider some ordering of the vertices. Define a split vertex to be a vertex $v$ such that both $succ(v)$ and $pred(v)$ are non-empty and define a non-split vertex to be any other vertex. Notice that in a 3-regular graph, the value of $|succ(v)| \times |pred(v)|$ is $0$ if $v$ is a non-split vertex and is $2$ otherwise.
Thus, $\sum\limits_{v\in V}|succ(v)|\times|pred(v)| = 2n_{\text{split}}$ where $n_{\text{split}}$ is the number of split vertices.
Therefore, in 3-regular graphs, solving your problem (minimizing the summation) is equivalent to choosing an ordering so as to minimize the number of split vertices.
The following theorem tells us that solving your problem in 3-regular graph $G$ is equivalent to solving Max Cut in $G$. Therefore, since Max Cut in 3-regular graphs is NP-hard, so is your problem.
Theorem If $G = (V, E)$ is a 3-regular graph with $n$ vertices then there exists an ordering of the vertices such that the number of split vertices is at most $k$ if and only if $G$ has a cut of size at least $\frac{3}{2}n - k$.
First direction
First, suppose there exists on ordering of the vertices such that the number of split vertices is at most $k$. Then we partition $V$ into two sets $A$ and $B$ as follows:
1. Place all non-split vertices (under the ordering) into $A$ and $B$ by putting the vertices with 3 successors into $A$ and the vertices with 3 predecessors into $B$.
2. Tentatively add all split vertices (under the ordering) into $A$.
3. Repeat steps 4 and 5 until step 4 fails:
4. Identify a split vertex $v$ that has more neighbors in its current set ($A$ or $B$) than in the other
5. Move $v$ to the other set
The partition of $V$ into $A$ and $B$ is a cut, and each occurrence of steps 4 and 5 increases the size of that cut, so this procedure will terminate. Below we compute the size of the resulting cut.
Notice also that each split-vertex $v$ in $A$ (or $B$) has at most one neighbor in $A$ (or $B$) because otherwise this vertex would have been identified in step 4 at the final iteration of the loop. Thus, the number of edges incident on split vertices that are within $A$ or within $B$ is bounded above by the number of split vertices (which we know is at most $k$).
Next consider any edge $(u, v)$ that is not incident on a split vertex. This edge must then be between two non-split vertices. Suppose wlog that $u$ is earlier than $v$ in the ordering. Then since $u$ is a non-split vertex with a successor, it must be the case that $u$ actually has 3 successors, and similarly $v$ must have 3 predecessors. We can conclude from this that $u$ is placed in $A$ and $v$ is placed in $B$, and therefore that $(u, v)$ is not an edge within $A$ or within $B$.
Thus, we conclude that the overall number of edges within $A$ or within $B$ is at most $k$. Then the size of the cut (i.e., the number of edges not within $A$ and not within $B$) is at least $|E| - k = \frac{3}{2}n - k$.
Second direction
Now suppose there exists a cut of size at least $\frac{3}{2}n - k$. Suppose this cut partitions $V$ into sets $A$ and $B$. Then we construct an ordering of the vertices in $V$ by putting the vertices in $A$ first and the vertices in $B$ later. We claim that the number of split vertices under this ordering is at most $k$.
Consider any edge incident on a vertex in $A$. If this edge is $(u, v)$ with $u \in A$ and $v \in B$ then $v$ is a successor of $u$ since all vertices in $B$ are after all vertices in $A$. Thus, edges between $A$ and $B$ do not contribute any predecessors to any vertices in $A$. The only other possible edge incident on a vertex in $A$ is an edge within $A$. Such an edge contributes a predecessor to exactly one of its endpoints. Thus, the total over all $v \in A$ of the number of predecessors of $v$ is at most the number of edges within $A$, and so the number of vertices in $A$ with a predecessor is at most the number of edges within $A$. Since every split-vertex in $A$ is a vertex in $A$ with a predecessor, the number of split-vertices in $A$ is at most the number of edges within $A$.
Analogous logic can be used to show that the number of split-vertices in $B$ is at most the number of edges within $B$. Then the total number of split vertices under this ordering is at most the number of edges within $A$ or within $B$. Since the cut has at least $\frac{3}{2}n - k = |E| - k$ edges between $A$ and $B$, it must have at most $k$ edges within $A$ or within $B$. We conclude that the number of split vertices under the constructed ordering is at most $k$.
$\endgroup$
6
• $\begingroup$ Do you have a reference for "Max Cut in 3-regular graphs is NP-hard" ? $\endgroup$ – user6973 May 23 '17 at 8:20
• 1
$\begingroup$ Sure. According to ISGCI (graphclasses.org/classes/gc_1100.html), cubic Max Cut is NP-complete due to the paper "On some tighter inapproximability results" by P. Berman and M. Karpinski. $\endgroup$ – Mikhail Rudoy May 23 '17 at 8:32
• $\begingroup$ I actually looked at that paper before commenting here. That paper says its "method uses randomized reductions" . Unless they use them in a very strange way, that means they only show hardness under randomized reductions. (... continued) $\endgroup$ – user6973 May 23 '17 at 9:38
• $\begingroup$ (continued ...) As further evidence towards that, section 3 specifies that in the first reduction, "every node (variable) is replaced with a wheel, a random graph that ..." . Are you aware of any proof of actual NP-hardness for the problem, rather than under randomized reductions? $\endgroup$ – user6973 May 23 '17 at 9:38
• $\begingroup$ Thanks for pointing that out! I have to say, I'm surprised and disappointed by ISGCI :(. I am not aware of a proof of hardness for Cubic Max Cut from the literature. However, I can think of a very simple reduction from Max-Degree-3 Max Cut to Cubic Max Cut and ISGCI says that Max-Degree-3 Max Cut is NP-complete. The source listed for this is "Node- and edge-deletion NP-complete problems" by "M. Yannakakis", but I have not yet verified that it contains a correct proof of NP-hardness for Max-Degree-3 Max Cut. If you think I should, I can edit my answer to reflect this state of things. $\endgroup$ – Mikhail Rudoy May 24 '17 at 5:29
4
$\begingroup$
To close a gap, I prove FNP-hardness (in fact, FETH-hardness)
of Max Cut in 3-regular graphs under deterministic reductions.
3-SAT to NAE-3SAT :
One variable is analogous to 3-coloring's palette. Similarly to how 3-coloring works,
flipping all truth-values does not affect NAE-satisfaction, so treat that constant as FALSE.
With that done, (x NAE y NAE local) and (z NAE false NAE not local) is a clause gadget.
NAE-3SAT to Max Cut in 3-regular graphs:
("parity" means evenness/oddness, whichever one the integer is.)
Each variable gadget is the result of taking [a pair of binary trees such that [the heights of the childless nodes in the pair of trees] all have the same parity] and putting an edge between
their roots. (For example, perfect binary trees whose heights have the same parity will work.)
Observe that those results are connected trees whose non-leaf nodes all have degree exactly 3.
In particular, they are connected and bipartite, so they have a unique maximum cut
and that cut includes every edge. For each such gadget, since the parities of the
distances from childless nodes to their root are the same and the roots are 1 away
from each other, that cut splits the leaves according to which root they are closer to.
Thus, with the choice between sides being arbitrary, the childless nodes closer to one
root are available as positive occurrences of the variable and the childless nodes
closer to the other root are available as negative occurrences of the variable.
In the resulting trees, the leaves are exactly the childless nodes, so those all have
degree exactly 1. Thus triangles using such nodes bring their degrees to exactly 3.
Notice that in a triangle, the size of a cut is 0 if the vertices are all on the same side and
is 2 otherwise. Finally, cap the remaining leaves like half (the halves are isomorphic)
of the 10-vertex cubic graph whose connect is 1. That makes the graph 3-regular.
Regardless of how the the rest of the graph is handled, for any given cap, cutting across
the 6 edges that are at less than two away from the leaf is the unique maximum extension
of the rest of the cut. Therefore, when T is [the number of edges in the variable gadgets]
plus [2 times the number of NAE-constraints] plus [6 times the number of caps],
no cut is larger than T, each NAE-satisfying assignment induces a cut of size T,
and for each cut of size T, the sides-of-the-cut of the
positive-side-of-variable-gadgets leaves form a NAE-satisfying assignment.
(One can use unbalanced and/or different-height trees to be more efficient, but even even when using perfect trees of the same height, the number of the number of leaves in a variable gadget
will be less than 4 times the number of occurrences of that variable, so the number of vertices
in a variable gadget will be less than 8 times the number of occurrences of that variable.
The constraint-gadgets do not change the number of vertices, the caps are constant-size,
and the number of them is at most the number of leaves. Therefore output size
is linear in input size, so the above reduction suffices for FETH-hardness too.)
$\endgroup$
1
$\begingroup$
The proof here also shows NP-hardness of the problem.
It shows NP-hardness of the less constrained problem where we look for an orientation of the edges instead of an ordering of the nodes (that is the orientation of the edges does not have to induce a DAG) by reduction to NAE3SAT.
Observe that in the constructed graph, a minimum edge orientation does not have any cycle when there is a NAE3SAT solution. That minimum edge orientation thus induces a DAG showing NP-hardness of the node ordering version.
$\endgroup$
0
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Question about primes and divisibility abstract algebra/number theory
1. Apr 12, 2009 #1
Can someone please tell me how to go about answering a question like this? I've been racking my brain for a long time and still don't have a clue...I guess because my background in algebra/number theory really isn't that strong.
"What is the greatest integer that divides [itex]p^4 - 1[/itex] for every prime number [itex]p[/itex] greater than 5?"
Thanks!
2. jcsd
3. Apr 12, 2009 #2
Hurkyl
User Avatar
Staff Emeritus
Science Advisor
Gold Member
There seems an obvious first thing to try:
Compute the greatest integer that divides [itex]p^4 - 1[/itex] for every prime number p in the range 5 < p < N
where N is whatever number you like. I'd probably start with 10 and then increase it a few times until I had an idea what was going on.
4. Apr 12, 2009 #3
Wouldn't it be [tex]p^4-1[/tex]? Maybe I'm not understanding the question.
5. Apr 12, 2009 #4
Sorry; this is a multiple choice question off of an old Math Subject GRE exam. There are five answer choices:
(A) 12
(B) 30
(C) 48
(D) 120
(E) 240
6. Apr 12, 2009 #5
This is what I have so far.
[tex]p^4-1= (p+1)(p-1)(p^2+1)[/tex]
p is odd so [tex]p = 1 \text{ or } 3 [/tex] (mod 4) so there are three 2's in (p+1) and (p-1) plus another in [tex](p^2+1)[/tex] so [tex]16|p^4-1[/tex]. Furthermore, 3 does not divide p (since p>5) so (p-1) or (p+1) does and so [tex]3|p^4-1[/tex]. Now it's between 240 and 48.
7. Apr 12, 2009 #6
HallsofIvy
User Avatar
Science Advisor
Yes, you are. The question is about a single number that divides [itex]p^4- 1[/itex] for all primes p> 5. It cannot depend on p.
8. Apr 12, 2009 #7
Alright I found the the last factor.
1^2 = 1 mod 5
2^2 = 4
3^2 = 4
4^2 = 1
So [tex]p^2[/tex] = 1 or 4 mod 5
[tex](p^2)^2 = 1[/tex] mod 5
[tex]p^4-1 = 0[/tex] mod 5
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Topic: Matheology § 176
Replies: 10 Last Post: Dec 11, 2012 12:33 AM
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[email protected]
Posts: 1,458
Registered: 2/15/09
Re: Matheology § 176
Posted: Dec 10, 2012 11:52 PM
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On Dec 10, 1:11 pm, Virgil <[email protected]> wrote:
> In article
> <c293798d-5b7e-4eba-a4df-6332461dc...@f19g2000vbv.googlegroups.com>,
>
>
>
>
>
>
>
>
>
> WM <[email protected]> wrote:
> > Matheology § 176
>
> > Here's a paradox of infinity noticed by Galileo in 1638. It seems that
> > the even numbers are as numerous as the evens and the odds put
> > together. Why? Because they can be put into one-to-one correspondence.
> > The evens and odds put together are called the natural numbers. The
> > first even number and the first natural number can be paired; the
> > second even and the second natural can be paired, and so on. When two
> > finite sets can be put into one-to-one correspondence in this way,
> > they always have the same number of members.
>
> > Supporting this conclusion from another direction is our intuition
> > that "infinity is infinity", or that all infinite sets are the same
> > size. If we can speak of infinite sets as having some number of
> > members, then this intuition tells us that all infinite sets have the
> > same number of members.
>
> > Galileo's paradox is paradoxical because this intuitive view that the
> > two sets are the same size violates another intuition which is just as
> > strong {{and as justified! If it is possible to put two sets A and B
> > in bijection but also to put A in bijection with a proper subset of B
> > and to put B in bijection with a proper subset of A, then it is insane
> > to judge the first bijection as more valid than the others and to talk
> > about equinumerousity of A and B.}}
>
> > [Peter Suber: "Infinite Reflections", St. John's Review, XLIV, 2
> > (1998) 1-59]
> >http://www.earlham.edu/~peters/writing/infinity.htm#galileo
>
> > Regards, WM
>
> Note that the part in {{ }} above is WM's addition, which runs totally
> counter to the Peter Suber's own conclusion which reads:
>
> "Conclusion
> Properly understood, the idea of a completed infinity is no longer a
> problem in mathematics or philosophy. It is perfectly intelligible and
> coherent. Perhaps it cannot be imagined but it can be conceived; it is
> not reserved for infinite omniscience, but knowable by finite humanity;
> it may contradict intuition, but it does not contradict itself. To
> conceive it adequately we need not enumerate or visualize infinitely
> many objects, but merely understand self-nesting. We have an actual,
> positive idea of it, or at least with training we can have one; we are
> not limited to the idea of finitude and its negation. In fact, it is at
> least as plausible to think that we understand finitude as the negation
> of infinitude as the other way around. The world of the infinite is not
> barred to exploration by the equivalent of sea monsters and tempests; it
> is barred by the equivalent of motion sickness. The world of the
> infinite is already open for exploration, but to embark we must unlearn
> our finitistic intuitions which instill fear and confusion by making
> some consistent and demonstrable results about the infinite literally
> counter-intuitive. Exploration itself will create an alternative set of
> intuitions which make us more susceptible to the feeling which Kant
> called the sublime. Longer acquaintance will confirm Spinoza's
> conclusion that the secret of joy is to love something infinite."
>
> http://www.earlham.edu/~peters/writing/infinity.htm#galileo
>
> --
Baruch Spinoza, a 17th century enlightenment thinker, sees the natural
integers as a continuum, of individua (thanks Eco, for the word).
http://en.wikipedia.org/wiki/Baruch_Spinoza
That's quite a fine quote, Suber's. Yet, we might not need the
infinite regress when reason gives us already the firmament, of the
continuum, as the individua, what it is.
And it's turtles: all the way down.
Regards,
Ross Finlayson
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-859,169,734,978,624,100 | Answers
Solutions by everydaycalculation.com
Answers.everydaycalculation.com » LCM of two numbers
What is the LCM of 378 and 20?
The LCM of 378 and 20 is 3780.
Steps to find LCM
1. Find the prime factorization of 378
378 = 2 × 3 × 3 × 3 × 7
2. Find the prime factorization of 20
20 = 2 × 2 × 5
3. Multiply each factor the greater number of times it occurs in steps i) or ii) above to find the LCM:
LCM = 2 × 2 × 3 × 3 × 3 × 5 × 7
4. LCM = 3780
MathStep (Works offline)
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7,914,949,172,330,691,000 | Случайной величиной называется величина, которая в результате опыта может принять то или иное значение, неизвестно заранее – какое именно. Различают случайные величины прерывного (дискретного) и непрерывного типа. Возможные значения прерывных величин могут быть заранее перечислены. Возможные значения непрерывных величин не могут быть заранее перечислены и непрерывно заполняют некоторый промежуток.
Примеры прерывных случайных величин:
1) число появлений герба при трех бросаниях монеты (возможные значения \(0, 1, 2, 3)\);
2) число попаданий в самолет, достаточное для вывода его из строя.
Примеры непрерывных случайных величин:
1) абсцисса точки попадания при выстреле;
2) ошибка измерителя высоты.
Рассмотрим непрерывную случайную величину \(X\) с возможными значениями \(x_{1}, x_{1}, ..., x_{n}\). Каждое из этих значений возможно, но не достоверно, и величина \(X\) может принять каждое из них с некоторой вероятность. В результате опыта величина \(X\) примет одно из этих значений, т.е. произойдет одно из полной группы несовместных событий: $$ \begin{cases}X=x_{1}\\......\\ X=x_{n}\end{cases} $$ Обозначим вероятность этих событий буквами \(p\) с соответствующими индексами:
\(P(X=x_{1})=p_{1};\) \(P(X=x_{2})=p_{2};\)... \(P(X=x_{n})=p_{n};\)
Так как несовместные события образуют полную группу, то
$$\sum_{i=1}^{n}{p_{i}}=1,$$
т.е. сумма вероятностей всех возможных значений случайной величины равна единице. Эта суммарная вероятность каким-то образом распределена между отдельными значениями. Случайная величина будет полностью описана с вероятностной точки зрения, если мы зададим это распределение, т.е. в точности укажем, какой вероятностью обладает каждое из событий. Этим мы установим так называемый закон распределения случайной величины.
Законом распределения случайной величины называется всякое соотношение, устанавливающее связь между возможными значениями случайной величины и соответствующими им вероятностями. Про случайную величину мы будем говорить, что она подчинена данному закону распределения.
Установим форму, в которой может быть задан закон распределения прерывной случайной величины \(X\). Простейшей формой задания этого закона является таблица, в которой перечислены возможные значения случайной величины и соответствующие им вероятности:
\(x_{i}\)
\(x_{1}\)
\(x_{2}\)
\(...\)
\(x_{n}\)
\(p_{i}\)
\(p_{1}\)
\(p_{2}\)
\(...\)
\(p_{n}\)
Такую таблицу мы будем называть рядом распределения случайной величины \(X\).
Чтобы придать ряду более наглядный вид, часто прибегают к его графическому изображению: по оси абсцисс откладываются возможные значения случайной величины, а по оси ординат – вероятности этих значений. Для наглядности полученные точки соединяются отрезками прямых. Такая фигура называется многоугольником распределения.
Многоугольник распределения, так же как и ряд распределения, полностью характеризует случайную величину; он является одной из форм закона распределения.
Иногда удобной оказывается так называемая «механическая» интерпретация ряда распределения. Представим себе, что некоторая масса, равна единице, распределена по оси абсцисс так, что в \(n\) отдельных точках \(x_{1}, x_{2},..., x_{n}\) сосредоточены соответственно массы \(p_{1}, p_{2},..., p_{n}\). Тогда ряд распределения интерпретируется как система материальных точек с какими-то массами, расположенных на оси абсцисс.
2012-11-07 • Просмотров [ 1596 ] | {
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-5,869,572,688,627,730,000 | Check sibling questions
Ex 6.1, 1 Class 12 Maths - Find rate of change of area of a circle
Ex 6.1, 1 - Chapter 6 Class 12 Application of Derivatives - Part 2
Ex 6.1, 1 - Chapter 6 Class 12 Application of Derivatives - Part 3 Ex 6.1, 1 - Chapter 6 Class 12 Application of Derivatives - Part 4
Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class
Transcript
Ex 6.1, 1 Find the rate of change of the area of a circle with respect to its radius r when (a) r = 3 cm (b) r = 4 cmLet Radius of circle = 𝑟 & Area of circle = A We need to find rate of change of Area w. r. t Radius i.e. we need to calculate 𝒅𝑨/𝒅𝒓 We know that Area of Circle = A = 〖𝜋𝑟〗^2 Finding 𝒅𝑨/𝒅𝒓 𝑑𝐴/𝑑𝑟 = (𝑑 (〖𝜋𝑟〗^2 ))/𝑑𝑟 𝑑𝐴/𝑑𝑟 = 𝜋 (𝑑〖(𝑟〗^2))/𝑑𝑟 𝑑𝐴/𝑑𝑟 = 𝜋(2𝑟) 𝒅𝑨/𝒅𝒓 = 𝟐𝝅𝒓 When r = 3 cm 𝑑𝐴/𝑑𝑟 = 2πr Putting r = 3 cm ├ 𝑑𝐴/𝑑𝑟┤|_(𝑟 = 3)= 2π × 3 ├ 𝑑𝐴/𝑑𝑟┤|_(𝑟 = 3) = 6π Since Area is in cm2 & radius is in cm 𝑑𝐴/𝑑𝑟 = 6π cm2/cm Hence, Area is increasing at the rate of 6π cm2/ cm when r = 3 cm (ii) When r = 4 cm 𝑑𝐴/𝑑𝑟 = 2πr Putting r = 4 cm 𝑑𝐴/𝑑𝑟 = 2π × 4 𝑑𝐴/𝑑𝑟 = 8π Since Area is in cm2 & radius is in cm 𝒅𝑨/𝒅𝒓 = 8π cm2/cm Hence, Area is increasing at the rate of 8π cm2/cm when r = 4 cm
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8,503,622,818,510,306,000 | 155765 est-il un nombre premier ? Quels sont les diviseurs de 155765 ?
155765 est-il un nombre premier ?
Non, 155765 n’est pas un nombre premier.
Par exemple, 155765 est divisible par 5 : 155765 / 5 = 31 153.
Pour que 155765 soit un nombre premier, il aurait fallu que 155765 ne soit divisible que par lui-même et par 1.
Parité du nombre 155765
155765 est un nombre impair, puisqu’il n’est pas divisible par 2.
155765 est-il un nombre carré parfait ?
Un nombre est un carré parfait si sa racine carrée est un nombre entier ; autrement dit, il est égal au produit d’un nombre entier par ce même nombre entier. Ici, la racine de 155765 est 394,671 environ.
Donc la racine carrée de 155765 n’est pas un nombre entier, et par conséquent 155765 n’est pas un carré parfait.
Quel est le carré de 155765 ?
Le carré d’un nombre (ici 155765) est le produit de ce nombre (155765) par lui-même (c’est-à-dire 155765 × 155765) ; le carré de 155765 est aussi parfois noté « 155765 à la puissance 2 ».
Le carré de 155765 est 24 262 735 225 car 155765 × 155765 = 1557652 = 24 262 735 225.
Par conséquent, 155765 est la racine carrée de 24 262 735 225.
Nombre de chiffres de 155765
155765 est un nombre à 6 chiffres.
Quels sont les multiples de 155765 ?
Les multiples de 155765 sont tous les nombres entiers divisibles par 155765, c’est-à-dire dont le reste de la division entière par 155765 est nul. Il existe une infinité de multiples du nombre 155765. Les plus petits multiples de 155765 sont :
• 0 : en effet, 0 est divisible par n’importe quel nombre entier, il est donc aussi un multiple de 155765 puisque 0 × 155765 = 0
• 155765 : en effet, 155765 est bien un multiple de lui-même, puisque 155765 est divisible par 155765 (on a 155765 / 155765 = 1, donc le reste de cette division est bien nul)
• 311 530 : en effet, 311 530 = 155765 × 2
• 467 295 : en effet, 467 295 = 155765 × 3
• 623 060 : en effet, 623 060 = 155765 × 4
• 778 825 : en effet, 778 825 = 155765 × 5
• etc.
Nombres contigus à 155765
Nombres premiers les plus proches de 155765
Trouver si un nombre entier est premier | {
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1,101,224,023,437,509,200 | Pages 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 181 182 183 184 185 186 187 188 189 190 191 192 193 194 195 196 197 198 199 200 201 202 203 204 205 206 207 208 209 210 211 212 213 214 215 216 217 218 219 220 221 222 223 224 225 226 227 228 229 230 231 232 233 234 235 236 237 238 239 240 241 242 243 244 245 246 247 248 249 250 251 252 253 254 255 256 257 258 259 260 261 262 263 264 265 266 267 268 269 270 271 272 273 274 275 276 277 278 279 280 281 282 283 284 285 286 287 288 289 290 291 292 293 294 295 296 297 298 299 300 301 302 303 304 305 306 307 308 309 310 311 312 313 314 315 316 317 318 319 320 321 322 323 324 325 326 327 328 329 330 331 332 333 334 335 336 337 338 339 340 341 342 343 344 345 346 347 348 349 350 351 352 353 354 355 356 357 358 359 360 361 362 363 364 365 366 367 368 369 370 371 372 373 374 375 376 377 378 379 380 381 382 383 384 385 386 387 388 389 390 391 392 393 394 395 396 397 398 399 400 401 402 403 404 405 406 407 408 409 410 411 412 413 414 415 416 417 418 419 420 421 422 423 424 425 426 427 428 429 430 431 432 433 434 435 436 437 438 439 440 441 442 443 444 445 446 447 448 449 450 451 452 453 454 455 456 457 458 459 460 461 462 463 464 465 466 467 468 469 470 471 472 473 474 475 476 477 478 9546 result(s) returned
Hardships of the Oregon Trail
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This video uses a model to point out each of the major human skeletal muscles using the proper pronunciation, although the student stumbles a bit over the names at times. It would be a good video for students to use as a review before a lab muscle practical. All student made video with sound.
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Writing a poem in iambic pentameter requires writing five metrical feet in a specific rhythm. Write a poem in iambic pentameter with tips from a produced playwright in this video. This video goes slowly enough so that students can grasp the concept. A practice activity would help the students understand better as it looks very easy in the video and some students may struggle without added practice. The video is in lecture format with an instructor standing in front of the camera. On occasi
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Calculus Functions
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The video works through strategies for solving 3 equations having 3 variables. White board in a class setting, some interaction, engaging, several examples of increasing complexity. The discussion is clear and understandable. Preview - full version at http://video.google.com/videoplay?docid=7858227191027997293&hl=en Produced by Robert Ahdoot, yaymath.org
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The video demonstrates synthetic division of equations. White board in a class setting, some interaction, engaging, several examples of increasing complexity. The discussion is clear and understandable. Preview - full version at http://video.google.com/videoplay?docid=4571296703082266244&hl=en Produced by Robert Ahdoot, yaymath.org
Author(s): No creator set
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Simplifying Cube Roots
Students learn to simplify a square root by setting up a factor tree for the number inside the radical. If a factor pairs up in the factor tree, then it comes out of the radical. If a factor does not pair up, then it stays inside. Students also learn to simplify a cube root by setting up a factor tree for the number inside the radical. If a factor is part of a group of three factors that are the same, then it comes out of the radical. If a factor is not part of a group of three factors that are
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Solving Radical Equations
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Calculus Limits: A Numerical Approach
In Calculus, the term limit is used to describe the value that a function approaches as the input of that function approaches a certain value. This video explains the two ways to demonstrate Calculus limits: a numerical approach or a graphical approach. In the numerical approach, we determine the point where the function is undefined and create a table of values to determine the value of the variable as it approaches that point. (1:45)
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This video explains one-sided limits. A limit is the value that a function approaches as the input of that function approaches a certain value. In Calculus, sometimes functions behave differently depending on what side of the function that they are on. By definition, a one-sided limit is the behavior on only one side of the value where the function is undefined. (3:02)
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Calculus Limits: A Graphical Approach
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This musical video is about measurement using non standard unit. No narration in this video, just text instructions. Non standard unit is the glass that being used to measure 1 kg of flour.
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Some 99.9 percent of all species that have ever lived on Earth are now extinct. This documentary explores why, then confronts a frightening notion: Are humans causing the next mass extinction -- the sixth in the history of life on Earth? This documentary is suitable for high school grades. Run time 02:52.
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8,460,468,986,288,839,000 | Prime numbers - 6th grade (11y) - math problems
Number of problems found: 115
• LCD 2
zlomky_6 The least common denominator of 2/5, 1/2, and 3/4
• Numbers
ten Determine the number of all positive integers less than 4183444 if each is divisible by 29, 7, 17. What is its sum?
• Dance group
dancers The dance group formed groups of 4, 5, and 6 members. Always one dancer remains. How many dancers were there in the whole group?
• Lcd3
lcd What is LCD of the equation of ? ? And what is x?
• Quotient
math Find quotient before the bracket - the largest divisor 51 a + 34 b + 68 121y-99z-33
• Florist
kvetinarka Florist has 84 red and 48 white roses. How many same bouquets can he make from them if he must use all roses?
• The King
gold_4 The King wants to divide his sons equally. He has 42 rubies and 45 diamonds. How many sons and how will they share them?
• MO Z6-6-1
kruhy_1 Write integers greater than 1 to the blanks in the following figure, so that each darker box was product of the numbers in the neighboring lighter boxes. What number is in the middle box?
• Largest squares
metals How many of the largest square sheets did the plumber cut the honeycomb from 16 dm and 96 dm?
• Class
pytagoras_class When Pythagoras asked how many students attend the school, said: "Half of the students studying mathematics, 1/4 music, seventh silent and there are three girls at school." How many students had Pythagoras at school?
• Game room
roulette Winner can took win in three types of jettons with value 3, 30 and 100 dollars. What is minimal value of win payable in this values of jettons?
• Chairs
stol In the two dining rooms in the recreational building, there are equally arranged chairs around the tables. A maximum of 78 people can dine in the first dining room and 54 people in the second. How many chairs can be around one table?
• Florist's
kvetiny The florist got 72 white and 90 red roses. How many bouquets can bind from all these roses when each bouquets should have the same number of white and red roses?
• Dining tables
stolik In the dining room are tables with 4 chairs, 6 chairs, 8 chairs. How many diners must be at least to be occupy all tables (chairs) and diners are more than 50?
• School
ratios_2 Headteacher think whether the distribution of pupils in race in groups of 4,5,6,9 or 10. How many pupils must have at least school at possible options?
• The smallest number
numbers_49 What is the smallest number that can be divided by both 5 and 7
• Three Titanics
Titanic Three steamers sailed from the same port on the same day. The first came back on the third day, fourth 4th day and the third returned sixth day. How many days after leaving the steamers met again in the harbor?
• Ornamental shrubs
delnik_3 Children committed to plant 240 ornamental shrubs. Their commitment however exceeded by 48 shrubs. Write ratio of actually planted shrubs and commitment by lowest possible integers a/b.
• Apples 2
jableka How many minimum apples are in the cart, if possible is completely divided into packages of 6, 14 and 21 apples?
• Matches
matches George poured out of the box matches and composing them triangles and no match was left. Then he tries squares, hexagons and octagons and no match was left. How many matches must be at least in the box?
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8,415,314,042,034,117,000 | Qnt 561 Week2 Assignment
Qnt 561 Week2 Assignment Words: 1076
Central Limit Theorem and Confidence Intervals Problem Sets Tiffany Blount QNT 561 September 7, 2010 Michelle Barnet University of Phoenix Central Limit Theorem and Confidence Intervals Problem Sets Chapter 8 Exercises: 21. What is sampling error? Could the value of the sampling error be zero? If it were zero, what would this mean? * Sampling error is the difference between the statistic estimated from a sample and the true population statistic. It is not impossible for the sampling error to not be zero. If the sampling error is zero then the population is uniform.
For example if I were evaluating the ethnicities of a populations and everyone is the population was Black then taking any sample would give me the true proportion of 100% Black. 22. List the reasons for sampling. Give an example of each reason for sampling. * The population size is too large and costly for making the study feasible in reasonable period. For example, if I want to know how watching the violent shows on television affects the behavior of children, it won’t be realistic to study each child in the population, so I would use sampling. * Only estimation of particular section of population is required
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For example, if I want to take an example of nation which is combined unit of states. I can choose the random samples of states which can be further divided into smaller units like cities. These cities can be clustered into smaller areas for observation. Researchers can define his pattern of selecting the sample data until data condition of observation is fully satisfied. * It is not possible to study the entire population and accessibility of them is time consuming and difficult For Example, if I wanted to prepare a list of all the customers from a chain of hardware stores. This would be a tedious task.
But it is convenient to choose a subset of stores in stage one of cluster sampling which can be used for interviewing the customers from those stores in the second stage of cluster sampling. 34. Information from the American Institute of Insurance indicates the mean amount of life insurance per household in the United States is $110,000. This distribution follows the normal distribution with a standard deviation of $40,000. A. If we select a random sample of 50 households, what is the standard error of the mean? * Standard error of the sample mean = ? /vn == 40000/v50 = 5656. 85
B. What is the expected shape of the distribution of the sample mean? * Since sample size is greater than 50, it should be normally distributed according to the Central Limit Theorem. C. What is the likelihood of selecting a sample with a mean of at least $112,000? * z = (X – ? ) / ? x, Where X is a normal random variable, ? is the mean, and ? is the standard deviation. P(X ; 112000) = P(z ; (112000 –110000)/5656. 85) = P(z;0. 3535) = 0. 5 – P (0;z;0. 3535) = 0. 5 – 0. 1368 = 0. 3632. D. What is the likelihood of selecting a sample with a mean of more than $100,000? P(X ; 100000) = P (z ; (100000 – 110000)/5656. 85) = P (z ; –1. 7677) = 0. 5+ P (–1. 7677 ; z ; 0) = 0. 5 + P (0 ; z ; 1. 7677) =0. 5 + 0. 4616 = 0. 9616. E. Find the likelihood of selecting a sample with a mean of more than $100,000 but less than $112,000. * P (100000 ; X ; 112000) = P(X ; 100000) – P(X ;112000) = 0. 9616 – 0. 3632 =0. 5984. Chapter 9 Exercises 32. A state meat inspector in Iowa has been given the assignment of estimating the mean net weight of packages of ground chuck labeled “3 pounds. ” Of course, he realizes that the weights cannot be precisely 3 pounds.
A sample of 36 packages reveals the mean weight to be 3. 01 pounds, with a standard deviation of 0. 03 pounds. a. What is the estimated population mean? * 3. 01. b. Determine a 95 percent confidence interval for the population mean. * 3. 01 ± 1. 96*0. 03/sqrt(36)= 3. 0002 , 3. 0198 34. A recent survey of 50 executives who were laid off from their previous position revealed it took a mean of 26 weeks for them to find another position. The standard deviation of the sample was 6. 2 weeks. Construct a 95 percent confidence interval for the population mean.
Is it reasonable that the population mean is 28 weeks? Justify your answer. z = 1. 96 (from a table) N = 50 sd = 6. 2 mean = 26 * 26 – 1. 96*6. 2/sqrt(50) to 26 + 1. 96*6. 2/sqrt(50)=24. 281 to 27. 719; The value of 28 weeks in not inside that confidence interval, so it is not reasonable that the population mean is 28 weeks. 46. As a condition of employment, Fashion Industries applicants must pass a drug test. Of the last 220 applicants 14 failed the test. Develop a 99 percent confidence interval for the proportion of applicants that fail the test.
Would it be reasonable to conclude that more than 10 percent of the applicants are now failing the test? In addition to the testing of applicants, Fashion Industries randomly tests its employees throughout the year. Last year in the 400 random tests conducted, 14 employees failed the test. Would it be reasonable to conclude that less than 5 percent of the employees are not able to pass the random drug test? 1st: z = 2. 5758 p = 14/220 p – z*sqrt(p*(1-p)/N) to p + z*sqrt(p*(1-p)/N) 14/220 – 2. 5758*sqrt(14/220*(1-14/220)/220) to 14/220 + 2. 5758*sqrt(14/220*(1-14/220)/220) * 0. 212 to 0. 1060; 10% is within that interval, so yes, it is reasonable 2nd: z = 2. 5758 p = 14/400 p – z*sqrt(p*(1-p)/N) to p + z*sqrt(p*(1-p)/N) 14/400 – 2. 5758*sqrt(14/400*(1-14/400)/400) to 14/220 + 2. 5758*sqrt(14/400*(1-14/400)/400) * 0. 01133 to 0. 05866899; No, it’s not reasonable to assume that less than 5% fail the test, since the interval goes higher than 5%. Discussion Question 5 Chapter 3 5. You have been approached by the editor of Gentlemen’s Magazine to carry out a research study. The magazine has been unsuccessful in attracting shoe manufacturers as advertisers.
When the sales force tried to secure advertising from shoe manufacturers, they were told men’s clothing stores are a small and dying segment of their business. Since Gentlemen’s Magazine goes chiefly to men’s clothing stores, the manufacturers reasoned that it was, therefore, not a good vehicle for their advertising. The editor believes that a survey (via mail questionnaire) of men’s clothing stores in the United States will probably show that these stores are important outlets for men’s shoes, and are not declining in importance as shoe outlets. He asks you to develop a proposal for the study and submit it to him.
Develop the management-research question hierarchy that will help you to develop a scientific proposal. * 1st determine our management dilemma. How can we get shoe manufacturers to purchase advertising from Gentlemen’s Magazine? * 2nd determine the management questions. Do men’s clothing stores provide an important outlet for men’s shoes? Do shoe manufacturers provide a profitable sales source? * 3rd determine the research questions. What volume of sales does men’s shoes provide to men’s clothing stores? What profits do men’s shoe manufacturers provide to Gentlemen’s Magazine?
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-6,700,083,104,827,173,000 | 克莱尼代数
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克莱尼代数(名稱源自于美国数学家逻辑学家 斯蒂芬·科尔·克莱尼)在数学中是下列两个事物之一:
定义[编辑]
在文献中给出了 Kleene 代数和相关结构的各种不等价的定义。总揽可见 [1]。下面给出当代最常用的定义。
Kleene 代数是带有分别写为 a + baba* 的,两个二元运算 + : A × AA 和 · : A × AA,和一个函数 * : AA集合 A,所以满足下列公理。
• + 和 · 的结合律: a + (b + c) = (a + b) + ca(bc) = (ab)c 对于 A 中所有的 a, b, c
• + 的交换律: a + b = b + a 对于 A 中所有的 a, b
• 分配律: a(b + c) = (ab) + (ac) 和 (b + c)a = (ba) + (ca) 对于 A 中所有的 a, b, c
• + 和 · 的单位元: 对于 A 中所有的 a 存在一个 A 中元素 0 使得: a + 0 = 0 + a = a。 对于 A 中所有的 a 存在一个 A 中元素 1 使得: a1 = 1a = a
• a0 = 0a = 0 对于 A 中所有的 a
上述公理定义了一个半环。我们进一步要求:
• + 是等幂的: a + a = a 对于 A 中所有的 a
现在可以定义在 A 上的偏序 ≤,通过设定 ab 当且仅当 a + b = b (或等价的: ab 当且仅当 A 存在一个 x 使得 a + x = b)。通过这个次序我们可以公式化关于运算 * 的最后两个公理:
• 1 + a(a*) ≤ a* 对于 A 中所有的 a
• 1 + (a*)aa* 对于 A 中所有的 a
• 如果 axA 中使得 axx,则 a*xx
• 如果 axA 中使得 xax,则 x(a*) ≤ x
在直觉上,我们应当把 a + b 当作"并"或 ab 的"最小上界",和把 ab 当作某种单调性的乘法,在 ab 蕴涵 axbx 的意义上。星号背后的想法是 a* = 1 + a + aa + aaa + ... 从编程理论的观点,你还可以把 + 解释所谓"选择",把 · 解释为"顺序",把 * 解释为"重复"。
例子[编辑]
设 Σ 是有限集合("字母表")并设 A 是在 Σ 上所有正则表达式的集合。我们认为两个正则表达式是相等的,如果它们描述同样的语言。则 A 形成一个 Kleene 代数。事实上,这是自由 Kleene 代数,在正则表达式上的任何等式都从 Kleene 代数的公理得出,并且因此在所有 Kleene 代数中都是有效的意义上。
再次设 Σ 是字母表。设 A 是在 Σ 上所有正则语言的集合(或在 Σ 上所有上下文无关语言的集合;或在 Σ 上所有递归语言的集合;或在 Σ 上所有语言的集合)。则 A 的两个元素的并集(写为 +)和串接(写为 ·)再次属于 A,并且 Kleene星号运算也适用于 A 的任何元素。我们获得了 0 为空集而 1 为只包含空字符串的集合的 Kleene 代数 A
M 是带有单位元 e幺半群,并设 AM 的所有子集的集合。对于两个这样的子集 ST,设 S + TST 的并集并设 ST = {st : sStT}。S* 被定义为 S 生成的 M 的子幺半群,它可以被描述为 {e} ∪ SSSSSS ∪ ... 则 A 形成 0 为空集而 1 为 {e} 的 Kleene 代数。对任何小范畴都可以进行类似的构造。
假设 M 是一个集合而 A 是在 M 上所有二元关系的集合。采用 + 为并,· 为复合,* 为自反传递凸包(hull),我们就得到了 Kleene 代数。
带有运算 ∨ 和 ∧ 的所有布尔代数成为 Kleene 代数,如果我们对 + 使用 ∨,对 ·使用 ∧,并对所有 a 设立 a* = 1。
在计算权重有向图中的最短路径的时候一个非常不同的 Kleene 代数是有用的: 设 A扩展的实数轴,采用 a + bab 的最小者,而 abab 的普通和(带有 +∞ 和 -∞ 的和并定义为 +∞)。a* 被定义对非负 a 为实数零对负 a 为 -∞。这是带有零元素 +∞ 和一元素是实数零的 Kleene 代数。
性质[编辑]
零是最小元素: 0 ≤ a 对于 A 中所有的 a
a + b 的和是 ab 的最小上界: 我们有 aa + bba + b 并且如果 xA 的一个元素有着 axbx,则 a + bx。类似的,a1 + ... + an 是元素 a1, ..., an 的最小上界。
乘法和加法是单调性的: 如果 ab,则 a + xb + xaxbxxaxb 对于 A 中所有的 x
关于 * 运算,我们有 0* = 1 和 1* = 1,* 是单调性的(ab 蕴涵 a* ≤ b*),而 ana* 对于所有自然数 n。进一步的,(a*)(a*) = a*、(a*)* = a* 和 ab* 当且仅当 a* ≤ b*。
如果 A 是 Kleene 代数而 n 是自然数,则你可以认为集合 Mn(A) 由带有 A 中条目的所有 n×n 矩阵构成。使用普通的矩阵加法和乘法概念,你可以定义一个唯一的 *-运算,所以 Mn(A) 成为一个 Kleene 代数。
历史[编辑]
Kleene 代数不是 Kleene 定义的;他介入了正则表达式并寻求一个完备的公理集合来允许在正则表达式上的所有等式的推导。首先约翰·何顿·康威正则代数的名义下研究了这个问题。Dexter Kozen 最先证明了 Kleene 代数的公理解决了这个问题。
参见[编辑]
引用[编辑]
1. Dexter Kozen: On Kleene algebras and closed semirings. In Rovan, editor, Proc. Math. Found. Comput. Sci., volume 452 of Lecture Notes in Computer Science, pages 26-47. Springer, 1990. Online at https://web.archive.org/web/20060923121313/http://www.cs.cornell.edu/~kozen/papers/kacs.ps | {
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Thread: expression manipulating
1. #1
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expression manipulating
I have done the first part and found the new limits but I can't do the last part.
Attached Thumbnails Attached Thumbnails expression manipulating-scan-1.jpg
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2. #2
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Re: expression manipulating
Quote Originally Posted by Stuck Man View Post
I have done the first part and found the new limits but I can't do the last part.
\displaystyle \begin{align*} y &= \frac{x}{1 + x^2} \\ \frac{y}{x} &= \frac{1}{1 + x^2} \\ \frac{x}{y} &= 1 + x^2 \\ \left(\frac{x}{y}\right)^2 &= (1 + x^2)^2 \\ \left(\frac{x}{y}\right)^2 &= 1 + 2x^2 + x^4 \\ \left(\frac{x}{y}\right)^2 - 2x^2 &= 1 + x^4 \end{align*}
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3. #3
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Re: expression manipulating
I've done that part.
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4. #4
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Re: expression manipulating
Quote Originally Posted by Stuck Man View Post
I've done that part.
You're welcome ><
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5. #5
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Re: expression manipulating
\displaystyle{y = \frac{x}{1 + x^2}}
And,
\displaystyle{\frac{dy}{dx} = \frac{1 - x^2}{(1 + x^2)^2}}
Therefore,
\displaystyle{\frac{1 - x^2}{(1 + x^2) \sqrt{1 + x^4}}}
=\displaystyle{\frac{1 - x^2}{(1 + x^2) \sqrt{(\frac{x}{y})^2 - 2x^2}}}
=\displaystyle{\frac{1 - x^2}{(1 + x^2) \sqrt{(\frac{x}{y})^2 (1 - 2y^2)}}}}
=\displaystyle{\frac{1 - x^2}{(1 + x^2)\ \frac{x}{y} \sqrt{1 - 2y^2}}}
=\displaystyle{\frac{1 - x^2}{x(1 + x^2)}\ \frac{y}{\sqrt{1 - 2y^2}}}
=\displaystyle{\frac{1 + x^2}{x}\ \frac{y}{\sqrt{1 - 2y^2}}\ \frac{1 - x^2}{(1 + x^2)^2}}}
=\displaystyle{\frac{1}{y}\ \frac{y}{\sqrt{1 - 2y^2}}\ \frac{dy}{dx}}}
=\displaystyle{\frac{1}{\sqrt{1 - 2y^2}}\ \frac{dy}{dx}}}
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+0
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avatar+65
A mother is 34 years old, and her son is 6 yers old. How old will her son be, if her mother gets two times older than her?
(Explanation will be nice)
Lunar May 10, 2016
Best Answer
#4
avatar
+5
34 - 6 = 28 years - mother's age when her son was born.
28 x 2 =56 mother's age when her son will be 28, or half her age.
Guys-Remember Occam's Razor Principle!.
Guest May 10, 2016
#1
avatar+13590
+5
34 = mothers age
6 = son's age
in 'x' years mom will be 34 + x son will be 6+x ALSO two times sons age is equal to mom's 34+x = 2(6+x) solve for x = 22 years
so sone will be 6 + 22 = 28 and mom will be 34 + 22 = 56
ElectricPavlov May 10, 2016
#2
avatar+91027
+5
Let y be the number of years until the mother is twice as old.....so we have......
The mother's age at that time = 2 times the son's age at that time
34 + y = 2 ( 6 + y) simplify
34 + y = 12 + 2y subtract y, 12 from both sides
22 = y
The mother will be twice as old in 22 more years
And in 22 years, the son will be 6 + 22 = 28 years old
Check :
34 + 22 = 56
6 + 22 = 28
And 56 is twice as old as 28 !!!!
cool cool cool
CPhill May 10, 2016
#3
avatar+65
0
Thanks to both of you!
Lunar May 10, 2016
#4
avatar
+5
Best Answer
34 - 6 = 28 years - mother's age when her son was born.
28 x 2 =56 mother's age when her son will be 28, or half her age.
Guys-Remember Occam's Razor Principle!.
Guest May 10, 2016
#5
avatar+13590
0
Ha! Good one 'Guest' !
ElectricPavlov May 10, 2016
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Thread: Find the volume of a solid using both the disk/washer and shell method.
1. #1
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Find the volume of a solid using both the disk/washer and shell method.
Let R be the region bounded by the following curves. Let S be the solid generated when R is revolved about the given axis. If possible, find the volume of S by both the disk/washer and shell methods. Check that your results agree.
I have attached an image with the functions and the correct answers.
I have tried integrating this both as f(x) and f(y), but both lead me to have to take anti-derivatives that seem impossible with the calculus i have learned so far (u-substitution would not work).
I tried to type some of my work here, but when it posted it came out in a bunch of random letters and numbers and such.
Thank you for the help!
Attached Thumbnails Attached Thumbnails Find the volume of a solid using both the disk/washer and shell method.-math.jpg
Last edited by tanagholat; Jun 23rd 2012 at 10:55 AM.
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2. #2
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Re: Find the volume of a solid using both the disk/washer and shell method.
I don't see why it would give difficult integrals. Rotating around the x-axis gives disks with radii y and so area \pi y^2. The square root, after squaring, will be gone so you just have integrals of logarithms. What is the integral of \int ln(ax) dx?
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3. #3
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Re: Find the volume of a solid using both the disk/washer and shell method.
Quote Originally Posted by tanagholat View Post
I have tried integrating this both as f(x) and f(y), but both lead me to have to take anti-derivatives that seem impossible with the calculus i have learned so far (u-substitution would not work).
For integrals involving logarithms, it's often useful to integrate by parts. For example, to integrate \int\ln x^n\,dx,\ n\geq1 let
u = \ln x^n\Rightarrow du=\frac{nx^{n-1}}{x^n}\,dx
\Rightarrow du=\frac nx\,dx,
and let dv = dx\Rightarrow v = x, which gives
\int\ln x^n\,dx = \int u\,dv
=uv - \int v\,du
=x\ln x^n - \int n\,dx
=x\left(\ln x^n - 1\right) + C
Now, to the problem at hand.
Washer Method:
First we determine where y=1 intersects the other two curves:
\sqrt{\ln\frac{x^2}{36}} = 1\Rightarrow x = 6\sqrt e (ignoring the negative root)
\sqrt{\ln\frac x6} = 1\Rightarrow x = 6e
So,
V = \pi\int_6^{6\sqrt e}\left[\left(\sqrt{\ln\frac{x^2}{36}}\right)^2 - \left(\sqrt{\ln\frac x6}\right)^2\right]\,dx + \pi\int_{6\sqrt e}^{6e}\left[1^2 - \left(\sqrt{\ln\frac x6}\right)^2\right]\,dx
= \pi\int_6^{6\sqrt e}\left(\ln\frac{x^2}{36} - \ln\frac x6\right)\,dx + \pi\int_{6\sqrt e}^{6e}\left(1 - \ln\frac x6}\right)\,dx
= \pi\left[x\left(\ln\frac{x^2}{36} - 2\right) - x\left(\ln\frac x6 - 1\right)\right]_6^{6\sqrt e} + \pi\left[x - x\left(\ln\frac x6 - 1\right)\right]_{6\sqrt e}^{6e}
=\pi\left[\left(6\sqrt e\right)(-1) - \left(6\sqrt e\right)\left(-\frac12\right) - 6(-2) + 6(-1) + 6e - 6e(0) - 6\sqrt e + \left(6\sqrt e\right)\left(-\frac12\right)\right]
=6\pi\left(e - 2\sqrt e + 1\right)
=6\pi\left(\sqrt e - 1\right)^2
Shell Method:
y = \sqrt{\ln\frac x6}\Rightarrow x = 6e^{y^2}
y = \sqrt{\ln\frac{x^2}{36}}\Rightarrow x = 6e^{y^2/2}
V = 2\pi\int_0^1y\left(6e^{y^2} - 6e^{y^2/2}\right)\,dy
This integral is relatively straightforward. I leave it to you to show that it produces the same result.
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Re: Find the volume of a solid using both the disk/washer and shell method.
Thank you very much to the both of you.
So we haven't learned how to do integration by parts with logarithms and such, so i guess that's where I got stuck, but other than that it's nice to know that my thought process was all correct. I did some research on how you got that and it makes some sense now. (yay)
Much appreciation to the both of you. I'm now a bit more confident with knowledge of this material.
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-5,034,519,813,395,545,000 | Inequação do 1º grau
Resolução da Inequação do 1º Grau
Resolução da Inequação do 1º Grau
Por Gabriel Alessandro de Oliveira
PUBLICIDADE
O primeiro estudo realizado com relação às expressões algébricas envolve a análise dos valores da incógnita que satisfazem uma determinada igualdade, ou seja, o estudo das equações. Nesse artigo faremos o estudo das inequações, ou seja, estudaremos os valores da incógnita que fazem com que a expressão algébrica possua um determinado valor (positivo ou negativo), pois as inequações consistem em desigualdades (≠, ≤, ≥, <, >). Caso você ainda possua dúvidas sobre os conceitos básicos da inequação, acesse o artigo “Inequação”.
As inequações do 1º grau consistem em desigualdades nas quais as expressões algébricas são expressões do 1º grau (maior expoente da incógnita é 1).
Os métodos para solucionar uma inequação do 1º grau são bem simples. Devemos isolar a incógnita e, caso façamos uma operação que envolva um número negativo, devemos inverter o sinal da desigualdade. As incógnitas são valores que estão no conjunto dos números reais, portanto, quando você obtiver a solução de uma inequação, faça a representação dessa solução nas retas dos reais. Por exemplo, quando você obtém a solução x > 1, em outras palavras você possui a informação de que para a expressão algébrica inicial, todos os valores maiores do que 1 irão satisfazer aquela desigualdade.
Vejamos alguns exemplos:
“Resolva a inequação a seguir: 3 (x+1) – 3 ≤ x+4”
Primeiramente devemos desenvolver a multiplicação dos parênteses, para poder eliminá-los.
Depois de feitas as operações necessárias, devemos isolar a incógnita em um dos membros da desigualdade e os termos constantes no outro. Isolemos, então, a incógnita no primeiro membro da desigualdade:
Por fim, divida os dois membros pelo valor que está acompanhando a incógnita x:
Com isso, obtemos os valores que satisfazem a inequação inicial, que consiste no nosso conjunto solução da inequação 3(x+1) – 3 ≤ x+4.
Nas retas dos reais teríamos:
Solução da inequação do 1º grau.
Por Gabriel Alessandro de Oliveira
DESTAQUES
Confira os destaques abaixo
..................................................
Soluções
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Olho humano
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.................................................. | {
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6,158,508,271,965,804,000 | Ask your own question, for FREE!
Mathematics
OpenStudy (anonymous):
Is 6% of 34 2.04? Also what is 2.75% of 36.04?
8 years ago
OpenStudy (shadowfiend):
Yes, 6% of 32 is 2.04. You can multiply 32 by 6 (204) and then divide by 100 (2.04) to find that. You can similar do a decimal multiplication of 2.75 * 36.04 and then divide by 100 (i.e. move the decimal point two places to the left) to find that percentage. Does that help?
8 years ago
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Results 1 to 4 of 4
Thread: Exact differential equation
1. #1
Newbie
Joined
Nov 2009
Posts
9
Exact differential equation
hello all
I am having problem understanding this:
i solved (a) :
(4x^3 y^2 + 3x^2 y^4)dx+(2x^4 y + 4x^3 y^3)dy=0
and got this:
x^4 y^2 +x^3 Y^4 = c
the question is (b) to solve for IC :
y(0)=0
and give all the solution. what is there beside x=0 ; y=0 ??
i think maybe x=-y^2 but I'm not sure
thanks for the help
mormor83
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2. #2
MHF Contributor Calculus26's Avatar
Joined
Mar 2009
From
Florida
Posts
1,271
From the IC:
c= 0
Therfore you would think
x^4 y^2 + x^3 y^4 = 0 is the solution
However
dy/dx = (4x^3 y^2 + 3x^2 y^4)/2x^4 y + 4x^3 y^3)
reduces to:
dy/dx = (4x^3 y + 3x^2 y^3)/2x^4 + 4x^3 y^2)
which has the equiilibrium solution y = 0 as well
Another way of looking at this is
x^4 y^2 + x^3 y^4 = 0
y^2( x^4 +x^3y^2) =0
which yields y = 0 or x = -y^2 as you suggested.
Note
dy/dx = (4x^3 y + 3x^2 y^3)/2x^4 + 4x^3 y^2)
does not satisfy the conditions of the uniqueness theorem.
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3. #3
Senior Member Peritus's Avatar
Joined
Nov 2007
Posts
397
This is the resulting equation after applying the IC:
$\displaystyle
x^3 y^2 \left( {x + y^2 } \right) = 0
$
so all the possible solutions are on the following curve:
$\displaystyle
y = \left\{ {\left. { \pm \sqrt { - x} } \right|x \geqslant 0} \right\}
$
which is a canonical parabola rotated 90 degrees to the left.
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4. #4
Newbie
Joined
Nov 2009
Posts
9
thank you very much for that. ( just as i thought but i was not sure)
by the way what do you use to write in "math font"?
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508,972,266,151,473,200 | How do you prove without calculation `int_-a^a (x^2*sin2x dx)/(x^2+1)` = 0?
1 Answer
sciencesolve's profile pic
sciencesolve | Teacher | (Level 3) Educator Emeritus
Posted on
You need to remember how to exploit the symmetry of a function when evaluating the definite integral of this function.
- if f(x) is odd, then `int_-a^a f(x) dx = 0`
Since the problem claims that the definite integral of symmetric function is cancelling, then f(x) must be an odd function.
You need to remember the behaviour of an odd function:
`f(-x) = -f(x)`
You need to plug -x in the equation of the given function such that:
`f(-x) = ((-x)^2*sin(-2x))/((-x)^2+1)`
`f(-x) = (x^2*(-sin 2x))/(x^2+1) =gt f(-x) =- (x^2*(sin 2x))/(x^2+1) =gt f(-x) = -f(x)`
This proves that the function `f(x) = (x^2*(sin 2x))/(x^2+1)` is odd, hence `int_-a^a f(x) dx = 0` . | {
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-2,523,661,614,348,639,000 | Prove that
[tex] \sqrt{5} [/tex]
is an irrational number
Prove that
[tex] \sqrt{5} [/tex]
is an irrational number
1 thought on “Prove that <br />[tex] \sqrt{5} [/tex]<br />is an irrational number”
1. [tex]\huge{\boxed{\mathfrak\blue{\boxed{\mathfrak\blue{\boxed{\mathfrak\blue{\boxed{\mathfrak\blue{\boxed{\mathfrak\blue{\boxed{\mathfrak\blue{\boxed{\mathfrak\blue{\boxed{\mathfrak\blue{\fcolorbox{red}{black}{\red{Answer}}}}}}}}}}}}}}}}}}}[/tex]
Let us assume√5 is rational.
√5=[tex]\frac{p}{q}[/tex] [p and q are co-prime]
p=√5q …(1)
p²=5q [Squaring both the sides]
[tex]\frac{p²}{5}=q²[/tex]…(2)
p² divides 5, p also divides 5.
p=5m [m is any integer]
From equation 1,
√5q=5m
q=[tex]\frac{5m}{√5}[/tex]
q=√5m
q²=5m² [Squaring both the sides]
5 divides both p and q.
But p and q are co-primes.
It means our assumption is wrong.
√5 is irrational
_______________________________________
Leave a Comment | {
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} | 0c090d63199a0a01e3b08e4a255778a0 |
-5,054,994,702,178,732,000 | Tuesday
March 3, 2015
Homework Help: trig
Posted by tony on Saturday, July 13, 2013 at 12:49pm.
use basic identities to simplify the expression.
csc θ cot θ / sec θ
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-7,080,900,902,107,718,000 | If the equation of plane passing
Question:
If the equation of plane passing through the mirror image of a point $(2,3,1)$ with respect to line $\frac{x+1}{2}=\frac{y-3}{1}=\frac{z+2}{-1}$ and containing the line $\frac{x-2}{3}=\frac{1-y}{2}=\frac{z+1}{1}$ is $\alpha x+\beta y+\gamma z=24$, then $\alpha+\beta+\gamma$ is equal to :
1. 20
2. 19
3. 18
4. 21
Correct Option: , 2
Solution:
Line $\frac{x+1}{2}=\frac{y-3}{1}=\frac{z+2}{-1}$
$\overrightarrow{\mathrm{PM}}=(2 \lambda-3, \lambda,-\lambda-3)$
$\overrightarrow{\mathrm{PM}} \perp(2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}})$
$4 \lambda-6+\lambda+\lambda+3=0 \Rightarrow \lambda=\frac{1}{2}$
$\therefore \mathbf{M} \equiv\left(0, \frac{7}{2}, \frac{-5}{2}\right)$
$\therefore$ Reflection $(-2,4,-6)$
Plane : $\left|\begin{array}{ccc}x-2 & y-1 & z+1 \\ 3 & -2 & 1 \\ 4 & -3 & 5\end{array}\right|=0$
$\Rightarrow(x-2)(-10+3)-(y-1)(15-4)+(z+1)(-1)=0$
$\Rightarrow-7 x+14-11 y+11-z-1=0$
$\Rightarrow 7 x+11 y+z=24$
$\therefore \alpha=7, \beta=11, \gamma=1$
$\alpha+\beta+\gamma=19 \quad$ Option (2)
Leave a comment | {
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