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1,166,422,517,531,652,900 | Thursday, March 7, 2013
Triangle & Parallelogram Families
Triangle and Parallelogram Families
What is a Triangle Family?
A triangle family is a group of triangles with the same base and height.
As you see in this picture, all of these triangles have the same base of 4cm and a height of 7cm. However, the perimeter changes and the area stays the same. The area stays the same because you are still using the same base and height. The formula for finding the area of a triangle is A =1/2 bh. You can make an infinite amount of triangles with the same base and height.
What is a Parallelogram Family?
A parallelogram family is a group of parallelograms with the same base and height.
In the picture above, you see that the base and height of these parallelograms stay the same but, the perimeter changes. The area of the parallelograms are the same because the formula for finding the area of parallelograms is A =bh. When you have a parallelogram family the height and base remain the same but the side lengths change which affect the perimeter. You can make an infinite amount of parallelograms with the same base and height.
Why does the area stay the same and the perimeter changes?
When finding the area of a parallelogram you do not use the side lengths only the base and height matter. The same is true for finding the area of a triangle. That is why the area remains the same in a family of parallelograms or triangles. The perimeter changes because the side lengths change.
Blog Post Written By: EO & SR
3 comments:
1. Thanks for sharing the information and its good to have all the topics under one place and I am here to give the definition and properties of quadrilaterals as they are figures with four sides and four angles. There are many different types of quadrilaterals, which are classified by their sides and angles here are some examples-rectangle, parallelogram, trapezoid, rhombus, square and kite.
How do you Find the Area of a Rectangle
ReplyDelete
2. Best ssc coaching centres in Bangalore
http://www.miqclasses.com/ssc-coaching-centers-in-bangalore/
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-3,312,804,495,743,479,300 | Lowest Common Multiples and Highest Common Factors
16-year-old Helena Muffly wrote exactly 100 years ago today:
Thursday, October 26, 1911: Have such difficult algebra problems. So much work it is to find the H.C. F. and L.C.M. Good bye for me if we happen to get one of these in examination.
Her middle-aged granddaughter’s comments 100 years later:
Important: If you aren’t into math—skip my comments today and come back tomorrow. Suffice it to say that Grandma was doing some fairly difficult algebra.
But, if you enjoy math here’s my take on what this diary entry is talking about–
First I’ll give an example of the L.C.M. (lowest common multiple) and H.C.F. (highest common factor) of two whole numbers (integers); then I’ll explain how it’s done for algebraic expressions.
Integers
The L.C.M. is the smallest integer that two whole numbers can be divided by. 1 would always be the L.C.M.
For example, for 8 and 12 the L.C.M. would be 1.
The H.C.F.(highest common factor) is the largest integer that two whole numbers can be divided by.
For the same two numbers (8 and 12), the H.C.F. would be 4.
If the H.C. F. is 1, it is a prime number.
Algebraic Expressions
The basic idea is the same as for algebraic expressions. For example, for H.C.F. of 2ab and 4a2b is 2a.
But it quickly gets complicated. I’m going to give you directions and examples from a 1911 algebra textbook below for H.C.F. [An aside: If you really want to understand this concept you might find the information on the algebrahelp.com website helpful.]
Now, here are the directions for finding the H.C.F. in Durrell’s School Algebra (1912):
The method of finding the H.C.F. is to:
Factor the given expressions, if necessary:
Take the H.C.F. of the numerical coefficients:
Annex the literal factors common to all of the expressions, giving to each factor the lowest exponent which it has in any expression.
Ex. 1: Find the H.C. F. of 6x2y – 12xy2 + 6y3 and 3x2y2 + 9xy3 – 12y4
6x2y – 12xy2 + 6y3 = 6y(x – y)2
3x2y2 + 9xy3 – 12y4 = 3y2(x2 + 3xy – 4y2) = 3y2(x + 4y)(x – y)
H.C.F. = 3y(x – y)
Whew, I’m getting a headache just typing these expressions. But if you’re still with me, here’s a couple problems you could try from the 1911 textbook:
Find the H.C.F.
1. 4a2b , 6ab2
2. x2 – 3x , x2 – 9
3. x2 + x , x2 – 1 , x2 – x – 2
4. 4a3x – 4ax3 , 8a2x3 – 8ax4 , 4a2x2(a – x)2
5. 3a2 – 10a + 3 , 9a – a3 , (3 – a)3
7 thoughts on “Lowest Common Multiples and Highest Common Factors
1. Haha…I feel for her. I never enjoyed LCM and HCF math problems either. They always seemed a bit pointless to me….
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-6,810,864,563,772,273,000 | login
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A135811 Number of coincidence-free length n lists of 5-tuples with all numbers 1,...,n in tuple position k, for k=1..5. 3
1, 0, 31, 7682, 7931709, 24843464324, 193342583284315, 3250662144028779654, 106536051676371091349113, 6291424280473807580386161416, 629175403160580417773688864819351 (list; graph; refs; listen; history; text; internal format)
OFFSET
0,3
COMMENTS
a(n) enumerates (ordered) lists of n 5-tuples such that every number from 1 to n appears once at each of the five tuple positions and the j-th list member is not the tuple (j,j,j,j,j), for every j=1,..,n. Called coincidence-free 5-tuple lists of length n. See the Charalambides reference for this combinatorial interpretation.
REFERENCES
Ch. A. Charalambides, Enumerative Combinatorics, Chapman & Hall/CRC, Boca Raton, Florida, 2002, p. 187, Exercise 13.(a), for r=5.
LINKS
G. C. Greubel, Table of n, a(n) for n = 0..100
FORMULA
a(n) = Sum_{j=0..n} ( ((-1)^(n-j))*binomial(n,j)*(j!)^5 ). See the Charalambides reference a(n)=B_{n,5}.
EXAMPLE
5-tuple combinatorics: a(1)=0 because the only list of 5-tuples composed of 1 is [(1,1,1,1,1)] and this is a coincidence for j=1.
5-tuple combinatorics: from the 2^5 possible 5-tuples of numbers 1 and 2 all except (1,1,1,1,1) appear as first members of the length 2 lists. The second members are the 5-tuples obtained by interchanging 1 and 2 in the first member. E.g. one of the a(2)=2^5-1 =31 lists is [(1,1,1,1,2),(2,2,2,2,1)]. The list [(1,1,1,1,1),(2,2,2,2,2) does not qualify because it has in fact two coincidences, those for j=1 and j=2.
MATHEMATICA
Table[Sum[(-1)^(n - k)*Binomial[n, k]*(k!)^5, {k, 0, n}], {n, 0, 25}] (* G. C. Greubel, Nov 23 2016 *)
CROSSREFS
Cf. A135810 (coincidence-free 4-tuples). A135812 (coincidence-free 6-tuples).
Sequence in context: A212858 A059384 A136676 * A119598 A139295 A261947
Adjacent sequences: A135808 A135809 A135810 * A135812 A135813 A135814
KEYWORD
nonn,easy
AUTHOR
Wolfdieter Lang, Jan 21 2008
STATUS
approved
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4,338,531,691,873,027,000 | Copyright
The sum of two numbers is 18 and their difference is two. What are the two numbers?
Question:
The sum of two numbers is 18 and their difference is two. What are the two numbers?
Elimination Method:
In the elimination method, we solve a system of two equations of two variables by adding or subtracting the equations. By doing so, we get a linear equation in one variable, which we can solve easily.
Answer and Explanation:
Let us assume the two numbers to be {eq}x {/eq} and {eq}y {/eq}.
The sum of the two numbers is 18. So we get: {eq}x+y=18\,\,\,\rightarrow (1) {/eq}.
The difference between the two numbers is 2. So we get: {eq}x-y=2\,\,\,\rightarrow (2) {/eq}
Adding (1) and (2):
$$x+y+x-y =18+2 \\ 2x = 20\\ \text{Dividing both sides by 2}, \\ \boxed{\mathbf{x=10}} $$
Substitute this in (1):
$$10+y=18 \\ \text{Subtracting 10 from both sides},\\ \boxed{\mathbf{y=8}} $$
Therefore, the numbers are 10 and 8.
Learn more about this topic:
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Elimination Method in Algebra: Definition & Examples
from High School Algebra II: Help and Review
Chapter 7 / Lesson 9
23K
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5,423,541,214,045,864,000 | Decimals and Operations STAAR Task Cards #1
Decimals and Operations STAAR Task Cards #1
Decimals and Operations STAAR Task Cards #1
Decimals and Operations STAAR Task Cards #1
Decimals and Operations STAAR Task Cards #1
Decimals and Operations STAAR Task Cards #1
Decimals and Operations STAAR Task Cards #1
Decimals and Operations STAAR Task Cards #1
Decimals and Operations STAAR Task Cards #1
Grade Levels
Common Core Standards
Product Rating
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File Type
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2 MB|32 pages
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Product Description
5th Grade STAAR Review Task Cards. STAAR review task cards focus on released STAAR questions from category 1 - operations, fractions, decimals, and numbers.
The STAAR task cards and printables are perfect for 5th grade math class practicing decimals and operations. They also help 6th and 7th STAAR remediation. Students will compare decimals, order decimals, use order of operations to find the value of an expression, and solve complex problems.
What's in these STAAR math task cards?
--16 half-page task cards
--16 duplicate full page cards
--Decimals to the thousandths
--Compare up to three decimals
--Compare the value of digits in numbers
--Use <, >, =
--Operations with parentheses
--Evaluating expressions
--Reading tables with decimals and real-life data
--Numerical reasoning
--Answer Key
++++++++++++++++++++++++++++++++++++++++++
Ideas for the STAAR Review:
- Give each pair of students one task card. Let them work the card and walk around to give feedback. Provide an incentive like a game for each correct answer.
-Provide individuals or small groups of students with copies task cards to complete.
-Work the full pages whole group, displaying the images on an interactive whiteboard or projector.
-Laminate the full pages or place in clear covers in a binder so students can do during math centers.
-Copy specific pages to use as an assessment. Answers keys are available for every page.
Make STAAR Prep fun, and students will learn more!!!
PLEASE PREVIEW THE PRODUCT BEFORE PURCHASING.
+++++++++++++++++++++++++++++++++++++++++++++
Great STAAR Math Task Cards and Games
3rd Grade Math STAAR Prep
3rd Grade Numbers & Fractions Task Card Bundle #1
⚫ 16 Cards Task Cards #1 STAAR Math Readiness Standards Category 1
⚫ 16 Cards Task Cards #2 Math 3 Numbers & Fractions
⚫ 32 Cards Task Cards #3 Math 3 Fractions & Place Value 26 Mixed Questions
4th Grade Math STAAR Prep
4th Grade Math STAAR Task Card Bundle (Includes Sets 1-3 below)
Task Cards #1 Category 1 Readiness
Task Cards #2 Fractions and Decimals
Task Cards #3 Fractions and Decimals
Task Cards #4 Category 2 Readiness (Grades 3-5)
5th grade STAAR Math...
Decimals & Fractions Task Cards #1
Decimals & Fractions Task Cards #2
MATH GAMES
Scavenger Hunt: Comparing Fractions (Grades 3-5)
Sum It Up: Area, Perimeter, and Volume (Grades 3-5)
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Please purchase the multiple license to share with your team (of no more than 4 teachers) on your campus.
NOTE: The TeamTom Education Store is NOT Affiliated with STAAR. All these products created are original and created to enhance teaching in preparation for standardized testing.
Total Pages
32 pages
Answer Key
Included
Teaching Duration
90 minutes
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8,891,553,280,226,902,000 | 4
$\begingroup$
The hint I was given was to simply prove that $y=xz$ is irrational given that $x$ is nonzero, $x$ is rational and $z$ is irrational. Here's how I did it:
Claim: $y=xz$ is irrational.
Proof: Assume $x\neq0$, $x$ is rational and $z$ is irrational.
By contradiction assume that $y=xz$ is rational. This means $y$ can be expressed as $m/n$, $m$ and $n$ being integers; $y$ can be expressed similarly as $p/q$, $p$ and $q$ being integers. By substitution, we have that $$ p/q=mz/n$$ and $$z=pn/qm, qm \neq 0.$$ Since $pn$ and $qm$ are integers $z$ has to be rational.
In addition to this it seems like there's a part 2 as follows:
Proof: Given an interval $(x,y)$ we will choose a positive irrational number, $z$, say. By density of the rationals there is a rational $p$ in the interval $(x/z, y/z)$ s.t. $$ x/z <p< y/z.$$ From this we see that $pz$ is irrational since it is the product of a rational and irrational number.
Is the $pz$ the $xz$ that we proved is irrational in the first proof? So ideally when presenting a full proof like this, should we do part 2 then part 1?
$\endgroup$
2
$\begingroup$
Great proof! You can apply your result about what you called $xz$ in part 1 to $pz$ in part 2 because it obeys your assumptions: $p \ne 0$, $p$ is rational and $z$ is irrational. So from part 1 we know that $pz$ is irrational.
We also know that $pz$ lies in the interval $(x,y)$, so irrationals are dense.
What is ideal to present when presenting a full proof depends on what your reader already knows. It sounds like in this case you want to present both part 1 and part 2, and that will be a complete proof. But in some other case your reader (or lecturer or marker) might not know or want to assume, for example, that rationals are dense in the real numbers, so you might have to include a part 1.5 that proves that.
EDIT: Oops, I think I misunderstood you. Yeah, I would present part 2 then part 1 when presenting this proof; I think the sequence of ideas flows better. But it's up to you!
Also, just to point out a few typos:
• I think you mean 'This means $x$ can be expressed as $m/n$' when you write "This means y can be expressed as m/n".
• I think you mean 'By substitution' when you write "By substation".
Hope that helps!
$\endgroup$
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-4,607,141,279,246,566,400 | View more editions
Calculus: Early Transcendental Functions
TEXTBOOK SOLUTIONS FOR
Calculus: Early Transcendental Functions 4th Edition
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SAMPLE SOLUTION
Chapter: Problem:
• Step 1 of 1
We are told to let .
(a)
We need to find L(1). So we find that
(b)
Now we need to find L'(x) and L'(1). So we find that
by the Second Fundamental Theorem of Calculus and L'(1) = 1.
(c)
We need to approximate the value of x (to three decimal places) for which L(x) = 1. So we have that
for
So then we obtain
(Note: the exact value of x is e, the base of the natural logarithm function)
(d)
Finally, we need to prove that L(x1x2) = L(x1) + L(x2) for all positive values of x1 and x2.
We first show that .
To see this, let . So then we get
So finally we obtain
Corresponding Textbook
Calculus 4th edition 9780618606245 0618606246
Calculus: Early Transcendental Functions | 4th Edition
9780618606245ISBN-13: 0618606246ISBN: Bruce E. Edwards, Ron Larson, Robert HostetlerAuthors: | {
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-9,206,269,352,144,804,000 |
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Math Help - Arc Length Parametrization
1. #1
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Feb 2011
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Arc Length Parametrization
Don't need too much help here, just a bit of clarification.
Suppose I have to find the arclength parametrization of t --> (cost,sint,cosht) : [-pi,pi] --> R^3 .
I take the norm of the derivitive of the function which ends up as cosh(t) and integrate it. Do I integrate cosh(s)ds (random dummy variable) with upper limit t and lower limit 0, or upper limit t and lower limit -pi ?
Unfortunately our notes don't explain the choice of limits.
Thanks.
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2. #2
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3. #3
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It doesn't matter. Your lower limit determines where you start measuring the arclength but any starting point will give you an arclength parameterization.
Specifically, s= \int_0^s |\vec{T}|dt gives a parameterization with s= 0 at (1, 0, 1), where t= 0. That way, values of t in [-\pi, 0) will have negative values of s and values of t in (0, \pi] will have positive values of s.
If you use, instead, s= \int_{-\pi}^s|\vec{T}|dt then s= 0 will correspond to the point (-1, 0, cosh(-1)) where t= -\pi and all points in [-\pi, \pi] will correspond to positive values of s.
(You do have "t" and "s" reversed in your integrals. If "s" represents arc length, your limits of integration should be "s" not t.)
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6,104,939,688,731,950,000 | How do you graph # y=2+1/2sin2(x-5)#?
1 Answer
Apr 5, 2017
Answer:
Please see below.
Explanation:
As we have #y=2+1/2sin2(x-5)# and any sine ratio has maximum value of #+1# and minimum value of #-1#,
maximum value of #y# will be #2+1/2=2 1/2# and minimum value will be #2-1/2=1 1/2#. As such there is no #x#-intercept.
Hence #y# will move between these two numbers.
Maxima #2 1/2# is there when #2(x-5)=(4n+1)/2pi# i.e. at #x=(4n+1)/4pi+5=npi+pi/4+5# and some values are #x={-0.4978,2.6438,5.7854,8.927,12.0686}#
Minima #1 1/2# is there when #2(x-5)=(4n-1)/2pi# i.e. at #x=(4n-1)/4pi+5=npi-pi/4+5# and some values are #x={-2.0686,1.073,4.2146,7.3562,10.4978}#
Mean value #2# appears at #2(x-5)=npi# i.e. #x=n/2pi+5# and some values are #x={0.2876,1.8584,3.4292,5,6.5708,8.1416,9.7124}#
Now when #x=0#, we have #y=2+1/2xxsin(-5)# ad considering it in radiansm #y=2+1/2xx0.9589=2.4795# and hence #y# intercept is #~=2.48#
and function appears as follows
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-589,395,437,717,755,400 | Thursday
June 30, 2016
Homework Help: Pre-calculus-check answers
Posted by Lucy on Saturday, August 23, 2008 at 1:32pm.
Write the polynomial equation of least degree that has the roots: -3i, 3i, i, and -i.
Answer:
(x+3i)(x-3i)(x-i)(x+i)=x^4+x^3-x^3i+9x^2+9x-9xi-9i
2)Determine if the expression 4m^5-6m^8+m+3 is a polynomial in one variable. If so, state the degree.
Answer: It is a polynomial in one variable; Degree = 8
Thanks
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814,478,909,997,155,300 | Wednesday
March 1, 2017
Homework Help: 6th grade math
Posted by Meg on Monday, November 8, 2010 at 10:31pm.
use compatible numbers to estimate the quotient:
7.35/6
answer: 1.20 is this wrong if so what did i do wrong? Thanks
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Let be $$6,8,12,15,20,24,30,35,42,… $$ a sequence of natural numbers. Guessing the recurrence then using generating functions I can prove that general term of sequence is $$a_n=\frac{1}{8}(2n^2+20n+45+3(-1)^n)$$ I think that my proof is too long. Any shorter more elementary proof.
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up vote 3 down vote accepted
Let $n=1$, the right side of your equation is $(1+20+45-3)/8=63/8$, so something's wrong.
Taking the odd index terms $6,12,20,30,42,\dots$, the first differences are $6,8,10,12,\dots$, the second differences are constant $2,2,2,\dots$, so the sequence is quadratic, $n^2+3n+2$.
The even index terms $8,15,24,35,\dots$, first difference $7,9,11,\dots$, second difference $2,2,\dots$, also quadratic, $n^2+4n+3$.
So we get $a_n=((n+1)/2)^2+3(n+1)/2+2$ if $n$ is odd, $a_n=(n/2)^2+2n+3$ if $n$ is even. Can you see how to combine those into a single formula?
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sequnce starts at n=0 not by n=1 so I am not wrong – Adi Dani Aug 14 '11 at 11:37
@Adi: right; so why didn't you mention that piece in the question in the first place? – Ｊ. Ｍ. Aug 14 '11 at 12:10
3
@Adi, your formula sure as heck was wrong; if it wasn't wrong, you wouldn't have edited it to change $n^2$ to $2n^2$. Anyway, I've given you a solution that doesn't use generating functions and is probably shorter than the one you found; do you have anything positive to say about it? If you want to start at zero, it's easy enough to make an appropriate change to the formulas I give. – Gerry Myerson Aug 14 '11 at 12:37
@Gerry I knew those formulas for even and odd terms, question is how to get a single formula for general term. – Adi Dani Aug 14 '11 at 13:17
1
@Adi, $4a_n=n^2+8n+15$ if $n$ is odd, $4a_n=n^2+8n+12$ if $n$ is even, so $4a_n=n^2+8n+(27/2)-(3/2)(-1)^n$. – Gerry Myerson Aug 15 '11 at 1:15
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• AND gate(.) – The AND gate gives an output of 1 if both the two inputs are 1, it gives 0 otherwise.
• OR gate(+) – The OR gate gives an output of 1 if either of the two inputs are 1, it gives 0 otherwise.
• NOT gate(‘) – The NOT gate gives an output of 1 input is 0 and vice-versa.
• XOR gate(\oplus) – The XOR gate gives an output of 1 if either both inputs are different, it gives 0 if they are same.
Three more logic gates are obtained if the output of above-mentioned gates is negated.
• NAND gate(\uparrow)- The NAND gate (negated AND) gives an output of 0 if both inputs are 1, it gives 1 otherwise.
• NOR gate(\downarrow)- The NOR gate (negated OR) gives an output of 1 if both inputs are 0, it gives 0 otherwise.
• XNOR gate(\odot)- The XNOR gate (negated XOR) gives an output of 1 both inputs are same and 0 if both are different.
Every Logic gate has a graphical representation or symbol associated with it. Below is an image which shows the graphical symbols and truth tables associated with each logic gate.
Universal Logic Gates –
Out of the seven logic gates discussed above, NAND and NOR are also known as universal gates since they can be used to implement any digital circuit without using any other gate. This means that every gate can be created by NAND or NOR gates only.
Implementation of three basic gates using NAND and NOR gates is shown below –
For the XOR gate, NAND and NOR implementation is –
• Implemented Using NAND –
• Implemented using NOR –
Note – For implementing XNOR gate, a single NAND or NOR gate can be added to the above circuits to negate the output of the XOR gate.
GATE CS Corner Questions
Practicing the following questions will help you test your knowledge. All questions have been asked in GATE in previous years or in GATE Mock Tests. It is highly recommended that you practice them.
1. GATE CS 2013, Question 21
2. GATE CS 2012, Question 10
3. GATE CS 2007, Question 33
4. GATE CS 2005, Question 15
Reference –
Digital Design, 5th edition by Morris Mano and Michael Ciletti
This article is contributed by Chirag Manwani. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to [email protected]. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
Attention reader! Don’t stop learning now. Get hold of all the important CS Theory concepts for SDE interviews with the CS Theory Course at a student-friendly price and become industry ready.
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1,859,762,120,412,378,000 | PercentageCalculator .pro
How much is one hundred percent of one over nine
Using this tool you can find any percentage in three ways. So, we think you reached us looking for answers like: What is one hundred (100%) percent (%) of one over nine (1/9)? Or may be: How much is one hundred percent of one over nine. See the solutions to these problems just after the percentage calculator below.
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How to work out percentages - Step by Step
Here are the solutions to the questions stated above:
1) What is 100% of 1/9?
Always use this formula to find a percentage:
% / 100 = Part / Whole replace the given values:
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100 x 1/9 = 100 x Part, or
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Part = 11.111111111111 / 100 = 1
2) 100 is what percent of 1/9?
Use again the same percentage formula:
% / 100 = Part / Whole replace the given values:
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4,928,345,985,507,041,000 | Liters to Centiliters Converter (l to cl)
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Liters to Centiliters
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Convert Centiliters to Liters (cl to l) ▶
Conversion Table
liters to centiliters
lcl
1 l 100 cl
2 l 200 cl
3 l 300 cl
4 l 400 cl
5 l 500 cl
6 l 600 cl
7 l 700 cl
8 l 800 cl
9 l 900 cl
10 l 1000 cl
11 l 1100 cl
12 l 1200 cl
13 l 1300 cl
14 l 1400 cl
15 l 1500 cl
16 l 1600 cl
17 l 1700 cl
18 l 1800 cl
19 l 1900 cl
20 l 2000 cl
How to convert
1 liter (l) = 100 centiliter (cl). Liter (l) is a unit of Volume used in Metric system. Centiliter (cl) is a unit of Volume used in Metric system.
Liters - A Unit of Volume
A liter (international spelling) or liter (American English spelling) (SI symbols L and l) is a unit of volume that is used in the metric system. It is defined as the volume of a cube with sides of 10 centimeters (0.1 meter) in length.
One liter is equal to 1000 cubic centimeters, 0.001 cubic meters, or 1 cubic decimeter.
How to Convert Liters
To convert liters to other units of volume, you need to multiply or divide by the appropriate conversion factor. Here are some common conversion factors and examples:
• To convert liters to cubic centimeters, multiply by 1000.
• Example: 2 L x 1000 = 2000 cm3
• To convert liters to cubic meters, multiply by 0.001.
• Example: 2 L x 0.001 = 0.002 m3
• To convert liters to cubic decimeters, multiply by 1.
• Example: 2 L x 1 = 2 dm3
• To convert liters to gallons (US liquid), multiply by 0.264.
• Example: 2 L x 0.264 = 0.528 gal
• To convert liters to bushels (US), multiply by 0.037.
• Example: 2 L x 0.037 = 0.074 bu
• To convert liters to barrels (oil), multiply by 0.0063.
• Example: 2 L x 0.0063 = 0.0126 bbl
To convert other units of volume to liters, you need to divide by the appropriate conversion factor. Here are some common conversion factors and examples:
• To convert cubic centimeters to liters, divide by 1000.
• Example: 2000 cm3 / 1000 = 2 L
• To convert cubic meters to liters, divide by 0.001.
• Example: 0.002 m3 / 0.001 = 2 L
• To convert cubic decimeters to liters, divide by 1.
• Example: 2 dm3 / 1 = 2 L
• To convert gallons (US liquid) to liters, divide by 0.264.
• Example: 0.528 gal / 0.264 = 2 L
• To convert bushels (US) to liters, divide by 0.037.
• Example: 0.074 bu / 0.037 = 2 L
• To convert barrels (oil) to liters, divide by 0.0063.
• Example: 0.0126 bbl / 0.0063 = 2 L
Centiliters - A Unit of Volume
A centiliter (international spelling) or centiliter (American English spelling) (SI symbols cL or cl) is a unit of volume that is used in the metric system. It is defined as one hundredth of a liter, or the volume of a cube with sides of 1 centimeter (0.01 meter) in length.
One centiliter is equal to 10 cubic centimeters, 0.00001 cubic meters, or 0.01 cubic decimeters.
How to Convert Centiliters
To convert centiliters to other units of volume, you need to multiply or divide by the appropriate conversion factor. Here are some common conversion factors and examples:
• To convert centiliters to cubic centimeters, multiply by 10.
• Example: 2 cL x 10 = 20 cm3
• To convert centiliters to cubic meters, multiply by 0.00001.
• Example: 2 cL x 0.00001 = 0.00002 m3
• To convert centiliters to cubic decimeters, multiply by 0.01.
• Example: 2 cL x 0.01 = 0.02 dm3
• To convert centiliters to gallons (US liquid), multiply by 0.00264.
• Example: 2 cL x 0.00264 = 0.00528 gal
• To convert centiliters to bushels (US), multiply by 0.000374.
• Example: 2 cL x 0.000374 = 0.000748 bu
• To convert centiliters to barrels (oil), multiply by 0.000063.
• Example: 2 cL x 0.000063 = 0.000126 bbl
To convert other units of volume to centiliters, you need to divide by the appropriate conversion factor. Here are some common conversion factors and examples:
• To convert cubic centimeters to centiliters, divide by 10.
• Example: 20 cm3 / 10 = 2 cL
• To convert cubic meters to centiliters, divide by 0.00001.
• Example: 0.00002 m3 / 0.00001 = 2 cL
• To convert cubic decimeters to centiliters, divide by 0.01.
• Example: 0.02 dm3 / 0.01 = 2 cL
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Main content
Introduction to one-dimensional motion with calculus
Straight-line motion can be modeled by giving position as a function of time. Calculus helps us learn about velocity, speed, and acceleration, all from our knowledge about the change in position.
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Video transcript
- [Instructor] What we're going to do in this video is start to think about how we describe position in one dimension as a function of time. So we could say our position, and we're gonna think about position on the x-axis as a function of time. And we could define it by some expression, let's say, in this situation, it is going to be our time to the third power minus three times our time squared plus five. And this is going to apply for our time being non-negative 'cause the idea of negative time, at least for now, is a bit strange. So let's think about what this right over here is describing. And to help us do that, we could set up a little bit of a table to understand that depending on what time we are, let's say that time is in seconds, what is going to be our position along our x-axis? So at time equals zero, x of zero is just going to be five. At time one, you're gonna have one minus three plus five. So that is going to be, let's see, one minus three is negative two, plus five is going to be, we're going to be at position three. And then at time two, we are going to be at eight minus 12 plus five, so we're going to be at position one. And then at time t equals three, it's gonna be 27 minus 27 plus five, we're gonna be back at five. And so, this can at least help us understand what's going on for the first three seconds. So let me draw our positive x-axis. So, say it looks something like that. And this is x equals zero. This is our x-axis. X equals one, two, three, four, and five. And now let's play out how this particle that's being described is moving along the x-axis. So we're gonna start here, and we're gonna go one, two, three. Let's do it again. We're going to go one, two, three, the way I just moved my mouse, if we assume that I got the time roughly right, is how that particle would move. And we can graph this as well. So for example, it would look like this. We are starting at time t equals zero. Our position, this is our vertical axis, our y-axis, but we're just saying y is going to be equal to our position along the x-axis. So that's a little bit counterintuitive because we're talking about our position in the left-right dimension, and here you're seeing it start off in the vertical dimension, but you see the same thing. At time t equals one, our position has gone down to three. Then it goes down further. At time equals two, our position is down to one. And then, we switch direction and then over the next, if we say that time is in seconds, over the next second, we get back to five. Now an interesting thing to think about in the context of calculus is, well, what is our velocity at any point in time? And velocity as you might remember, is the derivative of position. So let me write that down. So we're gonna be thinking about velocity as a function of time. And you could view velocity as the first derivative of position with respect to time which is just going to be equal to, or we're gonna apply the power rule and some derivative properties multiple times. If this is unfamiliar to you, I encourage you to review it. But this is going to be three t squared minus six t and then plus zero, and we're gonna restrict the domain as well for t is greater than or equal to zero. And then, we can plot that. It would look like that. Now let's see if this curve makes intuitive sense. We mentioned that one second, two seconds, three seconds. So we're starting moving to the left. And the convention is if you're moving to the left, you have negative velocity, and if you're moving to the right, you have positive velocity. And you can see here our velocity immediately gets more and more negative until we get to one second, and then it stays negative but it's getting less and less negative until we get to two seconds. And at two seconds, our velocity becomes positive. And that makes sense, because at two seconds was when our velocity switched directions to the rightward direction. So, our velocity is getting more and more negative, less and less negative, and then we switch directions and we go just like that. And we see it right over here. Now one thing to keep in mind when we're thinking about velocity as a function of time is that velocity and speed are two different things. Speed, speed, let me write it over here. Speed is equal to, if you think about it in one dimension, you could think about it as the absolute value of your velocity as a function of time or your magnitude of velocity as a function of time. So in the beginning, even though your velocity is becoming more and more negative, your speed is actually increasing. Your speed is increasing to the left, then your speed is decreasing, you slow down, and then your speed is increasing as we go to the right. And we'll do some worked examples that work through that a little bit more. Now the last concept we will touch on in this video is the idea of acceleration. And acceleration you can view as the rate of change of velocity with respect to time. So acceleration as a function of time is just going to be the first derivative of velocity with respect to time which is equal to the second derivative of position with respect to time. It's just going to be the derivative of this expression. So once again, using the power rule here, that's going to be six t. And then using the power rule here, minus six, and once again, we will restrict the domain. And we can graph that as well. And we would see right over here, this is y is equal to acceleration as a function of time. And you can see at time equals zero, our acceleration is quite negative. It is negative six. And then it becomes less and less and less negative. And then, our acceleration actually becomes positive at t equals one. Now, does that makes sense? Well, we're going one, two, three. You might say, "Wait, we didn't switch directions "until we get to the second second." But remember, after we get to the first second, our velocity in the negative direction becomes less negative which means that our acceleration is positive. If that's a little confusing, pause the video and really think through that. So, our acceleration is negative, then positive, and then positive continues. And so, this is just to give you an intuition. In the next few videos, we'll do several worked examples that help us dive deeper into this idea of studying motion and position, into this idea of studying motion in one dimension. | {
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-6,918,971,157,783,297,000 | Your device is currently offline. You can view downloaded files in My Downloads.
Lesson Plan
Understand functions in terms of inputs and outputs
teaches Common Core State Standards A.12B http://ritter.tea.state.tx.us/rules/tac/chapter111/index.html
teaches Common Core State Standards 2A.4F http://ritter.tea.state.tx.us/rules/tac/chapter111/index.html
teaches Common Core State Standards 2A.5D http://ritter.tea.state.tx.us/rules/tac/chapter111/index.html
teaches Common Core State Standards 7.11A http://ritter.tea.state.tx.us/rules/tac/chapter111/index.html
teaches Common Core State Standards MAFS.912.F-IF.1.2 http://www.cpalms.org/Public/search/Standard
teaches Common Core State Standards AI.F.4 http://www.doe.in.gov/standards/mathematics
teaches Common Core State Standards CCSS.Math.Content.HSF-IF.A.2 http://corestandards.org/Math/Content/HSF/IF/A/2
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In this lesson you will learn how to input values into a function to get output values by working with a function machine.
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8,895,394,428,486,513,000 | What is 16 to the 37th Power?
So you want to know what 16 to the 37th power is do you? In this article we'll explain exactly how to perform the mathematical operation called "the exponentiation of 16 to the power of 37". That might sound fancy, but we'll explain this with no jargon! Let's do it.
What is an Exponentiation?
Let's get our terms nailed down first and then we can see how to work out what 16 to the 37th power is.
When we talk about exponentiation all we really mean is that we are multiplying a number which we call the base (in this case 16) by itself a certain number of times. The exponent is the number of times to multiply 16 by itself, which in this case is 37 times.
16 to the Power of 37
There are a number of ways this can be expressed and the most common ways you'll see 16 to the 37th shown are:
• 1637
• 16^37
So basically, you'll either see the exponent using superscript (to make it smaller and slightly above the base number) or you'll use the caret symbol (^) to signify the exponent. The caret is useful in situations where you might not want or need to use superscript.
So we mentioned that exponentation means multiplying the base number by itself for the exponent number of times. Let's look at that a little more visually:
16 to the 37th Power = 16 x ... x 16 (37 times)
So What is the Answer?
Now that we've explained the theory behind this, let's crunch the numbers and figure out what 16 to the 37th power is:
16 to the power of 37 = 1637 = 356,811,923,176,489,970,264,571,492,362,373,784,095,686,656
Why do we use exponentiations like 1637 anyway? Well, it makes it much easier for us to write multiplications and conduct mathematical operations with both large and small numbers when you are working with numbers with a lot of trailing zeroes or a lot of decimal places.
Hopefully this article has helped you to understand how and why we use exponentiation and given you the answer you were originally looking for. Now that you know what 16 to the 37th power is you can continue on your merry way.
Feel free to share this article with a friend if you think it will help them, or continue on down to find some more examples.
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!
• "What is 16 to the 37th Power?". VisualFractions.com. Accessed on August 18, 2022. http://visualfractions.com/calculator/exponent/what-is-16-to-the-37th-power/.
• "What is 16 to the 37th Power?". VisualFractions.com, http://visualfractions.com/calculator/exponent/what-is-16-to-the-37th-power/. Accessed 18 August, 2022.
• What is 16 to the 37th Power?. VisualFractions.com. Retrieved from http://visualfractions.com/calculator/exponent/what-is-16-to-the-37th-power/.
Exponentiation Calculator
Want to find the answer to another problem? Enter your number and power below and click calculate.
Calculate Exponentiation
Random List of Exponentiation Examples
If you made it this far you must REALLY like exponentiation! Here are some random calculations for you: | {
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6,628,966,694,867,001,000 | CameraIcon
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Question
In the given figure, find the measure of ABC, if ABCD is a cyclic quadrilateral.
328065_977f8418a3134ed0bccf19225402d795.png
A
120
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B
118
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C
138
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D
128
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Solution
The correct option is D 128
Given that: ABCD is a cyclic quadrilateral
DCB=82°
DCA=46°
CYB=62°
ACB=DCBDCA=8246=36°
Consider CYB
CBY=180°BCYCYB=180°62°36°=82° (sum of three angles of a triangle is 180°)
Also, ACD=ABD=46° ......... (angles in the same segment)
Thus, ABC=ABD+CBY=(82+46)°=128°
Answer: ABC=128°
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6,820,894,188,790,960,000 | Skip to content
Koshe dekhi 8.2 class 9
Koshe dekhi 8.2 class 9
Koshe Dekhi 8.2 Class 9
Koshe Dekhi 8.2 Class 9
উৎপাদকে বিশ্লেষণ : কষে দেখি - 8.2
নিচের বহুপদী সংখ্যামালাগুলিকে উৎপাদকে বিশ্লেষণ করি :
1. \fn_cm {\color{Blue} \frac{x^{4}}{16}-\frac{y^{4}}{81}}
সমাধানঃ
\fn_cm \frac{x^{4}}{16}-\frac{y^{4}}{81}
=\left ( \frac{x^{2}}{4} \right )^{2}-\left ( \frac{y^{2}}{9} \right )^{2}
=\left [ \left ( \frac{x}{2} \right )^{2} \right ]^{2}-\left [ \left ( \frac{y}{3} \right )^{2} \right ]^{2}
=\left [\left ( \frac{x}{2} \right )^{2}+\left ( \frac{y}{3} \right )^{2} \right ]\left [\left ( \frac{x}{2} \right )^{2}-\left ( \frac{y}{3} \right )^{2} \right ]
[ সূত্রঃ {\color{Blue} a^{2}-b^{2}=\left ( a+b \right )\left ( a-b \right )} ]
=\left [\left ( \frac{x^{2}}{4} \right )+\left ( \frac{y^{2}}{9} \right ) \right ]\left [ \left (\frac{x}{2} + \frac{y}{3} \right ) \left (\frac{x}{2} - \frac{y}{3} \right ) \right ]
[ সূত্রঃ {\color{Blue} a^{2}-b^{2}=\left ( a+b \right )\left ( a-b \right )} ]
=\left ( \frac{x^{2}}{4}+\frac{y^{2}}{9} \right ) \left (\frac{x}{2} + \frac{y}{3} \right ) \left (\frac{x}{2} - \frac{y}{3} \right ) (উত্তর)
উৎপাদকে বিশ্লেষণ : কষে দেখি - 8.2
2. \fn_cm {\color{Blue} m^{2}+\frac{1}{m^{2}}+2-2m-\frac{2}{m}}
সমাধানঃ
\fn_cm m^{2}+\frac{1}{m^{2}}+2-2m-\frac{2}{m}
\fn_cm =\left (m \right )^{2}+\left (\frac{1}{m} \right )^{2}+2-2\left (m+\frac{1}{m} \right )
=\left [\left ( m+\frac{1}{m} \right )^{2}-2\times m\times \frac{1}{m} \right ]+2-2\left ( m+\frac{1}{m} \right )
[ সূত্রঃ {\color{Blue} a^{2}+b^{2}=\left ( a+b \right )^{2}-2ab} ]
=\left ( m+\frac{1}{m} \right )^{2}-2+2-2\left ( m+\frac{1}{m} \right )
=\left ( m+\frac{1}{m} \right )^{2}-2\left ( m+\frac{1}{m} \right )
=\left ( m+\frac{1}{m} \right )\left ( m+\frac{1}{m}-2 \right ) (উত্তর)
উৎপাদকে বিশ্লেষণ : কষে দেখি - 8.2
3. \fn_cm {\color{Blue} 9p^{2}-24pq+16q^{2}+3ap-4aq}
সমাধানঃ
9p^{2}-24pq+16q^{2}+3ap-4aq
=\left [\left (3p \right )^{2}-2\times 3p\times 4q+\left (4q \right )^{2} \right ]+a\left (3p-4q \right )
=\left ( 3p-4q \right )^{2}+a\left ( 3p-4q \right )
[ সূত্রঃ {\color{Blue} a^{2}-2ab+b^{2}=\left ( a-b \right )^{2}} ]
=\left ( 3p-4q \right )\left ( 3p-4q \right )+a\left ( 3p-4q \right )
=\left ( 3p-4q \right )\left [\left ( 3p-4q \right )+a \right ]
=\left ( 3p-4q \right )\left ( 3p-4q+a \right ) (উত্তর)
উৎপাদকে বিশ্লেষণ : কষে দেখি - 8.2
4. \fn_cm {\color{Blue} 4x^{4}+81}
সমাধানঃ
\fn_cm 4x^{4}+81
=\left ( 2x^{2} \right )^{2}+\left ( 9 \right )^{2}
=\left ( 2x^{2}+9 \right )^{2}-2\times 2x^{2}\times 9
[ সূত্রঃ {\color{Blue} a^{2}+b^{2}=\left ( a+b \right )^{2}-2ab} ]
=\left ( 2x^{2}+9 \right )^{2}-36x^{2}
=\left ( 2x^{2}+9 \right )^{2}-\left (6x \right )^{2}
=\left ( 2x^{2}+9+6x \right )\left ( 2x^{2}+9-6x \right )
[ সূত্রঃ {\color{Blue} a^{2}-b^{2}=\left ( a+b \right )\left ( a-b \right )} ]
=\left ( 2x^{2}+6x+9 \right )\left ( 2x^{2}-6x+9 \right ) (উত্তর)
উৎপাদকে বিশ্লেষণ : কষে দেখি - 8.2
5. \fn_cm {\color{Blue} x^{4}-7x^{2}+1}
সমাধানঃ
x^{4}-7x^{2}+1
=x^{4}+2x^{2}+1-9x^{2}
=\left [\left (x^{2} \right )^{2}+2x^{2}+1 \right ]-9x^{2}
=\left [\left (x^{2} \right )^{2}+2\times x^{2}\times 1+\left ( 1 \right )^{2} \right ]-9x^{2}
=\left [\left (x^{2}+1 \right ) \right ]^{2}-\left (3x \right )^{2}
[ সূত্রঃ {\color{Blue} a^{2}+2ab+b^{2}=\left ( a+b \right )^{2}} ]
=\left ( x^{2}+1+3x \right )\left ( x^{2}+1-3x \right )
[ সূত্রঃ {\color{Blue} a^{2}-b^{2}=\left ( a+b \right )\left ( a-b \right )} ]
=\left ( x^{2}+3x+1 \right )\left ( x^{2}-3x+1 \right ) (উত্তর)
উৎপাদকে বিশ্লেষণ : কষে দেখি - 8.2
6. \fn_cm {\color{Blue} p^{4}-11p^{2}q^{2}+q^{4}}
সমাধানঃ
\fn_cm p^{4}-11p^{2}q^{2}+q^{4}
=\left ( p^{2} \right )^{2}-2p^{2}q^{2}+\left ( q^{2} \right )^{2}-9p^{2}q^{2}
=\left ( p^{2}-q^{2} \right )^{2}-9p^{2}q^{2}
[ সূত্রঃ {\color{Blue} a^{2}-2ab+b^{2}=\left ( a-b \right )^{2}} ]
=\left ( p^{2}-q^{2} \right )^{2}-\left (3pq \right )^{2}
=\left ( p^{2}-q^{2}+3pq \right )\left ( p^{2}-q^{2}-3pq \right )
[ সূত্রঃ {\color{Blue} a^{2}-b^{2}=\left ( a+b \right )\left ( a-b \right )} ]
=\left ( p^{2}+3pq-q^{2} \right )\left ( p^{2}-3pq-q^{2} \right ) (উত্তর)
উৎপাদকে বিশ্লেষণ : কষে দেখি - 8.2
7. \fn_cm {\color{Blue} a^{2}+b^{2}-c^{2}-2ab}
সমাধানঃ
\fn_cm a^{2}+b^{2}-c^{2}-2ab
=\left [\left ( a \right )^{2}-2\times a\times b+\left ( b \right )^{2} \right ]-\left ( c \right )^{2}
=\left ( a-b \right )^{2}-\left ( c \right )^{2}
[ সূত্রঃ {\color{Blue} a^{2}-2ab+b^{2}=\left ( a-b \right )^{2}} ]
=\left ( a-b+c \right )\left ( a-b-c \right )
[ সূত্রঃ {\color{Blue} a^{2}-b^{2}=\left ( a+b \right )\left ( a-b \right )} ]
(উত্তর)
উৎপাদকে বিশ্লেষণ : কষে দেখি - 8.2
8. \fn_cm {\color{Blue} 3a\left ( 3a+2c \right )-4b\left ( b+c \right )}
সমাধানঃ
\fn_cm 3a\left ( 3a+2c \right )-4b\left ( b+c \right )
=9a^{2}+6ac-4b^{2}-4bc
=9a^{2}+6ac+c^{2}-4b^{2}-4bc-c^{2}
=\left [ \left ( 3a \right )^{2}+2\times 3a\times c+\left ( c \right )^{2} \right ]-\left [ \left ( 2b \right )^{2}+2\times 2b\times c+\left ( c \right )^{2} \right ]
=\left ( 3a+c \right )^{2}-\left ( 2b+c \right )^{2}
[ সূত্রঃ {\color{Blue} a^{2}+2ab+b^{2}=\left ( a+b \right )^{2}} ]
=\left ( 3a+c+2b+c \right )\left ( 3a+c-2b-c \right )
[ সূত্রঃ {\color{Blue} a^{2}-b^{2}=\left ( a+b \right )\left ( a-b \right )} ]
=\left ( 3a+2b+2c \right )\left ( 3a-2b \right ) (উত্তর)
উৎপাদকে বিশ্লেষণ : কষে দেখি - 8.2
9. \fn_cm {\color{Blue} a^{2}-6ab+12bc-4c^{2}}
সমাধানঃ
\fn_cm a^{2}-6ab+12bc-4c^{2}
=a^{2}-2\times a\times 3b+2\times 3b\times 2c-4c^{2}
=\left [\left (a \right )^{2}-2\times a\times 3b+\left ( 3b \right )^{2} \right ]-\left [ \left ( 3b \right )^{2}-2\times 3b\times 2c+\left ( 2c \right )^{2} \right ]
=\left ( a-3b \right )^{2}-\left ( 3b-2c \right )^{2}
[ সূত্রঃ {\color{Blue} a^{2}-2ab+b^{2}=\left ( a-b \right )^{2}} ]
=\left ( a-3b+3b-2c \right )\left ( a-3b-3b+2c \right )
[ সূত্রঃ {\color{Blue} a^{2}-b^{2}=\left ( a+b \right )\left ( a-b \right )} ]
=\left ( a-2c \right )\left ( a-6b+2c \right ) (উত্তর)
উৎপাদকে বিশ্লেষণ : কষে দেখি - 8.2
10. \fn_cm {\color{Blue} 3a^{2}+4ab+b^{2}-2ac-c^{2}}
সমাধানঃ
\fn_cm 3a^{2}+4ab+b^{2}-2ac-c^{2}
=4a^{2}+4ab+b^{2}-a^{2}-2ac-c^{2}
=\left [\left ( 2a \right )^{2}+2\times 2a\times b+\left ( b \right )^{2} \right ]-\left [ \left ( a \right )^{2}+2\times a\times c+\left ( c \right )^{2} \right ]
=\left ( 2a+b \right )^{2}-\left ( a+c \right )^{2}
[ সূত্রঃ {\color{Blue} a^{2}+2ab+b^{2}=\left ( a+b \right )^{2}} ]
=\left ( 2a+b+a+c \right )\left ( 2a+b-a-c \right )
[ সূত্রঃ {\color{Blue} a^{2}-b^{2}=\left ( a+b \right )\left ( a-b \right )} ]
=\left ( 3a+b+c \right )\left ( a+b-c \right ) (উত্তর)
উৎপাদকে বিশ্লেষণ : কষে দেখি - 8.2
11. \fn_cm {\color{Blue} x^{2}-y^{2}-6ax+2ay+8a^{2}}
সমাধানঃ
\fn_cm x^{2}-y^{2}-6ax+2ay+8a^{2}
=x^{2}-y^{2}-2\times x\times 3a+2\times a\times y+9a^{2}-a^{2}
=\left [ \left ( x \right )^{2}-2\times x\times 3a+\left ( 3a \right )^{2} \right ]-\left [\left ( a \right )^{2}-2\times a\times y+\left ( y \right )^{2} \right ]
=\left ( x-3a \right )^{2}-\left ( a-y \right )^{2}
[ সূত্রঃ {\color{Blue} a^{2}-2ab+b^{2}=\left ( a-b \right )^{2}} ]
=\left ( x-3a+a-y \right )\left ( x-3a-a+y \right )
[ সূত্রঃ {\color{Blue} a^{2}-b^{2}=\left ( a+b \right )\left ( a-b \right )} ]
=\left ( x-y-2a \right )\left ( x+y-4a \right ) (উত্তর)
উৎপাদকে বিশ্লেষণ : কষে দেখি - 8.2
12. \fn_cm {\color{Blue} a^{2}-9b^{2}+4c^{2}-25d^{2}-4ac+30bd}
সমাধানঃ
\fn_cm a^{2}-9b^{2}+4c^{2}-25d^{2}-4ac+30bd
=\left [a^{2}-2\times a\times 2c+\left ( 2c \right )^{2} \right ]-\left [ \left ( 3b \right )^{2}-2\times 3b\times 5d+\left ( 5d \right )^{2} \right ]
=\left ( a-2c \right )^{2}-\left ( 3b-5d \right )^{2}
[ সূত্রঃ {\color{Blue} a^{2}-2ab+b^{2}=\left ( a-b \right )^{2}} ]
=\left ( a-2c+3b-5d \right )\left ( a-2c-3b+5d \right )
[ সূত্রঃ {\color{Blue} a^{2}-b^{2}=\left ( a+b \right )\left ( a-b \right )} ]
=\left ( a+3b-2c-5d \right )\left ( a-3b-2c+5d \right ) (উত্তর)
উৎপাদকে বিশ্লেষণ : কষে দেখি - 8.2
13. \fn_cm {\color{Blue} 3a^{2}-b^{2}-c^{2}+2ab-2bc+2ca}
সমাধানঃ
\fn_cm 3a^{2}-b^{2}-c^{2}+2ab-2bc+2ca
=4a^{2}-a^{2}-b^{2}-c^{2}+2ab-2bc+2ca
=\left (2a \right )^{2}-\left (a^{2}+b^{2}+c^{2}-2ab+2bc-2ca \right )
=\left ( 2a \right )^{2}-\left [ \left ( -a \right )^{2}+\left ( b \right )^{2}+\left ( c \right )^{2} +2\left ( -a \right )b+2bc+2c\left ( -a \right )\right ]
=\left ( 2a \right )^{2}-\left ( -a+b+c \right )^{2}
[ সূত্রঃ {\color{Blue} x^{2}+y^{2}+z^{2}+2xy+2yz+2zx=\left ( x+y+z \right )^{2}} ]
=\left [ \left ( 2a \right )+\left ( -a+b+c \right ) \right ]\left [ \left ( 2a \right )-\left ( -a+b+c \right ) \right ]
[ সূত্রঃ {\color{Blue} a^{2}-b^{2}=\left ( a+b \right )\left ( a-b \right )} ]
=\left ( 2a-a+b+c \right )\left ( 2a+a-b-c \right )
=\left ( a+b+c \right )\left ( 3a-b-c \right ) (উত্তর)
উৎপাদকে বিশ্লেষণ : কষে দেখি - 8.2
14. \fn_cm {\color{Blue} x^{2}-2x-22499}
সমাধানঃ
\fn_cm x^{2}-2x-22499
=x^{2}-2x+1-22500
=\left ( x^{2}-2.x.1+1^{2} \right )-\left ( 150 \right )^{2}
=\left ( x-1 \right )^{2}-\left ( 150 \right )^{2}
[ সূত্রঃ {\color{Blue} a^{2}-2ab+b^{2}=\left ( a-b \right )^{2}} ]
=\left ( x-1+150 \right )\left ( x-1-150 \right )
[ সূত্রঃ {\color{Blue} a^{2}-b^{2}=\left ( a+b \right )\left ( a-b \right )} ]
=\left (x+149 \right )\left ( x-151 \right ) (উত্তর)
উৎপাদকে বিশ্লেষণ : কষে দেখি - 8.2
15. \fn_cm {\color{Blue} \left ( x^{2}-y^{2} \right )\left ( a^{2}-b^{2} \right )+4abxy}
সমাধানঃ
\fn_cm \left ( x^{2}-y^{2} \right )\left ( a^{2}-b^{2} \right )+4abxy
=a^{2}x^{2}-b^{2}x^{2}-a^{2}y^{2}+b^{2}y^{2}+4abxy
=a^{2}x^{2}-b^{2}x^{2}-a^{2}y^{2}+b^{2}y^{2}+2abxy+2abxy
=a^{2}x^{2}+2abxy+b^{2}y^{2}-\left [b^{2}x^{2}-2abxy+a^{2}y^{2} \right ]
=\left ( ax+by \right )^{2}-\left ( bx-ay \right )^{2}
[ সূত্রঃ {\color{Blue} m^{2}+2mn+n^{2}=\left ( m+n \right )^{2}} ও
{\color{Blue} m^{2}-2mn+n^{2}=\left ( m-n \right )^{2}} ]
=\left (ax+by+bx-ay \right )\left ( ax+by-bx+ay \right )
[ সূত্রঃ {\color{Blue} a^{2}-b^{2}=\left ( a+b \right )\left ( a-b \right )} ]
=\left ( ax-bx+ay+by\right )\left (ax+bx-ay+by \right ) (উত্তর)
Koshe dekhi 8.2 Class 9
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3,356,674,501,767,054,300 | Numberpedia 整数や数列に関するまとめサイト
整数100の性質や関連する数列について
100の性質
素数
100 は素数ではありません
100未満の最大の素数97
100より大きい最小の素数101
次の整数101
前の整数99
素因数分解
2 2 * 5 2
2 * 2 * 5 * 5
約数1, 2, 4, 5, 10, 20, 25, 50, 100
約数の個数9
約数の和217
平方根10.0
立方根4.641588833612778
自然対数4.605170185988092
常用対数2.0
sin(100)-0.5063656411097588
cos(100)0.8623188722876839
tan(100)-0.5872139151569291
100の表記
漢字
2進数表記1100100
8進数表記144
16進数表記64
36進数表記
(英字26文字+数字10文字)
2s
62進数表記
(大文字小文字英数字)
1C
その他の表現
100 = 11 * 32 + 1
100 = 73 - 35
100を含む数列
100 は 3 番目の10のべき乗数です
100 は 5 番目の十八角数です
100 は 7 番目のレイランド数です
100 は 11 番目の平方数です
100 は 14 番目の多冪数です
100 は 20 番目のハッピー数です
100 は 22 番目の過剰数です
100 は 23 番目の疑似完全数です
100 は 33 番目のハーシャッド数です
100 は 34 番目のハミング数です | {
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4,329,565,421,439,443,500 | 2018088期双色球开机号
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SAT数学真题精粹
2014-07-11 15:28 供稿单位: 新航道
出国英语考试有哪些 雅思6.5是什么水平 雅思阅读评分标准 托福阅读评分标准 雅思和托福的区别
sat数学考试中有一部分考题作为经典考题，每年都会重复出现1-2道。新航道小编整理历年sat数学考题之精粹，供大家参考。
1. If 2 x + 3 = 9, what is the value of 4 x – 3 ?
(A) 5 (B) 9 (C) 15 (D) 18 (E) 21
2. If 4(t + u) + 3 = 19, then t + u = ?
(A) 3 (B) 4 (C) 5 (D) 6 (E) 7
3. In the xy-coordinate (坐标) plane above, the line contains the points (0,0) and (1,2). If line M (not shown) contains the point (0,0) and is perpendicular (垂直) to L, what is an equation of M?
(A) y = -1/2 x (B) y = -1/2 x + 1 (C) y = - x (D) y = - x + 2 (E) y = -2x
4. If K is divisible by 2,3, and 15, which of the following is also divisible by these numbers?
(A) K + 5 (B) K + 15 (C) K + 20 (D) K + 30 (E) K + 45
5. There are 8 sections of seats in an auditorium. Each section contains at least 150 seats but not more than 200 seats. Which of the following could be the number of seats in this auditorium?
(A) 800 (B) 1,000 (C) 1,100 (D) 1,300 (E) 1,700
6. If rsuv = 1 and rsum = 0, which of the following must be true?
(A) r < 1 (B) s < 1 (C) u= 2 (D) r = 0 (E) m = 0
7. The least integer of a set of consecutive integers (连续整数) is –126. if the sum of these integers is 127, how many integers are in this set?
(A) 126 (B) 127 (C) 252 (D) 253 (E) 254
8. A special lottery is to be held to select the student who will live in the only deluxe room in a dormitory. There are 200 seniors, 300 juniors, and 400 sophomores who applied. Each senior’s name is placed in the lottery 3 times; each junior’s name, 2 time; and each sophomore’s name, 1 times. If a student’s name is chosen at random from the names in the lottery, what is the probability that a senior’s name will be chosen?
(A)1/8 (B) 2/9 (C) 2/7 (D) 3/8 (E) 1/2
更多sat数学信息可以参看新航道sat频道。
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责编:边媛媛
2018088期双色球开机号 | {
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7,730,149,588,994,825,000 | Sei sulla pagina 1di 11
I GRAFICI DELLE FUNZIONI
Davide Boldarin
[email protected]
a cura di DAVIDE BOLDARIN
FUNZIONI IRRAZIONALI La gura 1 mostra il graco della funzione f (x) = denita per valori 0)
x. (Si noti che x
Figura 1: Funzione irrazionale con radice pari
La gura 2 mostra il graco della funzione f (x) = funzione denita x R)
x. (Questa
Figura 2: Funzione irrazionale con radice dispari
a cura di DAVIDE BOLDARIN
FUNZIONI LOGARITMICHE La gura 3 mostra il graco della funzione ln x, vale a dire logaritmo con base e 2.718 numero di Nepero. Il graco si presenta di questo tipo (vale a dire crescente) per tutti i logaritmi con base a > 1.(Si noti che x denita per valori > 0)
Figura 3: Funzione logaritmica con a > 1
La gura 4 mostra il graco della funzione log0.5 x, vale a dire logaritmo con base a = 0.5. Il graco si presenta decrescente per tutti i logaritmi con base 0 < a < 1.(Si noti che x denita per valori > 0)
Figura 4: Funzione logaritmica con 0 < a < 1
a cura di DAVIDE BOLDARIN
ln x2 = ln x = ln x ln x
FUNZIONI ESPONENZIALI La gura 5 mostra il graco della funzione exp x, vale a dire il numero di Nepero elevato alla x. Il graco si presenta di questo tipo (vale a dire crescente) se a > 1.(Si noti come il codominio f (x) > 0)
Figura 5: Funzione esponenziale con a > 1
La gura 6 mostra il graco della funzione 0.5x . Il graco si presenta decrescente per 0 < a < 1.(Si noti come il codominio f (x) > 0)
Figura 6: Funzione esponenziale con 0 < a < 1
a cura di DAVIDE BOLDARIN
ln (exp x) = x , exp (ln x) = x loga ax = x , aloga x = x
IPERBOLE EQUILATERA (Funzione omograca)
+b y = ax cx+d con c = 0 (altrimenti sarebbe una retta) a Il centro delliperbole si trova: C = d c , c . Nelle coordinate del a centro (x = d c e y = c ) abbiamo un asintoto verticale e un asintoto orizzontale. Per capire in che quadrante sta liperbole diamo un valore b a x (ad esempio x = 0) e troviamo il corrispettivo f (x) = d . (liperbole sta sempre in due quadranti opposti tra loro)
x+1 Figura 7: Graco delliperbole y = 3 x+1
a cura di DAVIDE BOLDARIN
FUNZIONI TRIGONOMETRICHE
Figura 8: Graco della funzione y = cos x
Figura 9: Graco della funzione y = senx
a cura di DAVIDE BOLDARIN
Figura 10: Graco della funzione y = tan x
sin x tan x = cos x
cos2 x + sin2 x = 1 (Teorema fondamentale della trigonometria)
a cura di DAVIDE BOLDARIN
FUNZIONI DI FUNZIONI
1. y = |g (x)|
Figure 11: Graco della funzione y = |ln(x)| 2. y = g (|x|)
Figure 12: Graco della funzione y =ln |x|
a cura di DAVIDE BOLDARIN 3. y = g (x) + k
Figure 13: Graco della funzione y = ln(x) + 1 4. y = g (x + k )
Figure 14: Graco della funzione y = ln(x + 1)
a cura di DAVIDE BOLDARIN 5. y = c g (x)
Figure 15: Graco della funzione y = 2 cos x 6. y = h[g (x)]
Figure 16: Graco della funzione y = exp
3x+1 x+1
a cura di DAVIDE BOLDARIN
Per costruire questa funzione consideriamo innanzitutto il graco della x+1 nostra g (x), vale a dire 3x +1 (vedi gura 7). Lordinata di questo graco costituisce largomento della nostra funzione esponenziale. Lordinata va da 3 a +, e poi va da a 3. Quindi, applicando la funzione esponenziale a qusti valori troviamo che il graco andr da e3 20 a + 3 e = , e poi da e =0ae 20. N.B.Il punto (1, 0) non fa parte della funzione poich x = 1 non fa parte del campo di esistenza della funzione. | {
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A group of friends have decided to take a roadtrip. Whenever
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A group of friends have decided to take a roadtrip. Whenever they come to a fork in the road, they will flip a fair coin to decide whether to head right or left. If the coin is heads, they will head right. If the coin is tails, they will head left. If the friends have made seven decisions, what is the probability that they took exactly four lefts?
A. 1/16
B. 1/32
C. 21/25
D. 35/128
E. 4/7
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New post 02 Oct 2005, 00:21
d) 35/128
7C4*(1/2)^4*(1/2)^3 = 35/(16*8)=35/128
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New post 02 Oct 2005, 04:18
D too
I used bernoulli method too as chets did
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New post 02 Oct 2005, 04:27
OA is D.
Antmavel, could you please explain the bernoulli method?? :oops:
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New post 02 Oct 2005, 06:08
D) Same method ... applying the formula nCr*(p^r)*(q^(n-r))
When to apply Bernoulli's trials....
1) Only two possible outcomes for each trial (Yes or No, Heads or Tail, or simply p=Success & q= Failure. Note: p need not be same as q)
2) The probability of success (p) is same for each trial
3) Trials are independent (n is independent)
So in this problem, there are seven trials and you want to find the probability of success for exactly 4 of them.
For 1 trial Success is: p= 1/2 (its either Yes or No)
q = 1-p = 1/2
n =7
r=4
Applying BT, you get 35/128.
Added: I also found the following link on this website (see Appendix A) http://www.gmatclub.com/content/courses/quantitative/probability.php
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New post 02 Oct 2005, 20:36
GMATT73 wrote:
OA is D.
Antmavel, could you please explain the bernoulli method?? :oops:
vikramm, your post was perfect :) Nice explanation
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Convergence to pi^2/6 using Fourier Series and f(x) = x^2
1. Nov 22, 2009 #1
1. The problem statement, all variables and given/known data
Using the Fourier trigonometrical series for [tex]f(x) = {x^2},{\rm{ }}0 \le x < 2\pi [/tex], prove that [tex]\sum\limits_{n = 1}^\infty {\frac{1}{{{n^2}}}} = \frac{{{\pi ^2}}}{6}[/tex]
3. The attempt at a solution
This is more of a "what am I doing wrong question". First, because I'm not in the period defined by the trigonometrical coefficients, I have to change the limits using that:
[tex]0 \le x < 2\pi [/tex]
[tex] - \pi \le x - \pi < \pi [/tex]
[tex]\begin{array}{l}
- \pi \le t < \pi , \\
t = x - \pi , \\
f(t) = f(x - \pi ) \\
\end{array}[/tex]
Then I find the Fourier coefficients with this in hand:
[tex]{a_0} = \frac{1}{\pi }\int\limits_{ - \pi }^\pi {f(t).dt = } \frac{1}{\pi }\int\limits_0^{2\pi } {f(x).dx = } \frac{1}{\pi }\int\limits_0^{2\pi } {{x^2}.dx = } \frac{8}{3}{\pi ^2}[/tex]
[tex]{a_n} = \frac{1}{\pi }\int\limits_{ - \pi }^\pi {f(t).\cos (nt)dt = } \frac{1}{\pi }\int\limits_0^{2\pi } {f(x).\cos (n(x - \pi ))dx = } \frac{1}{\pi }\int\limits_0^{2\pi } {{x^2}.\left[ {\cos (nx)\cos (n\pi ) - \sin (nx)\sin (n\pi )} \right].dx = }
[/tex]
[tex]{a_n} = \frac{1}{\pi }\int\limits_0^{2\pi } {{x^2}.\cos (nx).{{( - 1)}^n}dx = } \frac{{{{( - 1)}^n}}}{\pi }\left( {\left. {\frac{{2x.\cos (nx)}}{{{n^2}}}} \right|_0^{2\pi }} \right) = \frac{{2{{( - 1)}^n}}}{{{n^2}}}[/tex]
[tex]{b_n} = \frac{1}{\pi }\int\limits_{ - \pi }^\pi {f(t).\sin (nt).dx = } \frac{1}{\pi }\int\limits_0^{2\pi } {f(x).\sin (n(x - \pi )).dx = } \frac{1}{\pi }\int\limits_0^{2\pi } {{x^2}.\left[ {\sin (nx)\cos (n\pi ) + \sin (n\pi )\cos (nx)} \right].dx = } [/tex]
[tex]{b_n} = \frac{{{{\left( { - 1} \right)}^n}}}{\pi }\int\limits_0^{2\pi } {{x^2}.\sin (nx).dx = } \frac{{{{\left( { - 1} \right)}^n}}}{\pi }\left( {\left. {(\frac{2}{{{n^3}}} - \frac{{{x^2}}}{n})\cos (nx)} \right|_0^{2\pi }} \right) = \frac{{4{{\left( { - 1} \right)}^{n + 1}}}}{n}[/tex]
Then, using Parseval's identity:
[tex]{\left\| {f(x)} \right\|^2} = \frac{{{a_0}^2}}{2} + \sum\limits_{n = 1}^\infty {{a_n}^2 + {b_n}^2} [/tex]
[tex]\frac{1}{\pi }\int\limits_0^{2\pi } {{{({x^2})}^2}.dx} = \frac{{{{\left( {\frac{8}{3}{\pi ^2}} \right)}^2}}}{2} + \sum\limits_{n = 1}^\infty {{{\left( {\frac{{2{{( - 1)}^n}}}{{{n^2}}}} \right)}^2} + {{\left( {\frac{{4{{( - 1)}^{n + 1}}}}{n}} \right)}^2}} [/tex]
[tex]\frac{{32}}{5}{\pi ^4} = \frac{{32{\pi ^4}}}{9} + \sum\limits_{n = 1}^\infty {\frac{4}{{{n^4}}} + \frac{{16}}{{{n^2}}}} [/tex]
[tex]\frac{{112}}{9}{\pi ^4} = \sum\limits_{n = 1}^\infty {\frac{{4 + 16{n^2}}}{{{n^4}}}} [/tex]
And here I reach a deadpoint. What do I do?
2. jcsd
3. Nov 22, 2009 #2
LCKurtz
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You are extending f(x) periodically with period [itex]2\pi[/itex]. You get the same thing when you integrate a periodic function over any period. That is, if g(x) has period P, then
[tex]\int_a^{a+P} g(x)\, dx = \int_b^{b+P} g(x)\, dx [/tex]
That means you don't have to use [itex]-\pi[/itex] to [itex]\pi[/itex] in your formulas for the coefficients; you can use 0 to [itex]2\pi[/itex], so you don't have to fiddle with changing the x2.
The other thing you are missing is using the Dirichlet theorem which tells you what the series will converge to at 0 and [itex]2\pi[/itex]. Notice that your periodic extension of your function has a jump discontinuity at those points.
4. Nov 22, 2009 #3
LCKurtz
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Also, one other question. Are you sure your function x2 wasn't given on [itex](-\pi,\pi)[/itex] in the first place?
5. Nov 23, 2009 #4
Let me see if I get it: you're saying that x^2.cos(nx) and x^2.sin(nx) are 2Pi periodic, so that change of limits that produced the (-1)^n was wrong? I didn't have to do anything, just integrate them between 0 and 2Pi?
And what do you want me to do with the borders? I know that they will converge to the lateral limits average but I don't know what you mean.
And yes, x^2 was defined in (0, 2Pi).
6. Nov 23, 2009 #5
LCKurtz
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No, I am not saying that x2cos(nx) and x2sin(nx) are periodic. I am saying that the periodic extensions of them are periodic, and that is what you are expanding in a FS. I didn't go through your steps to see if you had any mistakes an your calculation of an and bn. I am just saying you can use the usual formulas on any period you wish, and in the present case, it is much easier to use [itex](0,2\pi)[/itex].
The reason I asked this is because, if I recall correctly, evaluating the even extension of x2 on [itex](-\pi,\pi)[/itex] at 0 is the way I have seen the problem done.
7. Nov 23, 2009 #6
But still, the expressions for an and bn, etc, are made for (-[tex]\pi[/tex], [tex]\pi[/tex]), I still have to make some correction when I calculate them for (0, 2[tex]\pi[/tex]), right?
8. Nov 23, 2009 #7
LCKurtz
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No. You don't have to change anything but the limits. If f(x) has period [itex]2\pi[/itex] then
[tex]\frac 1 {\pi}\int_{-\pi}^{\pi} f(x)\cos{nx}\, dx = \frac 1 {\pi}\int_{0}^{2\pi} f(x)\cos{nx}\, dx[/tex]
and ditto for the sine. In the current problem the latter is much preferred because it is the only one in which you can use [itex]f(x) = x^2[/itex].
And looking ahead, I'm guessing you might want to use the even periodic extension of x2, which would be [itex]4\pi[/itex] periodic. As I mentioned before, the usual series for this is the periodic extension of x2 on [itex](-\pi,\pi)[/itex], but that is apparently not what you have been asked to do.
9. Nov 23, 2009 #8
So I use the even periodic extension for f(x): that is, f(x) = x2 for -2[tex]\pi[/tex]<x<2[tex]\pi[/tex], and then I make the following calculations?:
[tex]{a_0} = \frac{1}{{2\pi }}\int\limits_{ - 2\pi }^{2\pi } {{x^2}dx} [/tex]
[tex]{a_n} = \frac{1}{{2\pi }}\int\limits_{ - 2\pi }^{2\pi } {{x^2}\cos (\frac{1}{{2\pi }}nx)dx} [/tex]
[tex]{b_n} = \frac{1}{{2\pi }}\int\limits_{ - 2\pi }^{2\pi } {{x^2}\sin (\frac{1}{{2\pi }}nx)dx} [/tex]
Is that what you're saying?
10. Nov 23, 2009 #9
LCKurtz
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There shouldn't be any [itex]\pi[/itex] in your cosines and sines. [itex]2p=4\pi[/itex] so
[tex]\sin{\frac{n\pi x}{p}} = sin{\frac n 2 x}[/tex]
11. Nov 23, 2009 #10
But that's the way?
12. Nov 23, 2009 #11
LCKurtz
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And, once more, I will add that I suspect you are doing the wrong problem. Maybe there is a typo somewhere. Because if you take the periodic extension of x2 on [itex](-\pi,\pi)[/itex] and evaluate its series at [itex]x = \pi[/itex] you will get your desired answer.
13. Nov 23, 2009 #12
OK, thanks. One question, though, when do I know when to use the periodic extensions (even an odd extensions)?
14. Nov 23, 2009 #13
LCKurtz
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That's a good question. It depends on what you intend to do with the series, but generally, the smoother the periodic function is, the faster the series will converge. In this case, if you take the even extension of x2 you get a continuous function whereas if you take the odd extension you get jump discontinuities at the ends. And the even extension has no sine terms so that's half the work. Another hint in this problem would be the series you are asked about has terms like 1/n2 in it which suggests you need a continuous periodic function.
15. Nov 23, 2009 #14
Thank you.
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3,247,746,463,905,332,700 | Roman Numerals
Roman Numerals: CCCXCIV = 394
Convert Roman Numerals
Arabic numerals:
Roman numerals:
Arabic
numerals
0
1MCXI
2MMCCXXII
3MMMCCCXXXIII
4CDXLIV
5DLV
6DCLXVI
7DCCLXXVII
8DCCCLXXXVIII
9CMXCIX
The converter lets you go from arabic to roman numerals and vice versa. Simply type in the number you would like to convert in the field you would like to convert from, and the number in the other format will appear in the other field. Due to the limitations of the roman number system you can only convert numbers from 1 to 3999.
To easily convert between roman and arabic numerals you can use the table above. The key is to handle one arabic digit at a time, and translate it to the right roman number, where zeroes become empty. Go ahead and use the converter and observe how the table shows the solution in realtime!
Current date and time in Roman Numerals
2022-09-26 21:53:09
MMXXII-IX-XXVI XXI:LIII:IX
Here is the current date and time written in roman numerals. Since the roman number system doesn't have a zero, the hour, minute, and second component of the timestamps sometimes become empty.
The year 394
Here you can read more about what happened in the year 394.
The number 394
The number 394 is divisble by 2 and 197 and can be prime factorized into 2×197.
394 as a binary number: 110001010
394 as an octal number: 612
394 as a hexadecimal number: 18A
Numbers close to CCCXCIV
Below are the numbers CCCXCI through CCCXCVII, which are close to CCCXCIV. The right column shows how each roman numeral adds up to the total.
391 = CCCXCI = 100 + 100 + 100 + 100 − 10 + 1
392 = CCCXCII = 100 + 100 + 100 + 100 − 10 + 1 + 1
393 = CCCXCIII = 100 + 100 + 100 + 100 − 10 + 1 + 1 + 1
394 = CCCXCIV = 100 + 100 + 100 + 100 − 10 + 5 − 1
395 = CCCXCV = 100 + 100 + 100 + 100 − 10 + 5
396 = CCCXCVI = 100 + 100 + 100 + 100 − 10 + 5 + 1
397 = CCCXCVII = 100 + 100 + 100 + 100 − 10 + 5 + 1 + 1
About Roman Numerals
Roman numerals originate, as the name suggests, from the Ancient Roman empire. Unlike our position based system with base 10, the roman system is based on addition (and sometimes subtraction) of seven different values. These are symbols used to represent these values:
SymbolValue
I1
V5
X10
L50
C100
D500
M1000
For example, to express the number 737 in roman numerals you write DCCXXXVII, that is 500 + 100 + 100 + 10 + 10 + 10 + 5 + 1 + 1. However, for the numbers 4 and 9, subtraction is used instead of addition, and the smaller number is written in front of the greater number: e.g. 14 is written as XIV, i.e. 10 + 5 − 1, and 199 is expressed as CXCIX i.e. 100 + 100 − 10 + 10 − 1. It could be argued that 199 would be more easily written as CIC, but according to the most common definition you can only subtract a number that is one order of magnitude smaller than the numbers you're subtracting from, meaning that IC for 99 is incorrect.
Roman numerals are often used in numbered lists, on buildings to state the year they were built, and in names of regents, such as Louis XVI of France.
Feel free to link to this site if you find it useful. It's also possible to link directly to specific numbers, such as roman-numerals.info/XXXVII or roman-numerals.info/37. You can also link to intervals, for instance roman-numerals.info/1-100 or roman-numerals.info/1980-2020, to see the numbers in a list format. | {
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Question and answer
Q: Solve |2x - 5| = 4.
A: |2x - 5| = 4 2x - 5 = -4 or 2x - 5 = 4 2x - 5 = -4; 2x = -4 + 5; 2x = 1; x = 1/2; 2x - 5 = 4; 2x = 4 + 5; 2x = 9; x = 9/2; x = 1/2 or x = 9/2
Question
Asked 6/14/2011 9:13:14 AM
Updated 8/24/2014 3:46:36 PM
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Edited by andrewpallarca [8/24/2014 3:46:36 PM]
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|2x - 5| = 4
2x - 5 = -4 or 2x - 5 = 4
2x - 5 = -4;
2x = -4 + 5;
2x = 1;
x = 1/2;
2x - 5 = 4;
2x = 4 + 5;
2x = 9;
x = 9/2;
x = 1/2 or x = 9/2
Added 8/24/2014 2:47:21 PM
This answer has been confirmed as correct and helpful.
Confirmed by andrewpallarca [8/24/2014 3:46:37 PM]
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Results 1 to 2 of 2
Thread: Linear law problem
1. #1
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Exclamation Linear law problem
The diagram shows part of a straight line obtained by plotting log10 y against log10 x for two variables x and y. Find
a) log10 y in terms of log10 x
b) y in terms of x
c) the valve of x when y = 40
Can teach me how to do??Linear law problem-image-62-.jpg
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2. #2
Forum Admin topsquark's Avatar
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Re: Linear law problem
How far have you gotten on this?
Hint: The "log(y) and log(x)" act as y and x values. So your line on the graph will be of the form:
log_{10}(y) = m \cdot \log_{10}(x) + b, where m and b are the usual slope of the line, and b is the y-intercept.
-Dan
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3,696,688,904,816,947,000 | [幼儿园数学练习题]幼儿园数学课件整理
数学课件 2018-07-22 17:18:32 数学课件
[摘要]导语： 活动目的： 1、幼儿知道应用题的结构，初步学会看图列式，能根据不同的画面，学会口编8以内的加减法应用题。具有一定的推理能力。 2、懂得运用互换规律列出另一道算式，并列式运算。 活动课件
【www.cddlwy.com--数学课件】
导语：
活动目的：
1、幼儿知道应用题的结构，初步学会看图列式，能根据不同的画面，学会口编8以内的加减法应用题。具有一定的推理能力。
2、懂得运用互换规律列出另一道算式，并列式运算。
活动准备：
课件，幼儿每人一套数字卡片及加号、减号、等号，练习纸，铅笔。
活动过程：
一、复习8的分合。
1、 “老师带来了一蓝鲜花，要分给小朋友。” 教师点击课件
“数数看，有几朵鲜花？”“一共有8朵鲜花，分给小朋友一朵，另外一位小朋友是几朵鲜花？”用拍手、跺脚或体态动作来表示？说对的电脑给予鼓掌。
2、“老师又摘了几朵鲜花，数数看。”“分给小朋友二朵，另外一位小朋友是几朵鲜花？”
3、“老师又摘了几朵鲜花，数数看。”“分给小朋友三朵，另外一位小朋友是几朵鲜花？”
二、学习8的加减
1、 出示课件，看图列式，学习列加法算式，先让幼儿观察，知道两种不同颜色的气球可以列加法题。7＋1＝8，根据互换规律，找出另一道题1＋7＝8。
2、 师：应用题讲了一件事，（妈妈买气球）2个已知道的数（7和1），还提出一个问题？（一共有几个气球）这道应用题用什么方法运算？为什么说7+1=8？（7和1合起来是8）。
幼儿根据不同形状的树，列出加法算式。6＋2＝8，根据互换规律，找出另一道题2＋6＝8。师：刚才编的应用题讲了一件事？有哪两个已知道的数？还提出一个什么问题？（教师小结：编应用题有三个要求：要说出一件事情，有2个已知道的数；还提出一个问题）这道应用题用什么方法运算？为什么？怎样列式？为什么说2+6=8？对了，一共有8棵树。
3. 幼儿看图编减法应用题（点击课件）。
师：看谁能根据三个要求来编应用题，编得又快又完整（并用“三个要求”检查应用题对、错）。
出示课件，看图列式，学习列减法算式，让幼儿知道划去的符号表示减少的意思，可以列减法算式。8－1＝7，另一道题是8－7＝1。
看图汽车，列出算式8－2＝6，另一道题是8－6＝2。
三、幼儿动手操作活动
将老师给出的三个数字2、6、8和3、5、8，用卡片排出两道加法和两道减法算式，并将结果记录在练习纸上。引导幼儿根据生活经验编题。
四、游戏《找朋友》
幼儿根据自已卡片上的数字找合起来是8的朋友。
[page_break]
活动结束：
小朋友一起听音乐。
延伸阅读
发展历史
Mathematics (pinyin: shu xue; Greek: mu alpha theta eta mu alpha tau; (English: Mathematics), derived from the ancient Greek mu theta eta mu alpha (math), which has the meaning of learning, learning and science. The ancient Greek scholars regarded it as the starting point of philosophy, "the foundation of learning". There is also a narrower and more technical significance, "mathematical research". Even in its etymology, its adjective meaning has to do with learning and is used for exponential learning.
It is in the plural form of English, and in the plural form of French, plus es into mathematiques, which can be traced to the Latin neutral plural (Mathematica), which is translated from the Greek plural tao alpha mu alpha mu alpha theta eta mu alpha theta eta mu alpha tau theta mu alpha theta.
In ancient China, mathematics was called arithmetic, also called mathematics, and finally mathematics. The arithmetic of ancient China is one of six arts (six art is called "number").
Mathematics originated from the early production activities of human beings. Ancient babylonians have accumulated certain mathematical knowledge since ancient times and can apply practical problems. From the math itself, their knowledge of mathematics is only observation and experience, without comprehensive conclusions and proofs, but also full affirmation of their contribution to mathematics.
The knowledge and application of basic mathematics is an indispensable part in the life of a person and a group. Its basic concept of refining is long before ancient Egypt, Mesopotamia and ancient Indian ancient mathematical texts. Since then, its development has continued to have small progress. But algebra and geometry had long remained independent.
Algebra is arguably the most widely accepted "mathematics". It's fair to say that every single person starts learning the math when they are young, and the first mathematics that comes into contact with is algebra. Mathematics, as a study of "number", is also one of the most important parts of mathematics. Geometry was the first branch of mathematics to be studied.
It wasn't until the Renaissance of the 16th century that Descartes founded analytic geometry that brought together the algebra and geometry that were completely separated at the time. Since then, we can finally prove the theorems of geometry by computing. It can also represent abstract algebraic equations with graphic representation. And then it developed even more subtle calculus.
Mathematics now includes many branches. The French bourbaki school, founded in the 1930s, argued that mathematics, at least pure mathematics, was the theory of abstract structures. Structure is a deductive system based on initial concepts and axioms. They believe that mathematics has three basic maternal structures: algebraic structures (groups, loops, domains, and so on). ), sequence structure. ), topological structure (neighborhood, limit, connectivity, dimension... ).
Mathematics is applied in many different fields, including science, engineering, medicine and economics. The applications of mathematics in these fields are generally called applied mathematics, and sometimes they provoke new mathematical discoveries and lead to the development of new mathematical disciplines. Mathematicians also study pure mathematics, which is mathematics itself, without any practical application. Although there is a lot of work to start with pure mathematics, it may be possible to find suitable applications later.
Concrete, there are used to explore the links between math core to other areas of sub areas: by logic, set theory, mathematical basis, to different scientific experience in mathematics, applied mathematics, at a relatively modern research to uncertainty (chaos, fuzzy mathematics).
In terms of longitudinally, the exploration in the fields of mathematics is also deepened.
数学（汉语拼音：shù xué；希腊语：μαθηματικ；英语：Mathematics），源自于古希腊语的μθημα（máthēma），其有学习、学问、科学之意。古希腊学者视其为哲学之起点，“学问的基础”。另外，还有个较狭隘且技术性的意义——“数学研究”。即使在其语源内，其形容意义凡与学习有关的，亦会被用来指数学的。
其在英语的复数形式，及在法语中的复数形式+es成mathématiques，可溯至拉丁文的中性复数（Mathematica），由西塞罗译自希腊文复数τα μαθηματικ（ta mathēmatiká）。
在中国古代，数学叫作算术，又称算学，最后才改为数学。中国古代的算术是六艺之一（六艺中称为“数”）。
数学起源于人类早期的生产活动，古巴比伦人从远古时代开始已经积累了一定的数学知识，并能应用实际问题。从数学本身看，他们的数学知识也只是观察和经验所得，没有综合结论和证明，但也要充分肯定他们对数学所做出的贡献。
基础数学的知识与运用是个人与团体生活中不可或缺的一部分。其基本概念的精炼早在古埃及、美索不达米亚及古印度内的古代数学文本内便可观见。从那时开始，其发展便持续不断地有小幅度的进展。但当时的代数学和几何学长久以来仍处于独立的状态。
代数学可以说是最为人们广泛接受的“数学”。可以说每一个人从小时候开始学数数起，最先接触到的数学就是代数学。而数学作为一个研究“数”的学科，代数学也是数学最重要的组成部分之一。几何学则是最早开始被人们研究的数学分支。
直到16世纪的文艺复兴时期，笛卡尔创立了解析几何，将当时完全分开的代数和几何学联系到了一起。从那以后，我们终于可以用计算证明几何学的定理；同时也可以用图形来形象的表示抽象的代数方程。而其后更发展出更加精微的微积分。
现时数学已包括多个分支。创立于二十世纪三十年代的法国的布尔巴基学派则认为：数学，至少纯数学，是研究抽象结构的理论。结构，就是以初始概念和公理出发的演绎系统。他们认为，数学有三种基本的母结构：代数结构（群，环，域，格……）、序结构（偏序，全序……）、拓扑结构（邻域，极限，连通性，维数……）。
数学被应用在很多不同的领域上，包括科学、工程、医学和经济学等。数学在这些领域的应用一般被称为应用数学，有时亦会激起新的数学发现，并促成全新数学学科的发展。数学家也研究纯数学，也就是数学本身，而不以任何实际应用为目标。虽然有许多工作以研究纯数学为开端，但之后也许会发现合适的应用。
具体的，有用来探索由数学核心至其他领域上之间的连结的子领域：由逻辑、集合论（数学基础）、至不同科学的经验上的数学（应用数学）、以较近代的对于不确定性的研究（混沌、模糊数学）。
就纵度而言，在数学各自领域上的探索亦越发深入。
本文来源：https://www.cddlwy.com/kejian/123807.html
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1,011,828,970,028,254,300 | The following table shows the ages of the patients admitted in a hospital during a year :
Question:
The following table shows the ages of the patients admitted in a hospital during a year :
Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency.
Solution:
From the given data, we have the modal class 35-45.
$\{\because$ It has largest frequency among the given classes of the data $\}$
So, $\ell=35, \mathrm{f}_{\mathrm{m}}=23, \mathrm{f}_{1}=21, \mathrm{f}_{2}=14$ and $\mathrm{h}=10$.
Mode $=\ell+\left\{\frac{\mathbf{f}_{\mathbf{m}}-\mathbf{f}}{\mathbf{2 f}_{\mathbf{m}}-\mathbf{f}-\mathbf{f}_{\mathbf{2}}}\right\} \times \mathbf{h}$
$=35+\left\{\frac{\mathbf{2 3}-\mathbf{2 1}}{\mathbf{4 6}-\mathbf{2 1}-\mathbf{1 4}}\right\} \times 10=35+\frac{\mathbf{2 0}}{\mathbf{1 1}}=36.8$ years
Now, let us find the mean of the data :
$\mathrm{a}=30, \mathrm{~h}=10, \mathrm{n}=80$ and $\Sigma \mathrm{fu}_{\mathrm{i}} \mathrm{u}_{\mathrm{i}}=43$
Mean $=\mathrm{a}+\mathrm{h} \times \frac{\mathbf{1}}{\mathbf{n}} \times \Sigma \mathrm{f}_{\mathrm{i}} \mathrm{u}_{\mathrm{i}}=30+10 \times \frac{\mathbf{1}}{\mathbf{8 0}} \times 43$
$=30+5.37=35.37$ years
Thus, mode = 36.8 years and mean = 35.37 years.
So, we conclude that the maximum number of patients admitted in the hospital are of the age 36.8 years (approx), whereas on an average the age of a patient admitted to the hospital is 35.37 years.
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6,049,863,889,918,284,000 |
Results 1 to 6 of 6
Thread: proving sum
1. #1
Newbie darknight's Avatar
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proving sum
A+B+C = 180 (pi)
prove that,
cotA + cotB + cotC = cotA cotB cotC (1 + secAsecBsecC)
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2. #2
Senior Member pacman's Avatar
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cotA + cotB + cotC = cotA cotB cotC (1 + secAsecBsecC)
= cotA cotB cotC + (cotA cotB cotC)(secAsecBsecC)
= cotA cotB cotC + (cos A/sin A)(cos B/sin B)(cos C/sin C)(1/(cosA cos B cos C))
cotA + cotB + cotC = (cot A cot B cot C) + (csc A csc B csc C)
= (cot A cot B cot (pi - A - B)) + (csc A csc B csc (pi - A - B))
= ?
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3. #3
Newbie darknight's Avatar
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Quote Originally Posted by pacman View Post
cotA + cotB + cotC = cotA cotB cotC (1 + secAsecBsecC)
= cotA cotB cotC + (cotA cotB cotC)(secAsecBsecC)
= cotA cotB cotC + (cos A/sin A)(cos B/sin B)(cos C/sin C)(1/(cosA cos B cos C))
cotA + cotB + cotC = (cot A cot B cot C) + (csc A csc B csc C)
after that,
cotA + cotB + cotC
= (cot A cot B cot C) + (csc A csc B csc C)
= (cosAcosBcosC / sinAsinBsinC) + (1/sinAsinBsinC)
taking 1/sinAsinBsinC outside the bracket,
= 1/sinAsinBsinC (cosAcosBcosC + 1) ....................(1)
____________________________________________
Now,
cotA + cotB + cotC
= cosA/sinA + cosB/sinB + cosC/sinC
cosAsinBsinC + cosBsinCsinA + cosCsinBsinA
= _________________________________________......... ........................(2)
sinAsinBsinC
Equating 1 and 2,
cosAsinBsinC + cosBsinCsinA + cosCsinBsinA
________________________________________ = 1/sinAsinBsinC (cosAcosBcosC +1)
sinAsinBsinC
sinAsinBsinC cancel out.
cosAsinBsinC + cosBsinCsinA + cosCsinBsinA = cosAcosBcosC +1
cosAsinBsinC + cosBsinCsinA + cosCsinBsinA - cosAcosBcosC = 1
(cosAsinBsinC + cosBsinCsinA) + (cosCsinBsinA - cosAcosBcosC) = 1
sinC(sinBcosA + cosBsinA) + cosC(sinBsinA - cosAcosB) = 1
sinC(sin(A+B)) + cosC(-(cosAcosB - sinBsinA)) = 1
sinC (sin (pi - C)) + cosC (- (cos(A+B)) =1
sinC (sinC) + cosC (-(-cos(pi-C)) =1
sinC^2 + cosC (-(-cosC)) =1
sinC^2 + cosC^2 =1
1=1
Hence proved
Thanks a bunch for kick-starting my brain. Just worked on it for a minutes and I got it.
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4. #4
Senior Member pacman's Avatar
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Darknight, i believe it is wrong to prove an identity working on both sides because if you do, you are treating it as an equation not as an identity. Any piece of your thought on this?
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5. #5
Newbie darknight's Avatar
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Re:
Quote Originally Posted by pacman View Post
Darknight, i believe it is wrong to prove an identity working on both sides because if you do, you are treating it as an equation not as an identity. Any piece of your thought on this?
That's a good point, and you are correct: the proper way to
prove such identities is to begin on one side and algebraically
transform it into the form shown on the other side. Working on both
sides is technically incorrect because in doing so we are assuming
that the equality to be proven is already true.
However, in the sense that it is not mathematically incorrect - if
the identity is true, then the result of working on both sides will
eventually result in an equality that is "obviously" true.
That is, such a method won't lead to a wrong decision
on the truth or falseness of the identity in question. But in so far
as mathematical rigor, it is insufficient because of the reason
mentioned at the end of my first paragraph. Often, however, working on
both sides can help us form a rigorous proof of an identity. I will
illustrate this with the example.
Identity to be proven:
(1-TanA)/SecA + SecA/TanA = (1+TanA)/(SecATanA)
If we work on both sides, the first step is to multiply both sides by
Sec[a]Tan[a]:
(1-TanA)TanA + SecASecA = 1+TanA
Multiplying through, we find
TanA - TanA^2 + SecA^2 = 1+TanA,
or
-TanA^2 + SecA^2 = 1,
which is equivalent to the identity 1+TanA^2 = SecA^2, so we are
done. This is not a rigorous proof, but it leads to one, in which we
see the proper method is as follows:
(1-TanA)/SecA + SecA/TanA
= (1-TanA)TanA/ (SecATanA) + SecA^2/(SecATanA)
= (TanA - TanA^2 + SecA^2)/(SecATanA)
= (TanA - TanA^2 + 1 + TanA^2)/(SecATanA)
= (TanA + 1)/(SecATanA).
Therefore we have taken the left-hand side and transformed it into the
right-hand side using algebraic manipulation and other known
trigonometric identities. But it should not be too hard to see that
the basic "flow" of the proof is essentially the same procedure as the
"improper" method, just rearranged a bit.
I believe by retracing the steps it should be easy enough to prove my earlier identity, albeit it is much tougher than this simple example.
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6. #6
Senior Member pacman's Avatar
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448
thanks darknight for that insightful explaination . . . .
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-1,135,227,060,313,797,100 | 8
$\begingroup$
Does anyone knows a way to compute a list of all (p,q)-shuffles in mathematica?
For a definition of the shuffle permutations see for example http://ncatlab.org/nlab/show/shuffle
I'm dreaming of a function with p and q as arguments that gives me:
1.) The number of all (p,q)-shuffles.
2.) A list of the actual "shuffle" (p+q)-permutations
3.) The sign of each such permutation
$\endgroup$
6
• 1
$\begingroup$ Let's see if we can do better than this mathhelpforum.com/math-software/… $\endgroup$ May 4, 2012 at 4:24
• $\begingroup$ Ok. Since this was asked before, I guess there is no such thing. $\endgroup$ May 4, 2012 at 4:33
• $\begingroup$ But it was not asked here :D $\endgroup$ May 4, 2012 at 4:34
• $\begingroup$ Is there a package to work with general permutations? Then maybe I can use it as a starting point to write that function by myself. $\endgroup$ May 4, 2012 at 4:47
• $\begingroup$ Very painfully inefficient: With[{p = 4, q = 5}, Union[Join[Sort[Take[#, p]], Sort[Take[#, -q]]] & /@ Permutations[Range[p + q]]]] $\endgroup$ May 4, 2012 at 4:50
1 Answer 1
9
$\begingroup$
1) The number of all (p,q) shuffles is
Binomial[p+q,p]
since when you chose the first p elements, the whole thing (and its order) is given.
2)The actual shuffles are given by: (See JM's comment below*)
With[{x = Range@#1}, {#, Complement[x, #]} & /@ Subsets[x, {#2}]] &[p + q, p]
Example:
p = 3; q = 2;
With[{x = Range@#1}, {#, Complement[x, #]} & /@ Subsets[x, {#2}]] &[p + q, p]
(*
->
{{{1, 2, 3}, {4, 5}}, {{1, 2, 4}, {3, 5}}, {{1, 2, 5}, {3, 4}}, {{1, 3, 4}, {2, 5}},
{{1, 3, 5}, {2, 4}}, {{1, 4, 5}, {2, 3}}, {{2, 3, 4}, {1, 5}}, {{2, 3, 5}, {1, 4}},
{{2, 4, 5}, {1, 3}}, {{3, 4, 5}, {1, 2}}}
3) The sign of each permutation (for the above shuffles) is given by:
Signature/@ With[{x=Range@#1}, Join[#, Complement[x, #]] & /@ Subsets[x, {#2}]] &[p+q,p]
Example:
p = 3; q = 2;
Signature/@ With[{x=Range@#1}, Join[#, Complement[x, #]] & /@ Subsets[x, {#2}]] &[p+q,p]
(*
-> {1, -1, 1, 1, -1, 1, -1, 1, -1, 1}
*)
$\endgroup$
7
• $\begingroup$ mmm looking again ... all those p+q looks nasty. Lets see if I can improve $\endgroup$ May 4, 2012 at 5:00
• $\begingroup$ Signature[] is built-in; no need for an auxiliary package... $\endgroup$ May 4, 2012 at 5:01
• 1
$\begingroup$ @J.M. Thanks. I keep using Combinatorica ... old vice. Editing. $\endgroup$ May 4, 2012 at 5:02
• $\begingroup$ On that note: it might be a bit better if you don't generate Range[p + q] more than once: With[{pq = Range[p + q]}, Join[#, Complement[pq, #]] & /@ Subsets[pq, {p}]]. $\endgroup$ May 4, 2012 at 5:09
• $\begingroup$ @J.M. yep. I'm trying to keep it as a one liner function of p,q $\endgroup$ May 4, 2012 at 5:11
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-8,224,381,486,297,972,000 | Tuesday
October 21, 2014
Homework Help: Math
Posted by Anonymous on Monday, October 14, 2013 at 10:01pm.
find the coordinates of the point on the curve xy = 10 in the first quadrant such that the normal line passes through the origin
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Computer Science 111. Fundamentals of Programming I Developing and Testing Programs. Start with a User Request. Write a program that converts pounds to kilograms. Determine the Inputs and Outputs. The input will be the number of pounds, as an integer or floating-point number
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Computer science 111
Computer Science 111
Fundamentals of Programming I
Developing and Testing Programs
Start with a user request
Start with a User Request
Write a program that converts pounds to kilograms.
Determine the inputs and outputs
Determine the Inputs and Outputs
• The input will be the number of pounds, as an integer or floating-point number
• The output will be
• a prompt for the number of pounds
• A label, and the number of kilograms as a floating-point number
Example session with the program
Example Session with the Program
Enter the number of pounds: 330
The number of kilograms is 150.0
Gather information about the problem
Gather Information About the Problem
There are 2.2 pounds in a kilogram
Design the algorithm in pseudocode
Design the Algorithm in Pseudocode
Prompt the user for the number of pounds
Input pounds
Set kilograms to pounds / 2.2
Output the number of kilograms
Code as the convert script
Code as the convert Script
"""
File: convert.py
This program converts pounds to kilograms.
Input: the number of pounds, as an integer or float.
Output: the number of kilograms, as a float, suitably
labeled.
"""
convert.py
Start with the prefatory docstring, not as an afterthought!!!
Code as the convert script1
Code as the convert Script
pounds = float(input('Enter the number of pounds: '))
kilograms = pounds / 2.2
print('The number of kilograms is', kilograms)
convert.py
The Python shell
Test the program in IDLE first!
Test with several inputs
Test with Several Inputs
pounds = float(input('Enter the number of pounds: '))
kilograms = pounds / 2.2
print('The number of kilograms is', kilograms)
convert.py
The Python shell
Then test in a terminal window
Then Test in a Terminal Window
pounds = float(input('Enter the number of pounds: '))
kilograms = pounds / 2.2
print('The number of kilograms is', kilograms)
convert.py
The terminal
For monday
For Monday
Start Chapter 3
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7,403,172,732,716,474,000 | Vous êtes sur la page 1sur 3
La fonction Γ
1) Définition.
Z +∞
Soit x ∈ R. On pose Γ (x) = tx−1 e−t dt. La fonction f : t 7→ tx e−t est continue et positive sur ]0, +∞[.
0
1
Etude en +∞. D’après un théorème de croissances comparées, t2 × tx−1 e−t → 0 et donc tx−1 e−t = o . On
t→ +∞ t→ +∞ t2
en déduit que la fonction f est intégrable sur un voisinage de +∞.
Etude en 0. tx−1 e−t ∼ tx−1 et donc la fonction f est intégrable sur un voisinage de 0 si et seulement si x − 1 > −1
t→ +∞
ce qui équivaut à x > 0.
Finalement, γ(x) existe si et seulement si x > 0.
Z +∞
∀x > 0, Γ (x) = tx−1 e−t dt.
0
2) Relation fonctionnelle.
Soit x > 0. Soient a et A deux réels tels que 0 < a < A. Les deux fonctions t 7→ tx et t 7→ −e−t sont de classe C1 sur le
segment [a, A]. On peut donc effectuer une intégration par parties et on obtient
ZA ZA ZA
x −t
A
dt = −tx e−t a + x x−1 −t x −A x −a
tx−1 e−t dt
t e t e dt = −A e +a e +x
a a a
Puisque x > 0 et donc x + 1 > 0, quand a tend vers 0 et A tend vers +∞, on obtient Γ (x + 1) = xΓ (x).
∀x > 0, Γ (x + 1) = xΓ (x).
3) Quelques valeurs.
Z +∞
+∞
e−t dt = −e−t 0 = 1.
• En particulier, pour tout entier naturel n > 2, Γ (n) = (n − 1)Γ (n − 1). De plus, Γ (1) =
0
Par récurrence, on obtient alors
∀n ∈ N∗ , Γ (n) = (n − 1)!.
1
• Calculons aussi Γ . On pose u = t et donc t = u2 et dt = 2u du et on obtient
2
Z +∞ −t Z +∞ −u2 Z +∞
1 e e 2 √
Γ = √ dt = 2udu = 2 e−u du = π (intégrale de Gauss).
2 0 t 0 u 0
1 √
Γ = π.
2
1 1 1
La relation fonctionnelle du 2) permet encore d’écrire : ∀n ∈ N∗ , Γ n + = n− Γ n− et donc pour n ∈ N∗ ,
2 2 2
√
1 2n − 1 2n − 3 1 1 (2n) × (2n − 1) × . . . × 3 × 2 √ (2n)! π
Γ n+ = × × ...× × Γ = n π = 2n ,
2 2 2 2 2 2 (2n) × (2n − 2) × . . . × 2 2 n!
ce qui reste vrai quand n = 0.
√
1 (2n)! π
∀n ∈ N, Γ n + = 2n .
2 2 n!
4) Continuité. Soit a et A deux réels tels que 0 < a < A. Soit Φ : [a, A]×]0, +∞[ → R .
(x, t) 7→ tx−1 e−t
• Pour chaque x ∈ [a, A], la fonction t 7→ Φ(x, t) est continue par morceaux sur ]0, +∞[,
• pour chaque t ∈]0, +∞[, la fonction x 7→ Φ(x, t) est continue sur [a, A],
• Soit (x, t) ∈ [a, A]×]0, +∞[. Si 0 < t 6 1, alors |tx−1 e−t | = tx−1 e−t 6 ta−1 e−t et si t > 1, |tx−1 e−t | 6 tA−1 e−t . On en
déduit que
∀(x, t) ∈ [a, A]×]0, +∞[, |Φ(x, t)| 6 ta−1 e−t + tA−1 e−t = ϕ0 (t).
http ://www.maths-france.fr 1 c Jean-Louis Rouget, 2008. Tous droits réservés.
D’après le 1), la fonction ϕ0 est continue par morceaux et intégrable sur ]0, +∞[ en tant que somme de deux fonctions
continues par morceaux et intégrables sur ]0, +∞[.
D’après le théorème de continuité des intégrales à paramètres, la fonction Γ est continue sur [a, A]. Ceci étant vrai pour
tous réels a et A tels que 0 < a < A, on a montré que
La fonction Γ est continue sur ]0, +∞[.
5) Dérivation.
a) Dérivée première. On reprend les notations de 4).
• Pour chaque x de [a, A], la fonction t 7→ Φ(x, t) est continue par morceaux et intégrable sur ]0, +∞[.
• La fonction φ admet sur [a, A]×]0, +∞[ une dérivée partielle par rapport à sa première variable x définie par
∂Φ
∀(x, t) ∈ [a, A]×]0, +∞[, (x, t) = (ln t)tx−1 e−t .
∂x
De plus,
∂Φ
- pour chaque x de [a, A], la fonction t 7→ (x, t) est continue par morceaux sur ]0, +∞[,
∂x
∂Φ
- pour chaque t ∈]0, +∞[, la fonction x 7→ (x, t) est continue sur [a, A],
∂x
∂Φ
- pour chaque (x, t) ∈ [a, A]×]0, +∞[, (x, t) 6 (ln t)(ta−1 + tA−1 )e−t = ϕ1 (t).
∂x
Vérifions alors l’intégrabilité de la fonction ϕ1 sur ]0, +∞[. Pour cela, pour α > 0 donné, vérifions l’intégrabilité de
la fonction t 7→ (ln t)tα−1 e−t sur ]0, +∞[. Cette fonction est
∗ continue par morceaux sur ]0, +∞[,
1
∗ négligeable en +∞ devant 2 d’après un théorème de croissances comparées,
t
α α α
∗ négligeable en 0 devant t−1+ 2 avec −1 + > −1 car t1− 2 × (ln t)tα−1 e−t ∼ tα/2 (ln t) → 0 d’après un
2 t→ 0 t→ 0
théorème de croissances comparées.
On en déduit que la fonction t 7→ (ln t)tα−1 e−t est intégrable sur ]0, +∞[ et il en est de même de la fonction ϕ1 .
D’après le théorème de dérivation des intégrales à paramètres (théorème de Leibniz), la fonction Γ est de classe C1 sur
[a, A] et sa dérivée s’obtient par dérivation sous le signe somme. Ceci étant vrai pour tous réels a et A tels que 0 < a < A,
on a montré que
Z +∞
La fonction Γ est de classe C1 sur ]0, +∞[ et ∀x > 0, Γ ′ (x) = (ln t)tx−1 e−t dt.
0
b) Dérivées successives.
• Pour chaque x de [a, A], la fonction t 7→ Φ(x, t) est continue par morceaux et intégrable sur ]0, +∞[.
• La fonction φ admet sur [a, A]×]0, +∞[ des dérivées partielles à tout ordre par rapport à sa première variable x définies
par
∂k Φ
∀k ∈ N∗ , ∀(x, t) ∈ [a, A]×]0, +∞[, (x, t) = (ln t)k tx−1 e−t .
∂xk
De plus, pour chaque k ∈ N∗ ,
∂k Φ
- pour chaque x de [a, A], la fonction t 7→ (x, t) est continue par morceaux sur ]0, +∞[,
∂xk
k
∂ Φ
- pour chaque t ∈]0, +∞[, la fonction x 7→ (x, t) est continue sur [a, A],
k ∂xk
∂ Φ
- pour chaque (x, t) ∈ [a, A]×]0, +∞[, k (x, t) 6 (ln t)(ta−1 + tA−1 )e−t = ϕ1 (t).
∂x
Enfin, les fonctions ϕk , k ∈ N∗ , sont intégrables sur ]0, +∞[ pour les mêmes raisons que la fonction ϕ1 .
D’après une généralisation du théorème de dérivation des intégrales à paramètres, la fonction Γ est de classe C∞ sur [a, A]
et ses dérivées successives s’obtiennent par dérivation sous le signe somme. Ceci étant vrai pour tous réels a et A tels que
0 < a < A, on a montré que
Z +∞
La fonction Γ est de classe C∞ sur ]0, +∞[ et ∀k ∈ N∗ , ∀x > 0, Γ (k) (x) = (ln t)k tx−1 e−t dt.
0
6) Convexité.
Z +∞
D’après 5), la fonction Γ est deux fois dérivable sur ]0, +∞[ et ∀x > 0, Γ ′′ (x) = (ln t)2 tx−1 e−t dt > 0 (intégrable
0
d’une fonction continue positive et non nulle).
http ://www.maths-france.fr 2 c Jean-Louis Rouget, 2008. Tous droits réservés.
Donc
La fonction Γ est strictement convexe sur ]0, +∞[.
7) Variations.
Puisque la fonction Γ ′′ est strictement positive sur ]0, +∞[, la fonction Γ ′ est strictement croissante sur ]0, +∞[. De plus,
- la fonction Γ est continue sur [1, 2],
- la fonction Γ est dérivable sur ]1, 2[,
- Γ (1) = Γ (2) = 1,
et le théorème de Rolle permet d’affirmer qu’il existe x0 ∈]1, 2[ tel que Γ ′ (x0 ) = 0. Puisque la fonction Γ ′ est strictement
croissante sur ]0, +∞[, la fonction Γ ′ est strictement négative sur ]0, x0 [ et strictement positive sur ]x0 , +∞[. On a montré
que
∃x0 ∈]1, 2[/ la fonction Γ est strictement décroissante sur ]0, x0 ] et strictement croissante sur [x0 , +∞[.
8) Etude en +∞.
Puisque la fonction Γ est croissante sur 52, +∞[, pour x > 3, Γ (x) = (x − 1)Γ (x − 1) > (x − 1)Γ (2) = x − 1 et on en déduit
que lim Γ (x) = +∞.
x→ +∞
Γ (x) x−1
De plus, pour x > 1, = Γ (x − 1) → +∞. On en déduit que la courbe représentative de la fonction Γ admet
x x x→ +∞
en +∞ une branche parabolique de direction (Oy).
Γ (x)
lim Γ (x) = +∞ et lim = +∞.
x→ +∞ x→ +∞ x
9) Etude en 0.
Pour x > 0, xΓ (x) = Γ (x + 1) ⇒ Γ (1) = 1 par continuité de la fonction Γ en 1. Donc
x→ 0
1
lim Γ (x) = 0 et de plus Γ (x) ∼ .
x→ 0+ x→ 0+ x
10) Graphe.
6 b
5
(x)
y=Γ
2 b
b
1 b b
1 2 3 4
http ://www.maths-france.fr 3 c Jean-Louis Rouget, 2008. Tous droits réservés.
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282,650,881,091,895,550 | Very-Short-Answer Questions
Question:
Very-Short-Answer Questions
If one root of the quadratic equation $3 x^{2}-10 x+k=0$ is reciprocal of the other, find the value of $k$.
Solution:
Let $\alpha$ and $\beta$ be the roots of the equation $3 x^{2}-10 x+k=0$.
$\therefore \alpha=\frac{1}{\beta}$ (Given)
$\Rightarrow \alpha \beta=1$
$\Rightarrow \frac{k}{3}=1 \quad$ (Product of the roots $=\frac{c}{a}$ )
$\Rightarrow k=3$
Hence, the value of k is 3.
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-3,096,350,405,935,672,300 | Polar or Distance Form of a Straight Line Equation
Go back to 'Straight Lines'
\(\textbf{Art 10 :} \qquad \boxed{{\text{Polar / Distance form of a line}}}\)
Sometimes, it is very convenient to write the equation of a straight line in polar / distance form. Suppose we know that the line passes through the fixed point \(P(h,\,k)\) and is at an inclination of \(\theta :\)
For any point \(Q(x,\,y)\) at a distance r from along this line, we can write the simple relation
\[\boxed{{\frac{{x - h}}{{\cos \theta }} = \frac{{y - k}}{{\sin \theta }} = r}}\]
This is the required equation of the line. The point \(Q(x,\,y),\) at a distance r from P, has the coordinates
\[Q(x,\,y) \equiv (h + r\cos \theta ,\,k + r\sin \theta ).\]
\[\left\{ \begin{array}{l}{\text{Obviously, there will be another point, say }}Q'(x,y),{\text{ at a distance }}r{\text{ from }}P\\{\text{along this line but on the opposite side of }}Q{\text{; thus }}Q'(x,{\rm{ }}y){\text{
will have the }}\\{\text{coordinates }}Q'(x,\,y) \equiv (h - r\cos \theta ,\,\,\,k - r\sin \theta)\end{array} \right\}\]
Example – 15
A line through \(A( - 5,\, - 4)\) meets the lines \(x + 3y = 2,\,\,2x + y + 4 = 0\) and \(x - y - 5 = 0\) at the points B, C and D respectively. If
\[\begin{align}{\left( {\frac{{15}}{{AB}}} \right)^2} + {\left( {\frac{{10}}{{AC}}} \right)^2} = {\left( {\frac{6}{{AD}}} \right)^2},\end{align}\]
find the equation of the line.
Solution:
The figure above roughly sketches the situation described in the equation. Let B, C and D be at distances \({r_1},\,{r_2}\) and \({r_3}\) from A along the line \(L = 0,\) whose equation we wish to determine. Assume the inclination of L to be \(\theta .\) Thus, B, C and D have the coordinates (respectively):
\[\begin{align}B \equiv ( - 5 + {r_1}\cos \theta ,\,\,\,\,\, - 4 + {r_1}\sin \theta )\\C \equiv ( - 5 + {r_2}\cos \theta ,\,\,\,\,\, - 4 + {r_2}\sin \theta )\\D \equiv ( - 5 + {r_3}\cos \theta ,\,\,\,\,\, - 4 + {r_3}\sin \theta )\end{align}\]
Since these three points(respectively) satisfy the three given equations, we have :
Point B\(\begin{align}( - 5 + {r_1}\cos \theta ) + 3( - 4 + {r_1}\sin \theta ) + 2 = 0 \quad \Rightarrow \qquad {r_1} = \frac{{15}}{{\cos \theta + 3\sin \theta }}\end{align}\)
Point C : \(\begin{align}2( - 5 + {r_2}\cos \theta ) + ( - 4 + {r_2}\sin \theta ) + 4 = 0 \quad \Rightarrow \qquad {r_2} = \frac{{10}}{{2\cos \theta + \sin \theta }}\end{align}\)
Point D : \(\begin{align}( - 5 + {r_3}\cos \theta ) - ( - 4 + {r_3}\sin \theta ) - 5 = 0 \quad \Rightarrow \qquad {r_3} = \frac{6}{{\cos \theta - \sin \theta }}\end{align}\)
It is given that
\[\begin{align}&{\left( {\frac{{15}}{{AB}}} \right)^2} + {\left( {\frac{{10}}{{AC}}} \right)^2} = {\left( {\frac{6}{{AD}}} \right)^2}\\
\text{i.e;}\qquad \qquad \qquad &{\left( {\frac{{15}}{{{r_1}}}} \right)^2} + {\left( {\frac{{10}}{{{r_2}}}} \right)^2} = {\left( {\frac{6}{{{r_3}}}} \right)^2}\\ \Rightarrow \qquad &{(\cos \theta + 3\sin \theta )^2} + {(2\cos \theta + \sin \theta )^2} = {(\cos \theta - \sin \theta )^2}\\ \Rightarrow \qquad &4{\cos ^2}\theta + 9{\sin ^2}\theta + 12\sin \theta \cos \theta = 0\\ \Rightarrow \qquad &{(2\cos \theta + 3\sin \theta )^2} = 0\\ \Rightarrow \qquad &\tan \theta = \frac{{ - 2}}{3}\\ \Rightarrow \qquad &m = \frac{{ - 2}}{3}\end{align}\]
Thus, we obtain the slope of as \(\begin{align}\frac{{ - 2}}{3}. \end{align}\) The equation of L can now be easily written :
\[\begin{align}&L:y - ( - 4) = \frac{{ - 2}}{3}(x - ( - 5))\\ \Rightarrow \qquad &L:2x + 3y + 22 = 0\end{align}\]
TRY YOURSELF - I
Q1. A variable straight line drawn through the intersection of the lines\(\begin{align}\frac{x}{a} + \frac{y}{b} = 1\;\;and\;\;\frac{x}{b} + \frac{y}{a} = 1\end{align}\) and meets the axes in A and B. Show that the locus of the mid-point of AB is \(2xy(a + b) = ab(x + y)\)
Q2. The line \(bx + ay = ab\) cuts the axes in A and B. Another variable line cuts the axes in C and D such that \(OA + OB = OC + OD\) where O is the origin. Prove that the locus of the point of intersection of the lines AD and BC is the line\(\;x + y = a + b\)
Q3. A point P moves so that the square of its distance from (3, –2) is equal to its distance from the line \(5x - 12y = 13\) Find the locus of P.
Q4. A line intersects the x-axis in A(7, 0) and the y-axis in B(0, –5). A variable line perpendicular to AB intersects the x-axis in P and the y-axis in Q. If AQ and BP intersect in R, find the locus of R.
Q5. If the sum of the distances of a point from two perpendicular lines in a plane is 1, prove that its locus is a square.
Q6. A vertex of an equilateral triangle is (2, 3) and the opposite side is \(\;x + y = 2\). Find the equations of the other sides.
Q7. A ray of light along the line \(x - 2y - 3 = 0\) is incident upon the mirror-line \(3x - 2y - 5 = 0\) Find the equation of the reflected ray.
Q8. If the vertices of a triangle have integral coordinates, show that it cannot be equilateral.
Q9. Show using coordinate geometry that the angle bisectors of the sides of a triangle are concurrent.
Q10. The sides of a triangle are \(4x + 3y + 7 = 0\;,\;5x + 12y - 27 = 0\;\;and\;\;3x + 4y + 8 = 0\) and By explicitly evaluating the medians in this triangle, show that they are concurrent.
Q11. A rod APB of constant length meets the axes in A and B. If AP = b and PB = a and the rod slides between the axes, show that the locus of P is\(\;{b^2}{x^2} + {a^2}{y^2} = {a^2}{b^2}\)
Q12. If p is the length of the perpendicular from the origin to the line whose intercepts on the axes are a and b, show that \(\begin{align}\frac{1}{{{p^2}}} = \frac{1}{{{a^2}}} + \frac{1}{{{b^2}}}\end{align}\)
Q13. The lines \(3x + 4y - 8 = 0\;and\;5x + 12y + 3 = 0\) intersect in A. Find the equations of the lines passing through which intersect the given lines at B and C, such that \(AB = AC\).
Q14. The equal sides AB and AC of an isosceles triangle ABC are produced to the points P and Q such that \(BP.CQ = A{B^2}\) Prove that the line PQ always passes through a fixed point.
Q15. One side of a square is inclined to the x-axis at an angle and one of its extremities is at the origin; prove that the equations to its diagonals are
\[\begin{array}{l}
&y(\cos \alpha - \sin \alpha ) = x(\sin \alpha + \cos \alpha )\\\\
and\qquad &y(\sin \alpha + \cos \alpha ) + x(\cos \alpha - \sin \alpha ) = a
\end{array}\]
where a is the length of the side of the sqaure.
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-8,163,465,306,737,080,000 | 902 Percent As A Decimal
Let's find how to write 902% as a decimal.
Answer : 902% is equal to 9.02
How to convert 902% to decimal number form
Let's see how do you write 902% as a decimal.
Converting a percentage to a decimal is a simple process. Let's break it down step by step:
Step 1: Understand that the word "percent" means "per hundred." So, 902% is the same as 902 per hundred.
Step 2: To convert a percentage to a decimal, you divide it by 100.
Step 3: For 902%, you can write it as a fraction: 902/100.
Step 4: Now, simplify the fraction if possible. In this case, both 902 and 100 can be divided by 100, resulting in 9.02/1.
Step 5: The final step is to write the simplified fraction as a decimal. In this case, 9.02/1 is equivalent to 9.02 .
So, 902% as a decimal is 9.02.
Question :Answer :
Express 902 out of 100 as a decimal.9.02
Convert the percentage 902% to its decimal form.9.02
What is the decimal representation of 902 percent?9.02
If you have 902 parts out of a total of 100, what is the corresponding decimal?9.02
Write 902% as a decimal.9.02
If you divide 902 by 100, what decimal do you get?9.02
Determine the decimal equivalent of the fraction 902/100.9.02
When you convert 902 per hundred to decimal form, what is the result?9.02
Percentage To Decimal Number Converter :
%
Make new percentage calculation from decimal number to percentage.
Let's learn how to write fraction as a percent, decimal as a percent and percentage to decimal.
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-1,770,523,372,671,653,600 | Factoring Examples
Video Transcript
Let's factor: $$x^2+\textcolor{#800080}{5}x+\textcolor{#5cb85c}{4}$$ Factoring means we want something like: $$(x+\textcolor{#d9534f}{\_})(x+\textcolor{#2d6da3}{\_})$$
What numbers go in the blanks?
Let's put a and b in the blanks and use FOIL to multiply it out. $$(x+\textcolor{#d9534f}{a})(x+\textcolor{#2d6da3}{b})$$ $$=x^2+\textcolor{#2d6da3}{b}x+\textcolor{#d9534f}{a}x+\textcolor{#d9534f}{a}\textcolor{#2d6da3}{b}$$ $$=x^2+\textcolor{#800080}{(a+b)}x+\textcolor{#5cb85c}{ab}$$
We see that a and b need to be two numbers that...
Multiply together to get 4: $$\textcolor{#5cb85c}{ab=4}$$
And add together to get 5: $$\textcolor{#800080}{a+b=5}$$
Can you think of the two numbers?
Let's think of pairs of numbers that multiply together to get 4. $$1\textcolor{#5cb85c}{*}4=\textcolor{#5cb85c}{4}$$ $$2\textcolor{#5cb85c}{*}2=\textcolor{#5cb85c}{4}$$
For each pair. let's add the two numbers together to see which pair adds up to 5.
We see that: $$1\textcolor{#800080}{+}4=\textcolor{#800080}{5}$$
So 1 and 4 are the numbers we were looking for.
Let's go back and fill in the blanks with 1 and 4 to get... $$(x+\textcolor{#d9534f}{1})(x+\textcolor{#2d6da3}{4})$$ Let's check our answer using FOIL to multiply it out. We get: $$(x+\textcolor{#d9534f}{1})(x+\textcolor{#2d6da3}{4})$$ $$=x^2+\textcolor{#2d6da3}{4}x+\textcolor{#d9534f}{1}x+(\textcolor{#d9534f}{1})(\textcolor{#2d6da3}{4})$$ $$=x^2+5x+4$$ so we know our answer is correct.
Answer
Our final answer is: $$(x+\textcolor{#d9534f}{1})(x+\textcolor{#2d6da3}{4})$$
Calculator Examples
Here are more examples of how to factor expressions in the Factoring Calculator. Feel free to try them now.
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8,289,272,711,100,581,000 | GRE Math : How to find a ratio
Study concepts, example questions & explanations for GRE Math
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Example Questions
← Previous 1
Example Question #1 : How To Find A Ratio
1 : 1
2 : 3
3 : 4
1 : 3
There are 28 students in a room. The ratio of boys to girls cannot be which of the above.
Possible Answers:
2 : 3
1 : 1
1 : 3
3 : 4
Correct answer:
2 : 3
Explanation:
When selecting ratios for two variables (boys and girls) the two sides of the ratio must add up to be a factor of the total student count. The factors of 28 include 14, 7, 4, and 2. (1 + 1 = 2), (2 + 3 = 5), (3 + 4 = 7), and (1 + 3 = 4). 5 is the only nonfactor and cannot be the ratio of boys to girls, thus making 2 : 3 the correct answer.
Example Question #1 : How To Find A Ratio
For All Sweets Bakery, the daily sales ratio of bread to cakes is 5:2. If the bakery sells 12 more loaves of bread on Tuesday than its daily sale of 40 loaves, how many cakes were sold on Tuesday? (round to the nearest integer)
Possible Answers:
Correct answer:
Explanation:
Since 40 loaves of bread are sold daily, and the ratio of bread to cakes is 5:2, then cakes are sold daily.
Using the ratio in the same way, we can find the additional amount of cakes sold:
cakes with approximation.
Thus, the total amount of cakes sold on Tuesday is cakes.
Example Question #3 : How To Find A Ratio
You are making a cake that requires, by volume, three times as much flour as sugar, twice as much sugar as milk, eight times more milk than baking powder and twice as much baking powder as salt. If you start with a teaspoon of salt, how many cups of flour do you need (there are 48 teaspoons in one cup)?
Possible Answers:
\dpi{100} \small 48
\dpi{100} \small 8
\dpi{100} \small 32
\dpi{100} \small 96
\dpi{100} \small 2
Correct answer:
\dpi{100} \small 2
Explanation:
One teaspoon of salt requires 2 teaspoons of baking powder, which requires 16 teaspoons of milk and 32 teaspoons of sugar. 32 teaspoons of sugar requires 96 teaspoons of flour, which equals two cups of flour.
Example Question #1 : How To Find A Ratio
A bakery stocks 3 cookies for every 2 cupcakes and 6 pastries for every 5 cookies. What is the ratio of cupcakes to pastries?
Possible Answers:
Correct answer:
Explanation:
First, you have to set up the given ratios, which is 3 cookies : 2 cupcakes and 5 cookies : 6 pastries. Then, you find a common multiple of cookies (i.e. 15) and convert the ratios to 15 cookies : 10 cupcakes and 15 cookies : 18 pastries. Since both ratios now have 15 cookies, you can infer that the ration of cupcakes to pastries is 10:18 or 5:9.
Example Question #5 : How To Find A Ratio
The ratio of male students to female students in a class is 13 to 19. If there are 224 people in the class, including one teacher, one administrator, and thirty evaluators, how many people in the class are male students?
Possible Answers:
133
80
78
91
114
Correct answer:
78
Explanation:
Begin by eliminating people in the class who are not students. Subtracting 1 teacher, 1 administrator and 30 evaluators leaves 192 students.
We also need to determine the ratio of male students to total students from the information in the question. Out of every thirty-two students, thirteen are male.
Now we can set up a proportion, and solve for the number of male students.
Example Question #6 : How To Find A Ratio
If the length of a rectangle is increased by 50% and its width is decreased by 20%, what is the ratio of the area of the new rectangle to the original rectangle?
Possible Answers:
Correct answer:
Explanation:
First, pick a number for the original length and width. To make it easy, you can choose a length of 1 and width of 1, which would give it an area of 1. If we increased the length by 50% and decreased the width by 20%, then the dimensions of the new rectangle would be , which would give it an area of 1.2. Thus, the ratio of the new rectangle to the original rectangle is 6:5
Example Question #1 : How To Find A Ratio
There are 3,500 people in group A and 5,000 people in group B:
Car Type
% in Group A Who Own
% in Group B Who Own
Motorbike
4
9
Sedan
35
25
Minivan
22
15
Van
9
12
Coupe
3
6
What is the ratio of the number of people in group A with motorbikes and vans to the total number of people across both groups with motorbikes and vans?
Possible Answers:
Correct answer:
Explanation:
First find the total number that own both motorbikes and vans of each group separately:
Group A:
Group B:
Now we have a ratio that should be:
Example Question #1 : How To Find A Ratio
A dessert is made using 2 parts cake and 3 parts icing. The cake contains 4 parts sugar, 5 parts milk, and 11 parts of other ingredients. The icing contains 3 parts sugar, 2 parts milk, and 15 parts of other ingredients. Which quantity is greater?
Quantity A: Parts sugar in the dessert
Quantity B: Parts milk in the dessert
Possible Answers:
The two quantities are equal.
Quantity B is greater.
The relationship cannot be determined from the information given.
Quantity A is greater.
Correct answer:
Quantity A is greater.
Explanation:
Quantity A: To determine the parts of sugar in the dessert, we use the following process. Let's first figure out the amount of sugar in the cake. This is 4/20. Next, find the amount of sugar in the icing topping: 3/20. Then, we need to account for the amount of cake and icing in the dessert. Using the fact that there are 2 parts cake and 3 parts icing, we can say that 2/5 of the dessert is cake and 3/5 is icing. Combining this information with the amount of sugar in both the cake and the icing, we obtain: 2/5 * 4/20 + 3/5 * 3/20 = 17/100. So, there are 17 parts of sugar in the dessert.
Quantity B: Use the same method to find the amount of milk: 2/5 * 5/20 + 3/5 * 2/20 = 16/100. So there are 16 parts milk in the dessert. Thus, Quantity A is larger.
Example Question #1 : How To Find A Ratio
Gre9
The ratio of the number of financial employees who remained in the same role for 2 to 9 years to the number of construction employees who remained in the same role for 0 to 4 years is closest to which of the following?
Possible Answers:
Correct answer:
Explanation:
For this problem, we need to find the number of employees who fall into the categories described, keeping in mind that multiple portions of the pie chart must be accommodated for. Then, we can fit them into a ratio:
For the "2 to 9 years" portion of the financial industry, include
(0.2 + 0.18)(12,000,000) = 4,560,000 workers.
For the "0 to 4 years" portion of the construction industry, include
(0.15 + 0.2)(8,000,000) = 2,800,000 workers.
Now divide and simplify to find the ratio:
4,560,000/2,800,000 = 8/5.
Example Question #12 : Proportion / Ratio / Rate
The ratio of to is to , while the ratio of to is to .
What is the ratio of to ?
Possible Answers:
Correct answer:
Explanation:
Since the ratios are fixed, regardless of the actual values of , , or , we can let and
In order to convert to a form where we can relate to , we must set the coefficient of of each ratio equal such that the ratio can be transferred. This is done most easily by finding a common multiple of and (the ratio of to and , respectively) which is
Thus, we now have and .
Setting the values equal, we get , or a ratio of
← Previous 1
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For a positive integer k written in decimal notation, #(k)
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For a positive integer k written in decimal notation, #(k) [#permalink] New post 27 Sep 2006, 01:35
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For a positive integer k written in decimal notation, #(k) is defined as the product of the digits of k. For instance, #(333)=27 and #(142)=8. If Y is the set of four digit whole numbers w such that #(w)=60, what percent of the elements of Y are greater than 5000?
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[#permalink] New post 27 Sep 2006, 02:11
25%
The four digist are the factors of 60..so they could be (5322) or (1652) or (1453).
The numbers in each case that are greater than 5000 are 3 of 12, 4 of 24 and 4 of 24...so total 15 of 60 or 25%
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[#permalink] New post 27 Sep 2006, 21:38
i am getting 21/60
numbers could be permutation of 5322 or 1652 or 1453.
total numbers = !3/!2 + 2*3*2*1 + !3 = 21
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[#permalink] New post 27 Sep 2006, 22:02
10/21
Permutations of 4351 2651 , 2353 = 3 * 4!
Total possible combinations = 9P4 (we have to ignore 0)
3*4!/9P4 * 100 = 10/21
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[#permalink] New post 27 Sep 2006, 22:39
Solution please :o I thought 25% as well.
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[#permalink] New post 27 Sep 2006, 22:59
lanuk wrote:
25%
The four digist are the factors of 60..so they could be (5322) or (1652) or (1453).
The numbers in each case that are greater than 5000 are 3 of 12, 4 of 24 and 4 of 24...so total 15 of 60 or 25%
Hey lanuk,
it must be 3 of 12, 6 of 24 and 6 of 24.........so total 15 out of 60 ie 25%.
I too got the same answer. So let's wait for the OA
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[#permalink] New post 27 Sep 2006, 23:43
kevincan wrote:
cicerone wrote:
lanuk wrote:
25%
The four digist are the factors of 60..so they could be (5322) or (1652) or (1453).
The numbers in each case that are greater than 5000 are 3 of 12, 4 of 24 and 4 of 24...so total 15 of 60 or 25%
Hey lanuk,
it must be 3 of 12, [color=red]6
of 24 and 6 of 24.........so total 15 out of 60 ie 25%.
I too got the same answer. So let's wait for the OA[/color]
Why only 6 of 24?
Hey kevin i got it...............
it is 21/60 ie 35%
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[#permalink] New post 28 Sep 2006, 01:12
Yup Cicerone..you're right..should be 21/60...
3 +12(Both starting with 5 and 6 are greater than 5000) +6
Silly mistake :oops:
[#permalink] 28 Sep 2006, 01:12
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Finding Probability Distribution Parameters from Percentiles
, 26 Jun 2014
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How to determine the parameters of a probability distribution given two percentile constraints
Introduction
Suppose there's a 10% chance of something being less than 30 and a 90% chance of it being less than 60. Once you pick a probability distribution family (normal, gamma, etc.) you need a way of determining what parameters will satisfy your two requirements.
More precisely, suppose a random variable X has a two-parameter distribution. The problem is to find values of those parameters so that Pr(X < x1) = p1 and Pr(X < x2) = p2.
This article will show how to compute these parameters for normal, Cauchy, Weibull, gamma, and inverse gamma distributions using Python's SciPy library.
The problem of finding parameters to satisfy two percentile equations is practical. It comes up, for example, in determining prior distributions in Bayesian statistics. This article also serves as a brief introduction to using SciPy.
Background
The mathematical solution to the problem posed here is described in the technical report Determining distribution parameters from quantiles. Here we explain how to implement in Python the calculations in that report. Here is an article giving more motivation for the problem.
Using the Code
The code described here is very simple to call. Suppose you want to find the mean and standard deviation for a normal distribution. You simply call normal_parameters with the appropriate arguments. It returns the mean and standard deviation as a pair.
(mean, stdev) = normal_parameters(x1, p1, x2, p2)
The functions for other distributions are similar:
• cauchy_parameters
• weibull_parameters
• gamma_parameters
• inverse_gamma_parameters
• beta_parameters
All functions take the same four arguments and all return two parameters. However, the meaning of the parameters is different for each distribution. For the normal distribution, the problem is finding μ and σ so that F(x1) = p1 and F(x2) = p2 where F(x) is defined by:
F(x) = \frac{1}{\sqrt{2\pi}\sigma} \int_{-\infty}^x \exp\left( -\frac{(t-\mu)^2}{2\sigma^2}\right) \, dt
For the Cauchy distribution, the problem is similar except now:
F(x) = \frac{1}{\pi\sigma} \int_{-\infty}^x \frac{dt}{ 1 + \left(\frac{t-\mu}{\sigma} \right)^2 }}
For the Weibull distribution, the problem is finding γ and β where:
F(x) = \frac{\gamma}{\beta^\gamma} \int_0^x t^{\gamma-1} \exp\left(-(t/\beta)^\gamma \right)\, dt
For the gamma distribution, the problem is finding α and β where:
F(x) = \frac{1}{\Gamma(\alpha)\,\beta^\alpha} \int_0^x t^{\alpha-1} \exp(-x/\beta)\, dt
For the inverse gamma distribution, the problem is finding α and β where:
F(x) = \frac{\beta^\alpha}{\Gamma(\alpha)} \int_0^x t^{-\alpha-1}\exp(-\beta/t)\, dt
For the beta distribution, the problem is finding a and b where:
F(x) = \frac{\Gamma(a+b)}{\Gamma(a)\,\Gamma(b)} \int_0^x t^{a-1}(1-t)^{b-1}\, dt
Looking Inside the Code
The code presented here depends critically on SciPy, a large library for scientific computing in Python. For an introduction to SciPy, see the CodeProject article Getting started with the SciPy (Scientific Python) library.
For the normal and Cauchy distributions, the location parameter is given by:
\frac{x_1 F^{-1}(p_2) - x_2 F^{-1}(p_1)}{F^{-1}(p_2) - F^{-1}(p_1)}
and the scale parameter is given by:
\frac{x_2 - x_1}{F^{-1}(p_2) - F^{-1}(p_1)}
where F(x) is the CDF of the normal or Cauchy distribution as in the previous section. The inverse of the CDF is given by the ppt method in SciPy. Here "ppt" stands for "percentile point function." Other libraries may call this the quantile function. The probability distribution classes are located in scipy.stats.
The parameters for the Weibull distribution can be given by a simple formula not requiring any SciPy functionality. The inverse gamma parameters are also easy to find since the inverse gamma problem can be reduced to the problem of finding parameters for the gamma distribution.
The gamma distribution parameters cannot be obtained so simply. The code to find these parameters illustrates SciPy and the functools module.
def gamma_parameters(x1, p1, x2, p2):
# Standardize so that x1 < x2 and p1 < p2
if p1 > p2:
(p1, p2) = (p2, p1)
(x1, x2) = (x2, x1)
# function to find roots of for gamma distribution parameters
def objective(alpha):
return stats.gamma.ppf(p2, alpha) / stats.gamma.ppf(p1, alpha) - x2/x1
# The objective function we're wanting to find a root of is decreasing.
# We need to find an interval over which is goes from positive to negative.
left = right = 1.0
while objective(left) < 0.0:
left /= 2
while objective(right) > 0.0:
right *= 2
alpha = optimize.bisect(objective, left, right)
beta = x1 / stats.gamma.ppf(p1, alpha)
return (alpha, beta)
Notice the auxiliary function objective defined inside the gamma_parameters function. A nice feature of Python is that you can easily define little functions like this using closures. Only the gamma_parameters needs this function so we define it inside gamma_parameters to keep from cluttering the outer scope.
To find where the function objective is 0, we use one of the root-finding methods in the scipy.optimize. The code here uses the bisect method, an algorithm that is slow compared to other root-finding algorithms but is very reliable.
For the beta distribution, we have to direct method for solving for the distribution parameters. Instead we solve an optimization problem using the SciPy function fmin. We search for the parameters which come closest to satisfying the percentile constraints. In fact these parameters will satisfy the constraints exactly.
def beta_parameters(x1, p1, x2, p2):
"Find parameters for a beta random variable X
so that P(X > x1) = p1 and P(X > x2) = p2."
def square(x):
return x*x
def objective(v):
(a, b) = v
temp = square( stats.beta.cdf(x1, a, b) - p1 )
temp += square( stats.beta.cdf(x2, a, b) - p2 )
return temp
# arbitrary initial guess of (3, 3) for parameters
xopt = optimize.fmin(objective, (3, 3))
return (xopt[0], xopt[1])
The approach used to calculate the beta distribution parameters could be used for other distributions as well. In fact, we could have used this approach for all distributions. However, the specialized methods used for the other distributions are more accurate and more efficient.
History
• 3rd February, 2010: Initial version
• 4th February, 2010: Article updated
• 26th July, 2010: Beta distribution added
License
This article, along with any associated source code and files, is licensed under A Public Domain dedication
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About the Author
John D. Cook
United States United States
I am an independent consultant in software development and applied mathematics. I help companies learn from their data to make better decisions.
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Comments and Discussions
QuestionLooking for a C# version Pinmembervictorbos21-Oct-11 4:15
Greetings John.
Given your impressive expertise on this subject, I thought you might have answers/suggestions for two problems I am grappling with:
1) Where might I find C# equivalent for this problem, i.e., estimating parameters for a distribution?
2) Where I might find pseudo code (or better yet C# code) for an efficient way two get the distribution of the sum of two discrete "ill-behaved" (i.e., no clear distribution type) real values? Not being a whiz at math or numerical techniques, I have so far tackled it the dumb brute force way, which is very inefficient.)
Any tips/pointers would be greatly appreciated. Thanks in advance.
--VVX
AnswerRe: Looking for a C# version PinmemberJohn D. Cook21-Oct-11 4:36
GeneralRe: Looking for a C# version Pinmembervictorbos21-Oct-11 5:02
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5,881,561,207,375,128,000 | Answers
Solutions by everydaycalculation.com
Answers.everydaycalculation.com » Compare fractions
Compare 70/60 and 3/70
1st number: 1 10/60, 2nd number: 3/70
70/60 is greater than 3/70
Steps for comparing fractions
1. Find the least common denominator or LCM of the two denominators:
LCM of 60 and 70 is 420
Next, find the equivalent fraction of both fractional numbers with denominator 420
2. For the 1st fraction, since 60 × 7 = 420,
70/60 = 70 × 7/60 × 7 = 490/420
3. Likewise, for the 2nd fraction, since 70 × 6 = 420,
3/70 = 3 × 6/70 × 6 = 18/420
4. Since the denominators are now the same, the fraction with the bigger numerator is the greater fraction
5. 490/420 > 18/420 or 70/60 > 3/70
MathStep (Works offline)
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2,530,844,880,918,872,600 |
Results 1 to 9 of 9
Thread: variable in the numerator and denominator
1. #1
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variable in the numerator and denominator
Say I have an equation that goes something like this:
(200.8kg*2+3333kg+15S+78kg*14+.04kg*1000kg+20kg*28)*(9.806m/s^2) / (0.5*(1.22kg/m^3)*(1.6*52m/s)*S)=0.92
How do I go about solving for S?
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2. #2
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Quote Originally Posted by strangiatotheme View Post
Say I have an equation that goes something like this:
(200.8kg*2+3333kg+15S+78kg*14+.04kg*1000kg+20kg*28)*(9.806m/s^2) / (0.5*(1.22kg/m^3)*(1.6*52m/s)*S)=0.92
How do I go about solving for S?
First, simplify where you can: (dropping the units for convenience)
(5426.6 + 15*S)*9.806/(50.752*S) = 0.92
Multiply both sides by 50.752*S/9.806:
5426.6 + 15*S = 0.92*50.752*S/9.806
5426.6 + 15*S = 4.76155823
15*S = -5421.83844177
S = -361.455896118
The unit for S is "kg" since we have to have the same unit for any two terms we add. This also checks when I put "kg" in the denominator for S. Why S is a negative number though, I can't say. Obviously a negative mass makes no sense. What kind of problem is this from?
-Dan
Last edited by topsquark; October 17th 2006 at 04:26 PM. Reason: Missed the g factor!
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3. #3
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Quote Originally Posted by topsquark View Post
Ummm...looking at your units, they don't make any sense. Supposedly S is a mass, since each term in the numerator has to have the same unit (though its being negative I have a hard time explaining). Given that unit for S the units in your denominator seem to claim that the 0.92 has some units tacked onto it but you didn't write them. I'm very confused by what the units are doing. What kind of problem is this from?
-Dan
Whoops, sorry...the .92 is unitless, it's the coefficient of lift. S is surface area. So I'm still a bit confused, after you multiply 50.752*S on both sides, what happes to that S?
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4. #4
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Quote Originally Posted by strangiatotheme View Post
Whoops, sorry...the .92 is unitless, it's the coefficient of lift. S is surface area. So I'm still a bit confused, after you multiply 50.752*S on both sides, what happes to that S?
C***. I dropped the S on the RHS. Let me run this again:
(5426.6 + 15*S)*9.806/(50.752*S) = 0.92
5426.6 + 15*S = 0.92*50.752*S/9.806
5426.6 + 15*S = 4.76155823*S
5426.6 = (4.76155823 - 15)*S
5426.6 = -10.23844177*S
S = -530.022060183
Shoot. I was hoping that would get rid of the negative sign.
Also, the units clearly indicate that S should be a mass, unless the "15" coefficient is supposed to be in kg?
Also, assuming S is in m^2 and the summation part of the numerator is in kg*m^2, when I do the unit division I get that the LHS has units of m^3/s, not unitless. There appears to be something wrong with the units here.
-Dan
Last edited by ThePerfectHacker; October 17th 2006 at 07:20 PM.
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5. #5
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Looking over the assignment sheet again, it looks like I messed up typing the first equation. It should be 1.6*52m/s^2. Also, the 15 should be kg/m^2. It looks like I end up with just m...?
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6. #6
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Quote Originally Posted by strangiatotheme View Post
Looking over the assignment sheet again, it looks like I messed up typing the first equation. It should be 1.6*52m/s^2. Also, the 15 should be kg/m^2. It looks like I end up with just m...?
Now I'm getting:
(kg/m^2)*m^2*m/s^2
(kg/m^3)*(m/s^2)
which is the same as m^3, a volume.
What specifically is the equation you are using? Maybe that will help.
-Dan
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7. #7
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CL=Weight/Lift
where weight is defined as: ME*2+MF+15S+MC+MP+MB
There are 5 different situations. These are the values for 1 of them:
ME=200.8kg
MF=3333kg
MC=400kg
MP=1092kg
MB=560kg
S is in m^2
lift is defined as: 0.5*P*V^2*S
Again, these are the values for the same situation above:
P=1.22kg/m^3
V=1.6(52m/s)
S is in m^2
and CL is 0.92
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8. #8
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Quote Originally Posted by strangiatotheme View Post
CL=Weight/Lift
where weight is defined as: ME*2+MF+15S+MC+MP+MB
There are 5 different situations. These are the values for 1 of them:
ME=200.8kg
MF=3333kg
MC=400kg
MP=1092kg
MB=560kg
S is in m^2
lift is defined as: 0.5*P*V^2*S
Again, these are the values for the same situation above:
P=1.22kg/m^3
V=1.6(52m/s)
S is in m^2
and CL is 0.92
For what you've got here I'm now getting CL in m^2. However I note that you are saying here that you have V^2, which you didn't specify in the initial problem; this will change the numbers, which I'll get to in a minute. I'm looking at your equation and you are calling the denominator the "lift," and the units are in N as they should be. Now, if your "m"s for some reason are areal mass densities (kg/m^2) I note that the numerator is now also N. Is this a possibility? This is the possibility that makes the most sense to me and would give the numerator the interpretation of being the weight that you are trying to lift. (It's the only reason I can think of that you would multiply a "mass" by a surface area.)
(200.8*2+3333+15S+78*14+.04*1000+20*28)*(9.806) / (0.5*(1.22)*[(1.6*52)]^2*S)=0.92
(5426.6 + 15*S)*9.806/(4222.5664*S) = 0.92
(5426.6 + 15*S)/(430.610483378*S) = 0.92
5426.6 + 15*S = 396.161644707*S
5426.6 = 381.161644707*S
S = 14.237004366
which (finally!) is not negative.
-Dan
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9. #9
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Quote Originally Posted by topsquark View Post
(200.8*2+3333+15S+78*14+.04*1000+20*28)*(9.806) / (0.5*(1.22)*[(1.6*52)]^2*S)=0.92
(5426.6 + 15*S)*9.806/(4222.5664*S) = 0.92
(5426.6 + 15*S)/(430.610483378*S) = 0.92
5426.6 + 15*S = 396.161644707*S
5426.6 = 381.161644707*S
S = 14.237004366
which (finally!) is not negative.
-Dan
Oh man do I feel stupid. All I had to do was subtract! Thanks Dan, I'm working out the other 4 problems right now...
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-4,485,885,260,354,565,000 | What is the percentage increase/decrease from 7281 to 1557?
Quickly work out the percentage increase or decrease from 7281 to 1557 in this step-by-step percentage calculator tutorial. (Spoiler alert: it's -78.62%!)
So you want to work out the percentage increase or decrease from 7281 to 1557? Fear not, intrepid math seeker! Today, we will guide you through the calculation so you can figure out how to work out the increase or decrease in any numbers as a percentage. Onwards!
In a rush and just need to know the answer? The percentage decrease from 7281 to 1557 is -78.62%.
What is the % change from to
Percentage increase/decrease from 7281 to 1557?
An increase or decrease percentage of two numbers can be very useful. Let's say you are a shop that sold 7281 t-shirts in January, and then sold 1557 t-shirts in February. What is the percentage increase or decrease there? Knowing the answer allows you to compare and track numbers to look for trends or reasons for the change.
Working out a percentage increase or decrease between two numbers is pretty simple. The resulting number (the second input) is 1557 and what we need to do first is subtract the old number, 7281, from it:
1557 - 7281 = -5724
Once we've done that we need to divide the result, -5724, by the original number, 7281. We do this because we need to compare the difference between the new number and the original:
-5724 / 7281 = -0.78615574783684
We now have our answer in decimal format. How do we get this into percentage format? Multiply -0.78615574783684 by 100? Ding ding ding! We have a winner:
-0.78615574783684 x 100 = -78.62%
We're done! You just successfully calculated the percentage difference from 7281 to 1557. You can now go forth and use this method to work out and calculate the increase/decrease in percentage of any numbers.
Head back to the percentage calculator to work out any more calculations you need to make or be brave and give it a go by hand. Hopefully this article has shown you that it's easier than you might think! | {
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-5,637,663,676,652,201,000 | Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.
I came across the following problem and I just can't solve it.
Suppose that $x_1,...,x_k \in \Bbb N$ and $c_1,...,c_k \in \Bbb N $ are such that $\sum_{j=1}^k c_j x_j$ is a multiple of $\operatorname{lcm}(x_1,...,x_k)$. Show that there exist integers $0 \leq d_j \leq c_j$ such that $\sum_{j=1}^k d_j x_j =\operatorname{lcm}(x_1,...,x_k)$.
share|improve this question
Where did you come across it? – Matthew Conroy Nov 1 '13 at 21:43
A problem in commutative algebra seems to boil down to this question. It is about generators in a graded algebra. I wanted to write down that problem initially, but I was able to solve that with a different(perhaps more standard method). But I am still interested in seeing a solution or counter-example of this seemingly indeoendent question. – Hammerhead Nov 2 '13 at 5:06
It might be possible with $| d_j | \le c_j$. – NovaDenizen Dec 8 '13 at 23:14
@NovaDenizen Interesting; do you have a proof of that? – Post No Bulls Dec 9 '13 at 7:01
@Post No Bills nope. – NovaDenizen Dec 10 '13 at 14:48
1 Answer 1
up vote 2 down vote accepted
+100
The lcm of $\{3,4,15,24,30,40,60\}$ is 120.
$$3+4+15+2\times24+30+2\times40+60=2\times120$$
Now suppose $3a+4b+15c+24d+30e+40f+60g=120$ with the appropriate bounds on $a,b,\dots,g$. Looking at it modulo 10, we need $a=b=c=1$, $d=2$, so we get $$30e+40f+60g=50$$ which is impossible.
EDIT: A smaller example. The lcm of $\{1,6,10,15\}$ is 30. $$1+4\times6+2\times10+15=2\times30$$ But no subsum of the left side can give 30; to get a multiple of 5, you need the 1 and all four sixes, which makes 25, so no way to get to 30.
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7,979,182,211,079,554,000 | Interesting
How many tickets do I need to buy to guarantee winning the jackpot?
How many tickets do I need to buy to guarantee winning the jackpot?
The five white balls have possible numbers ranging between 1 and 70, while the Mega Ball can be between 1 and 25. That means you need to cover 302,575,350 combinations to guarantee a jackpot.
How do you calculate odds of winning multiple tickets?
Understand the calculations involved. To find the odds of winning any lottery, divide the number of winning lottery numbers by the total number of possible lottery numbers. If the numbers are chosen from a set and the order of the numbers doesn’t matter, use the formula. r ! ( n − r ) !
READ: Should I take 2 spares Grade 12?
How do you calculate the odds of winning a giveaway?
Odds, are given as (chances for success) : (chances against success) or vice versa. If odds are stated as an A to B chance of winning then the probability of winning is given as PW = A / (A + B) while the probability of losing is given as PL = B / (A + B).
How do you calculate odds?
1. Odds are most simply calculated as the number of events divided by the number of non-events.
2. The formal way to describe the odds is as the probability of the event divided by the probability of the non-event.
3. So odds are the ratio of two fractions:
4. If event occurs 1 of 5 times, probability = 0.2.
Do odds increase with more tickets?
The odds of winning the lottery do not increase by playing frequently, rather, you’d do better by purchasing more tickets for the same drawing. Although there is no guarantee in the stock market, the likelihood of getting a return on your investment is far better than your chances of winning the lottery.
READ: Is Thanos afraid of Darkseid?
How do I calculate odds?
How can I increase my chances of winning a scratch off?
Top Tips To Improve Your Chances Of Winning Scratch Cards
1. Don’t Buy The Cheapest Ones.
2. Check The Small Print.
3. Buy In Bulk.
4. Play Them Like Slots.
5. Keep Your Old Tickets.
6. Submit All Losing Tickets.
7. Study The Scratch Cards.
8. Be Strict With Your Budget.
What does 1 to 500 odds are for winning mean?
A 1 in 500 chance of winning, or probability of winning, is entered into this calculator as “1 to 500 Odds are for winning”. You may also see odds reported simply as chance of winning as 500:1. This most likely means “500 to 1 Odds are against winning” which is exactly the same as “1 to 500 Odds are for winning.”.
What are the odds of winning on a scratch ticket?
The overall odds of winning on a scratch ticket apply to each ticket purchased in a game, not a series of tickets. So, if a scratch game has overall odds of winning of about 1:4, that doesn’t mean that if you buy four tickets that one will automatically be a winner.
READ: How do I get rid of bugs in my house without killing them?
What are odds in sports betting and wagering?
Simply put, positive odds show you what you win for every hundred bucks you wager, and negative odds show you how much to bet to get $100. Let’s look at a more concrete example. Say the Patriots are at +200 to beat the Falcons for Monday Night Football. This means for every $100 you bet, you win $200.
What are decimal odds in gambling?
With decimal odds, you will multiply your wager by the decimal shown, with the answer being what you will receive in winnings, including your original wager. Betting $10 on a 5.00 bet will net you $50 in winnings (including your returned bet). | {
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-8,996,531,618,572,444,000 |
+0
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n+10=40,then what is the value of the expression n-25 n=plug in the n value found and solve
Guest Mar 13, 2017
#1
avatar+7153
+2
n + 10 = 40
Subtract 10 from both sides.
n = 40 - 10
n = 30
Replace "n" with 30, since n = 30.
n - 25
30 - 25 = 5
hectictar Mar 13, 2017
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-7,774,879,492,842,431,000 | Normal distribution
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The normal or Gaussian probability distribution is actually a family of distributions of the same general form, differing only in their location and scale parameters, commonly called the mean and standard deviation.
The shape of a graph of the distribution consists of a central bulge centered on the mean with about 99.7% of the area under the density curve between the mean plus and minus three standard deviations. Its resemblance to the shape of a bell has led to the shape of the normal distribution being called the "bell curve".
The probability distribution function for the normal distribution with mean μ and standard deviation σ is
p(x) = exp(-(x-μ)2/2σ2) / (2π)1/2σ
One reason that this distribution occurs so often in statistical work is the Central Limit Theorem. Simply stated, this theorem says that if you add up a lot of little things, the resulting distribution will resemble the normal distribution. More precisely: if you have n independent identically distributed random variables with mean 0 and standard deviation 1, then n-1/2 times their sum converges in distribution to the normal distribution with mean 0 and standard deviation 1.
Beware! There are random variables which do not have both a mean and a standard deviation. (The Cauchy distribution is a famous example.) Sums of such unfriendly random variables need not tend to normality.
Links:
Back to Probability -- Statistics
/Talk | {
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8,444,222,844,441,534,000 | ILE Home Intuitionistic Logic Explorer < Previous Next >
Nearby theorems
Mirrors > Home > ILE Home > Th. List > en0 GIF version
Theorem en0 6275
Description: The empty set is equinumerous only to itself. Exercise 1 of [TakeutiZaring] p. 88. (Contributed by NM, 27-May-1998.)
Assertion
Ref Expression
en0 (𝐴 ≈ ∅ ↔ 𝐴 = ∅)
Proof of Theorem en0
Dummy variable 𝑓 is distinct from all other variables.
StepHypRef Expression
1 bren 6228 . . 3 (𝐴 ≈ ∅ ↔ ∃𝑓 𝑓:𝐴1-1-onto→∅)
2 f1ocnv 5139 . . . . 5 (𝑓:𝐴1-1-onto→∅ → 𝑓:∅–1-1-onto𝐴)
3 f1o00 5161 . . . . . 6 (𝑓:∅–1-1-onto𝐴 ↔ (𝑓 = ∅ ∧ 𝐴 = ∅))
43simprbi 260 . . . . 5 (𝑓:∅–1-1-onto𝐴𝐴 = ∅)
52, 4syl 14 . . . 4 (𝑓:𝐴1-1-onto→∅ → 𝐴 = ∅)
65exlimiv 1489 . . 3 (∃𝑓 𝑓:𝐴1-1-onto→∅ → 𝐴 = ∅)
71, 6sylbi 114 . 2 (𝐴 ≈ ∅ → 𝐴 = ∅)
8 0ex 3884 . . . 4 ∅ ∈ V
98enref 6245 . . 3 ∅ ≈ ∅
10 breq1 3767 . . 3 (𝐴 = ∅ → (𝐴 ≈ ∅ ↔ ∅ ≈ ∅))
119, 10mpbiri 157 . 2 (𝐴 = ∅ → 𝐴 ≈ ∅)
127, 11impbii 117 1 (𝐴 ≈ ∅ ↔ 𝐴 = ∅)
Colors of variables: wff set class
Syntax hints: wb 98 = wceq 1243 wex 1381 c0 3224 class class class wbr 3764 ccnv 4344 1-1-ontowf1o 4901 cen 6219
This theorem was proved from axioms: ax-1 5 ax-2 6 ax-mp 7 ax-ia1 99 ax-ia2 100 ax-ia3 101 ax-in1 544 ax-in2 545 ax-io 630 ax-5 1336 ax-7 1337 ax-gen 1338 ax-ie1 1382 ax-ie2 1383 ax-8 1395 ax-10 1396 ax-11 1397 ax-i12 1398 ax-bndl 1399 ax-4 1400 ax-13 1404 ax-14 1405 ax-17 1419 ax-i9 1423 ax-ial 1427 ax-i5r 1428 ax-ext 2022 ax-sep 3875 ax-nul 3883 ax-pow 3927 ax-pr 3944 ax-un 4170
This theorem depends on definitions: df-bi 110 df-3an 887 df-tru 1246 df-fal 1249 df-nf 1350 df-sb 1646 df-eu 1903 df-mo 1904 df-clab 2027 df-cleq 2033 df-clel 2036 df-nfc 2167 df-ral 2311 df-rex 2312 df-v 2559 df-dif 2920 df-un 2922 df-in 2924 df-ss 2931 df-nul 3225 df-pw 3361 df-sn 3381 df-pr 3382 df-op 3384 df-uni 3581 df-br 3765 df-opab 3819 df-id 4030 df-xp 4351 df-rel 4352 df-cnv 4353 df-co 4354 df-dm 4355 df-rn 4356 df-res 4357 df-ima 4358 df-fun 4904 df-fn 4905 df-f 4906 df-f1 4907 df-fo 4908 df-f1o 4909 df-en 6222
This theorem is referenced by: nneneq 6320 php5 6321 snnen2oprc 6323 php5dom 6325 ssfiexmid 6336 fin0 6342 fin0or 6343 diffitest 6344 findcard 6345 findcard2 6346 findcard2s 6347 diffisn 6350
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-2,268,672,260,921,190,000 | Você está na página 1de 7
A UA UL L AA
25
25
A fórmula da equação
do 2º grau
Introdução N
para resolver a equação do 2º grau.
esta aula vamos encontrar uma fórmula
ax² + bx + c = 0 (com a ¹ 0)
ax
Você poderá naturalmente perguntar por que será necessária tal fórmu-
la, já que conseguimos, na aula anterior, resolver equações sem usar fórmulas.
Diremos então que a fórmula torna a resolução mais rápida e permite o uso mais
eficiente da máquina de calcular para obter as raízes da equação. Ainda
observando a fórmula, vamos descobrir quando uma equação do 2º grau possui
soluções ou não.
Nossa aula Inicialmente, vamos resolver uma equação do 2º grau para recordar o
método que desenvolvemos na aula passada. Observe cuidadosamente todos
os passos porque eles serão os mesmos que utilizaremos no caso geral.
Resolução da equação 3x² + 5x + 1 = 0
EXEMPLO 1
Solução
Solução:
1 º passo
passo: Como o coeficiente de x ² é 3, dividimos todos os termos da
equação por 3.
5 1
x2 + x + = 0
3 3
2 º passo
passo: Passamos o termo independente para o outro lado.
5 1
x2 + x = -
3 3
3 º passo
passo: Agora, vamos acrescentar aos dois lados da equação um número A U L A
capaz de transformar o lado esquerdo em um quadrado perfeito. Para fazer
isso, pegamos a metade do coeficiente de x:
1 5 5
25
× =
2 3 6
æ5ö² 25
e elevamos ao quadrado: =
è6ø 36
5 æ5 ö² 1 æ5ö²
Temos, então, x² + x+ =- +
3 è6 ø 3 è6ø
ou, ainda, x² + 2 . ²5 x + æ5ö² = - 1 + 25
6 è6ø 3 36
Observe agora que o lado esquerdo é um quadrado perfeito e que podemos
reunir as duas frações do lado direito igualando seus denominadores.
æx + 5 ö² 12 25
=- +
è 6ø 36 36
æ
x + ö² =
5 13
è 6ø 36
4 º passo
passo: Extraímos a raiz quadrada dos dois lados.
5 13
x+ =±
6 6
5 º passo
passo: Deixamos a letra x isolada do lado esquerdo para obter as duas
soluções.
5 13
x=- ± ou
6 6
-5 ± 13
x=
6
O caso geral: a solução da equação ax² + bx + c = 0
Desejamos agora que você acompanhe a dedução da fórmula, observando
que os passos são exatamente os mesmos.
1 º passo
passo: Como o coeficiente de x ² é a , dividimos todos os termos da
equação por a .
b c
x2 + x+ =0
a a
A U L A 2 º passo
passo: Passamos o termo independente para o outro lado.
25
b c
x2 + x=-
a a
3 º passo
passo: Para transformar o lado esquerdo em um quadrado perfeito,
pegamos a metade do coeficiente de x :
1 b b
× =
2 a 2a
æ b ö² b²
e o elevamos ao quadrado: =
è 2a ø 4a²
Depois, acrescentamos esse número aos dois lados:
b æ b ö² c æ b ö²
x² + 2 . x +è = - +
2a 2a ø a è 2aø
æ
x² + 2 . b x + b ö² = - + b²
c
2a è 2aø a 4a²
Observando que o lado esquerdo é agora um quadrado perfeito e que
podemos reunir as duas frações do lado direito igualando seus denomina-
dores, temos
æ b ö² c 4a b²
x+ =- . +
è 2a ø a 4a 4a²
æ b ö² 4ac b²
èx + 2aø = - 4a² + 4a²
æ ö²
x + b = b² - 4ac
è 2aø 4a²
4 º passo
passo: Extraímos a raiz quadrada dos dois lados.
b ± b2 - 4ac
x+ =
2a 2a
5 º passo
passo: Deixamos x isolado do lado esquerdo.
b b2 - 4ac
x=- ± ou
2a 2a
- b ± b2 - 4ac
x=
2a
E aí está nossa fórmula.
Quando uma equação do 2º grau possui solução? A U L A
Na fórmula que encontramos para a solução da equação do 2º grau, vemos
que, dentro da raiz quadrada, existe o número b ² - 4ac
4ac. Esse número é, em geral, 25
representado pela letra grega D (delta) e chama-se discriminante . Usando essa
nova letra, temos que as raízes da equação axax² + bx + c = 0 são:
-b + D -b - D
D
x=- e x=-
2a 2a
onde D = b2 - 4ac
Veja agora que, se o número D for positivo , encontramos duas raízes
diferentes.
Se, entretanto, D for zero , encontramos um só valor para a raiz. Se D for
negativo a equação não terá solução.
EXEMPLO 2
Resolver a equação 2x
2x² - 7x + 3 = 0
Solução
Solução: Vamos resolvê-la usando a fórmula:
- b ± b2 - 4ac
x=
2a
Na nossa equação, a = 22, b = − 7 e c = 33. Substituindo, temos:
x = - (- 7) ± Ö (- 7)² - 4 . 2 . 3
2.2
7 ± 49 - 24
x=
4
7 ± 25
x=
4
7±5
x=
4
As soluções são, portanto: 7 + 5 12
x= = =3
4 4
7-5 2 1
x= = =
4 4 2
A U L A Veja que, nesse exemplo, o discriminante é 25, que possui raiz quadrada
exata. Mas, isso nem sempre acontece. No exemplo do início desta aula,
25 encontramos, para raízes da equação 3x² + 5x + 1 = 00, os valores:
-5 + 13 -5 - 13
x= e x=
2 2
Para obter valores aproximados desses números, podemos utilizar a má-
quina de calcular. É o que veremos a seguir.
Usando a máquina de calcular
Consideremos, mais uma vez, a equação 3x
3x² + 5x + 1 = 00.
Vamos resolvê-la outra vez, usando agora a fórmula.
Temos a = 33, b = 5 e c = 11. Substituindo, temos:
x = - 5 ± Ö 5² - 4 . 3 . 1
2.3
-5 ± 25 - 12
x=
6
-5 ± 13
x=
6
Rapidamente encontramos as soluções. Para obter valores aproximados
dessas duas raízes, começamos calculando 13 e guardando o resultado na
memória.
Digitamos, então:
VISOR
1 3 M+ _ 3,6055512
O resultado que aparece no visor está guardado. Podemos então limpá-lo
apertando a tecla
ON/C
Para obter a 1ª solução, digitamos.
VISOR
− 5 + MR ¸ 6 = _ − 0,2324081
Para obter a 2ª solução, digitamos:
VISOR
- 5 - MR ¸ 6 = _ − 1,4342585
Concluímos então que, com duas casas decimais, as raízes da
equação 3 x ² + 5 x + 1 = 0 são, aproximadamente, − 0 , 2 3 e − 1 , 4 33.
Casos particulares A U L A
Na equação ax
ax² + bx + c = 00, quando encontramos b = 0 ou c = 0, não há
vantagem em utilizar a fórmula. Observe os exemplos seguintes. 25
EXEMPLO 3
Resolva a equação 2x2 − 32 = 00.
Solução
Solução: Para resolver essa equação, passamos o termo independente
para o outro lado e, em seguida, dividimos os dois lados por 2 (o
coeficiente de x ²²).
2x2 = 32
2x2 32
=
2 2
x2 = 16
Extraindo a raiz quadrada, temos x = ± 44.
EXEMPLO 4
Resolva a equação 2 x ² − 5x = 00.
Solução: Para resolver essa equação (que possui c = 00), o procedimento é
Solução
diferente. Inicialmente colocamos a letra x em evidência:
x . (2x - 5) = 0
Temos então um produto de dois números que dá zero. Isto só é possível
se um desses números for zero. Como primeiro caso, podemos ter x = 00.
Como segundo caso, podemos ter:
2x - 5 = 0
2x = 5
5
x=
2
5
Assim, as duas raízes de 2x² − 5x = 0 são x = 0 e x = .
2
Exercícios
A U L A Exercício 1
25 Resolva as equações:
a) x² − 9 = 0
b) x² + 5 = 0
c) x² − 3 = 0
Exercício 2
Resolva as equações:
a) x² − 3x = 0
b) 3x² + 12x = 0
Exercício 3
Resolva as equações:
a) x² − 5x + 6 = 0
b) x² − 3x − 10 = 0
c) x² − 3x + 1 = 0
d) x² − 6x + 9 = 0
e) x² + 2x + 3 = 0
Exercício 4
Resolva as equações seguintes e use a máquina de calcular para obter
valores aproximados das raízes (duas casas decimais são suficientes).
a) 2x² + 3x − 4 = 0
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9,101,217,222,177,719,000 |
Results 1 to 3 of 3
Thread: Rearranging help?
1. #1
Newbie
Joined
Sep 2010
Posts
1
Rearranging help?
Alright, this has stumped me.
How can I rearrange:
2x^3 - x - 4 = 0
to be:
x = ROOT(2/x + 1/2) ?
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2. #2
MHF Contributor
Joined
Dec 2009
Posts
3,120
Thanks
4
Quote Originally Posted by Drummerdude93 View Post
Alright, this has stumped me.
How can I rearrange:
2x^3 - x - 4 = 0
to be:
x = ROOT(2/x + 1/2) ?
Hi Drummerdude,
2x^3-x-4=0
2x^3-x=4
Factor...
x\left(2x^2-1\right)=4
2x^2-1=\frac{4}{x}
2x^2=\frac{4}{x}+1
x^2=\frac{2}{x}+\frac{1}{2}
Finally, take the square root.
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3. #3
MHF Contributor
skeeter's Avatar
Joined
Jun 2008
From
North Texas
Posts
15,875
Thanks
3524
Quote Originally Posted by Drummerdude93 View Post
Alright, this has stumped me.
How can I rearrange:
2x^3 - x - 4 = 0
to be:
x = ROOT(2/x + 1/2) ?
note that the transformation is incorrect ... it should be x = \pm \sqrt{\frac{2}{x} + \frac{1}{2}} \, ; x \ne 0
what problem are you solving that led you to ask about this?
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-8,698,942,502,784,733,000 | Sunday
April 20, 2014
Homework Help: Significant Figures
Posted by Britt on Sunday, January 27, 2008 at 5:58pm.
Report the following to two significant figures
a. 3.1459
b. 6.003
c. 42
d. 6.0233 x 10^23
e. 14.456 + 3.87
f. 9.9999995 - 0.0108
g. 67.3 x 6.667
h. 4.93 / 3.946
Can someone please check my answers
a. 3.1
b. 6.0
c. 42
d. 6.0 x 10^23
e. 18
f. 10.
g. 4.5 x 10^2
h. 1.2
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2,969,570,275,000,852,000 | Friday
December 9, 2016
Homework Help: math
Posted by Gorege on Thursday, December 16, 2010 at 9:17pm.
to identical jars are filled with equal amount of marbles. the marbles are coloured red or white. the ratio of red to white marblesis 7:1 in jar 1 and 9:1 in jar 2 if there are 90 marbles altogether, determin the number of red marbles in jar 2.
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2,118,990,234,266,384,000 | Analytische Geometrie und Lineare Algebra 1 by Ina Kersten
By Ina Kersten
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Zu zeigen: u = 0 Es ist u=u+0=0+u nach 2. =⇒ u = 0 3 =⇒ 1 Sei u1 + u2 = 0. Zu zeigen u1 = u2 = 0 Da u1 + u 2 = 0 =⇒ =⇒ u1 = −u2 ∈ U2 u1 = 0 =⇒ =⇒ u1 ∈ U1 ∩ U2 u2 = 0 Definition. • Seien U1 , U2 Teilr¨aume eines K-Vektorraumes. Dann heißt die Summe U1 + U2 die (innere) direkte Summe von U1 und U2 , falls eine der Bedingungen (und damit alle) aus obigem Satz erf¨ ullt sind. 8 Ubungsaufgaben 7 – 11 33 • Seien V1 , V2 beliebige K-Vektorr¨aume. Wir definieren die (¨ außere) direkte Summe als V1 ⊕ V2 := {(v1 , v2 ) | v1 ∈ V1 , v2 ∈ V2 } mit komponentenweiser Addition und Skalarmultiplikation.
0) , e2 = (0, 1, 0, . . , 0) , . . , en = (0, 0, . . , 0, 1) eine Basis von K n . Sie heißt die Standardbasis des K n . 6 gezeigt, ist B ein Erzeugendensystem von ❘2 . Zu zeigen bleibt die lineare Unabh¨ angigkeit. 3 sehen werden. 6 wissen wir, dass B = {(−3, 3), (1, −1)} kein Erzeugendensystem und damit auch keine Basis des ❘2 ist. Die Vektoren v1 und v2 sind wegen v1 +3v2 = 0 linear abh¨angig. 4 werden wir sehen, dass zwei linear unabh¨angige Vektoren in ❘2 stets eine Basis von ❘2 bilden.
Wt } eine Basis von U2 bilden folgt λi = 0 f¨ ur 1 ur 1 j t. Damit ist B linear unabh¨angig und bildet i r und µj = 0 f¨ eine Basis von U1 + U2 . Es folgt dimK (U1 + U2 ) = r + s + t = (r + s) + (r + t) − r = dimK U1 + dimK U2 − dimK (U1 ∩ U2 ) Lernerfolgstest. 2 f¨ ur m = 2 ? • Wie wird die Existenz einer Basis und der Basiserg¨ anzungssatz bewiesen? • Referieren Sie, wieso in einem endlich erzeugten K-Vektorraum je zwei Basen dieselbe Anzahl von Elementen haben. • Aufgaben 12–19, im Aufgabenpool Abschnitt 3.
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2,946,909,441,162,578,400 | How to reduce entropy?
This is not necessarily a research question, since I do not know, if someone is working on this or not, but I hope to gain some insight by asking it here:
The idea behind this question is to attach to a natural number in a “natural” way an entropy, such that “multiplication increases entropy”. (Of course one can attach very different entropies to natural numbers such that multiplication reduces entropy, but I will try to give an argument, why this choice is natural.) Let $ n$ be a composite number $ \ge 2$ , $ \phi$ the Euler totient function. Suppose a factorization algorithm $ A$ outputs a number $ X$ , $ 1 \le X \le n-1$ with equal probability $ \frac{1}{n-1-\phi(n)}$ such that $ 1 < \gcd(X,n) < n$ . Then we can view $ X$ as a random variable and attach the entropy to it $ H(X_n):=H(n):= \log_2(n-1-\phi(n))$ .
The motivation behind this definition comes from an analogy to physics: I think that “one-way-functions” correspond to the “arrow of time” in physics. Since “the arrow of time” increases entropy, so should “one-way-functions”, if they exist. Since integer factorization is known to be a candidate for owf, my idea was to attach an entropy which would increase when multiplying two numbers.
It is proved here ( https://math.stackexchange.com/questions/3275096/does-entropy-increase-when-multiplying-two-numbers ) that:
$ H(mn) > H(n) + H(m)$ for all composite numbers $ n \ge 2, m \ge 2$
The entropy of $ n=pq$ , $ p<q<2p$ will be $ H(pq) = \log_2(p+q-1) > \log_2(2p-1) \approx \log_2(2p) = 1+\log_2(p)$ .
At the beginning of the factorization, the entropy of $ X_n$ , $ n=pq$ will be $ \log_2(p+q-1)$ since it is “unclear” which $ X=x$ will be printed. The algorithm must output $ X=x$ as described above. But knowing the value of $ X$ , this will reduce the entropy of $ X$ to $ 0$ . So the algorithm must reduce the entropy from $ \log_2(p+q-1)$ to $ 0$ . From physics it is known, that reducing entropy can be done with work. Hence the algorithm “must do some work” to reduce the entropy.
My question is, which functions do reduce entropy? (So I am thinking about how many function calls will the algorithm at least make to reduce the entropy by the amount described above?)
Thanks for your help!
Does multiplication increase entropy?
Does multiplication increase entropy?
The Shannon entropy of a number $ k$ in binary digits is defined as $ $ H = -\log(\frac{a}{l})\cdot\frac{a}{l} – \log(1-\frac{a}{l})\cdot (1-\frac{a}{l})$ $ where $ l = \text{ floor }(\frac{\log(k)}{\log(2)})$ is the number of binary digits of $ k$ and $ a$ is the number of $ 1$ -s in the binary expansion of $ k$ . So we view the number $ k$ as a “random variable”.
Suppose that $ n,m$ are uniformly randomly chosen in the interval $ 1 \le N$ .
Hypothesis 1):
$ H_{m \cdot n}$ is “significantly” larger then $ H_n$ .
Hypothesis 2):
$ H_{m + n}$ is not “significantly” larger then $ H_n$ .
Here is some empirical statistical test indicating that multiplication increases entropy, but addition does not:
def entropyOfCounter(c): S = 0 for k in c.keys(): S += c[k] prob = [] for k in c.keys(): prob.append(c[k]/S) H = -sum([ p*log(p,2) for p in prob]).N() return H def HH(l): return entropyOfCounter(Counter(l)) N = 10^4 HN = [] HmXn = [] HmPn = [] for k in range(N): n = randint(1,17^50) m = randint(1,17^50) Hn = HH(Integer(n).digits(2)) Hm = HH(Integer(m).digits(2)) HmXn.append(HH(Integer(n*m).digits(2))) HmPn.append(HH(Integer(n+m).digits(2))) HN.append(Hn) X = mean(HN) Y = mean(HmPn) Z = mean(HmXn) n = len(HN) m = n SX2 = variance(HN) SY2 = variance(HmPn) SZ2 = variance(HmXn) SXY2 = ((n-1)*SX2 + (m-1)*SY2)/(n+m-2) SXZ2 = ((n-1)*SX2 + (m-1)*SZ2)/(n+m-2) TXY = sqrt((m*n)/(n+m)).N()*(X-Y)/sqrt(SXY2).N() TXZ = sqrt((m*n)/(n+m)).N()*(X-Z)/sqrt(SXZ2).N() print TXY,TXZ,n+m-2 Output: -1.43265218355297 -32.5323306851490 19998
The second case (multiplication) increases entropy significantly. The first case ( addition) does not.
Is there a way to give a heuristic explanation why this is so in general (if it is), or is this empirical obervation in general $ 1 \le N$ wrong?
Related: https://physics.stackexchange.com/questions/487780/increase-in-entropy-and-integer-factorization-how-much-work-does-one-have-to-do
Minimising an Integrated Relative Entropy Functional
Suppose I am given
• A probability distribution on $ \mathbf R^d$ , with density $ \pi (x)$ .
• A family of transition kernels $ \{ q^0 (x \to \cdot) \}_{x \in \mathbf R^d}$ on $ \mathbf R^d$ , with densities $ q^0 (x \to y)$ .
I now want to find a new family of transition kernels $ \{ q^1 (x \to \cdot) \}_{x \in \mathbf R^d}$ such that
• $ q^1$ is in detailed balance with respect to $ \pi$ , i.e.
\begin{align} \pi (x) q^1 ( x \to y ) = \pi (y) q^1 ( y \to x) \end{align}
• $ q^1$ are as close as possible to $ q^0$ in KL divergence, averaged across $ x \sim \pi$ .
To this end, I introduce the following functional
\begin{align} \mathbb{F} [ q^1 ] &= \int_{x \in \mathbf R^d} \pi (x) \cdot KL \left( q^1 (x \to \cdot ) || q^0 (x \to \cdot ) \right) dx\ &= \int_{x \in \mathbf R^d} \int_{y \in \mathbf R^d} \pi (x) \cdot q^1 (x \to y) \log \frac{ q^1 (x \to y) }{ q^0 (x \to y) } \, dx dy \ &= \int_{x \in \mathbf R^d} \int_{y \in \mathbf R^d} \pi (x) \cdot \left\{ q^1 (x \to y) \log \frac{ q^1 (x \to y) }{ q^0 (x \to y) } – q^1 (x \to y) + q^0 (x \to y) \right \} \, dx dy. \end{align}
I thus wish to minimise $ \mathbb{F}$ over all transition kernels $ \{ q^1 (x \to \cdot) \}_{x \in \mathbf R^d}$ satisfying the desired detailed balance condition. Implicitly, there are also the constraints that the $ q^1$ are nonnegative and normalised.
I would like to solve this minimisation problem, or at least derive e.g. an Euler-Lagrange equation for it.
Currently, I can show that the first variation of $ \mathbb{F}$ is given by
\begin{align} \left( \frac{d}{dt} \vert_{t = 0} \right) \mathbb{F} [q^1 + t h] = \int_{x \in \mathbf R^d} \int_{y \in \mathbf R^d} \pi (x) \cdot \log \frac{ q^1 (x \to y) }{ q^0 (x \to y) } \cdot h (x, y) \, dx dy. \end{align}
Moreover, my constraints stipulate that any admissible variation $ h$ must satisfy the following two conditions:
1. $ \forall x, y, \quad \pi (x) h (x, y) = \pi (y) h (y, x)$
2. $ \forall x, \quad \int_{y \in \mathbf R^d} h (x, y) dy = 0$
I have not been able to translate these conditions into an Euler-Lagrange equation. I acknowledge that since the functional involves no derivatives of $ q^1$ , the calculus of variations approach may be ill-suited. If readers are able to recommend alternative approaches, this would also be appreciated. Anything which would allow for a more concrete characterisation of the optimal $ q^1$ would be ideal.
Cross Entropy Function
I’ve seen two versions of the cross entropy cost function, and conflicting information about it. \begin{equation}J(\theta) = -\frac{1}{N} \sum_{n=1}^N\sum_{i=1}^C y_{ni}\log \hat{y}_{n_i} (\theta)\end{equation} $ $ C(\theta) = – \frac{1}{N}\sum_{n=1}^N \sum_{i=1}^{C}[ y_{ni}\log (\hat{y}_{ni})+ (1-y_{ni}) \log(1-\hat{y}_{ni})] $ $
Some are saying that the second one is equivalent to the first for the case where there are only two classes, which makes sense. But couldn’t the second one also be used for more than two classes? For example, say we have three class classification, with the ground truth $ \vec{y} = \begin{bmatrix} 1 \ 0 \ 0 \end{bmatrix}$ and $ \vec{\hat{y}} =\begin{bmatrix} 0.7 \ 0.1 \ 0.2 \end{bmatrix} $ . Then with the first equation, cross entropy would simply be $ -\log(0.7)$ . But couldn’t we also use the second equation and calculate cross entropy as $ -\log(0.7) – \log(0.9)-\log(0.8)?$ When would we use the first equation and when would we use the second one, and why?
Determining the entropy of a string if each character has a slight bias
Let’s say I need to generate a 32-character secret comprised of ASCII characters from the set ‘0’..’9′. Here’s one way of doing it:
VALID_CHARS = '0123456789' generate_secret_string() { random = get_crypto_random_bytes(32) secret = '' for (i = 0; i < 32; i++) { secret += VALID_CHARS[random[i] % 10] } return secret }
My concern is that my character selection is biased. Because 10 doesn’t divide evenly into 256, the first 6 VALID_CHARS are slightly more likely to occur.
The secret space is 1032, but my generated secrets have less entropy than that. How can I calculate precisely how much entropy I actually have?
How can ECDSA offer stronger security with the same key length (same amount of entropy)?
It’s always been told that ECDSA is more secure “per bit of key size”, such that it offers same security with a shorter key, or offers stronger security with the same key length.
However, per my understanding, if the length of the key is fixed, it means that the amount of entropy is also fixed (e.g. 160-bit ECDSA key contains no more entropy than 2^160, and a 192-bit AES key contains no more entropy than 2^192). So why is ECDSA considered “more secure” at the same level of entropy?
entropy of base map of skew product
Let $ (M, \mathcal{B}, \mu)$ be a probability space and $ f:M\rightarrow M$ be a continuous. Let $ T:M\rightarrow S^{1}$ be a measurable function. The $ \mathit{skew\hspace{0.1cm}product}$ defined by $ T$ over $ f$ is the transformation
$ $ F:M\times S^{1}\rightarrow M\times S^{1}.$ $ Let $ \nu$ is a stationary measure for $ F$ . That means $ \nu(D)=\int\nu(T^{-1}(x)(D))d\mu(x)$ for $ D\subset S^{1}$ .
We consider $ m=\mu \times \nu$ is $ F$ -invarinat measure.
Question:
Let $ \nu$ is non-atomic(i.e. $ \nu(V)=0$ for $ V\subset S^{1}$ ). Can we prove measure theoretic entropy $ h_{\mu}(f)=0$ for any invarinat measure $ \mu$ ?
By disintegration theorem, we can easily see $ m(\{(x, t_{x}), t_{x}\in D\subset S^{1}\})=0$ . So question said, can we prove $ h_{\mu}(f)=0$ for any invarinat measure $ \mu$ , if graph of function has zero measure?
Topological entropy of logistic map $f(x) = \mu x (1-x)$, $f:[0,1] \to [0,1]$ for $\mu \in (1,3)$
As stated in the question, I want to find the topological entropy of the logistic map on the interval $ [0,1]$ for a “nice” range of the parameter $ \mu$ , namely $ \mu \in (1,3)$ . I think the fact that $ f:[0,1] \to [0,1]$ is a very important additional condition here which simplifies things.
I’ve tried something, which I’m not sure is the right way to approach the problem, but I’ll outline it here anyway.
I know a theorem that states $ h_{top}(f) = h_{top}(f|_{NW(f)})$ , where $ NW(f)$ is the set of non-wandering points of $ f$ , so I wanted to find that set. By drawing a lot of little pictures, I concluded that for $ x \notin \{0,1\}$ , we should have $ \lim_{n\to \infty} f^{n}(x) = 1-\frac{1}{\mu}$ , which is the other fixed point of $ f$ . Also, the convergence seems fairly straightforward (i.e. it gets closer with every iteration), so I somehow got it into my head that I should have $ NW(f) = \{0, 1- \frac{1}{\mu}$ .
To confirm this, Wikipedia says:
By varying the parameter r, the following behavior is observed:
With r between 0 and 1, the population will eventually die, independent of the initial population. With r between 1 and 2, the population will quickly approach the value r − 1/r, independent of the initial population. With r between 2 and 3, the population will also eventually approach the same value r − 1/r, but first will fluctuate around that value for some time. The rate of convergence is linear, except for r = 3, when it is dramatically slow, less than linear (see Bifurcation memory).
However, I haven’t been able to find a proof of these claims. Can anyone show me how to prove this, or give me a reference where the proof is clearly written out?
Also, if there is an easier way of finding the topological entropy (again, I emphasize that $ f:[0,1] \to [0,1]$ ; I’ve lost a lot of time reading about Mandelbrot sets by conjugating $ f$ to $ g(z) = z^2 + c$ and looking at formulas for the entropy of $ g$ which exist, but with domains $ \mathbb{C}$ or some variant), I’d be very happy to hear it.
Entropy of endpoints of a random walk in a dense graph
Let $ p\in[0,1]$ be a constant and let $ G$ be a graph with $ n$ vertices and $ \approx p\binom{n}{2}$ edges. If you’d like, consider $ p=1/2$ .
Let $ X$ be a random vertex of $ G$ chosen proportional to its degree (i.e. the probability that $ X=v$ is $ d(v)/2|E(G)|$ ). That is, $ X$ is chosen according to the stationary distribution of a random walk. Now, generate two standard random walks starting at $ X$ denoted $ A_0,A_1,\dots$ and $ B_0,B_1,\dots$ where $ A_0=B_0=X$ .
For $ k\geq1$ , define $ \alpha_k$ so that $ H(A_k,B_k) =\log_2(\alpha_kn^2)$ where $ H$ denotes the binary entropy. That is, $ \alpha_k:=2^{H(A_k,B_k)}/n^2$ . The way that entropy works is that $ \alpha_k\approx1$ if and only if the distribution on $ (A_k,B_k)$ is roughly uniformly on $ V(G)^2$ .
Question: Given the value of $ \alpha_i$ , what kind of lower bound can one prove on $ \alpha_j$ for $ j>i$ ?
I am mainly interested in constant values of $ i,j$ . For example, I would be quite interested to know whether all such graphs satisfy any (non-trivial) bound of the form $ $ \alpha_2\geq f(\alpha_1)$ $ for some function $ f$ . I’d be very happy to know if there are any existing results related to this; please excuse my lack of knowledge on random walks… | {
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8,042,972,960,143,143,000 | Isotropic quadratic form
From Wikipedia, the free encyclopedia
Jump to: navigation, search
In mathematics, a quadratic form over a field F is said to be isotropic if there is a non-zero vector on which the form evaluates to zero. Otherwise the quadratic form is anisotropic. More precisely, if q is a quadratic form on a vector space V over F, then a non-zero vector v in V is said to be isotropic if q(v) = 0. A quadratic form is isotropic if and only if there exists a non-zero isotropic vector for that quadratic form.
Suppose that (V, q) is quadratic space and W is a subspace. Then W is called an isotropic subspace of V if some vector in it is isotropic, a totally isotropic subspace if all vectors in it are isotropic, and an anisotropic subspace if it does not contain any (non-zero) isotropic vectors. The isotropy index of a quadratic space is the maximum of the dimensions of the totally isotropic subspaces.[1]
A quadratic form q on a finite-dimensional real vector space V is anisotropic if and only if q is a definite form:
• either q is positive definite, i.e. q(v) > 0 for all non-zero v in V ;
• or q is negative definite, i.e. q(v) < 0 for all non-zero v in V.
More generally, if the quadratic form is non-degenerate and has the signature (a, b), then its isotropy index is the minimum of a and b.
Hyperbolic plane[edit]
Not to be confused with Hyperbolic plane in geometry.
Let V = F2 with elements (x, y). Then the quadratic forms q = xy and r = x2y2 are equivalent since there is a linear transformation on V that makes q look like r, and vice versa. Evidently, (V, q) and (V, r) are isotropic. This example is called the hyperbolic plane in the theory of quadratic forms. A common instance has F = real numbers in which case \scriptstyle \lbrace x \isin V : q(x)\ =\ \text{nonzero constant} \rbrace and \scriptstyle \lbrace x \isin V : r(x)\ =\ \text{nonzero constant} \rbrace are hyperbolas. In particular, \scriptstyle \lbrace x \isin V : r(x) = 1 \rbrace is the unit hyperbola. The notation \scriptstyle \langle 1 \rangle \oplus \langle -1 \rangle has been used by Milnor and Huseman[2] for the hyperbolic plane as the signs of the terms of the bivariate polynomial r are exhibited.
Split quadratic space[edit]
A space with quadratic form is split (or metabolic) if there is a subspace which is equal to its own orthogonal complement: equivalently, the index of isotropy is equal to half the dimension.[1] The hyperbolic plane is an example, and over a field of characteristic not equal to 2, every split space is a direct sum of hyperbolic planes.[3]
Relation with classification of quadratic forms[edit]
From the point of view of classification of quadratic forms, anisotropic spaces are the basic building blocks for quadratic spaces of arbitrary dimensions. For a general field F, classification of anisotropic quadratic forms is a nontrivial problem. By contrast, the isotropic forms are usually much easier to handle. By Witt's decomposition theorem, every inner product space over a field is an orthogonal direct sum of a split space and an anisotropic space.[4]
Field theory[edit]
• If F is an algebraically closed field, for example, the field of complex numbers, and (V, q) is a quadratic space of dimension at least two, then it is isotropic.
• If F is a finite field and (V, q) is a quadratic space of dimension at least three, then it is isotropic.
• If F is the field Qp of p-adic numbers and (V, q) is a quadratic space of dimension at least five, then it is isotropic.
See also[edit]
References[edit]
1. ^ a b Milnor & Husemoller (1973) p.57
2. ^ Milnor & Husemoller (1973) page 9
3. ^ Milnor & Husemoller (1973) pp.12–13
4. ^ Milnor & Husemoller (1973) p.56 | {
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2,718,983,261,468,823,000 | Games / Video Games Non Featured
How To Learn Betting Systems in Roulette
ClickHowTo Team
Written by ClickHowTo Team
If you’re like any gambler, you have a system on how to play roulette, but for some of those who don’t have a system – maybe you didn’t even know there was a system, you just pick your favorite numbers, like your birthday and favorite day of the week, and because we know that’s won you millions, you might want to read this article.
Roulette has different types of systems in how you can play. For example if you want to play
all on black or all on red, it might work. Chances are maybe you’ll win something, but there are a few different strategies that you may want to consider as you play the odds. We’ll explore these and hopefully you’ll think about betting on a strategy and not just an anniversary the next time you gamble so you can increase your money.
DOUBLE STREET QUAD STRATEGY
You can use this type of maneuver, where you make your bet on two sets of long lines of a number that you want to pick and you also place a corner bet. This will help you to actually place bets on 17 different numbers, so your bet may be six or seven chips down two long lines. So for example, you might have chips on 10, 11, 12, 13, 14, 15 and then your next row might be 28, 29, 30, 31, 32, 33. Next, bet on the corners and you have one bet left. As you can see, this has nothing to do with birthdays or anniversaries, its an actual system to trying to map out as many areas to play as you can because with this particular method, your betting based on odds. In regards to this type of bet, your odds may be five to one. The benefits of using this bet:
• You’re betting on 17 lines
• You’re increasing your odds
• You’re covering a more strategic area of the board
5 QUAD METHOD
Another method that is similar to this is considered the 5 quad method. With the five quad method, you make a bet on actual rows of numbers as your bets. That is, you would be betting on 21 different numbers. And with this particular method of play, your bets on 5 lines of play would total 21. For example, your numbers might look like:
• 5, 6, 8, 9 and 10,
• 11, 12, 13, 14, and then 17
• 18, 19, 20, 21, and then 25
• 26, 27, 28, 29, and then 32,
• 33, 34, 35, and 36
But if you’re unsure, you can also just stick with your original plan and bet on grandma’s birthday. But keep in mind, your odds go up 8 to 1 with this type of bet.
A few other ways that you can bet on roulette include different ways that you system bet. For example, you could try:
• A Martindale, as you bet on consistent numbers like 1 2 3 4 5 6 & 7
• There is also a Grand Martindale, where you do the same thing but increase a bet, as you would with the standard martingale, but then you would actually add to your bet by placing a wager on red or on black, in essence doubling your bet that you just placed by adding another unit to it
• And other systems if you ever decide you want to look them up and research them further include the cancellation system, one that’s considered Oscar’s grind, the shot well system, the red system, betting on the biased wheel, and there’s even one that’s considered the dealer signature. Or if you’re unsure, just flip the coin that’s in your pocket and if it lands on heads, bet on red, tails, bet on black and hope you get it all back!
Ultimately, you want to have as much fun as possible and if you can strategize that’s even better but if you can’t find a strategy that works for you, just have fun and enjoy the table games! Always remember that when you bet online you want to utilize features through mobile roulette sites as these are enhanced gaming platforms that are set up to ensure consistent play and reliability, when you play real money gaming in mobile casinos.
You might also be interested to explore good bingo sites, too – you can find many good bonuses with bingo and roulette alike.
About the author
ClickHowTo Team
ClickHowTo Team | {
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7,934,176,198,918,880,000 | Measuring objects in different ways
Little Chicks have been measuring different objects using varied resources.
Using straws to measure Using wooden blocks to measure Using unifix to measure
Little Chicks were imaginative at finding different ways to measure objects.
to view gallery follow…
Category: Little Chicks
measure activemaths
read more...
Number bond party for World Maths Day
Today, it was World Maths Day. We decided to have a number bond party to celebrate! Number bonds are pairs of numbers that make a particular total. We had fun playing lots of games that make 10! We had to find a partner with a numbered party hat that would match with our own to make 10. We rolled…
Category: Ducklings
maths activemaths number
read more...
Maths
Little Chicks have been learning mathamatical language and have been measuring different quantities.
Each Little Chick has been measuring the unifix into containers they had to look at the picture cue card which showed either a full,empty, half empty, nearly full and nearly empty…
Category: Little Chicks
weight activemaths Size
read more...
Counting Actions
In mathematics this week, we have been matching numbers to their cardinal value. We have picked numbers out of a bag and then completed that many actions such as star jumps, hops, nods of the head, runs around the garden, etc. Some of us were able to do the correct number or actions for 1 more or…
Category: Ducklings
maths activemaths
read more...
Little Chicks number 9
Little Chicks have been learning the number 9 and counting out objects to nine.
Counting out pictures to 9.
Counting the baubles out in different amounts to nine.
Category: Little Chicks
activemaths Counting number
read more...
Christmas
Little Chicks have a new role play and are learning all about Christmas
.
Little Chicks are busy Elves making toys. Santa and Ruldolph oversee the work shop. Little Chicks are learning the Nativity story.
Little Chicks thought about what Christmas meant to them and made a spider…
Category: Little Chicks
Christmas ICT activemaths Role-play Nativity
read more...
Number 7
Little Chicks are learning the number 7.
Little Chicks count with using their fingers to show 7.
Little Chicks counted out the objects to 7. They used the interactive board to trace over 7.
They looked around the classroom to find number 7 in their role…
Category: Little Chicks
activemaths Counting
read more...
Old McDonald had a farm
This week Little Chicks nursery rhyme is Old McDonald had a farm. Little Chicks will be learning the rhyme and looking at the what happens on a farm.
Follow the link https://www.youtube.com/hashtag/nurseryrhymes to learn song.
The Little Chicks have been putting the farm…
Category: Little Chicks
Rhyme activemaths language
read more...
Number 6
Little Chicks are learning number 6. We count out objects, find number six in the environment and use the interactive screen to count and find 6.
activemaths Counting
read more...
Nursery Rhyme Shop
Little chicks have a new role play Nursery Rhyme Shop. As part of our learning Little Chicks are learning Nursery Rhymes.
Little Chicks engage in role play being the shopkeeper or the customer. Counting out money, making order forms and delivering goods.
Little Chicks…
Category: Little Chicks
Rhyme activemaths Singing social
read more...
Number 3
Little Chicks have been learning the number three this week. They have traced over number three on the screen and attempted to draw three apples.
The rhyme we sing as we trace.
Round the tree,
Round the tree,
Now I've made the number three
The children have used chalk boards to…
Category: Little Chicks
writing fine motor activemaths
read more...
Making Clocks
In mathematics the Ducklings have been making their own clocks and showing different o'clock and even half-past times.
Category: Ducklings
maths time activemaths Creative learning
read more... | {
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5,232,147,167,641,713,000 | Answers
Solutions by everydaycalculation.com
Answers.everydaycalculation.com » Compare fractions
Compare 7/2 and 3/5
1st number: 3 1/2, 2nd number: 3/5
7/2 is greater than 3/5
Steps for comparing fractions
1. Find the least common denominator or LCM of the two denominators:
LCM of 2 and 5 is 10
Next, find the equivalent fraction of both fractional numbers with denominator 10
2. For the 1st fraction, since 2 × 5 = 10,
7/2 = 7 × 5/2 × 5 = 35/10
3. Likewise, for the 2nd fraction, since 5 × 2 = 10,
3/5 = 3 × 2/5 × 2 = 6/10
4. Since the denominators are now the same, the fraction with the bigger numerator is the greater fraction
5. 35/10 > 6/10 or 7/2 > 3/5
MathStep (Works offline)
Download our mobile app and learn to work with fractions in your own time:
Android and iPhone/ iPad
Related:
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-1,665,602,064,693,557,800 | Photographer's Future
How Statistics Homework Determines The Photographer’s Future
Remember the first time your kids took some shots with your camera phone. It was their first time and of course, they are not professional photographers. But guess what these are signs they will make good photographers in the future.
It might sound absurd but statistics does determine the future of the photographer. A good photographer will do well with the sound knowledge of statistics or other subjects involving calculations. This is why people can pay for statistics homework just to see the results. Let’s find out how.
1. Understanding the Rule of the third
A good photographer would obviously need a sound knowledge of statistics to excel. Take a look at the photos you have at home. Aren’t they looking great? There is one formula that makes it possible to determine if something is great to the human eye. It is known as the rule of thirds.
The rule in question is simple to understand and also needs some statistical knowledge. You can have a pleasing composition of the image if this rule is applied well. The photographer’s duty is to make sure that points of interest are placed along the lines of a grid of nine. This is also the area where the picture is shared into three main parts; vertically and horizontally. The view should also be divided fast.
The complex calculations in statistics are one of the reasons students dump statistics homework. Most of the calculations are abstract and difficult to understand. But if your dream is to become a professional photographer in the future, there is no point in running away from your statistics homework. It is very easy to find online help with statistics no matter how complex the homework is.
Writing service providers like Myhomeworkdone.com have made things a lot easier for students to pass statistics assignment and broaden their understanding. They are written by experts who have sound knowledge and qualification on statistics.
1. Making Use of the Camera
Pressing the green circle on the screen of the phone is what is used to take pictures. But there is more a professional photographer need to understand. The Single-lens reflex camera (SLR) is what the professional photographer uses. And to be effective, they need to know a thing or two about the depth of field and shutter speeds which will also involve sound statistical or mathematical knowledge. The shutter speed is used to determine how fast the camera blinks. The aperture, on the other hand, controls the amount of light that comes in.
As a photographer, one also needs to have a basic understanding of geometric sequence to determine the correct shutter speed. Calculations on shutter speed are in fractions of seconds. So photographers should be sound with calculations.
Most statistics homework is complex and difficult to understand. But students can seek statistics online help whenever they encounter challenges with their homework. There are dozens of professionals online who are willing to assist students to get good grades on their statistics assignment. So, if you want to become a professional photographer in the future, embrace statistics.
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-3,943,368,596,061,109,000 | Unit fraction
Summary
A unit fraction is a rational number written as a fraction where the numerator is one and the denominator is a positive integer. A unit fraction is therefore the reciprocal of a positive integer, 1/n. Examples are 1/1, 1/2, 1/3, 1/4, 1/5, etc.
Elementary arithmeticEdit
Multiplying any two unit fractions results in a product that is another unit fraction:
However, adding, subtracting, or dividing two unit fractions produces a result that is generally not a unit fraction:
Modular arithmeticEdit
Unit fractions play an important role in modular arithmetic, as they may be used to reduce modular division to the calculation of greatest common divisors. Specifically, suppose that we wish to perform divisions by a value x, modulo y. In order for division by x to be well defined modulo y, x and y must be relatively prime. Then, by using the extended Euclidean algorithm for greatest common divisors we may find a and b such that
from which it follows that
or equivalently
Thus, to divide by x (modulo y) we need merely instead multiply by a.
Finite sums of unit fractionsEdit
Any positive rational number can be written as the sum of unit fractions, in multiple ways. For example,
The ancient Egyptian civilisations used sums of distinct unit fractions in their notation for more general rational numbers, and so such sums are often called Egyptian fractions. There is still interest today in analyzing the methods used by the ancients to choose among the possible representations for a fractional number, and to calculate with such representations.[1] The topic of Egyptian fractions has also seen interest in modern number theory; for instance, the Erdős–Graham conjecture and the Erdős–Straus conjecture concern sums of unit fractions, as does the definition of Ore's harmonic numbers.
In geometric group theory, triangle groups are classified into Euclidean, spherical, and hyperbolic cases according to whether an associated sum of unit fractions is equal to one, greater than one, or less than one respectively.
Series of unit fractionsEdit
Many well-known infinite series have terms that are unit fractions. These include:
• The harmonic series, the sum of all positive unit fractions. This sum diverges, and its partial sums
closely approximate the natural logarithm of plus the Euler–Mascheroni constant. Changing every other addition to a subtraction produces the alternating harmonic series, which sums to the natural logarithm of 2:
• The Leibniz formula for π is
• The Basel problem concerns the sum of the square unit fractions:
Similarly, Apéry's constant is an irrational number, the sum of the cubed unit fractions.
• The binary geometric series is
Matrices of unit fractionsEdit
The Hilbert matrix is the matrix with elements
It has the unusual property that all elements in its inverse matrix are integers.[2] Similarly, Richardson (2001) defined a matrix with elements
where Fi denotes the ith Fibonacci number. He calls this matrix the Filbert matrix and it has the same property of having an integer inverse.[3]
Adjacent fractionsEdit
Fractions with tangent Ford circles differ by a unit fraction
Two fractions and (in lowest terms) are called adjacent if , which implies that their difference is a unit fraction. For instance, and are adjacent: and . However, some pairs of fractions whose difference is a unit fraction are not adjacent in this sense: for instance, and differ by a unit fraction, but are not adjacent, because for them . The terminology comes from the study of Ford circles, circles that are tangent to the number line at a given fraction and have the squared denominator of the fraction as their diameter: fractions and are adjacent if and only if their Ford circles are tangent circles.[4]
Unit fractions in probability and statisticsEdit
In a uniform distribution on a discrete space, all probabilities are equal unit fractions. Due to the principle of indifference, probabilities of this form arise frequently in statistical calculations.[5] Additionally, Zipf's law states that, for many observed phenomena involving the selection of items from an ordered sequence, the probability that the nth item is selected is proportional to the unit fraction 1/n.[6]
Unit fractions in physicsEdit
The energy levels of photons that can be absorbed or emitted by a hydrogen atom are, according to the Rydberg formula, proportional to the differences of two unit fractions. An explanation for this phenomenon is provided by the Bohr model, according to which the energy levels of electron orbitals in a hydrogen atom are inversely proportional to square unit fractions, and the energy of a photon is quantized to the difference between two levels.[7]
Arthur Eddington argued that the fine-structure constant was a unit fraction, first 1/136 then 1/137. This contention has been falsified, given that current estimates of the fine structure constant are (to 6 significant digits) 1/137.036.[8]
See alsoEdit
ReferencesEdit
1. ^ Guy, Richard K. (2004), "D11. Egyptian Fractions", Unsolved problems in number theory (3rd ed.), Springer-Verlag, pp. 252–262, ISBN 978-0-387-20860-2.
2. ^ Choi, Man Duen (1983), "Tricks or treats with the Hilbert matrix", The American Mathematical Monthly, 90 (5): 301–312, doi:10.2307/2975779, MR 0701570.
3. ^ Richardson, Thomas M. (2001), "The Filbert matrix" (PDF), Fibonacci Quarterly, 39 (3): 268–275, arXiv:math.RA/9905079, Bibcode:1999math......5079R
4. ^ Ford, L. R. (1938), "Fractions", The American Mathematical Monthly, 45 (9): 586–601, doi:10.1080/00029890.1938.11990863, JSTOR 2302799, MR 1524411
5. ^ Welsh, Alan H. (1996), Aspects of statistical inference, Wiley Series in Probability and Statistics, vol. 246, John Wiley and Sons, p. 66, ISBN 978-0-471-11591-5.
6. ^ Saichev, Alexander; Malevergne, Yannick; Sornette, Didier (2009), Theory of Zipf's Law and Beyond, Lecture Notes in Economics and Mathematical Systems, vol. 632, Springer-Verlag, ISBN 978-3-642-02945-5.
7. ^ Yang, Fujia; Hamilton, Joseph H. (2009), Modern Atomic and Nuclear Physics, World Scientific, pp. 81–86, ISBN 978-981-283-678-6.
8. ^ Kilmister, Clive William (1994), Eddington's search for a fundamental theory: a key to the universe, Cambridge University Press, ISBN 978-0-521-37165-0.
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-5,579,471,371,188,833,000 | The Unapologetic Mathematician
Mathematics for the interested outsider
Matrix notation
I just spent all day on the road back to NOLA to handle some end-of-month business, clean out my office, and so on. This one will have to do for today and tomorrow.
It gets annoying to write out matrices using the embedded LaTeX here, but I suppose I really should, just for thoroughness’ sake.
In general, a matrix is a collection of field elements with an upper and a lower index. We can write out all these elements in a rectangular array. The upper index should list the rows of our array, while the lower index should list the columns. The matrix \left(t_i^j\right) with entries t_i^j for i running from {1} to m and n running from {1} to n is written out in full as
\displaystyle\begin{pmatrix}t_1^1&t_2^1&\cdots&t_m^1\\t_1^2&t_2^2&\cdots&t_m^2\\\vdots&\vdots&\ddots&\vdots\\t_1^n&t_2^n&\cdots&t_m^n\end{pmatrix}
We call this an n\times m matrix, because the array is n rows high and m columns wide.
There is a natural isomorphism V\cong\hom(\mathbb{F},V). This means that every vector in dimension m, written out in the components relative to a given basis, can be seen as an m\times1 “column vector”:
\displaystyle\begin{pmatrix}v^1\\v^2\\\vdots\\v^m\end{pmatrix}
Similarly, a linear functional on an n-dimensional space can be written as a 1\times n “row vector”:
\displaystyle\begin{pmatrix}\mu_1&\mu_2&\cdots&\mu_n\end{pmatrix}
Notice that evaluation of linear transformations is now just a special case of matrix multiplication! Let’s practice by writing out the composition of a linear functional \mu\in\left(\mathbb{F}^n\right)^*, a linear map T:\mathbb{F}^m\rightarrow\mathbb{F}^n, and a vector v\in\mathbb{F}^m.
\displaystyle\begin{pmatrix}\mu_1&\mu_2&\cdots&\mu_n\end{pmatrix}\begin{pmatrix}t_1^1&t_2^1&\cdots&t_m^1\\t_1^2&t_2^2&\cdots&t_m^2\\\vdots&\vdots&\ddots&\vdots\\t_1^n&t_2^n&\cdots&t_m^n\end{pmatrix}\begin{pmatrix}v^1\\v^2\\\vdots\\v^m\end{pmatrix}
A matrix product makes sense if and only if the number of columns in the left-hand matrix is the same as the number of rows in the right-hand matrix. That is, an m\times n and an n\times p can be multiplied. The result will be an m\times p matrix. We calculate it by taking a row from the left-hand matrix and a column from the right-hand matrix. Since these are the same length (by assumption) we can multiply corresponding elements and sum up.
In the example above, the n\times m matrix \left(t_i^j\right) and the m\times1 matrix \left(v^i\right) can be multiplied. There is only one column in the latter to pick, so we simply choose row j out of n on the left: \begin{pmatrix}t_1^j&t_2^j&\cdots&t_m^j\end{pmatrix}. Multiplying corresponding elements and summing gives the single field element t_i^jv^i (remember the summation convention). We get n of these elements — one for each row — and we arrange them in a new n\times1 matrix:
\displaystyle\begin{pmatrix}t_1^1&t_2^1&\cdots&t_m^1\\t_1^2&t_2^2&\cdots&t_m^2\\\vdots&\vdots&\ddots&\vdots\\t_1^n&t_2^n&\cdots&t_m^n\end{pmatrix}\begin{pmatrix}v^1\\v^2\\\vdots\\v^m\end{pmatrix}=\begin{pmatrix}t_i^1v^i\\t_i^2v^i\\\vdots\\t_i^nv^i\end{pmatrix}
Then we can multiply the row vector \begin{pmatrix}\mu_1&\mu_2&\cdots&\mu_n\end{pmatrix} by this column vector to get the 1\times1 matrix:
\displaystyle\begin{pmatrix}\mu_1&\mu_2&\cdots&\mu_n\end{pmatrix}\begin{pmatrix}t_1^1&t_2^1&\cdots&t_m^1\\t_1^2&t_2^2&\cdots&t_m^2\\\vdots&\vdots&\ddots&\vdots\\t_1^n&t_2^n&\cdots&t_m^n\end{pmatrix}\begin{pmatrix}v^1\\v^2\\\vdots\\v^m\end{pmatrix}=\begin{pmatrix}\mu_jt_i^jv^i\end{pmatrix}
Just like we slip back and forth between vectors and m\times1 matrices, we will usually consider a field element and the 1\times1 matrix with that single entry as being pretty much the same thing.
The first multiplication here turned an m-dimensional (column) vector into an n-dimensional one, reflecting the source and target of the transformation T. Then we evaluated the linear functional \mu on the resulting vector. But by the associativity of matrix multiplication we could have first multiplied on the left:
\displaystyle\begin{pmatrix}\mu_1&\mu_2&\cdots&\mu_n\end{pmatrix}\begin{pmatrix}t_1^1&t_2^1&\cdots&t_m^1\\t_1^2&t_2^2&\cdots&t_m^2\\\vdots&\vdots&\ddots&\vdots\\t_1^n&t_2^n&\cdots&t_m^n\end{pmatrix}=\begin{pmatrix}\mu_jt_1^j&\mu_jt_2^j&\cdots&\mu_jt_m^j\end{pmatrix}
turning the linear functional on \mathbb{F}^n into one on \mathbb{F}^m. But this is just the dual transformation T^*:\left(\mathbb{F}^n\right)^*\rightarrow\left(\mathbb{F}^m\right)^*! Then we can evaluate this on the column vector to get the same result: \mu_jt_i^jv^j.
There is one slightly touchy thing we need to be careful about: Kronecker products. When the upper index is a pair (i_1,i_2) with 1\leq i_1\leq n_1 and 1\leq i_2\leq n_2 we have to pick an order on the set of such pairs. We’ll always use the “lexicographic” order. That is, we start with (1,1), then (1,2), and so on until (1,n_2) before starting over with (2,1), (2,2), and so on. Let’s write out a couple examples just to be clear:
\displaystyle\begin{pmatrix}s_1^1&s_2^1\\s_1^2&s_2^2\end{pmatrix}\boxtimes\begin{pmatrix}t_1^1&t_2^1&t_3^1\\t_1^2&t_2^2&t_3^2\end{pmatrix}=\begin{pmatrix}s_1^1t_1^1&s_1^1t_2^1&s_1^1t_3^1&s_2^1t_1^1&s_2^1t_2^1&s_2^1t_3^1\\s_1^1t_1^2&s_1^1t_2^2&s_1^1t_3^2&s_2^1t_1^2&s_2^1t_2^2&s_2^1t_3^2\\s_1^2t_1^1&s_1^2t_2^1&s_1^2t_3^1&s_2^2t_1^1&s_2^2t_2^1&s_2^2t_3^1\\s_1^2t_1^2&s_1^2t_2^2&s_1^2t_3^2&s_2^2t_1^2&s_2^2t_2^2&s_2^2t_3^2\end{pmatrix}
\displaystyle\begin{pmatrix}t_1^1&t_2^1&t_3^1\\t_1^2&t_2^2&t_3^2\end{pmatrix}\boxtimes\begin{pmatrix}s_1^1&s_2^1\\s_1^2&s_2^2\end{pmatrix}=\begin{pmatrix}t_1^1s_1^1&t_1^1s_2^1&t_2^1s_1^1&t_2^1s_2^1&t_3^1s_1^1&t_3^1s_2^1\\t_1^1s_1^2&t_1^1s_2^2&t_2^1s_1^2&t_2^1s_2^2&t_3^1s_1^2&t_3^1s_2^2\\t_1^2s_1^1&t_1^2s_2^1&t_2^2s_1^1&t_2^2s_2^1&t_3^2s_1^1&t_3^2s_2^1\\t_1^2s_1^2&t_1^2s_2^2&t_2^2s_1^2&t_2^2s_2^2&t_3^2s_1^2&t_3^2s_2^2\end{pmatrix}
So the Kronecker product depends on the order of multiplication. But this dependence is somewhat illusory. The only real difference is reordering the bases we use for the tensor products of the vector spaces involved, and so a change of basis can turn one into the other. This is an example of how matrices can carry artifacts of our choice of bases.
May 30, 2008 Posted by | Algebra, Linear Algebra | 7 Comments
Matrices IV
Like we saw with the tensor product of vector spaces, the dual space construction turns out to be a functor. In fact, it’s a contravariant functor. That is, if we have a linear transformation T:U\rightarrow V we get a linear transformation T^*:V^*\rightarrow U^*. As usual, we ask what this looks like for matrices.
First, how do we define the dual transformation? It turns out this is the contravariant functor represented by \mathbb{F}. That is, if \mu:V\rightarrow\mathbb{F} is a linear functional, we define T^*(\mu)=\mu\circ T:U\rightarrow\mathbb{F}. In terms of the action on vectors, \left[T^*(\mu)\right](v)=\mu(T(v))
Now let’s assume that U and V are finite-dimensional, and pick bases \left\{e_i\right\} and \left\{f_k\right\} for U and V, respectively. Then the linear transformation T has matrix coefficients t_i^k. We also get the dual bases \left\{\epsilon^j\right\} of U^* and \left\{\phi^l\right\} of V^*.
Given a basic linear functional \phi^l on V, we want to write T^*(\phi^l) in terms of the \epsilon^j. So let’s evaluate it on a generic basis vector e_i and see what we get. The formula above shows us that
\left[T^*(\phi^l)\right](e_i)=\phi^l(T(e_i))=\phi^l(t_i^kf_k)=t_i^k\delta_k^l=t_i^l
In other words, we can write T^*(\phi^l)=t_j^l\epsilon^j. The same matrix works, but we use its indices differently.
In general, given a linear functional \mu with coefficients \mu_l we find the coefficients of T^*(\mu) as t_j^l\mu_l. The value \left[T^*(\mu)\right](v)=\mu(T(u)) becomes \mu_lt_i^lu^i. Notice that the summation convention tells us this must be a scalar (as we expect) because there are no unpaired indices. Also notice that because we can use the same matrix for two different transformations we seem to have an ambiguity: is the lower index running over a basis for U or one for U^*? Luckily, since every basis gives rise to a dual basis, we don’t need to care. Both spaces have the same dimension anyhow.
May 28, 2008 Posted by | Algebra, Linear Algebra | 2 Comments
Dual Spaces
Another thing vector spaces come with is duals. That is, given a vector space V we have the dual vector space V^*=\hom(V,\mathbb{F}) of “linear functionals” on V — linear functions from V to the base field \mathbb{F}. Again we ask how this looks in terms of bases.
So let’s take a finite-dimensional vector space V with basis \left\{e_i\right\}, and consider some linear functional \mu\in V^*. Like any linear function, we can write down matrix coefficients \mu_i=\mu(e_i). Notice that since our target space (the base field \mathbb{F}) is only one-dimensional, we don’t need another index to count its basis.
Now let’s consider a specially-crafted linear functional. We can define one however we like on the basis vectors e_i and then let linearity handle the rest. So let’s say our functional takes the value {1} on e_1 and the value {0} on every other basis element. We’ll call this linear functional \epsilon^1. Notice that on any vector we have
\epsilon^1(v)=\epsilon^1(v^ie_i)=v^i\epsilon^1(e_i)=v^1
so it returns the coefficient of e_1. There’s nothing special about e_1 here, though. We can define functionals \epsilon^j by setting \epsilon^j(e_i)=\delta_i^j. This is the “Kronecker delta”, and it has the value {1} when its two indices match, and {0} when they don’t.
Now given a linear functional \mu with matrix coefficients \mu_j, let’s write out a new linear functional \mu_j\epsilon^j. What does this do to basis elements?
\mu_j\epsilon^j(e_i)=\mu_j\delta_i^j=\mu_i
so this new transformation has exactly the same matrix as \mu does. It must be the same transformation! So any linear functional can be written uniquely as a linear combination of the \epsilon^j, and thus they form a basis for the dual space. We call \left\{\epsilon^j\right\} the “dual basis” to \left\{e_i\right\}.
Now if we take a generic linear functional \mu and evaluate it on a generic vector v we find
\mu(v)=\mu_j\epsilon^j(v^ie_i)=\mu_jv^i\epsilon^j(e_i)=\mu_jv^i\delta_i^j=\mu_iv^i
Once we pick a basis for V we immediately get a basis for V^*, and evaluation of a linear functional on a vector looks neat in terms of these bases.
May 27, 2008 Posted by | Algebra, Linear Algebra | 30 Comments
Matrices III
Given two finite-dimensional vector spaces U and V, with bases \left\{e_i\right\} and \left\{f_j\right\} respectively, we know how to build a tensor product: use the basis \left\{e_i\otimes f_j\right\}.
But an important thing about the tensor product is that it’s a functor. That is, if we have linear transformations S:U\rightarrow U' and T:V\rightarrow V', then we get a linear transformation S\otimes T:U\otimes V\rightarrow U'\otimes V'. So what does this operation look like in terms of matrices?
First we have to remember exactly how we get the tensor product S\otimes T. Clearly we can consider the function S\times T:U\times V\rightarrow U'\times V'. Then we can compose with the bilinear function U'\times V'\rightarrow U'\otimes V' to get a bilinear function from U\times V to U'\otimes V'. By the universal property, this must factor uniquely through a linear function U\otimes V\rightarrow U'\otimes V'. It is this map we call S\otimes T.
We have to pick bases \left\{e_k'\right\} of U' and \left\{f_l'\right\} of V'. This gives us a matrix coefficients s_i^k for S and t_j^l for T. To calculate the matrix for S\otimes T we have to evaluate it on the basis elements e_i\otimes f_j of U\otimes V. By definition we find:
\left[S\otimes T\right](e_i\otimes f_j)=S(e_i)\otimes T(f_j)=\left(s_i^ke_k'\right)\otimes\left(t_j^lf_l'\right)=s_i^kt_j^le_k'\otimes f_l'
that is, the matrix coefficient between the index pair (i,j) and the index pair (k,l) is s_i^kt_j^l.
It’s not often taught anymore, but there is a name for this operation: the Kronecker product. If we write the matrices (as opposed to just their coefficients) \left(s_i^k\right) and \left(t_j^l\right), then we write the Kronecker product \left(s_i^k\right)\boxtimes\left(t_j^l\right)=\left(s_i^kt_j^l\right).
May 26, 2008 Posted by | Algebra, Linear Algebra | 6 Comments
Tensor Products and Bases
Since we’re looking at vector spaces, which are special kinds of modules, we know that \mathbf{Vec}(\mathbb{F}) has a tensor product structure. Let’s see what this means when we pick bases.
First off, let’s remember what the tensor product of two vector spaces U and V is. It’s a new vector space U\otimes V and a bilinear (linear in each of two variables separately) function B:U\times V\rightarrow U\otimes V satisfying a certain universal property. Specifically, if F:U\times V\rightarrow W is any bilinear function it must factor uniquely through B as F=\bar{F}\circ B. The catch here is that when we say “linear” and “bilinear” we mean that the functions preserve both addition and scalar multiplication. As with any other universal property, such a tensor product will be uniquely defined up to isomorphism.
So let’s take finite-dimensional vector spaces U and V, and bases \left\{e_i\right\} of U and \left\{f_j\right\} of V. I say that the vector space with basis \left\{e_i\otimes f_j\right\}, and with the bilinear function B(e_i,f_j)=e_i\otimes f_j is a tensor product. Here the expression e_i\otimes f_j is just a name for a basis element of the new vector space. Such elements are indexed by the set of pairs (i,j), where i indexes a basis for U and j indexes a basis for V.
First off, what do I mean by the bilinear function B(e_i,f_j)=e_i\otimes f_j? Just as for linear functions, we can define bilinear functions by defining them on bases. That is, if we have u=u^ie_i\in U and v=v^jf_j\in V, we get the vector
B(u,v)=B(u^ie_i,v^je_j)=u^iv^jB(e_i,f_j)=u^iv^je_i\otimes f_j
in our new vector space, with coefficients u^iv^j.
So let’s take a bilinear function F and define a linear function \bar{F} by setting
\bar{F}(e_i\otimes f_j)=F(e_i,f_j)
We can easily check that F does indeed factor as desired, since
\bar{F}(B(e_i,f_j))=\bar{F}(e_i\otimes f_j)=F(e_i,f_j)
so F=\bar{F}\circ B on basis elements. By linearity, they must agree for all pairs (u,v). It should also be clear that we can’t define \bar{F} any other way and hope to satisfy this equation, so the factorization is unique.
Thus if we have bases \left\{e_i\right\} of U and \left\{f_j\right\} of V, we immediately get a basis \left\{e_i\otimes f_j\right\} of U\otimes V. As a side note, we immediately see that the dimension of the tensor product of two vector spaces is the product of their dimensions.
May 23, 2008 Posted by | Algebra, Linear Algebra | 5 Comments
Matrices II
With the summation convention firmly in hand, we continue our discussion of matrices.
We’ve said before that the category of vector space is enriched over itself. That is, if we have vector spaces U and V over the field \mathbb{F}, the set of linear transformations \hom(U,V) is itself a vector space over \mathbb{F}. In fact, it inherits this structure from the one on V. We define the sum and the scalar product
\left[S+T\right](u)=S(u)+T(u)
\left[cT\right](u)=cT(u)
for linear transformations S and T from U to V, and for a constant c\in\mathbb{F}. Verifying that these are also linear transformations is straightforward.
So what do these structures look like in the language of matrices? If U and V are finite-dimensional, let’s pick bases \left\{e_i\right\} of U and \left\{f_j\right\} of V. Now we get matrix coefficients s_i^j and t_i^j, where i indexes the basis of U and j indexes the basis of V. Now we can calculate the matrices of the sum and scalar product above.
We do this, as usual, by calculating the value the transformations take at each basis element. First, the sum:
\left[S+T\right](e_i)=S(e_i)+T(e_i)=s_i^jf_j+t_i^jf_j=(s_i^j+t_i^j)f_j
and now the scalar product:
\left[cT\right](e_i)=cT(e_i)=(ct_i^j)f_j
so we calculate the matrix coefficients of the sum of two linear transformations by adding the corresponding matrix coefficients of each transformation, and the matrix coefficients of the scalar product by multiplying each coefficient by the same scalar.
May 22, 2008 Posted by | Algebra, Linear Algebra | 1 Comment
The Einstein Summation Convention
Look at the formulas we were using yesterday. There’s a lot of summations in there, and a lot of big sigmas. Those get really tiring to write over and over, and they get tiring really quick. Back when Einstein was writing up his papers, he used a lot of linear transformations, and wrote them all out in matrices. Accordingly, he used a lot of those big sigmas.
When we’re typing nowadays, or when we write on a pad or on the board, this isn’t a problem. But remember that up until very recently, publications had to actually set type. Actual little pieces of metal with characters raised (and reversed!) on them would get slathered with ink and pressed to paper. Incidentally, this is why companies that produce fonts are called “type foundries”. They actually forged those metal bits with letter shapes in different styles, and sold sets of them to printers.
Now Einstein was using a lot of these big sigmas, and there were pages that had so many of them that the printer would run out! Even if they set one page at once and printed them off, they just didn’t have enough little pieces of metal with big sigmas on them to handle it. Clearly something needed to be done to cut down on demand for them.
Here we note that we’re always summing over some basis. Even if there’s no basis element in a formula — say, the formula for a matrix product — the summation is over the dimension of some vector space. We also notice that when we chose to write some of our indices as subscripts and some as superscripts, we’re always summing over one of each. We now adopt the convention that if we ever see a repeated index — once as a superscript and once as a subscript — we’ll read that as summing over an appropriate basis.
For example, when we wanted to write a vector v\in V, we had to take the basis \{f_j\} of V and write up the sum
\displaystyle v=\sum\limits_{j=1}^{\dim(V)}v^jf_j
but now we just write v^jf_j. The repeated index and the fact that we’re talking about a vector in V means we sum for j running from {1} to the dimension of V. Similarly we write out the value of a linear transformation on a basis vector: T(f_j)=t_j^kg_k. Here we determine from context that k should run from {1} to the dimension of W.
What about finding the coefficients of a linear transformation acting on a vector? Before we wrote this as
\displaystyle T(v)^k=\sum\limits_{j=1}^{\dim(V)}t_j^kv^j
Where now we write the result as t_j^kv^j. Since the v^j are the coefficients of a vector in V, j must run from {1} to the dimension of V.
And similarly given linear transformations S:U\rightarrow V and T:V\rightarrow W represented (given choices of bases) by the matrices with components s_i^j and t_j^k, the matrix of their product is then written s_i^jt_j^k. Again, we determine from context that we should be summing j over a set indexing a basis for V.
One very important thing to note here is that it’s not going to matter what basis for V we use here! I’m not going to prove this quite yet, but built right into this notation is the fact that the composite of the two transformations is completely independent of the choice of basis of V. Of course, the matrix of the composite still depends on the bases of U and W we pick, but the dependence on V vanishes as we take the sum.
Einstein had a slightly easier time of things: he was always dealing with four-dimensional vector spaces, so all his indices had the same range of summation. We’ve got to pay some attention here and be careful about what vector space a given index is talking about, but in the long run it saves a lot of time.
May 21, 2008 Posted by | Fundamentals, Linear Algebra | 18 Comments
Matrices I
Yesterday we talked about the high-level views of linear algebra. That is, we’re discussing the category \mathbf{Vec}(\mathbb{F}) of vector spaces over a field \mathbb{F} and \mathbb{F}-linear transformations between them.
More concretely, now: we know that every vector space over \mathbb{F} is free as a module over \mathbb{F}. That is, every vector space has a basis — a set of vectors so that every other vector can be uniquely written as an \mathbb{F}-linear combination of them — though a basis is far from unique. Just how nonunique it is will be one of our subjects going forward.
Now if we’ve got a linear transformation T:V\rightarrow W from one finite-dimensional vector space to another, and if we have a basis \{f_j\}_{j=1}^{\mathrm{dim}(V)} of V and a basis \{g_k\}_{k=1}^{\mathrm{dim}(W)} of W, we can use these to write the transformation T in a particular form: as a matrix. Take the transformation and apply it to each basis element of V to get vectors T(f_j)\in W. These can be written uniquely as linear combinations
\displaystyle T(f_j)=\sum\limits_{k=1}^{\mathrm{dim}(W)}t_j^kg_k
for certain t_j^k\in\mathbb{F}. These coefficients, collected together, we call a matrix. They’re enough to calculate the value of the transformation on any vector v\in V, because we can write
\displaystyle v=\sum\limits_{j=1}^{\mathrm{dim}(V)}v^jf_j
We’re writing the indices of the components as superscripts here, just go with it. Then we can evaluate T(v) using linearity
\displaystyle T(v)=T\left(\sum\limits_{j=1}^{\mathrm{dim}(V)}v^jf_j\right)=\sum\limits_{j=1}^{\mathrm{dim}(V)}v^jT(f_j)=
\displaystyle=\sum\limits_{j=1}^{\mathrm{dim}(V)}v^j\sum\limits_{k=1}^{\mathrm{dim}(W)}t_j^kg_k=\sum\limits_{k=1}^{\mathrm{dim}(W)}\left(\sum\limits_{j=1}^{\mathrm{dim}(V)}t_j^kv^j\right)g_k
So the coefficients v^j defining the vector v\in V and the matrix coefficients t_j^k together give us the coefficients defining the vector T(v)\in W.
If we have another finite-dimensional vector space U with basis \{e_i\}_{i=1}^{\mathrm{dim}(U)} and another transformation S:U\rightarrow V then we have another matrix
\displaystyle S(e_i)=\sum_{j=1}^{\mathrm{dim}(V)}s_i^jf_j
Now we can compose these two transformations and calculate the result on a basis element
\displaystyle \left[T\circ S\right](e_i)=T\left(S(e_i)\right)=T\left(\sum_{j=1}^{\mathrm{dim}(V)}s_i^jf_j\right)=\sum_{j=1}^{\mathrm{dim}(V)}s_i^jT(f_j)=
\displaystyle=\sum_{j=1}^{\mathrm{dim}(V)}s_i^j\sum\limits_{k=1}^{\mathrm{dim}(W)}t_j^kg_k=\sum\limits_{k=1}^{\mathrm{dim}(W)}\left(\sum_{j=1}^{\mathrm{dim}(V)}t_j^ks_i^j\right)g_k
This last quantity in parens is then the matrix of the composite transformation T\circ S. Thus we can represent the operation of composition by this formula for matrix multiplication.
May 20, 2008 Posted by | Algebra, Linear Algebra | 7 Comments
Linear Algebra
Here we begin a discussion of linear algebra. There are three views on what this is all about.
The mid-level view is that we’re studying the properties of linear maps — homomorphisms — between abelian groups, and particularly between modules or vector spaces, which are just modules over a field. In particular we’ll focus on vector spaces over some arbitrary (but fixed!) field \mathbb{F}.
The high-level view is that what we’re really studying is the category of \mathbb{F}-modules. Categories are all about how morphisms between their objects interact, so this is what we’re really after here. And it turns out we already know a lot about these sorts of categories! Specifically, they’re abelian categories. In fact, since we’re working over a field \mathbb{F} (which is a commutative ring) the properties of \hom-functors tell us that \mathbb{F}-\mathbf{mod} is enriched over itself.
So this tells us that our category of vector spaces has a biproduct — the direct sum — and in particular a zero object — the trivial {0}-dimensional vector space \mathbf{0}. It also has a tensor product, which makes this a monoidal category, using the one-dimensional vector space \mathbb{F} itself as monoidal identity. We also know that kernels and cokernels exist, which then (along with biproducts) give us all finite limits and colimits.
The third viewpoint is that we’re talking about solving systems of linear equations, and that’s where “linear algebra” comes from. The connection between these abstract formulations and that concrete one is a bit mysterious at first blush, but we’ll start making it tomorrow.
May 19, 2008 Posted by | Algebra, Linear Algebra | 4 Comments
Commutativity in Series III
Okay, here’s the part I promised I’d finish last Friday. How do we deal with rearrangements that “go to infinity” more than once? That is, we chop up the infinite set of natural numbers into a bunch of other infinite sets, add each of these subseries up, and then add the results up. If the original series was absolutely convergent, we’ll get the same answer.
First of all, if a series \sum_{k=0}^\infty a_k converges absolutely, then so does any subseries \sum_{j=0}^\infty a_{p(j)}, where p is an injective (but not necessarily bijective!) function from the natural numbers to themselves. For instance, we could let p(j)=2j and add up all the even terms from the original series.
To see this, notice that at any finite n we have a maximum value N=\max\limits_{0\leq j\leq n}p(j). Then we find
\displaystyle\left|\sum\limits_{j=0}^na_{p(j)}\right|\leq\sum\limits_{j=0}^n\left|a_{p(j)}\right|\leq\sum\limits_{k=0}^N\left|a_k\right|\leq\sum\limits_{k=0}^\infty\left|a_k\right|
So the new sequence of partial sums of absolute values is increasing and bounded above, and thus converges.
Now let’s let p_0, p_1, p_2, and so on be a countable collection of functions defined on the natural numbers. We ask that
• Each p_n is injective.
• The image of p_n is a subset P_k\subseteq\mathbb{N}.
• The collection \left\{P_0,P_1,P_2,...\right\} is a partition of \mathbb{N}. That is, these subsets are mutually disjoint, and their union is all of \mathbb{N}.
If \sum_{k=0}^\infty a_k is an absolutely convergent series, we define \left(b_n\right)_j=a_{p_n(j)} — the subseries defined by p_n. Then from what we said above, each \sum_{j=0}^\infty\left(b_n\right)_j is an absolutely convergent series whose sum we call s_n. We assert now that \sum_{n=0}^\infty s_n is an absolutely convergent series whose sum is the same as that of \sum_{k=0}^\infty a_k.
Let’s set t_m=\sum_{n=0}^m\left|s_n\right|. That is, we have
\displaystyle t_m\leq\sum\limits_{j=0}^\infty\left|\left(b_1\right)_j\right|+...+\sum\limits_{j=0}^\infty\left|\left(b_m\right)_j\right|=\sum\limits_{j=0}^\infty\left(\left|\left(b_1\right)_j\right|+...+\left|\left(b_m\right)_j\right|\right)
But this is just the sum of a bunch of absolute values from the original series, and so is bounded by \sum_{k=0}^\infty\left|a_k\right|. So the series of absolute values of s_n has bounded partial sums, and so \sum_{n=0}^\infty s_n converges absolutely. That it has the same sum as the original is another argument exactly analogous to (but more complicated than) the one for a simple rearrangement, and for associativity of absolutely convergent series.
This pretty much wraps up all I want to say about calculus for now. I’m going to take a little time to regroup before I dive into linear algebra in more detail than the abstract algebra I covered before. But if you want to get ahead, go back and look over what I said about rings and modules. A lot of that will be revisited and fleshed out in the next sections.
May 12, 2008 Posted by | Analysis, Calculus | 3 Comments
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2,899,967,657,269,498,000 | You may also like
Geoboards
This practical challenge invites you to investigate the different squares you can make on a square geoboard or pegboard.
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If you had 36 cubes, what different cuboids could you make?
Tea Cups
Age 7 to 14
Challenge Level
pic 2
Aunt Jane had been to a jumble sale and bought a whole lot of cups and saucers - she's having many visitors these days and felt that she needed some more. You are staying with her and when she arrives home you help her to unpack the cups and saucers.
There are four sets: a set of white, a set of red, a set of blue and a set of green. In each set there are four cups and four saucers. So there are sixteen cups and sixteen saucers altogether.
Just for the fun of it, you decide to mix them around a bit so that there are sixteen different-looking cup/saucer combinations laid out on the table in a very long line.
So, for example:
a) there is a red cup on a green saucer but not another the same, although there is a green cup on a red saucer;
b) there is a red cup on a red saucer but that's the only one like it.
There are these sixteen different cup/saucer combinations on the table and you think about arranging them in a big square. Because there are sixteen, you realise that there are going to be four rows with four in each row (or if you like, four rows and four columns).
So here is the challenge to start off this investigation:
Place these sixteen different combinations of cup/saucer in this four by four arrangement with the following rules:-
1) In any row there must only be one cup of each colour;
2) In any row there must only be one saucer of each colour;
3) In any column there must only be one cup of each colour;
4) In any column there must be only one saucer of each colour.
Remember that these sixteen cup/saucers are all different so, for example, you CANNOT have a red cup on a green saucer somewhere and another red cup on a green saucer somewhere else.
There are a lot of different ways of approaching this challenge.
When you think you have completed it, check it through very carefully. It's even a good idea to get a friend who has seen the rules to check it also.
This challenge is also found at nrich.maths.org.uk/7397
Printable Roadshow.pdf resource
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1,483,385,459,345,568,500 | Démontrer qu’un triangle est rectangle isocèle
Merci !
Annales corrigées
Classe(s) : Tle S | Thème(s) : Nombres complexes et applications
Type : Exercice | Année : 2012 | Académie : Inédit
Unit 1 - | Corpus Sujets - 1 Sujet
Démontrer qu’un triangle est rectangle isocèle
Nombres complexes et applications
Corrigé
24
Ens. spécifique
matT_1200_00_48C
Sujet inédit
Exercice • 5 points
PARTIE A
On considère, dans l’ensemble des nombres complexes, l’équation (E) :
.
> 1. Démontrer que 2 est une solution de (E). (0,75 point)
> 2. Démontrer que (E) peut s’écrire sous la forme
a, b et c sont trois réels que l’on déterminera. (1 point)
> 3. En déduire les solutions de l’équation (E) sous forme algébrique, puis sous forme trigonométrique avec la notation exponentielle. (0,5 point)
PARTIE B
Le plan complexe est muni d’un repère orthonormé .
> 1. Placer les points A, B et D d’affixes respectives :
, et . (0,75 point)
> 2. Calculer l’affixe du point C tel que le quadrilatère ABCD soit un parallélogramme. Placer C. (1 point)
> 3. Démontrer que le triangle OAD est rectangle isocèle. (1 point)
Durée conseillée : 40 min.
Le thème en jeu
Nombres complexes.
Les conseils du correcteur
Partie A
> 1. Remplacez par la valeur 2 dans l’expression .
> 2. D’après la question 1., vous pouvez factoriser l’expression par . À l’aide d’un développement et d’une identification de coefficients, déduisez-en les valeurs de a, b et c.
> 3. Une fois les valeurs de a, b et c trouvées, résolvez dans une équation du second degré. → fiche C36 B
É crivez ensuite les nombres complexes sous forme trigonométrique, avec la notation exponentielle. → fiche C35
Partie B
> 1. Remarquez que deux des nombres complexes sont des nombres conjugués et que le troisième est un nombre réel. → fiche C33
> 2. Si ABCD est un parallélogramme, que peut-on en déduire vectoriellement ? Puis pour les affixes ? → fiche C37
> 3. Remarquez que zA, zB et zD sont les solutions de l’équation (E) de la partie A. Vous connaissez donc leurs modules et leurs arguments grâce à leurs notations exponentielles. Démontrez alors que à l’aide des modules, puis déterminez une mesure de l’angle en utilisant la relation de Chasles. → fiche C37 B | {
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1,164,291,681,980,212,700 | How to Plot and Name Points on a Coordinate Plane (Graph)
Use a ruler and pencil to create a coordinate plane and graph your slope.
••• ruler and pencil image by Hao Wang from Fotolia.com
A very common task in math class is to plot and name points on a rectangular coordinate plane, more commonly known as a four-quadrant graph. While this is not at all difficult, many students have a hard time with this task, which leads to difficulty in later math topics which depend on this basic skill. Learning how to read the graph simplifies this task.
Study the coordinate plane. Note that the plane (flat, 2D surface) is divided into four quadrants by way of two axes, one vertical and one horizontal. The horizontal axis is called the x-axis, and the vertical axis is the y-axis. Each axis has a positive and negative side. For the x-axis, the positive side is to the right of the y-axis and the negative side is to the left. For the y-axis, the positive side is above the x-axis and the negative side is below it.
Look where the two axes cross. This point is the origin; its coordinates are (0, 0). This means that its "address" is 0 on the x-axis, and 0 on the y-axis. Each increment away from the axes is marked by another grid line. The grid lines often indicate a value of 1, for example, the line one increment up from the x-axis has a y-value of 1, the line two increments up has a y-value of 2. For some purposes, however, each increment might have a value of 10, 100 or 1,000.
Whether the value corresponds with the x- or y-axis is determined by the direction of movement. If your point is vertical, either up or down, this represents a y-value. If your point is horizontal, left or right, this represents an x-value.
Determine first how far left or right to go to be in line with the point, then determine how far up or down to go. The format of a coordinate is always (x, y), written like that in parentheses, separated by a comma. The x-coordinate is always listed before the y-coordinate. You can remember that they are in alphabetical order.
Therefore, to plot the point (-2, 5), move two units to the left relative to the x-axis, and five units up relative to the y-axis.
To name a point on a graph, count how many units from the origin (0,0) to the left or right you go for the x-value, then count how many units up or down you go to get the y-value.
Tips
• If the point rests on the y-axis, it has an x-value of zero. If it rests on the x-axis, it has a y-value of zero.
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-4,663,823,277,789,317,000 | Edit Article
How to Calculate a Discount
Three Methods:Calculating the Discount and Sale PriceEstimating the Discount and Sale PriceCompleting Sample ProblemsCommunity Q&A
Calculating a discount is one of the most useful math skills you can learn. You can apply it to tips at a restaurant, sales in stores, and setting rates for your own services. The basic way to calculate a discount is to multiply the original price by the decimal form of the percentage. To calculate the sale price of an item, subtract the discount from the original price. You can do this using a calculator, or you can round the price and estimate the discount in your head.
1
Calculating the Discount and Sale Price
1. 1
Convert the percentage discount to a decimal. To do this, think of the percent number with a decimal to the right of the last digit. Move the decimal point two places to the left to get the converted decimal.[1] You can also use the sign on a calculator.
• For example, you might want to calculate the sale price of a pair of shoes that is regularly $69.95. If the shoes are 25% off, you need to convert 25% to a decimal by thinking .
2. 2
Multiply the original price by the decimal. You can multiply the decimal by hand, or use a calculator. This will tell you the discount, or what value is being taken off the original price.[2]
• For example, to find the 25% discount on a pair of $69.95 shoes, you would calculate .
3. 3
Subtract the discount from the original price. To subtract decimals, line up the decimal points and subtract as you would whole numbers. Be careful to drop the decimal point down into your answer. You can also use a calculator. The difference will be the sale price of the item.[3]
• For example, if a pair of shoes that are originally $69.95 have a discount of $17.49, calculate the sale price by subtracting:. So, the shoes are on sale for $52.46.
2
Estimating the Discount and Sale Price
1. 1
Round the original price to the nearest ten. Use normal rounding rules to round up or down. Doing this will make it easier to calculate the percent discount of the number.[4]
• For example, if the original price of a shirt is $47.89, round the price up to $50.00
2. 2
Calculate 10 percent of the rounded price. To mentally calculate 10% of a price, think of the price written as dollars and cents with a decimal point. Then, move the decimal point one place to the left. This will show you the number that is equal to 10%.[5]
• For example, to calculate 10% of $50, think . So, 5 is 10% of 50.
3. 3
Determine the number of tens in the percent off. To figure out the number of tens, divide by the percentage by 10 using normal division rules. Don’t worry about fives in the percentage for now.
• For example, if a shirt is 35% off, you would need to know how many tens are in 35. Since , there are 3 tens in 35.
4. 4
Multiply 10% of the rounded price by the appropriate factor. The factor is determined by the number of tens in the percent off. Since you determined what 10% of the price is, find a larger percent by multiplying by the number of tens.
• For example, if you found that 10% of $50 is 5, to find out how much 30% of 50 is, you would multiply $5 by 3, since there are 3 tens in 30: . So, 30% of $50 is $15.
5. 5
Calculate 5% of the rounded price, if necessary. You will need to do this step if the percent off discount ends in a 5 rather than a 0 (for example, 35% or 55% off). It is easy to calculate 5% by simply dividing 10% of the original price by 2, since 5% is half of 10%.
• For example, if 10% of $50 is $5, then 5% of $50 is $2.50, since $2.50 is half of $5.
6. 6
Add the remaining 5% to the discount, if necessary. This will give you the total estimated discount of the item.
• For example, if a shirt is 35% off, you first found 30% of the original price was $15. Then you found that 5% of the original price was $2.50. So adding the values of 30% and 5%, you get . So, the estimated discount of the shirt is $17.50.
7. 7
Subtract the discount from the rounded price. This will give you an estimate of the sale price of the item.
• For example, if the rounded price of a shirt is $50, and you found the 35% discount to be $17.50, you would calculate . So, a $47.89 shirt that is 35% off is about $32.50 on sale.
3
Completing Sample Problems
1. 1
Calculate the exact sale price. A television is originally priced at $154.88. It now has a 40% discount.
• Convert the percentage discount to a decimal by moving the decimal two places to the left: .
• Multiply the original price by the decimal: .
• Subtract the discount from the original price: . So, the sale price of the television is $92.93.
2. 2
Find the exact sale price of a camera that is 15% off. The original price is $449.95.
• Convert the percentage discount to a decimal by moving the decimal two places to the left: .
• Multiply the original price by the decimal: .
• Subtract the discount from the original price: . So, the sale price of the camera is $382.46.
3. 3
Estimate the sale price. A tablet is regularly $199.99. On sale, it is 45% off.
• Round the original price to the nearest ten. Since $199.99 is only 1 cent away from $200, you would round up.
• Calculate 10% of the rounded price. Moving the decimal one place to the left, you should see that 10% of $200.00 is $20.00.
• Determine the number of tens in the percent off. Since , you know that there are 4 tens in 45%.
• Multiply 10% of the rounded price by the appropriate factor. Since the percentage off is 45%, you would multiply 10% of the rounded price by 4:
• Calculate 5% of the rounded price. This is half of 10%, which is $20. So half of $20 is $10.
• Add the remaining 5% to the discount. 40% is $80, and 5% is $10, so 45% is $90.
• Subtract the discount from the rounded price: . So the estimated sale price is $110.
Community Q&A
Search
Add New Question
• Question
How do I calculate a discount if the rate is unknown?
Judy McGuirk
Community Answer
Assuming you know the original price and the sale price of an item, subtract sale price from original price to determine the discount amount. Next, divide the discount amount by original price. Convert this decimal amount into a percentage. This percent is the discount rate. For an example, a lamp shows a discount price of $30 with an original price of $50. $50 - $30 = $20 20 / 50 = 0.40 0.40 = 40%
• Question
How would I calculate a discount of 2.5 percent on an item?
wikiHow Contributor
Community Answer
if something costs $100, multiply the total amount being 100 by the amount of 2.5 percent after moving the decimal over two spaces to the left. So, it should look like $100 times .025 equals $2.50.
• Question
How do I calculate the marked price before discount?
wikiHow Contributor
Community Answer
Divide by the percent discounted, taken from 100%. Say an item on 20% discount has a price tag reading £40. 100% - 20% is 80%. £40 / 0.8 = £50. (The answer, 80% of £50 is £40).
• Question
A product that regularly sells for $425 is marked down to $318.75. What is the discount rate?
Lorejane Castillo
Community Answer
If the product regular price is $425 and the discounted price is $318.75. Divide the disc. price to orig. price EX. 318.75 / 425= 0.75, then multiply the 0.75 into 100 and 100 is the total % of an item Ex. 0.75x100= 75, now subtract the 100 to 75 . Ex. 100- 75= 25%. So the discount rate is = 25%.
• Question
Can I convert the decimal to a fraction, then multiply by the original price?
wikiHow Contributor
Community Answer
Yes, as long as your conversion is right.
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Quick Summary
To calculate a discount, start by converting the discount percentage to a decimal. Think of the percent number with a decimal to the right of the last digit, then move the decimal point two places to the left to convert it. Next, multiply the original price by the decimal either by hand or by using a calculator. This gives you the discount amount, or what value is being taken off the original price. Then, subtract the discount amount from the original price to figure out the discounted price.
Did this summary help you?
Tips
• Download a dedicated discount calculator on your smartphone. Search for “discount calculator” in Google Play or the App Store. Then, open it and click the button to set the percentage discount and type in your item price. Press “Calculate” to find your discount.
• You can also calculate the sale price automatically, instead of subtracting the discount from the original price. Subtract the discount percentage from 100. For example, if your discount is 30 percent, your remaining price will be 70 percent of its original price. Then, use the same methods to calculate 70 percent of the original price to find your new price.[6]
Article Info
Categories: Economics
In other languages:
Nederlands: De korting op een product berekenen, ไทย: คำนวณส่วนลด, العربية: حساب الخصومات على الأسعار
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-7,354,054,533,388,445,000 | 25814
25,814 (twenty-five thousand eight hundred fourteen) is an even five-digits composite number following 25813 and preceding 25815. In scientific notation, it is written as 2.5814 × 104. The sum of its digits is 20. It has a total of 2 prime factors and 4 positive divisors. There are 12,906 positive integers (up to 25814) that are relatively prime to 25814.
Basic properties
• Is Prime? No
• Number parity Even
• Number length 5
• Sum of Digits 20
• Digital Root 2
Name
Short name 25 thousand 814
Full name twenty-five thousand eight hundred fourteen
Notation
Scientific notation 2.5814 × 104
Engineering notation 25.814 × 103
Prime Factorization of 25814
Prime Factorization 2 × 12907
Composite number
ω(n) Distinct Factors 2
Total number of distinct prime factors
Ω(n) Total Factors 2
Total number of prime factors
rad(n) Radical 25814
Product of the distinct prime numbers
λ(n) Liouville Lambda 1
Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n)
μ(n) Mobius Mu 1
Returns:
• 1, if n has an even number of prime factors (and is square free)
• −1, if n has an odd number of prime factors (and is square free)
• 0, if n has a squared prime factor
Λ(n) Mangoldt function 0
Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 25,814 is 2 × 12907. Since it has a total of 2 prime factors, 25,814 is a composite number.
Divisors of 25814
1, 2, 12907, 25814
4 divisors
Even divisors 2
Odd divisors 2
4k+1 divisors 1
4k+3 divisors 1
τ(n) Total Divisors 4
Total number of the positive divisors of n
σ(n) Sum of Divisors 38724
Sum of all the positive divisors of n
s(n) Aliquot Sum 12910
Sum of the proper positive divisors of n
A(n) Arithmetic Mean 9681
Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n))
G(n) Geometric Mean 160.66735822811
Returns the nth root of the product of n divisors
H(n) Harmonic Mean 2.6664600764384
Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 25,814 can be divided by 4 positive divisors (out of which 2 are even, and 2 are odd). The sum of these divisors (counting 25,814) is 38,724, the average is 9,681.
Other Arithmetic Functions (n = 25814)
1 φ(n) n
φ(n) Euler Totient 12906
Total number of positive integers not greater than n that are coprime to n
λ(n) Carmichael Lambda 12906
Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n
π(n) Prime Pi ≈ 2842
Total number of primes less than or equal to n
r2(n) Sum of 2 squares 0
The number of ways n can be represented as the sum of 2 squares
There are 12,906 positive integers (less than 25,814) that are coprime with 25,814. And there are approximately 2,842 prime numbers less than or equal to 25,814.
Divisibility of 25814
m 2 3 4 5 6 7 8 9
n mod m 0 2 2 4 2 5 6 2
The number 25,814 is divisible by 2.
Classification of 25814
By Arithmetic functions
• Arithmetic
• Semiprime
• Deficient
Expressible via specific sums
• Polite
• Non-hypotenuse
By Powers
• Square Free
Base conversion (25814)
Base System Value
2 Binary 110010011010110
3 Ternary 1022102002
4 Quaternary 12103112
5 Quinary 1311224
6 Senary 315302
8 Octal 62326
10 Decimal 25814
12 Duodecimal 12b32
16 Hexadecimal 64d6
20 Vigesimal 34ae
36 Base36 jx2
Basic calculations (n = 25814)
Multiplication
n×i
n×2 51628
n×3 77442
n×4 103256
n×5 129070
Division
ni
n2 12907.000
n3 8604.666
n4 6453.500
n5 5162.800
Exponentiation
ni
n2 666362596
n3 17201484053144
n4 444039109347859216
n5 11462425568705637801824
Nth Root
i√n
2√n 160.66735822811
3√n 29.554147415235
4√n 12.675462840785
5√n 7.6273040846131
25814 as geometric shapes
Circle
Radius = n
Diameter 51628
Circumference 162194.14551953
Area 2093439836.2206
Sphere
Radius = n
Volume 72053407909599
Surface area 8373759344.8825
Circumference 162194.14551953
Square
Length = n
Perimeter 103256
Area 666362596
Diagonal 36506.508899099
Cube
Length = n
Surface area 3998175576
Volume 17201484053144
Space diagonal 44711.159546583
Equilateral Triangle
Length = n
Perimeter 77442
Area 288543468.13387
Altitude 22355.579773291
Triangular Pyramid
Length = n
Surface area 1154173872.5355
Volume 2027214336741.7
Height 21077.042740068
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5
Proof the universe doesn't exist?
Reply Sat 21 Jul, 2012 07:18 pm
Assume the universe exists. Then everything in it can be put into a set U such that U contains, by definition of U, everything in the universe and nothing else. Since U already contains everything in the universe, no set contains more elements in the universe than U. But by Cantor's Theorem, the powerset of any set S always contains more elements than S. So, the powerset of U contains more elements than U. But each element of the powerset of U forms a part of the universe and is thus also in the universe. Thus, the powerset of U contains more elements in the universe than U. This is a contradiction. Therefore by proof by contradiction, our assumption that the universe exists must be false. Therefore, the universe does not exist.
What do you think of this?
• Topic Stats
• Top Replies
• Link to this Topic
Type: Discussion • Score: 5 • Views: 12,920 • Replies: 132
Ding an Sich
2
Reply Sat 21 Jul, 2012 08:36 pm
@browser32,
browser32 wrote:
Assume the universe exists. Then everything in it can be put into a set U such that U contains, by definition of U, everything in the universe and nothing else. Since U already contains everything in the universe, no set contains more elements in the universe than U. But by Cantor's Theorem, the powerset of any set S always contains more elements than S. So, the powerset of U contains more elements than U. But each element of the powerset of U forms a part of the universe and is thus also in the universe. Thus, the powerset of U contains more elements in the universe than U. This is a contradiction. Therefore by proof by contradiction, our assumption that the universe exists must be false. Therefore, the universe does not exist.
What do you think of this?
You know, for once you're on to something. This argument is a very potent argument against actualists, who construe possible worlds as maximally consistent sets of propositions. A neo-Cantorian argument is applicable to their position as well.
But, refuting other philosophical positions aside, just because we assume or know that the universe exists does not entail that we have to couch it, ontologically, within a set-theoretic framework. One may take the notion of a world or universe as primitive, that is, it cannot be defined any further. You either know what a world is or you don't.
But you do have something here, which is nice. :-)
fresco
2
Reply Sun 22 Jul, 2012 01:46 am
@Ding an Sich,
Laughing
Once again, the assumption that we can treat "existence" as understood is the central issue. The "logic" is secondary to that.
fresco
3
Reply Sun 22 Jul, 2012 02:09 am
@browser32,
The application of logic to ontology is might be likened to trying to stir a hot drink with a spoon made out of ice !
0 Replies
Ding an Sich
2
Reply Sun 22 Jul, 2012 08:11 am
@fresco,
fresco wrote:
Laughing
Once again, the assumption that we can treat "existence" as understood is the central issue. The "logic" is secondary to that.
I think we can treat existence as understood; depends on the language game.
The issue is whether we can treat existence and logic as equivalent. Can we treat the entities we are dealing with as, ontologically, mathematical? Seems Pythagorean to me, and it leads to numerous problems, one of which browser pointed out. That's why I'm not an actualist when it comes to possible worlds talk; instead I'm a modal dimensionalist .
I would argue that the entities are not themselves mathematical, but can be "mathematized". A different position, and I believe far more fruitful one.
0 Replies
Fil Albuquerque
2
Reply Sun 22 Jul, 2012 09:25 am
@browser32,
I think for a proper definition, a Universe is a powerset on itself, you cannot treat it as a simple set...this is why I believe we need a system which is not dynamic, or rather has a simulated dynamic into it...on that regard, dynamics bring no emergent novelty if not from the primitive set...a cyclic dynamic boundless Universe can this way I believe be treated as a powerset and still be qualitatively finite in phenomena even if repeating...
Does this make sense ? (I don't have but a basic intuition on powersets knowledge for tool)
Fil Albuquerque
2
Reply Sun 22 Jul, 2012 09:32 am
@Fil Albuquerque,
...I very much feel like trying to "box" and contain infinity to render it harmless... Laughing
0 Replies
Icon
1
Reply Tue 24 Jul, 2012 12:52 pm
@browser32,
The problem with Cantor's Theorem is that it assumes a finite base set. If the basic set is infinite, it is impossible to have a power set greater , numerically, than infinity.
∞ = ∞ +1 = ∞√ (n² +x²) = ...
Fil Albuquerque
2
Reply Tue 24 Jul, 2012 01:14 pm
@Icon,
Semantics, as according to Cantor there are infinity's bigger then others...
Icon
2
Reply Tue 24 Jul, 2012 02:23 pm
@Fil Albuquerque,
Technically, he only proves that not all infinite sets are equal in infinity. This is assuming linear distribution of parts in a set. The Universe is not a linear model. Where it is true that, on a micro/molecular scale, no object is exactly identical to another, it is also true that no object can exist without space. Another way to think of it is like this:
P(X) = { {a }, { a, b }, { b, c, e }, { a, c }, { e }, . . . }
In your argument, you are assuming X is the universe and P would be the power set of the Universe. In reality, to use set theory to apply to the Universe you must realize that X is a single object in the universe: matter, for instance.
We would need to look at this more like this:
U ≡ (P(X) = { {a }, { a, b }, { b, c, e }, { a, c }, { e }, . . . }) ⁿ < Øⁿ
Where n = ∞
So U is the equivalent of all sets, to the infinite power, less than infinite nothing. The reason for this is that the Universe includes all sets less than infinite nothing as well as the infinite nothing (a measurement of space) in which all sets must exist. Since you cannot have a set of nothing, we are representing nothing (empty space) as a null set.
Cantor's Theorem can only apply to that which can be counted, even infinitely, within a space. Since space cannot be enumerated, it cannot be included in a set as it is the container of all sets and the Universe contains all space.
Fil Albuquerque
1
Reply Tue 24 Jul, 2012 03:20 pm
@Icon,
As I see it the universe is the powerset with all the space time info, a collection of sets of local instants in contact or not, as any object or collection of objects is not just in the here and now but distributed trough vectors of space time...I imagine a motionless ensemble of all space times frames along an axis, a set itself with a circular repeating order...whether it suffices or not...
Icon
1
Reply Tue 24 Jul, 2012 10:19 pm
@Fil Albuquerque,
I feel I may have misrepresented something in my previous post. Allow me to clarify. I am not speaking of distance and motion(space time). I am referring to the space that this measurement is taken in. Space exists without motion while distance does not. Space time frames are still enumerable. Space cannot be observed, thus cannot be enumerated and placed in a set. I imagine space would be the brackets around the set. Not part of the set itself, but a fundamental defining point or key that translates the set for us.
fresco
1
Reply Wed 25 Jul, 2012 12:41 am
@Icon,
Quote:
Space cannot be observed, thus cannot be enumerated and placed in a set. I imagine space would be the brackets around the set. Not part of the set itself, but a fundamental defining point or key that translates the set for us.
These spoons made of ice are good fun ! Laughing
Now try observing the observer !
Fil Albuquerque
1
Reply Wed 25 Jul, 2012 04:01 am
@Icon,
...oh pleeeease...It works like an ocean, Einstein thought of aether at the time, lately some are turning to it again...space is such an empty set is full of energy and virtual pairs of particles coming in and out of existence anywhere even intergalactic space not to mention it can be deformed...but even it was the case you can work spacetime as a function all you need is a set of dynamics correlated with every particle ever existed...in fact you experience reality from a certain layer of code/language programming, from a certain layer of correlated ratios which give rise to correlated ratios of effects which you "observe"/measure...as far as I know the whole thing can be reduced to binary coding in a huge algorithm, matter energy space dark energy dark matter n all the remaining ****...yeah even movement, trajectory's and dynamics, are but a collection of ratios...a collection of "objects"...you are far off and green in your conjectures dude...space is an empty set ? oh dear... Rolling Eyes
Ding an Sich
1
Reply Wed 25 Jul, 2012 09:07 am
@fresco,
These observers observing observers are good fun! Laughing
Now try something less constructivist! Twisted Evil
0 Replies
Icon
1
Reply Wed 25 Jul, 2012 10:04 am
@Fil Albuquerque,
It is true that space is filled with "things". All the time. And those "things" exist and can be classified into sets. What I am talking about is the nothing in which they exist. Space time can be deformed, this is true, but space cannot because there is nothing to deform and no way to measure the deformation. Space time exists within space. Space is the void in which all things exist but is not, in and of itself, a measurable/observable anything. There is no equation for it; no theorem or measurement to define it. It is simply the "place" where all observable actions happen. It is the void in which an infinite number of possibilities occur simultaneously.
In multi-verse theory, it is the void in which all universes simultaneously exist without bumping into each other. The standard model of physics has no true definition for it as the standard model focuses on observable reactions. It's a bit like explaining the color red to a man born blind (verbally, without the assistance of retinal receptor manipulation).
And don't "oh pleeeease" me. It just shows ignorance and unwillingness to consider unconventional ideas and does not further the conversation or provide rational, credible argument. I am trying to be amicable while discussing your ideas. Please return the respect.
Icon
1
Reply Wed 25 Jul, 2012 10:08 am
@fresco,
I can do one better, I can observe the observer observing. Shocked
0 Replies
Fil Albuquerque
1
Reply Wed 25 Jul, 2012 10:19 am
@Icon,
You see but you are totally wrong as space is a shape around and in between objects as much objects are a shape in space...so there is this shadow geometry between form and its absence, the form of things, and the form which things themselves give to the absence that can be translated in information...through time space, dynamics is dependent on this relations...what is the shape of a void between objects in a given local sequence of space time ? It does matter...I can only think of music to give you an example on voids being important...the whole binary sequencing is yet another one...
the "oh pleeeease" thing was not ill intended only spontaneous as expecting the counter you offer...in fact I believe you reason quite well, even if I disagree...
Icon
1
Reply Wed 25 Jul, 2012 12:39 pm
@Fil Albuquerque,
I believe we are arguing over a definition at this point. Let's see if we can get past semantics and move forward. I think I have a solution. Let's change a few definitions and see what we come up with.
Let us assume that the Universe can be expressed as a set, space and everything in it, just as you say. Still, the universe must exist (or NOT exist) somewhere. That somewhere must have the potential to contain everything and nothing simultaneously in order to provide a platform in which our current observations can hold true.
If we are to assume that this has any physical attributes of any type, we create a paradox and would generally break every model of physics that we have come to understand.
So let us call the platform C, the Universe U, and n will be ∞ as before.
C ≡ (P(U) = { {a}, {b, c} ,...})ⁿ ∈ Cⁿ ∴ C ≡ Øⁿ ∈ Cⁿ
So the "platform" C is the equivalent of infinite power sets of infinite sets of Universe as elements of infinite C and therefore also the equivalent of infinite nothing as element of Infinite C. This is where we create the paradox in that C ≡ Cⁿ ≡ C(C)ⁿ(...). So C is not a set or power set but contains the potential of all sets, no sets, and itself infinitely.
There must be a platform to define potential or there could be nothing, including any experiences.
In short, the potential for something defines the existence of something and potential is constant where observation is possible.
cicerone imposter
1
Reply Wed 25 Jul, 2012 12:47 pm
Semantics and definitions seem to create the problems for humans.
Simply put, we're here. Anybody need proof?
Related Topics
1. Forums
2. » Proof the universe doesn't exist?
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-1,948,885,390,623,154,700 | My All 2020 Mathematics A to Z: Zero Divisor
Jacob Siehler had several suggestions for this last of the A-to-Z essays for 2020. Zorn’s Lemma was an obvious choice. It’s got an important place in set theory, it’s got some neat and weird implications. It’s got a great name. The zero divisor is one of those technical things mathematics majors have deal with. It never gets any pop-mathematics attention. I picked the less-travelled road and found a delightful scenic spot.
Color cartoon illustration of a coati in a beret and neckerchief, holding up a director's megaphone and looking over the Hollywood hills. The megaphone has the symbols + x (division obelus) and = on it. The Hollywood sign is, instead, the letters MATHEMATICS. In the background are spotlights, with several of them crossing so as to make the letters A and Z; one leg of the spotlights has 'TO' in it, so the art reads out, subtly, 'Mathematics A to Z'.
Art by Thomas K Dye, creator of the web comics Projection Edge, Newshounds, Infinity Refugees, and Something Happens. He’s on Twitter as @projectionedge. You can get to read Projection Edge six months early by subscribing to his Patreon.
Zero Divisor.
3 times 4 is 12. That’s a clear, unambiguous, and easily-agreed-upon arithmetic statement. The thing to wonder is what kind of mathematics it takes to mess that up. The answer is algebra. Not the high school kind, with x’s and quadratic formulas and all. The college kind, with group theory and rings.
A ring is a mathematical construct that lets you do a bit of arithmetic. Something that looks like arithmetic, anyway. It has a set of elements. (An element is just a thing in a set. We say “element” because it feels weird to call it “thing” all the time.) The ring has an addition operation. The ring has a multiplication operation. Addition has an identity element, something you can add to any element without changing the original element. We can call that ‘0’. The integers, or to use the lingo Z , are a ring (among other things).
Among the rings you learn, after the integers, is the integers modulo … something. This can be modulo any counting number. The integers modulo 10, for example, we write as Z_{10} for short. There are different ways to think of what this means. The one convenient for this essay is that it’s the integers 0, 1, 2, up through 9. And that the result of any calculation is “how much more than a whole multiple of 10 this calculation would otherwise be”. So then 3 times 4 is now 2. 3 times 5 is 5; 3 times 6 is 8. 3 times 7 is 1, and doesn’t that seem peculiar? That’s part of how modulo arithmetic warns us that groups and rings can be quite strange things.
We can do modulo arithmetic with any of the counting numbers. Look, for example, at Z_{5} instead. In the integers modulo 5, 3 times 4 is … 2. This doesn’t seem to get us anything new. How about Z_{8} ? In this, 3 times 4 is 4. That’s interesting. It doesn’t make 3 the multiplicative identity for this ring. 3 times 3 is 1, for example. But you’d never see something like that for regular arithmetic.
How about Z_{12} ? Now we have 3 times 4 equalling 0. And that’s a dramatic break from how regular numbers work. One thing we know about regular numbers is that if a times b is 0, then either a is 0, or b is zero, or they’re both 0. We rely on this so much in high school algebra. It’s what lets us pick out roots of polynomials. Now? Now we can’t count on that.
When this does happen, when one thing times another equals zero, we have “zero divisors”. These are anything in your ring that can multiply by something else to give 0. Is, zero, the additive identity, always a zero divisor. … That depends on what the textbook you first learned algebra from said. To avoid ambiguity, you can write a “nonzero zero divisor”. This clarifies your intentions and slows down your copy editing every time you read “nonzero zero”. Or call it a “nontrivial zero divisor” or “proper zero divisor” instead. My preference is to accept 0 as always being a zero divisor. We can disagree on this. What of zero divisors other than zero?
Your ring might or might not have them. It depends on the ring. The ring of integers Z , for example, doesn’t have any zero divisors except for 0. The ring of integers modulo 12 Z_{12} , though? Anything that isn’t relatively prime to 12 is a zero divisor. So, 2, 3, 6, 8, 9, and 10 are zero divisors here. The ring of integers modulo 13 Z_{13} ? That doesn’t have any zero divisors, other than zero itself. In fact any ring of integers modulo a prime number, Z_{p} , lacks zero divisors besides 0.
Focusing too much on integers modulo something makes zero divisors sound like some curious shadow of prime numbers. There are some similarities. Whether a number is prime depends on your multiplication rule and what set of things it’s in. Being a zero divisor in one ring doesn’t directly relate to whether something’s a zero divisor in any other. Knowing what the zero divisors are tells you something about the structure of the ring.
It’s hard to resist focusing on integers-modulo-something when learning rings. They work very much like regular arithmetic does. Even the strange thing about them, that every result is from a finite set of digits, isn’t too alien. We do something quite like it when we observe that three hours after 10:00 is 1:00. But many sets of elements can create rings. Square matrixes are the obvious extension. Matrixes are grids of elements, each of which … well, they’re most often going to be numbers. Maybe integers, or real numbers, or complex numbers. They can be more abstract things, like rotations or whatnot, but they’re hard to typeset. It’s easy to find zero divisors in matrixes of numbers. Imagine, like, a matrix that’s all zeroes except for one element, somewhere. There are a lot of matrices which, multiplied by that, will be a zero matrix, one with nothing but zeroes in it. Another common kind of ring is the polynomials. For these you need some constraint like the polynomial coefficients being integers-modulo-something. You can make that work.
In 1988 Istvan Beck tried to establish a link between graph theory and ring theory. We now have a usable standard definition of one. If R is any ring, then \Gamma(R) is the zero-divisor graph of R . (I know some of you think R is the real numbers. No; that’s a bold-faced \mathbb{R} instead. Unless that’s too much bother to typeset.) You make the graph by putting in a vertex for the elements in R . You connect two vertices a and b if the product of the corresponding elements is zero. That is, if they’re zero divisors for one other. (In Beck’s original form, this included all the elements. In modern use, we don’t bother including the elements that are not zero divisors.)
Drawing this graph \Gamma(R) makes tools from graph theory available to study rings. We can measure things like the distance between elements, or what paths from one vertex to another exist. What cycles — paths that start and end at the same vertex — exist, and how large they are. Whether the graphs are bipartite. A bipartite graph is one where you can divide the vertices into two sets, and every edge connects one thing in the first set with one thing in the second. What the chromatic number — the minimum number of colors it takes to make sure no two adjacent vertices have the same color — is. What shape does the graph have?
It’s easy to think that zero divisors are just a thing which emerges from a ring. The graph theory connection tells us otherwise. You can make a potential zero divisor graph and ask whether any ring could fit that. And, from that, what we can know about a ring from its zero divisors. Mathematicians are drawn as if by an occult hand to things that let you answer questions about a thing from its “shape”.
And this lets me complete a cycle in this year’s A-to-Z, to my delight. There is an important question in topology which group theory could answer. It’s a generalization of the zero-divisors conjecture, a hypothesis about what fits in a ring based on certain types of groups. This hypothesis — actually, these hypotheses. There are a bunch of similar questions about invariants called the L2-Betti numbers can be. These we call the Atiyah Conjecture. This because of work Michael Atiyah did in the cohomology of manifolds starting in the 1970s. It’s work, I admit, I don’t understand well enough to summarize, and hope you’ll forgive me for that. I’m still amazed that one can get to cutting-edge mathematics research this. It seems, at its introduction, to be only a subversion of how we find x for which (x - 2)(x + 1) = 0 .
And this, I am amazed to say, completes the All 2020 A-to-Z project. All of this year’s essays should be gathered at this link. In the next couple days I plan t check that they actually are. All the essays from every A-to-Z series, going back to 2015, should be at this link. I plan to soon have an essay about what I learned in doing the A-to-Z this year. And then we can look to 2021 and hope that works out all right. Thank you for reading.
Reading the Comics, March 12, 2019: Back To Sequential Time Edition
Since I took the Pi Day comics ahead of their normal sequence on Sunday, it’s time I got back to the rest of the week. There weren’t any mathematically-themed comics worth mentioning from last Friday or Saturday, so I’m spending the latter part of this week covering stuff published before Pi Day. It’s got me slightly out of joint. It’ll all be better soon.
Mark Anderson’s Andertoons for the 11th is the Mark Anderson’s Andertoons for this week. That’s nice to have. It’s built on the concept of story problems. That there should be “stories” behind a problem makes sense. Most actual mathematics, even among mathematicians, is done because we want to know a thing. Acting on a want is a story. Wanting to know a thing justifies the work of doing this calculation. And real mathematics work involves looking at some thing, full of the messiness of the real world, and extracting from it mathematics. This would be the question to solve, the operations to do, the numbers (or shapes or connections or whatever) to use. We surely learn how to do that by doing simple examples. The kid — not Wavehead, for a change — points out a common problem here. There’s often not much of a story to a story problem. That is, where we don’t just want something, but someone else wants something too.
On the classroom chalkboard: 'Today's number story. 2 dogs are at the park. 3 more dogs arrive. How many dogs are there?' Non-Wavehead student: 'Really? That's it? I don't mean anything by it, but where the conflict? Where's the drama? OK, try this ... a cat shows up. Or a mailman! Ooh, a cat mailman! Now we're talking!'
Mark Anderson’s Andertoons for the 11th of March, 2019. Essays which discuss topics raised by Andertoons can be found at this link. Also at this link, nearly enough.
Parker and Hart’s The Wizard of Id for the 11th is a riff on the “when do you use algebra in real life” snark. Well, no one disputes that there are fields which depend on advanced mathematics. The snark comes in from supposing that a thing is worth learning only if it’s regularly “useful”.
King: 'Wiz! We need your help! There's a giant meteor heading towards earth!' Wizard: 'oh boy! Oh boy!' (He runs off and pulls out a chalkboard full of mathematics symbols.) King: 'What are you doing?' Wizard: 'Finally! This is where algebra saves lives!'
Parker and Hart’s The Wizard of Id for the 11th of March, 2019. This is part of the current run of The Wizard of Id, such as you might find in a newspaper. Both the current and 1960s-vintage reruns get discussion at this link.
Rick Detorie’s One Big Happy for the 12th has Joe stalling class to speak to “the guy who invented zero”. I really like this strip since it’s one of those cute little wordplay jokes that also raises a legitimate point. Zero is this fantastic idea and it’s hard to imagine mathematics as we know it without the concept. Of course, we could say the same thing about trying to do mathematics without the concept of, say, “twelve”.
Teacher: 'Are there any questions before the test? Yes, Joe?' Joe: 'I would like to say a few words to the guy who invented zero. Thanks for NOTHING, dude! ... I think we can all enjoy a little math humor at a time like this.'
Rick Detorie’s One Big Happy for the 12th of March, 2019. This is part of the current run of One Big Happy, such as you might find in a newspaper. At GoComics.com strips from several years ago are reprinted. Both runs of One Big Happy get their discussion at this link.
We don’t know who’s “the guy” who invented zero. It’s probably not all a single person, though, or even a single group of people. There are several threads of thought which merged together to zero. One is the notion of emptiness, the absense of a measurable thing. That probably occurred to whoever was the first person to notice a thing wasn’t where it was expected. Another part is the notion of zero as a number, something you could add to or subtract from a conventional number. That is, there’s this concept of “having nothing”, yes. But can you add “nothing” to a pile of things? And represent that using the addition we do with numbers? Sure, but that’s because we’re so comfortable with the idea of zero that we don’t ponder whether “2 + 1” and “2 + 0” are expressing similar ideas. You’ll occasionally see people asking web forums whether zero is really a number, often without getting much sympathy for their confusion. I admit I have to think hard to not let long reflex stop me wondering what I mean by a number and why zero should be one.
And then there’s zero, the symbol. As in having a representation, almost always a circle, to mean “there is a zero here”. We don’t know who wrote the first of that. The oldest instance of it that we know of dates to the year 683, and was written in what’s now Cambodia. It’s in a stone carving that seems to be some kind of bill of sale. I’m not aware whether there’s any indication from that who the zero was written for, or who wrote it, though. And there’s no reason to think that’s the first time zero was represented with a symbol. It’s the earliest we know about.
Lemont: 'My son asked what my favorite number is. I thought, who has a favorite number? Then remembered *I* did. When I was a kid I loved the number eight, because it reminded me of infinity.' C Dog: 'Nuh-uh, Big L. You tol' me you loved it 'cause it was the only number that reminded you of food.' Lemont: 'You sure? I remember being a much more profound kid.' C Dog: 'People always re-writing e'reything.'
Darrin Bell’s Candorville for the 12th of March, 2019. Essays featuring some topic raised by of Candorville should appear here.
Darrin Bell’s Candorville for the 12th has some talk about numbers, and favorite numbers. Lemont claims to have had 8 as his favorite number because its shape, rotated, is that of the infinity symbol. C-Dog disputes Lemont’s recollection of his motives. Which is fair enough; it’s hard to remember what motivated you that long ago. What people mostly do is think of a reason that they, today, would have done that, in the past.
The ∞ symbol as we know it is credited to John Wallis, one of that bunch of 17th-century English mathematicians. He did a good bit of substantial work, in fields like conic sections and physics and whatnot. But he was also one of those people good at coming up with notation. He developed what’s now the standard notation for raising a number to a power, that x^n stuff, and showed how to define raising a number to a rational-number power. Bunch of other things. He also seems to be the person who gave the name “continued fraction” to that concept.
Wallis never explained why he picked ∞ as a shape, of all the symbols one could draw, for this concept. There’s speculation he might have been varying the Roman numeral for 1,000, which we’ve simplified to M but which had been rendered as (|) or () and I can see that. (Well, really more of a C and a mirror-reflected C rather than parentheses, but I don’t have the typesetting skills to render that.) Conflating “a thousand” with “many” or “infinitely many” has a good heritage. We do the same thing when we talk about something having millions of parts or costing trillions of dollars or such. But, Wallis never explained (so far as we’re aware), so all this has to be considered speculation and maybe mnemonic helps to remembering the symbol.
Colin: 'These math problems are impossible!' Dad: 'Calm down, I'll help you think it through. [ Reading ] Dot has 20 gumballs. Jack has twice as many, and Joe has a third as many as Jack. If Dot gives Jack half her gumballs, and Jack chews half the new amount, how many will the teacher have to take away for all the kids' gumballs to be equal?' ... Now I see why people buy the answers to these things on the Internet!' Colin: 'I'm never going to harvard, am I?'
Terry Laban and Patty LaBan’s Edge City rerun for the 12th of March, 2019. (The strip has ended.) This comic originally ran in 2004. Edge City inspires discussions in the essays at this link.
Terry LaBan and Patty LaBan’s Edge City for the 12th is another story problem joke. Curiously the joke seems to be simply that the father gets confused following the convolutions of the story. The specific story problem circles around the “participation awards are the WORST” attitude that newspaper comics are surprisingly prone to. I think the LaBans just wanted the story problem to be long and seem tedious enough that our eyes glazed over. Anyway you could not pay me to read whatever the comments on this comic are. Sorry not sorry.
I figure to have one more Reading the Comics post this week. When that’s posted it should be available at this link. Thanks for being here.
Reading the Comics, February 10, 2018: I Meant To Post This Thursday Edition
Ah, yes, so, in the midst of feeling all proud that I’d gotten my Reading the Comics workflow improved, I went out to do my afternoon chores without posting the essay. I’m embarrassed. But it really only affects me looking at the WordPress Insights page. It publishes this neat little calendar-style grid that highlights the days when someone’s posted and this breaks up the columns. This can only unnerve me. I deserve it.
Tom Thaves’s Frank and Ernest for the 8th of February is about the struggle to understand zero. As often happens, the joke has a lot of truth to it. Zero bundles together several ideas, overlapping but not precisely equal. And part of that is the idea of “nothing”. Which is a subtly elusive concept: to talk about the properties of a thing that does not exist is hard. As adults it’s easy to not notice this anymore. Part’s likely because mastering a concept makes one forget what it took to understand. Part is likely because if you don’t have to ponder whether the “zero” that’s “one less than one” is the same as the “zero” that denotes “what separates the count of thousands from the count of tens in the numeral 2,038” you might not, and just assume you could explain the difference or similarity to someone who has no idea.
John Zakour and Scott Roberts’s Maria’s Day for the 8th has maria and another girl bonding over their hatred of mathematics. Well, at least they’re getting something out of it. The date in the strip leads me to realize this is probably a rerun. I’m not sure just when it’s from.
Zach Weinersmith’s Saturday Morning Breakfast Cereal for the 8th proposes a prank based on mathematical use of the word “arbitrarily”. This is a word that appears a lot in analysis, and the strip makes me realize I’m not sure I can give a precise definition. An “arbitrarily large number”, for example, would be any number that’s large enough. But this also makes me realize I’m not sure precisely what joke Weinersmith is going for. I suppose that if someone were to select an arbitrarily large number they might pick 53, or a hundred, or million billion trillion. I suppose Weinersmith’s point is that in ordinary speech an arbitrarily made choice is one selection from all the possible alternatives. In mathematical speech an arbitrarily made choice reflects every possible choice. To speak of an arbitrarily large number is to say that whatever selection is made, we can go on to show this interesting stuff is true. We’d typically like to prove the most generically true thing possible. But picking a single example can be easier to prove. It can certainly be easier to visualize. 53 is probably easier to imagine than “every number 52 or larger”, for example.
Quincy: 'Someday I'm gonna write a book, Gran.' Grandmom: 'Wonderful. Will you dedicate it to me?' Quincy: 'Sure. In fact, if you want, I'll dedicate this math homework to you.'
Ted Shearer’s Quincy for the 16th of December, 1978 and reprinted the 9th of February, 2018. I’m not sure just what mathematics homework Quincy could be doing to inspire him to write a book, but then, it’s not like my mind doesn’t drift while doing mathematics either. And book-writing’s a common enough daydream that most people are too sensible to act on.
Ted Shearer’s Quincy for the 16th of December, 1978 was rerun the 9th of February. It just shows Quincy at work on his mathematics homework, and considering dedicating it to his grandmother. Mathematics books have dedications, just as any other book does. I’m not aware of dedications of proofs or other shorter mathematics works, but there’s likely some. There’s often a note of thanks, usually given to people who’ve made the paper’s writers think harder about the subjects. But I don’t think there’s any reason a paper wouldn’t thank someone who provided “mere” emotional support. I just don’t have examples offhand.
Jef Mallet’s Frazz for the 9th looks like one of those creative-teaching exercises I sometimes see in Mathematics Education Twitter: the teacher gives answers and the students come up with story problems to match. That’s not a bad project. I’m not sure how to grade it, but I haven’t done anything that creative when I’ve taught. I’m sorry I haven’t got more to say about it since the idea seems fun.
Redeye: 'C'mon, Pokey. Time for your lessons. Okay, what do you get when you divide 5,967,342 by 973 ... ?' Pokey: 'A headache!'
Gordon Bess’s Redeye for the 30th of September, 1971 and reprinted the 10th of February, 2018. I realized I didn’t know the father’s name and looked it up, and Wikipedia revealed to me that he’s named Redeye. You know, like the comic strip implies right there in the title. Look, I just read the comics, I can’t be expected to think about the comics too.
Gordon Bess’s Redeye for the 30th of September, 1971 was rerun the 10th. It’s a bit of extremely long division and I don’t blame Pokey for giving up on that problem. Starting from 5,967,342 divided by 973 I’d say, well, that’s about six million divided by a thousand, so the answer should be near six thousand. I don’t think the last digits of 2 and 3 suggest anything about what the final digit should be, if this divides evenly. So the only guidance I have is that my answer ought to be around six thousand and then we have to go into actually working. It turns out that 973 doesn’t go into 5,967,342 a whole number of times, so I sympathize more with Pokey. The answer is a little more than 6,132.9311.
Reading the Comics, June 24, 2017: Saturday Morning Breakfast Cereal Edition
Somehow this is not the title of every Reading The Comics review! But it is for this post and we’ll explore why below.
Piers Baker’s Ollie and Quentin for the 18th is a Zeno’s Paradox-based joke. This uses the most familiar of Zeno’s Paradoxes, about the problem of covering any distance needing infinitely many steps to be done in a finite time. Zeno’s Paradoxes are often dismissed these days (probably were then, too), on the grounds that the Ancient Greeks Just Didn’t Understand about convergence. Hardly; they were as smart as we were. Zeno had a set of paradoxes, built on the questions of whether space and time are infinitely divisible or whether they’re not. Any answer to one paradox implies problems in others. There’s things we still don’t really understand about infinity and infinitesimals and continuity. Someday I should do a proper essay about them.
Dave Coverly’s Speed Bump for the 18th is not exactly an anthropomorphic-numerals joke. It is about making symbols manifest in the real world, at least. The greater-than and less-than signs as we know them were created by the English mathematician Thomas Harriot, and introduced to the world in his posthumous Artis Analyticae Praxis (1631). He also had an idea of putting a . between the numerals of an expression and the letters multiplied by them, for example, “4.x” to mean four times x. We mostly do without that now, taking multiplication as assumed if two meaningful quantities are put next to one another. But we will use, now, a vertically-centered dot to separate terms multiplied together when that helps our organization. The equals sign we trace to the 16th century mathematician Robert Recorde, whose 1557 Whetsone of Witte uses long but recognizable equals signs. The = sign went into hibernation after that, though, until the 17th century and it took some time to quite get well-used. So it often is with symbols.
Mr Tanner: 'Today we'll talk about where numbers come from. Take zero, for instance ... Quincy, do you know who invented the zero?' Quincy: 'I'm not sure, Mr Tanner, but from the grades I get it must have been one of my teachers.'
Ted Shearer’s Quincy for the 25th of April, 1978 and rerun the 19th of June, 2017. The question does make me wonder how far Mr Tanner was going to go with this. The origins of zero and one are great stuff for class discussion. Two, also. But what about three? Five? Ten? Twelve? Minus one? Irrational numbers, if the class has got up to them? How many students are going to be called on to talk about number origins? And how many truly different stories are there?
Ted Shearer’s Quincy for the 25th of April, 1978 and rerun the 19th of June, starts from the history of zero. It’s worth noting there are a couple of threads woven together in the concept of zero. One is the idea of “nothing”, which we’ve had just forever. I mean, the idea that there isn’t something to work with. Another is the idea of the … well, the additive identity, there being some number that’s one less than one and two less than two. That you can add to anything without changing the thing. And then there’s symbols. There’s the placeholder for “there are no examples of this quantity here”. There’s the denotation of … well, the additive identity. All these things are zeroes, and if you listen closely, they are not quite the same thing. Which is not weird. Most words mean a collection of several concepts. We’re lucky the concepts we mean by “zero” are so compatible in meaning. Think of the poor person trying to understand the word “bear”, or “cleave”.
John Deering’s Strange Brew for the 19th is a “New Math” joke, fittingly done with cavemen. Well, numerals were new things once. Amusing to me is that — while I’m not an expert — in quite a few cultures the symbol for “one” was pretty much the same thing, a single slash mark. It’s hard not to suppose that numbers started out with simple tallies, and the first thing to tally might get dressed up a bit with serifs or such but is, at heart, the same thing you’d get jabbing a sharp thing into a soft rock.
Guy Gilchrist’s Today’s Dogg for the 19th I’m sure is a rerun and I think I’ve featured it here before. So be it. It’s silly symbol-play and dog arithmetic. It’s a comic strip about how dogs are cute; embrace it or skip it.
Zach Weinersmith’s Saturday Morning Breakfast Cereal is properly speaking reruns when it appears on GoComics.com. For whatever reason Weinersmith ran a patch of mathematics strips there this past week. So let me bundle all that up. On the 19th he did a joke mathematicians get a lot, about how the only small talk anyone has about mathematics is how they hated mathematics. I’m not sure mathematicians have it any better than any other teachers, though. Have you ever known someone to say, “My high school gym class gave me a greater appreciation of the world”? Or talk about how grade school history opened their eyes to the wonders of the subject? It’s a sad thing. But there are a lot of things keeping teachers from making students feel joy in their subjects.
For the 21st Weinersmith makes a statisticians joke. I can wrangle some actual mathematics out of an otherwise correctly-formed joke. How do we ever know that something is true? Well, we gather evidence. But how do we know the evidence is relevant? Even if the evidence is relevant, how do we know we’ve interpreted it correctly? Even if we have interpreted it correctly, how do we know that it shows what we want to know? Statisticians become very familiar with hypothesis testing, which amounts to the question, “does this evidence indicate that some condition is implausibly unlikely”? And they can do great work with that. But “implausibly unlikely” is not the same thing as “false”. A person knowledgeable enough and honest turns out to have few things that can be said for certain.
The June 23rd strip I’ve seen go around Mathematics Twitter several times, as see above tweet, about the ways in which mathematical literacy would destroy modern society. It’s a cute and flattering portrait of mathematics’ power, probably why mathematicians like passing it back and forth. But … well, how would “logic” keep people from being fooled by scams? What makes a scam work is that the premise seems logical. And real-world problems — as opposed to logic-class problems — are rarely completely resolvable by deductive logic. There have to be the assumptions, the logical gaps, and the room for humbuggery that allow hoaxes and scams to slip through. And does anyone need a logic class to not “buy products that do nothing”? And what is “nothing”? I have more keychains than I have keys to chain, even if we allow for emergencies and reasonable unexpected extra needs. This doesn’t stop my buying keychains as souvenirs. Does a Penn Central-logo keychain “do nothing” merely because it sits on the windowsill rather than hold any sort of key? If so, was my love foolish to buy it as a present? Granted that buying a lottery ticket is a foolish use of money; is my life any worse for buying that than, say, a peanut butter cup that I won’t remember having eaten a week afterwards? As for credit cards — It’s not clear to me that people max out their credit cards because they don’t understand they will have to pay it back with interest. My experience has been people max out their credit cards because they have things they must pay for and no alternative but going further into debt. That people need more money is a problem of society, yes, but it’s not clear to me that a failure to understand differential equations is at the heart of it. (Also, really, differential equations are overkill to understand credit card debt. A calculator with a repeat-the-last-operation feature and ten minutes to play is enough.)
Some Mathematical Tweets To Read
Can’t deny that I will sometimes stockpile links of mathematics stuff to talk about. Sometimes I even remember to post it. Sometimes it’s a tweet like this, which apparently I’ve been carrying around since April:
I admit I do not know whether the claim is true. It’s plausible enough. English has many variants in England alone, and any trade will pick up its own specialized jargon. The words are fun as it is.
From the American Mathematical Society there’s this:
I talk a good bit about knot theory. It captures the imagination and it’s good for people who like to doodle. And it has a lot of real-world applications. Tangled wires, protein strands, high-energy plasmas, they all have knots in them. Some work by Paul Sutcliffe and Fabian Maucher, both of Durham University, studies tangled vortices. These are vortices that are, er, tangled together, just like you imagine. Knot theory tells us much about this kind of vortex. And it turns out these tangled vortices can untangle themselves and smooth out again, even without something to break them up and rebuild them. It gives hope for power cords everywhere.
Nerds have a streak which compels them to make blueprints of things. It can be part of the healthier side of nerd culture, the one that celebrates everything. The side that tries to fill in the real-world things that the thing-celebrated would have if it existed. So here’s a bit of news about doing that:
I like the attempt to map Sir Thomas More’s Utopia. It’s a fun exercise in matching stuff to a thin set of data. But as mentioned in the article, nobody should take it too seriously. The exact arrangement of things in Utopia isn’t the point of the book. More probably didn’t have a map for it himself.
(Although maybe. I believe I got this from Simon Garfield’s On The Map: A Mind-Expanding Exploration Of The Way The World Looks and apologize generally if I’ve got it wrong. My understanding is Robert Louis Stevenson drew a map of Treasure Island and used it to make sure references in the book were consistent. Then the map was lost in the mail to his publishers. He had to read his text and re-create it as best he could. Which, if true, makes the map all the better. It makes it so good a lost-map story that I start instinctively to doubt it; it’s so colorfully perfect, after all.)
And finally there’s this gem from the Magic Realism Bot:
Happy reading.
Reading the Comics, July 13, 2016: Catching Up On Vacation Week Edition
I confess I spent the last week on vacation, away from home and without the time to write about the comics. And it was another of those curiously busy weeks that happens when it’s inconvenient. I’ll try to get caught up ahead of the weekend. No promises.
Art and Chip Samson’s The Born Loser for the 10th talks about the statistics of body measurements. Measuring bodies is one of the foundations of modern statistics. Adolphe Quetelet, in the mid-19th century, found a rough relationship between body mass and the square of a person’s height, used today as the base for the body mass index.Francis Galton spent much of the late 19th century developing the tools of statistics and how they might be used to understand human populations with work I will describe as “problematic” because I don’t have the time to get into how much trouble the right mind at the right idea can be.
No attempt to measure people’s health with a few simple measurements and derived quantities can be fully successful. Health is too complicated a thing for one or two or even ten quantities to describe. Measures like height-to-waist ratios and body mass indices and the like should be understood as filters, the way temperature and blood pressure are. If one or more of these measurements are in dangerous ranges there’s reason to think there’s a health problem worth investigating here. It doesn’t mean there is; it means there’s reason to think it’s worth spending resources on tests that are more expensive in time and money and energy. And similarly just because all the simple numbers are fine doesn’t mean someone is perfectly healthy. But it suggests that the person is more likely all right than not. They’re guides to setting priorities, easy to understand and requiring no training to use. They’re not a replacement for thought; no guides are.
Jeff Harris’s Shortcuts educational panel for the 10th is about zero. It’s got a mix of facts and trivia and puzzles with a few jokes on the side.
I don’t have a strong reason to discuss Ashleigh Brilliant’s Pot-Shots rerun for the 11th. It only mentions odds in a way that doesn’t open up to discussing probability. But I do like Brilliant’s “Embrace-the-Doom” tone and I want to share that when I can.
John Hambrock’s The Brilliant Mind of Edison Lee for the 13th of July riffs on the world’s leading exporter of statistics, baseball. Organized baseball has always been a statistics-keeping game. The Olympic Ball Club of Philadelphia’s 1837 rules set out what statistics to keep. I’m not sure why the game is so statistics-friendly. It must be in part that the game lends itself to representation as a series of identical events — pitcher throws ball at batter, while runners wait on up to three bases — with so many different outcomes.
'Edison, let's discuss stats while we wait for the opening pitch.' 'Statistics? I have plenty of those. A hot dog has 400 calories and costs five dollars. A 12-ounce root beer has 38 grams of sugar.' 'I mean *player* stats.' 'Oh'. (To his grandfather instead) 'Did you know the average wait time to buy nachos is eight minutes and six seconds?'
John Hambrock’s The Brilliant Mind of Edison Lee for the 13th of July, 2016. Properly speaking, the waiting time to buy nachos isn’t a player statistic, but I guess Edison Lee did choose to stop talking to his father for it. Which is strange considering his father’s totally natural and human-like word emission ‘Edison, let’s discuss stats while we wait for the opening pitch’.
Alan Schwarz’s book The Numbers Game: Baseball’s Lifelong Fascination With Statistics describes much of the sport’s statistics and record-keeping history. The things recorded have varied over time, with the list of things mostly growing. The number of statistics kept have also tended to grow. Sometimes they get dropped. Runs Batted In were first calculated in 1880, then dropped as an inherently unfair statistic to keep; leadoff hitters were necessarily cheated of chances to get someone else home. How people’s idea of what is worth measuring changes is interesting. It speaks to how we change the ways we look at the same event.
Dana Summers’s Bound And Gagged for the 13th uses the old joke about computers being abacuses and the like. I suppose it’s properly true that anything you could do on a real computer could be done on the abacus, just, with a lot ore time and manual labor involved. At some point it’s not worth it, though.
Nate Fakes’s Break of Day for the 13th uses the whiteboard full of mathematics to denote intelligence. Cute birds, though. But any animal in eyeglasses looks good. Lab coats are almost as good as eyeglasses.
LERBE ( O O - O - ), GIRDI ( O O O - - ), TACNAV ( O - O - O - ), ULDNOA ( O O O - O - ). When it came to measuring the Earth's circumference, there was a ( - - - - - - - - ) ( - - - - - ).
David L Hoyt and Jeff Knurek’s Jumble for the 13th of July, 2016. The link will be gone sometime after mid-August I figure. I hadn’t thought of a student being baffled by using the same formula for an orange and a planet’s circumference because of their enormous difference in size. It feels authentic, though.
David L Hoyt and Jeff Knurek’s Jumble for the 13th is about one of geometry’s great applications, measuring how large the Earth is. It’s something that can be worked out through ingenuity and a bit of luck. Once you have that, some clever argument lets you work out the distance to the Moon, and its size. And that will let you work out the distance to the Sun, and its size. The Ancient Greeks had worked out all of this reasoning. But they had to make observations with the unaided eye, without good timekeeping — time and position are conjoined ideas — and without photographs or other instantly-made permanent records. So their numbers are, to our eyes, lousy. No matter. The reasoning is brilliant and deserves respect.
Reading the Comics, November 21, 2015: Communication Edition
And then three days pass and I have enough comic strips for another essay. That’s fine by me, really. I picked this edition’s name because there’s a comic strip that actually touches on information theory, and another that’s about a much-needed mathematical symbol, and another about the ways we represent numbers. That’s enough grounds for me to use the title.
Samson’s Dark Side Of The Horse for the 19th of November looks like this week’s bid for an anthropomorphic numerals joke. I suppose it’s actually numeral cosplay instead. I’m amused, anyway.
Zach Weinersmith’s Saturday Morning Breakfast Cereal for the 19th of November makes a patent-law joke out of the invention of zero. It’s also an amusing joke. It may be misplaced, though. The origins of zero as a concept is hard enough to trace. We can at least trace the symbol zero. In Finding Zero: A Mathematician’s Odyssey to Uncover the Origins of Numbers, Amir D Aczel traces out not just the (currently understood) history of Arabic numerals, but some of how the history of that history has evolved, and finally traces down the oldest known example of a written (well, carved) zero.
Tony Cochrane’s Agnes for the 20th of November is at heart just a joke about a student’s apocalyptically bad grades. It contains an interesting punch line, though, in Agnes’s statement that “math people are dreadful spellers”. I haven’t heard that before. It might be a joke about algebra introducing letters into numbers. But it does seem to me there’s a supposition that mathematics people aren’t very good writers or speakers. I do remember back as an undergraduate other people on the student newspaper being surprised I could write despite majoring in physics and mathematics. That may reflect people remembering bad experiences of sitting in class with no idea what the instructor was going on about. It’s easy to go from “I don’t understand this mathematics class” to “I don’t understand mathematics people”.
Steve Sicula’s Home and Away for the 20th of November is about using gambling as a way to teach mathematics. So it would be a late entry for the recent Gambling Edition of the Reading The Comics posts. Although this strip is a rerun from the 15th of August, 2008, so it’s actually an extremely early entry.
Ruben Bolling’s Tom The Dancing Bug for the 20th of November is a Super-Fun-Pak Comix installment. And for a wonder it hasn’t got a Chaos Butterfly sequence. Under the Guy Walks Into A Bar label is a joke about a horse doing arithmetic that itself swings into a base-ten joke. In this case it’s suggested the horse would count in base four, and I suppose that’s plausible enough. The joke depends on the horse pronouncing a base four “10” as “ten”, when the number is actually “four”. But the lure of the digits is very hard to resist, and saying “four” suggests the numeral “4” whatever the base is supposed to be.
Mark Leiknes’s Cow and Boy for the 21st of November is a rerun from the 9th of August, 2008. It mentions the holographic principle, which is a neat concept. The principle’s explained all right in the comic. The idea was first developed in the late 1970s, following the study of black hole thermodynamics. Black holes are fascinating because the mathematics of them suggest they have a temperature, and an entropy, and even information which can pass into and out of them. This study implied that information about the three-dimensional volume of the black hole was contained entirely in the two-dimensional surface, though. From here things get complicated, though, and I’m going to shy away from describing the whole thing because I’m not sure I can do it competently. It is an amazing thing that information about a volume can be encoded in the surface, though, and vice-versa. And it is astounding that we can imagine a logically consistent organization of the universe that has a structure completely unlike the one our senses suggest. It’s a lasting and hard-to-dismiss philosophical question. How much of the way the world appears to be structured is the result of our minds, our senses, imposing that structure on it? How much of it is because the world is ‘really’ like that? (And does ‘really’ mean anything that isn’t trivial, then?)
I should make clear that while we can imagine it, we haven’t been able to prove that this holographic universe is a valid organization. Explaining gravity in quantum mechanics terms is a difficult point, as it often is.
Dave Blazek’s Loose Parts for the 21st of November is a two- versus three-dimensions joke. The three-dimension figure on the right is a standard way of drawing x-, y-, and z-axes, organized in an ‘isometric’ view. That’s one of the common ways of drawing three-dimensional figures on a two-dimensional surface. The two-dimension figure on the left is a quirky representation, but it’s probably unavoidable as a way to make the whole panel read cleanly. Usually when the axes are drawn isometrically, the x- and y-axes are the lower ones, with the z-axis the one pointing vertically upward. That is, they’re the ones in the floor of the room. So the typical two-dimensional figure would be the lower axes.
The Set Tour, Part 6: One Big One Plus Some Rubble
I have a couple of sets for this installment of the Set Tour. It’s still an unusual installment because only one of the sets is that important for my purposes here. The rest I mention because they appear a lot, even if they aren’t much used in these contexts.
I, or J, or maybe Z
The important set here is the integers. You know the integers: they’re the numbers everyone knows. They’re the numbers we count with. They’re 1 and 2 and 3 and a hundred million billion. As we get older we come to accept 0 as an integer, and even the negative integers like “negative 12” and “minus 40” and all that. The integers might be the easiest mathematical construct to know. The positive integers, anyway. The negative ones are still a little suspicious.
The set of integers has several shorthand names. I is a popular and common one. As with the real-valued numbers R and the complex-valued numbers C it gets written by hand, and typically typeset, with a double vertical stroke. And we’ll put horizontal serifs on the top and bottom of the symbol. That’s a concession to readability. You see the same effect in comic strip lettering. A capital “I” in the middle of a word will often be written without serifs, while the word by itself needs the extra visual bulk.
The next popular symbol is J, again with a double vertical stroke. This gets used if we want to reserve “I”, or the word “I”, for some other purpose. J probably gets used because it’s so very close to I, and it’s only quite recently (in historic terms) that they’ve even been seen as different letters.
The symbol that seems to come out of nowhere is Z. It comes less from nowhere than it does from German. The symbol derives from “Zahl”, meaning “number”. It seems to have got into mathematics by way of Nicolas Bourbaki, the renowned imaginary French mathematician. The Z gets written with a double diagonal stroke.
Personally, I like Z most of this set, but on trivial grounds. It’s a more fun letter to write, especially since I write it with the middle horizontal stroke that. I’ve got no good cultural or historical reason for this. I just picked it up as a kid and never set it back down.
In these Set Tour essays I’m trying to write about sets that get used often as domains and ranges for functions. The integers get used a fair bit, although not nearly as often as real numbers do. The integers are a natural way to organize sequences of numbers. If the record of a week’s temperatures (in Fahrenheit) are “58, 45, 49, 54, 58, 60, 64”, there’s an almost compelling temperature function here. f(1) = 58, f(2) = 45, f(3) = 49, f(4) = 54, f(5) = 58, f(6) = 60, f(7) = 64. This is a function that has as its domain the integers. It happens that the range here is also integers, although you might be able to imagine a day when the temperature reading was 54.5.
Sequences turn up a lot. We are almost required to measure things we are interested in in discrete samples. So mathematical work with sequences uses integers as the domain almost by default. The use of integers as a domain gets done so often that it often becomes invisible, though. Someone studying my temperature data above might write the data as f1, f2, f3, and so on. One might reasonably never even notice there’s a function there, or a domain.
And that’s fine. A tool can be so useful it disappears. Attend a play; the stage is in light and the audience in darkness. The roles the light and darkness play disappear unless the director chooses to draw attention to this choice.
And to be honest, integers are a lousy domain for functions. It’s achingly hard to prove things for functions defined just on the integers. The easiest way to do anything useful is typically to find an equivalent problem for a related function that’s got the real numbers as a domain. Then show the answer for that gives you your best-possible answer for the original question.
If all we want are the positive integers, we put a little superscript + to our symbol: I+ or J+ or Z+. That’s a popular choice if we’re using the integers as an index. If we just want the negative numbers that’s a little weird, but, change the plus sign to a minus: I.
Now for some trouble.
Sometimes we want the positive numbers and zero, or in the lingo, the “nonnegative numbers”. Good luck with that. Mathematicians haven’t quite settled on what this should be called, or abbreviated. The “Natural numbers” is a common name for the numbers 0, 1, 2, 3, 4, and so on, and this makes perfect sense and gets abbreviated N. You can double-brace the left vertical stroke, or the diagonal stroke, as you like and that will be understood by everybody.
That is, everybody except the people who figure “natural numbers” should be 1, 2, 3, 4, and so on, and that zero has no place in this set. After all, every human culture counts with 1 and 2 and 3, and for that matter crows and raccoons understand the concept of “four”. Yet it took thousands of years for anyone to think of “zero”, so how natural could that be?
So we might resort to speaking of the “whole numbers” instead. More good luck with that. Besides leaving open the question of whether zero should be considered “whole” there’s the linguistic problem. “Whole” number carries, for many, the implication of a number that is an integer with no fractional part. We already have the word “integer” for that, yes. But the fact people will talk about rounding off to a whole number suggests the phrase “whole number” serves some role that the word “integer” doesn’t. Still, W is sitting around not doing anything useful.
Then there’s “counting numbers”. I would be willing to endorse this as a term for the integers 0, 1, 2, 3, 4, and so on, except. Have you ever met anybody who starts counting from zero? Yes, programmers for some — not all! — computer languages. You know which computer languages. They’re the languages which baffle new students because why on earth would we start counting things from zero all of a sudden? And the obvious single-letter abbreviation C is no good because we need that for complex numbers, a set that people actually use for domains a lot.
There is a good side to this, if you aren’t willing to sit out the 150 years or so mathematicians are going to need to sort this all out. You can set out a symbol that makes sense to you, early on in your writing, and stick with it. If you find you don’t like it, you can switch to something else in your next paper and nobody will protest. If you figure out a good one, people may imitate you. If you figure out a really good one, people will change it just a tiny bit so that their usage drives you crazy. Life is like that.
Eric Weisstein’s Mathworld recommends using Z* for the nonnegative integers. I don’t happen to care for that. I usually associate superscript * symbols with some operations involving complex-valued numbers and with the duals of sets, neither of which is in play here. But it’s not like he’s wrong and I’m right. If I were forced to pick a symbol right now I’d probably give Z0+. And for the nonpositive itself — the negative integers and zero — Z0- presents itself. I fully understand there are people who would be driven stark raving mad by this. Maybe you have a better one. I’d believe that.
Let me close with something non-controversial.
These are some sets that are too important to go unmentioned. But they don’t get used much in the domain-and-range role I’ve been using as basis for these essays. They are, in the terrain of these essays, some rubble.
You know the rational numbers? They’re the things you can write as fractions: 1/2, 5/13, 32/7, -6/7, 0 (think about it). This is a quite useful set, although it doesn’t get used much for the domain or range of functions, at least not in the fields of mathematics I see. It gets abbreviated as Q, though. There’s an extra vertical stroke on the left side of the loop, just as a vertical stroke gets added to the C for complex-valued numbers. Why Q? Well, “R” is already spoken for, as we need it for the real numbers. The key here is that every rational number can be written as the quotient of one integer divided by another. So, this is the set of Quotients. This abbreviation we get thanks to Bourbaki, the same folks who gave us Z for integers. If it strikes you that the imaginary French mathematician Bourbaki used a lot of German words, all I can say is I think that might have been part of the fun of the Bourbaki project. (Well, and German mathematicians gave us many breakthroughs in the understanding of sets in the late 19th and early 20th centuries. We speak with their language because they spoke so well.)
If you’re comfortable with real numbers and with rational numbers, you know of irrational numbers. These are (most) square roots, and pi and e, and the golden ratio and a lot of cosines of angles. Strangely, there really isn’t any common shorthand name or common notation for the irrational numbers. If we need to talk about them, we have the shorthand “R \ Q”. This means “the real numbers except for the rational numbers”. Or we have the shorthand “Qc”. This means “everything except the rational numbers”. That “everything” carries the implication “everything in the real numbers”. The “c” in the superscript stands for “complement”, everything outside the set we’re talking about. These are ungainly, yes. And it’s a bit odd considering that most real numbers are irrational numbers. The rational numbers are a most ineffable cloud of dust the atmosphere of the real numbers.
But, mostly, we don’t need to talk about functions that have an irrational-number domain. We can do our work with a real-number domain instead. So we leave that set with a clumsy symbol. If there’s ever a gold rush of fruitful mathematics to be done with functions on irrational domains then we’ll put in some better notation. Until then, there are better jobs for our letters to do.
Quick Little Calculus Puzzle
fluffy, one of my friends and regular readers, got to discussing with me a couple of limit problems, particularly, ones that seemed to be solved through L’Hopital’s Rule and then ran across some that don’t call for that tool of Freshman Calculus which you maybe remember. It’s the thing about limits of zero divided by zero, or infinity divided by infinity. (It can also be applied to a couple of other “indeterminate forms”; I remember when I took this level calculus the teacher explaining there were seven such forms. Without looking them up, I think they’re \frac00, \frac{\infty}{\infty}, 0^0, \infty^{0}, 0^{\infty}, 1^{\infty}, \mbox{ and } \infty - \infty but I would not recommend trusting my memory in favor of actually studying for your test.)
Anyway, fluffy put forth two cute little puzzles that I had immediate responses for, and then started getting plagued by doubts about, so I thought I’d put them out here for people who want the recreation. They’re both about taking the limit at zero of fractions, specifically:
\lim_{x \rightarrow 0} \frac{e^x}{x^e}
\lim_{x \rightarrow 0} \frac{x^e}{e^x}
where e here is the base of the natural logarithm, that is, that number just a little high of 2.71828 that mathematicians find so interesting even though it isn’t pi.
The limit is, if you want to be exact, a subtly and carefully defined idea that took centuries of really bright work to explain. But the first really good feeling that I really got for it is to imagine a function evaluated at the points near but not exactly at the target point — in the limits here, where x equals zero — and to see, if you keep evaluating x very near zero, are the values of your expression very near something? If it does, that thing the expression gets near is probably the limit at that point.
So, yes, you can plug in values of x like 0.1 and 0.01 and 0.0001 and so on into \frac{e^x}{x^e} and \frac{x^e}{e^x} and get a feeling for what the limit probably is. Saying what it definitely is takes a little more work. | {
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3,626,502,576,268,588,500 | Take the 2-minute tour ×
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I have a question about constructing projective plane over $\mathbb{F}_3$. We first establish seven equivalence classes $P= \{ [0,0,1], [0,1,0], [1,0,0], [0,1,1], [1,1,0], [1,0,1], [1,1,1] \}$.
Given a triple $(a_0, a_1, a_2) \in \mathbb{F}^3_3 \setminus (0, 0, 0)$ we define the line $L(a_0, a_1, a_2)$ as follows: $L(a_0, a_1, a_2) := \{ [x_0; x_1; x_2] \in P : a_0x_0 + a_1x_1 + a_2x_2 = 0 \}$.
It is quite easy for $L(0,0,1), L(0,1,0), L(1,0,0)$, because here we take the points which have zero on the first, second and third coordinate.
It gets more difficult for me for $L(0,1,1)$, because here we need to have $x_1+x_2=0$. Do we treat the coordinates of the points as elements of $\mathbb{F}_2$ or $\mathbb{F}_3$?
There are $26$ nonzero triples in $\mathbb{F}^3_3 $.
Do we check all 26 $L(x_0, x_1, x_2) $ sets?
Please help, because I really want to understand it.
Thank you.
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2 Answers 2
up vote 6 down vote accepted
You are describing a method for getting the Fano plane, which by definition is the projective plane over $\mathbb F_2$.
For getting the projective plane over $\mathbb F_3$ in a similar way, do the following:
Select a set of projective representatives of the nonzero elements of $\mathbb F_3^3$. One way to do this is to select only the vectors whose first nonzero entry is a $1$. Since $\mathbb F_3$ has $2$ units, you end up with $(3^3 - 1) / 2 = 13$ vectors. Those vectors are called coordinate vectors and give you the $13$ points of the projective plane.
You can also use the coordinate vectors for the description of the lines: Each coordinate vector $v$ corresponds to the line containing all the points $w$ such that the scalar product of $v$ and $w$ equals $0$ (so $\langle v,w\rangle = 0$). In this way, there are $13$ lines containing $4$ points each.
Typically, the projective plane over a field $K$ is defined in this way:
The subspaces of $K^3$ of dimension $1$ are the points, and the subspaces of $K^3$ of dimension $2$ are the lines. It may be worth to convince yourself that the above construction is compatible with this description.
share|improve this answer
What does first nonzero entry mean? There are 9 vectors whose first entry is 1, so it must mean something else. – Sandy May 26 '13 at 9:07
1
The first nonzero entry of $(0,1,0)$ is $1$, too! – azimut May 26 '13 at 9:08
1
You should understand that finite projective planes are primarily combinatorial objects. Their main feature is: For each pair of distinct points, there is exactly one line connecting those two points. Of course it is nice to visualize those planes, but it gets harder as he field size increases. The plane over $\mathbb F_3$ can still be visualized in a reasonable way, see geometrie.tuwien.ac.at/havlicek/img/pg23_240x240.gif for a common way to do it. – azimut May 26 '13 at 9:31
1
Yes, exactly. The projective plane over $\mathbb{F}_4$ hast 21 points and 21 lines. – azimut May 26 '13 at 12:23
2
In general, in the projective plane over $\mathbb F_q$, each line contains $q+1$ points, and through each point there are $q+1$ lines. The number of points and lines is $(q^3 - 1)/(q-1) = q^2 + q + 1$. Google brought up this picture of the plane over $\mathbb F_4$: maa.org/editorial/mathgames/21lines.gif You see, this is already quite involved. – azimut May 26 '13 at 13:15
The points and lines are not triples in $\mathbb {F}^3_3$, but they are elements of $\mathbb{F}^3_2$. So the coordinates of points and lines are in $\mathbb{F}_2$.
For example, your line $L(0,1,1)$ contains the points $[x_0,x_1,x_2]$ with $x_1+x_2=0$. Thus $x_1=x_2=1$ or $x_1=x_2=0$. So the possibilities are: $[0,1,1]$ , $[1,1,1]$ , $[1,0,0]$.
Edit: First the question was about the Fano plane, I provided my answer in that case.
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8,752,326,983,664,865,000 | Презентация "Выполнение вычислений по табличным данным в MS Word" 10 класс скачать
Презентация "Выполнение вычислений по табличным данным в MS Word" 10 класс
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«Выполнение вычислений по табличным данным в MS Word»
Учитель информатики
Лесконог Елена Викторовна
Златоустовская ОШ I-III ступеней
Основные принципы работы с полем Формула
• В раскрывающемся списке Формат числа (Number format) укажите числовой формат результата вычислений. Например, для отображения чисел в виде процентов выберите 0%.
Если над курсором расположены ячейки с числами, то в поле Формула (Formula) Word предложит формулу =SUM(ABOVE),
по которой производится суммирование чисел, расположенных выше в ячейках этого столбца. Если ячейки с числами расположены левее ячейки с курсором, то Word предложит формулу =SUM(LEFT).
Отметим, что суммирование производится до первой пустой ячейки. Чтобы просуммировать всю строку или весь столбец, вставьте в пустые ячейки нули.
Если Word предлагает неподходящую формулу, удалите ее из поля Формула (Formula) и из списка Вставить функцию (Paste function) выберите формулу, по которой будут проводиться вычисления. Сведения о доступных функциях приведены в таблице.
• Для вставки закладки выберите помеченный закладкой диапазон ячеек, который следует использовать в вычислениях, или введите его самостоятельно в поле Формула.
Стандартные функции, которые можно ввести в поле Формула
Функция
Возвращаемое значение
ABS(x)
Абсолютное значение числа или формулы (без знака).
AND(x;y)
1 (истина), если оба логические выражения х и у истинны, или 0 (ложь), если хотя бы одно из них ложно.
AVERAGE( )
Среднее значений, включенных в список.
COUNT( )
Число элементов в списке.
DEFINED(x)
1 (истина), если выражение х допустимо, или 0 (ложь), если оно не может быть вычислено.
FALSE
0 (нуль).
IF(x;y;z)
у, если условие х истинно, или z, если оно ложно.
INT(x)
Целая часть числа или значения формулы х.
MIN( )
Наименьшее значение в списке.
MAX()
Наибольшее значение в списке.
MOD(x;y)
Остаток от деления х на у.
Функция
Возвращаемое значение
NOT(x)
0 (ложь), если логическое выражение х истинно, или 1 (истина), если оно ложно.
OR(x;y)
1 (истина), если хотя бы одно из двух логических выражений х и у истинно, или 0 (ложь), если оба они ложны.
PRODUCT( )
Произведение значений, включенных в список. Например, функция { = PRODUCT (1;3;7;9) } возвращает значение 189.
ROUND(x;y)
Значение х, округленное до указанного десятичного разряда (у), х может быть числом или значением формулы.
SIGN(x)
Знак числа: 1 (если х > 0) или —1 (если х < 0).
SUM()
Сумма значений или формул, включенных в список.
TRUE
1
Функция
Возвращаемое значение
Для функций с пустыми скобками допустимо любое число аргументов, разделенных точками с запятыми (;). В скобки могут вводиться ссылки на ячейки таблицы, в которых находятся данные, вставляемые в формулу. Аргументами также могут быть числа и формулы. Для обновления поля выделите его и нажмите клавишу F9. Для обновления всех полей таблицы выделите всю таблицу и нажмите клавишу F9.
Задание 1 1. Создайте и заполните ячейки таблицы по образцу. Переход из ячейки в ячейку нажатием клавиши Tab.
Отформатируйте таблицу по образцу
Отделы
Канцелярские товары
Расходные материалы на оргтехнику
Отдел снабжения
46
200
Отдел маркетинга
200
230
Плановый
120
400
Бухгалтерия
340
560
Отдел кадров
20
140
Технический отдел
373
749
Преобразовать рассматриваемую таблицу следующим образом:
• Вставить столбцы Сумма и НДС и оформить их как вычисляемые.
• Вставить строку ИТОГО и подсчитать в ней суммы по каждому столбцу.
Добавление столбца
Выделите крайний столбец Расходные материалы на оргтехнику Таблица – Вставить – Столбцы слева /p>
Оформление столбца Сумма .
Значение этого столбца вычисляется как сумма столбцов
Канцелярские товары и Расходные материалы на оргтехнику с помощью команды Таблица - Формула.
В появившемся окне в строке Формула нужно набрать
= SUM ( LEFT ), что означает суммирование числовых ячеек, расположенных слева.
Формулу нужно набирать для каждой ячейки столбца.
Оформление столбца НДС.
Значение НДС определяется как 5% от значений столбца Сумма.
В этом случае столбцы таблицы имеют имена A,B,C,D ...., строки нумеруются 1,2,3,..., ячейки имеют адреса из имени столбца и номера строки, на пересечении которых они расположены - Al, B3, D4 и т.д., как в табличном процессоре.
Для подсчета значения в первой строке столбца НДС в окне команды Таблица, Формула набираем формулу = D 2*5%, для второй строки - = D 3*5% и т.д.
Оформление строки ИТОГО
Для вычисления суммы по столбцу Канцелярские товары нужно в окне команды Таблица, Формула набрать формулу = SUM ( ABOVE ). Аналогично вычисляются суммы по остальным столбцам.
Задание 2 1. Создайте и заполните ячейки таблицы по образцу. Переход из ячейки в ячейку нажатием клавиши Tab.
2. Выполнить необходимые вычисления по формулам, согласно данным в таблице.
3. Оформить таблицу по образцу.
4. Выполнить заливку ячеек (цвета выбрать произвольно).
5. Оформить границы в таблице:
1
2
3
4
5
6
7
8
9
Min
2
3
4
5
6
7
8
9
Max
Sum
AVERAGE
COUNT
Спасибо за урок !!!
Список используемых источников
http://www.klyaksa.net/htm/kopilka/index.htm
Информатика. 9 класс. Босова Л.Л., Босова А.Ю.
Информатика: 7-11 класс. Гаевский А.Ю. | {
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4,365,226,658,390,539,000 | IBM Research | Ponder This | August 2007 solutions
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August 2007
<<July August September>>
Ponder This Solution:
The sequence {f(n)/f(n-1)} for positive integers n-s covers all the positive rational numbers in reduced form.
If n is odd (=2m+1), from the definition of f(n) (modulo 2) we get that 1 (=2^0) appears exactly once and hence
(1a) f(2m+1) = f(m).
If n is even (=2m), 1 either does not appear at all or appears twice. In the first case, we get f(m) possibilities and in the second we get f(m-1). Therefore,
(1b) f(2m)=f(m)+f(m-1).
Clearly f(n)>0 and therefore
(2) f(2n-1) = f(n-1) < f(n-1)+f(n) = f(2n).
Claim 1: gcd(f(n),f(n-1))=1
Proof: By induction on n. The base (n=1) is trivial: gcd(f(1),f(0))=gcd(1,1)=1
Suppose the induction hypothesis holds for k<n, then from (1a) & (1b):
gcd(f(n),f(n-1)) = gcd(f(2m+1),f(2m)) = gcd(f(m), f(m)+f(m-1)) = gcd(f(m),f(m-1)) = 1
or
gcd(f(n),f(n-1)) = gcd(f(2m),f(2m-1)) = gcd(f(m)+f(m-1), f(m-1)) = gcd(f(m),f(m-1)) = 1
qed.
Claim 2: f(n+1)/f(n) = f(n’+1)/f(n’) implies that n=n’
Proof: Again, by induction on max(n,n’). The base (max(n,n’)=0) is trivial since n=n’=0.
Assume the induction hypothesis holds for k<n, then from (2) we get that
f(n)/f(n-1) = f(n’)/f(n’-1) can only hold when n=n’(mod 2) and from (1a)&(1b) we get
f(m)/f(m-1) = f(m’)/f(m’-1) and by induction m=m’ yields n=n’.
qed.
Claim 3: For every positive rational number r, there exists a positive integer n such that r = f(n)/f(n-1).
Proof: Write r as a quotient of two relative prime natural numbers p and q.
We prove by induction on max(p,q).
The base (p=q=1) is trivial: n=1.
Assume the induction hypothesis holds for all k<n.
If p<q (q<p), then define r’ as p’/q’ where p’=p & q’=q-p (p’=q & q’=p-q). r’ satisfies the induction hypothesis and therefore r’ = f(m)/f(m-1). Hence r = f(2m+1)/f(2m) (f(2m)/f(2m-1))
qed.
If you have any problems you think we might enjoy, please send them in. All replies should be sent to: [email protected]
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-5,019,921,934,973,796,000 | Recipe Ratios 7.RP.A.1.
Recipe Ratios 7.RP.A.1.
Recipe Ratios 7.RP.A.1.
Recipe Ratios 7.RP.A.1.
Recipe Ratios 7.RP.A.1.
Recipe Ratios 7.RP.A.1.
Recipe Ratios 7.RP.A.1.
Recipe Ratios 7.RP.A.1.
Recipe Ratios 7.RP.A.1.
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Product Description
Recipe Ratios 7.RP.A.1.
• 12 Problems
• 2 Recipes (Complete Recipes Now Included!)
• 3 Pages
• Answer Keys
................................................................................................................................
How can you use this resource?
• Use as a packet (3 worksheets) or 3 separate worksheets!
• There are 4 problems on each page (12 problems total)
• Use these practice problems to re-enforce necessary skills related to 7.RP.A.1.
• CCSS.MATH.CONTENT.7.RP.A.1Compute unit rates associated with ratios of fractions, including ratios of lengths, areas and other quantities measured in like or different units. For example, if a person walks 1/2 mile in each 1/4 hour, compute the unit rate as the complex fraction 1/2/1/4 miles per hour, equivalently 2 miles per hour. http://www.corestandards.org/Math/Content/7/RP/
• Use these practice problems when practicing dividing a fraction by a fraction.
................................................................................................................................
Use the resource as a…
• Quiz
• Homework Assignment
• Ticket Out the Door
• Warm-Up
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Related Products:
Dividing a Fraction by a Fraction.
Adding and Subtracting Fractions Animated PowerPoint.
Ratios and Proportional Relationships Middle School Math Stations.
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-7,495,060,229,928,434,000 | Jean-Paul CIPRIA
————— PHYSICS – SPACE – ALGORITHMICS ——-
Navigation 13 – Spherical Position and Local Plan – Spherical Trajectories ?
Written By: Jean-Paul Cipria - Août• 11•17
Spherical Trajectory - ©Jean-Paul Cipria 2017
Spherical Trajectory – Matlab Personal Algorithms – ©Jean-Paul Cipria 2017
How do we set a Position on Earth and how we determine Local Coordinates and Perpendicular Plan ?
And then how do we find trajectories point by point on Earth ? With and … without Latitude and Longitude, only with an Absolute Rotation Angle ?
Done ... Un Café ?
Done … Un Café ?
Complicated but Not Difficult - Bachelord Level
Complicated but Not Difficult – Bachelord Level
Created :2017-08-11 16:09:53. – Modified : 2018-04-01 17:35:45.
We want to « navigate » on earth with direct trajectory from P point to B point but merely (simplement ?) use latitude and longitude ? When we are in local coordinates we use for a small instant an absolute directionnal vector. When all coordinates are aligned on a known axes meridian for exemple we can « trace » trajectory easily but … if we turn our spherical earth or our route as we can see on following pictures then problems begin.
Some web sites and bachelord courses describe local plan as we do on the first paragraph but how do we navigate without use d\varphi and d\theta like changes of tack on sailboat (virements de bord d’un voilier) ?
The same applies for telescopes, radars, robotic arms [CIPRIA-Robotics-2015], or satellites trajectories. We can’t do some thousands small iterations on elevation and azimut motors ! Did you see the NASA trajectories panels at TV when they launched spatial engines ? Here are the beginning of understanding how to … think and, as an arrow indicate direction and sens, how to do …
Local Plan
Is Local Plan Perpendicular to Point Position ?
We can see a projected point P from position (0,0,0) to latitude and longitude at radius R. We calculate for this position e_\varphi and e_\theta local Plan perpendicular to the P Point.
Sphère et Plan Local 2 - ©Jean-Paul Cipria 2017
Sphère et Plan Local 2 – ©Jean-Paul Cipria 2017
e_\varphi and e_\theta vectors define the local plan by definition. Then Local plan is perpendicular to \vec{OP} vector.
Do Local Plan display Real Measurements ?
Vectors e_\varphi and e_\theta are like our LOCAL North-South and Est-West directions. Generaly they don’t match with latitude and longitude metrics. On the following example e_\varphi seems to follow the Greenwitch meridian. It is right only on the P Point.
Sphère et Plan Local 2 - ©Jean-Paul Cipria 2017
Sphère et Plan Local 2 – ©Jean-Paul Cipria 2017
Local plan don’t display correctly real metrics measurements except in P point.
Matlab : How do we plot a so beautiful blue Earth ?
Matlab Begin.
max=2*pi;
pas = max/50;
% A so beautiful blue earth !
% --------------------------
figure % Don't forget it otherwise plot, plot ... nothing !
hold on % Don't forget it otherwise you got only one point ... the last !
for lat=-pi:pas:+pi
for long=0:pas:max
x = R*cos(lat)*cos(long); % Don't use colat and colong but change sin(colat) to cos(lat) !!!
y = R*cos(lat)*sin(long);
z = R*sin(lat);
plot3(x,y,z,'.','color',[0.4 0.6 1],'markersize',12);
end
end
Matlab End.
Sorry for the missing of indentation on program writing but the object langage used to « concept » the php display « absorb », eliminate, « sont phagocytés » as said frenches, the blank or tabulation characters. We do some script in … 1990 to « remediate » those little bugs. But C++ managers forgot them for thirty years ? These are the invisible part of the object langage iceberg inconsistencies.
.
Do we use Quaternions Concepts to Solve Trajectories ?
Yes we did – 3D View ?
Quaternion Trajectory - ©Jean-Paul Cipria 2017
Quaternion Trajectory – ©Jean-Paul Cipria 2017
As we can see \vec{N} is perpendicular for \overrightarrow{OP} and \overrightarrow{OB} plan.
We can consider the green arrow under earth trajectory is a bad one (Not confuse with the green dots Greewitch referential). Is it like neutrino famous experiment from France to Italy [CIPRIA-Light Speed-2011] ?
Quaternion Trajectory - ©Jean-Paul Cipria 2017
Quaternion Trajectory – ©Jean-Paul Cipria 2017
CQFD
How do we do ?
The Famous Trigonometrical Formula ?
First : We determine the angle sin(\alpha) (or distance in radian) between P and B on the Spherical Distance like all web sites and french courses displayed it and ? … work is finish for them …
Matlab Begin
% Cosinus angle_distance entre OP et OB
% --------------------------------------
cosinus_angle = sin(latP)*sin(latB)+cos(latP)*cos(latB)*cos(longB-longP);
% Distance en Radian.
% --------------------------------
angle_distance = acos(cosinus_angle);
Matlab End.
Here below we determinate the TRAJECTORY and it is more difficult as :
Perpendicular Vector from Plan Trajectory ?
Second : We determinate perpendicular plan vector \vec{N} for \overrightarrow{OP} and \overrightarrow{OB} by crossing this two vectors.
• \vec{N}=\overrightarrow{OP}\wedge\overrightarrow{OB}=|\vec{N}|.\vec{e}_N . With \vec{e}_N is unitary vector perpendicular to OP and OB plan.
We know also that we can write this vector as :
• \vec{N}=|\overrightarrow{OP}|\wedge|\overrightarrow{OB}|=R.R.sin(\alpha).\vec{e}_N
We divide vector \vec{N} by R because the two vectors \overrightarrow{OP} and \overrightarrow{OB} have R length. To avoid \vec{N} vector to have R^2 lenght and also by sin(\alpha) to have a unitary vector with R lenght.
• \vec{N_{unitR}}=\frac{\overrightarrow{OP}\wedge\overrightarrow{OB}}{R.sin(\alpha)}=R.\vec{e}_N .
Rotate by Quaternion Concepts ?
Third : We turn \overrightarrow{OP} over \vec{N_{unitR}} by Quaternion Concept [CIPRIA-Quaternions-2017] by angle previously determined.
Quaternions de l'axe k - Rotations de vues 3d
Quaternions by 45° rotations Steps around the k (or Z or North) Vector.
What is a quaternion ?
As we detailled previous [CIPRIA-Quaternions-2017] quaternion is the same as complex number but with 3 ? 4 dimensions ? « Normal » complex number is 2 dimensions as : a+ib then a is the real part and b is the imaginary part projected in the i perpendicular one dimension space. Then in quaternion we add another two dimensions j and k and … ? A Spherical Angle \varphi ! Then quaternion is a 4 dimensions number. Example :
• q=a+bi+cj+dk
In this complex form to « see » the quaternion we are able to do some calculations with some multiplication and addition new rules. But it is another form more explicit to understand quaternion. We display the Spherical Angle on it :
• cos(\varphi)+sin(\varphi).u_x.i+sin(\varphi).u_y.j+sin(\varphi).u_z.k
We can see here that there are 4 variables : A Spherical Angle \varphi and a three dimensions Vector : u_x ,u_y , u_z .
Because an angle is INSIDE the quaternion we have define, by definition, a vector with its three coordinates AND a rotation in a four dimensions space. It is a great subtility. Then because we have a vector AND a spherical rotation position this quaternion seems to be a super vector or like a rotation matrix ? Yes it do.
After that we use for example the \vec{OP} vector to turn around the q quaternion as the method defined in previous article [CIPRIA-Quaternions-2017].
Navigate throught the Calculated Trajectory ?
Quarto and lasto : We can specify like as we do fractional angle d\alpha by a satellite speed and determinate for all s and ds positions the « right » trajectory to follow. Then we can then perfom servo-command for our navigation … And do not change of tack on sailboat at every 100 nautical milles (idem virements de bord d’un voilier) as every body explained us. Then we can do automatismes and pilote drone !
MP4 Video (French Explanations)
Click for MP4 Film
Earth Trajectory Paris New-York by Quaternion Rotations – ©Jean-Paul Cipria 2017
Local Plan for Paris New-York Trajectory
- ©Jean-Paul Cipria 2017
Local Plan For Paris New-York Trajectory – ©Jean-Paul Cipria 2017
.
Thank You Olinde Rodrigues [Rodrigues-1815 ].
That’s all Folks ! 😉
Result
Numeric Values
Paris Starting Route = -68.6°
New-York Ending Route = -54.1°
Latitude Paris = 48.8534100°
Longitude Paris = 2.3488000°
Latitude New-York = 40.5931°
Longitude New-York = -73.5686°
Distance = 5811 km.
Vertex Maximum Latitude = 52.22°
Trajectories Animation
Spherical Trajectory - ©Jean-Paul Cipria 2017
Spherical Trajectory – ©Jean-Paul Cipria 2017
Trajectory Explanation
Spherical Trajectory 2 - ©Jean-Paul Cipria 2017
Spherical Trajectory 2 – ©Jean-Paul Cipria 2017
References
Did Neutrinos exceed the speed of light in 2011?
• [CIPRIA-Light Speed-2011] : Jean-Paul Cipria, « Light Speed Measurements with neutrino experiments in 2011. Measurements Error in Physics. How do we made mistakes and how we correct them ? « – 28/11/2016.
Les neutrinos ont-ils dépassé la vitesse de la lumière en 2011 ?
Les quaternions
• [CIPRIA-Quaternions-2017] : Jean-Paul Cipria – Quaternions or real or mathematics space rotation ? How do we easily calculate with satelitte computer a rotation around a priviligied axe without evaluate it with rotation matrix ? French description.
Navigation 12 – Les Quaternions ou Rotations dans l’Espace Réel ou … Mathématique ?
Robotics Trajectories
• [CIPRIA-Robotics-2015] : Cipria, Jean-Paul : « Here a small film on how to Program a Trajectory on a Defined Plan … then we 3D Change Referential Position and we perform the same Program. Whats Up, Doc? »
Robotics – Industrials 4
Scientifics
• [Rodrigues-1815] : Olinde Rodrigues : « De l’attraction des sphéroïdes » – Thèse de Doctorat – 28 juin 1815
.
Jean-Paul Cipria
©11/08/2017
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6,477,084,202,833,221,000 | Categories
Development
Given n integer a1, a2, …, an and k. How many ways to choose k consecutive elements that those elements can construct a triangle? [closed]
Consecutive elements can construct a triangle if they can be divided into 3 sets as 3 edges (sum of elements in a set is the weight of 1 edge). (n <= 10^5, 0 <= a[i] <= 10^9, each element is distinct).
For example:
n = 6, k = 3
a = [1, 3, 4, 2, 5, 9]
There are 2 ways:
1) 1, 3, 4
2) 3, 4, 2
I have tried with this:
With every segment (i, i + k – 1), I divide it into 3 subsets and calculate the sum each set.
Let a, b, c be the sum of elements of each set respectively. I check:
a+b>c
a+c>b
b+c>a
With this algorithm, it takes O(n*(3^k)) time.
I read by this by this algorithm take O(n) time but I dont’t understand why it works:
With every segment (i, i + k – 1), assume that we sorted them so we have: ai < a(i+1) < … < an.
Let:
x = ai + a(i + 2) + … + a(n – 2)
y = a(i+1) + a(i + 3) + … + a(n-1)
z = an
We have y + z > x and x + z > y (can be easily proved),
we just need to check if:
x + y > z
<=> x + y + z > 2z
So we don’t need to sort the segment, just check if sum of all elements in the segment greater than 2 times the maximum element.
We can calculate the sum by prefix sum and maximum element by Deque in O(n).
I don’t understand that in a segment, (x, y, z) could be another way to divide into 3 sets, so why is this solution true?
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7,748,187,496,510,280,000 | Home > Games > Math > Subtract and FInd the DIfference Game
Subtract and FInd the DIfference Game
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6,308,429,820,109,640,000 | Sunday
February 7, 2016
Homework Help: math
Posted by Holly on Saturday, March 13, 2010 at 7:19pm.
Can someone help me solve the following....using elimination methods
1)
5x+7y=43
-4x+y=25
2)9x-9y=36
7y-3x=-14
I think the answers are
would the answer to number 1 be
(-4.26,1.52)?
would the answer to number 2 be
(-0.5,-6.5)?
Can someone please verify or tell me what I did wrong if they are not right!
Thanks
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Labeling an n-cube
1. Aug 11, 2004 #1
I need some closure on this one:
In how many ways can the vertices of an n-cube be labeled 0, … ,2ⁿ - 1 so that there is an edge between two vertices if and only if the binary representation of their labels differs in exactly one bit?
Let G be an appropriately labeled n-cube. Pick an arbitrary vertex v in G. How many ways can one change labels of the vertices appart from v in G and still mantain the properties of the n-cube? Because of the nature of G, there are n incident edges on v with n adjacent vertices. The labels of these vertices differ from the label of v by one bit. If one swaps the labels of two of these vertices and makes the appropriate swaps elsewhere so as to maintain the n-cube (which can be done due to the symmetry of G), one obtains a new labeling. The number of combination of adjacent vertices that can be swapped is n(n - 1)/2, so the number of different labelings of the vertices in G without changing the label of v is n(n - 1)/2.
The label of vertex v is arbitrary. Therefore, for every possible label of v there are n(n - 1)/2 labelings of the vertices in G. Because there are 2ⁿ - 1 possible labels for v, the number of possible labelings for the vertices in G is (2ⁿ - 1)(n - 1)n/2.
2. jcsd
3. Aug 12, 2004 #2
4. Aug 12, 2004 #3
Interesting link. I just realized the formula I gave is pure BS (for one thing, the possible labels for a vertex v in G is 2ⁿ and even with this mod. the formula doesn't work). I'll have to look at this again...
Last edited: Aug 12, 2004
5. Aug 12, 2004 #4
OK. I think I have it know. Suppose G is an n-cube, appropriately labeled. Let v be a vertex in G. In how many ways can I change the labels of the vertices in G without changing the label of v? There are n adjecent vertices to v. I can move the labels of these vertices around (making the appropriate changes to rest of G) and obtain a new labeling of the vertices in G in n! ways. Since the label of v can be one of 2ⁿ possibilities, the number of different vertex-labelings in G is n!2ⁿ.
The sequence generated seems to be the same as that in:
http://www.research.att.com/cgi-bin/access.cgi/as/njas/sequences/eismum.cgi
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-9,218,939,806,214,877,000 | Uniswap v3 流动性数学 [翻译]
为 Uniswap v3 流动性数学提供进一步阐释
本文翻译自：https://atiselsts.github.io/pdfs/uniswap-v3-liquidity-math.pdf
1.前言
Uniswap v3 作为 Uniswap 协议的最新版本，引入了集中流动性等许多新功能。它允许流动性提供者将其流动性集中在特定的价格范围内，从而提高资本效率。然而，头寸的流动性、资产数量及其价格范围之间的数学关系变得较为复杂。Uniswap v3 白皮书定义了流动性数学的核心概念，然而其提供的信息十分简洁，与 Uniswap 用户和开发人员提出的问题之间存在差距，例如：
• 给定头寸的总流动性和价格范围，该头寸在特定价格 $P$ 下拥有多少资产 $X$ 和资产 $Y$ ？
• 给定价格范围、当前价格 $P$ 和资产 $X$ 的数量 $x$ ，应向其供应多少资产 $Y$ 以涵盖这个价格范围？
• 给定要存入流动性池的 $x$ 数量的资产 $X$ 和 $y$ 数量的资产 $Y$ ，以及当前价格 $P$ 和预期价格范围的下限，预期价格范围的上限是多少？
本文旨在弥合这一差距，为 Uniswap v3 流动性数学提供进一步阐释。
2.计算
2.1 计算流动性和资产数量
白皮书中的方程 6.5 和 6.6 似乎给出了计算 $L$ ， $x$ 和 $y$ 的简单方法：
但是，这里的 $x$ 和 $y$ 是代币的虚拟数量，而不是真实数量。计算实际数量 $x$ 和 $y$ 的数学公式在白皮书的最后给出，具体见方程 6.29 和 6.30。
这些方程可以从白皮书中的关键方程 2.2 推导出来：
尝试直接解方程 2.2 以得到 $L$ 的结果会得到一个非常复杂的答案。相反，我们可以注意到在价格范围之外，流动性完全由单一资产提供，要么是 $X$ ，要么是 $Y$ ，具体取决于当前价格在价格范围的哪一侧。我们有以下三个选择：
1. 假设 $P ≤ p_a$ ，流动性的 position 完全由代币构成，因此 $y = 0$ ：
该头寸的流动性为：
1. 假设 $P ≥ p_b$ ，流动性的 position 同样完全由代币构成，因此 $x = 0$ ：
该头寸的流动性为：
当前价格位于范围内： $p_a < P < p_b$ 。理解这一点的方式是考虑在最优位置，两种资产将平等地贡献流动性。换句话说，在范围的一侧 $(P,p_b)$ 由资产 $x$ 提供的流动性 $Lx$ 必须等于在范围的另一侧 $(p_a,P)$ 由资产 $y$ 提供的流动性 $Ly$ 。
根据方程 (5) 和 (9)，可以得到如何计算单资产范围的流动性。当 $P$ 位于范围 $(p_a,p_b)$ 时，我们可以将 $(P,p_b)$ 视为 $X$ 提供流动性的子范围，将 $(p_a,P)$ 视为 $Y$ 提供流动性的子范围。将此代入方程 (5) 和 (9)，并令 $Lx (P,p_b) = Ly (p_a,P)$ ，可以得到：
方程式 (10) 非常重要，因为它可以解出 $x$ 、 $y$ 、 $P$ 、 $p_a$ 、 $p_b$ 中的任意一项，而无需参考流动性。然而，对于 $x$ 和 $y$ 来说，这并非必要；简单修改方程 (4) 和 (8) 就足够了：
总结：
• 如果 $P ≤ p_a$ ，则 $y = 0$ ，而 $x$ 可以通过方程 (4) 计算。
• 如果 $P ≥ p_b$ ，则 $x = 0$ ，而 $y$ 可以通过方程 (8) 计算。
• 否则，当 $p_a < P < p_b$ 时， $x$ 和 $y$ 可以分别通过方程 (11) 和 (12) 计算。
从概念上讲，这个结果只是以稍微不同的形式重新陈述了白皮书中的方程 6.29 和 6.30。然而，白皮书中对 $∆$ 的使用可能会对新用户造成困惑 —— $∆$ 的差异是什么？如果这是一个全新的头寸呢？为保持简单，方程 (4) - (12) 避免了以上提及的任何差异。当然，深入研究时可以明确，白皮书中的方程 6.29 和 6.30 甚至适用于新头寸：在这种情况下，我们只需要将 $∆L = L$ 。
2.2 根据资产数量计算价格范围
2.2.1 价格范围边界
在给定资产数量和当前价格 $P$ 的情况下，不可能同时推断出两个边界 $p_a$ 和 $p_b$ 。因为只有一个方程（方程(10)），因此当存在两个未知变量时，结果是不确定的。对于任何给定的资产比例，都存在无穷多个相应的价格范围。
然而，以下问题可以解决：
• $P, x, y$ 和 $p_a$ 已知； $p_b$ 的值是多少？
• $P, x, y$ 和 $p_b$ 已知； $p_a$ 的值是多少？
• $P, x$ 和 $y$ 已知；如果价格从当前价格增加 z%，则价格将达到 $p_b$ 。 $p_a$ 对应的价格下跌百分比是多少？换句话说，给定 $p_b/P$ 的比率， $p_a/P$ 是多少？
计算这些答案的一种方法是首先计算价格一侧的流动性。这里开始，我们假设价格在一个非空的范围内，并且不等于范围的端点；形式上， $p_a < P < p_b$ 。一旦 $L$ 已知，就可以分别使用方程 (12) 和 (11) 计算 $p_a$ 和 $p_b$ 。使用平方根能极大简化解，使用起来更方便计算。
或者，可以跳过流动性计算，直接解方程 (10) 得到 $p_a$ 和 $p_b$ 的平方根：
2.2.2 价格区间比例
用户还可以根据当前价格的增加或减少来考虑价格范围。我们引入新的符号 $c$ 和 $d$ 来 表示，令 $c^{2}\cdot P$ 等于范围的上限， $d^{2}\cdot P$ 等于下界：
显然可以得到以下内容：
现在可以使用方程 (10) 来表示它们之间的相互关系。
现在，例如已知 $p_a$ 对应于当前价格的 70%（ $c^2$ 等于 70% = 0.7），可以通过使用方程 (24) 计算 $d^2$ 来计算对应于 $p_b$ 的当前价格的百分比。
3 应用
3.1 实施细节
在将这些数学应用于源代码时，请注意以下几个方面：
• Ticks: Uniswap 中的价格范围具有称为 tick 的离散边界。第 i 个 tick 的价格定义为 $p(i) = 1.0001^i$ 。只有与初始化的 tick 对应的特定价格点可以作为价格范围的边界。
• 刻度间隔: 并非所有刻度都可以初始化。可以索引的确切刻度索引取决于池的费用水平。1% 的池具有最宽的刻度间隔，为 200 个刻度，0.3% 的池为 60 个刻度，0.05% 的池最小，为 10 个刻度。
• Fixed point math: Uniswap v3 使用 fixed point math。这是因为 Solidity 不支持浮点数，而且定点数学有助于最小化舍入误差。这种数学带来了新的挑战，例如需要将带有定点基数的数字相乘或相除以保持其正确的范围。
• Decimal: ERC20 代币定义了一个称为decimal的字段。在以太坊生态系统内部，加密货币的数量以整数表示。为了将 ERC20 代币的数量从这种内部表示转换为人类可读的值，需要除以 $10^{decimal}$ 。不同的 ERC20 合同为其代币定义了不同数量的小数点；例如，DAI 的 decimal 为 18，而 USDC 有 6 个。1 USDC 的人类可读价格约为 1 DAI；但从流动性计算的角度来看，1 USDC 相对于 DAI 的价格约为 $10^{(18-6)} = 10^{12}$ 。
3.2 没有实施细节的示例
在这个示例子节中，为了增加解释的清晰度，我们忽略了上面提到的实施细节。请注意，与从 Uniswap 代码和 UI 获得的结果相比，这可能导致小的数值误差。
3.2.1 示例 1：从给定价格范围计算资产金额
问题：一个用户拥有 x = 2 ETH，并希望在 ETH/USDC 池中建立一个流动性仓位。ETH 的当前价格为 $P$ = 2000 USDC，目标价格范围为 $p_a$ = 1500 到 $p_b$ = 2500 USDC。他们需要多少 USDC（ $y$ ）？
解决方案：首先，使用方程 (5) 计算范围顶半部分的流动性（将 $p_a$ 替换为 $P$ 以获得顶半部分）：
然后，使用 $L$ 的值通过方程 (8) 计算 $y$ ：
答案是 $y$ = 5076.10 USDC。
3.2.2 示例 2：从给定资产金额计算范围
问题：用户有 x = 2 ETH 和 y = 4000 USDC，并希望使用每 ETH $p_b$ = 3000 USDC 作为流动性提供上界。为确保已开仓位使用其全部资金，做市范围的下界（ $p_a$ ）是多少？
解决方案：可以使用与上述示例 1 中相同的半范围技巧计算头寸的流动性，然后计算做市范围的下界。然而， $p_a$ 也可以通过方程 (15) 直接计算：
答案是 $p_a$ = 1333.33 USDC。较低的价格是当前价格的三分之二，而当前价格是较高价格的三分之二：这是因为 USDC 和 ETH 的初始值相等。
3.2.3 示例 3：价格变动后的资产
问题：使用示例 2 中创建的流动性头寸，当价格变为 $P$ = 2500USDC/ETH 时，资产余额是多少？
解决方案：首先计算头寸的流动性，使用之前计算出的 $p_a$ = 1333.33：
使用 64 位浮点数运算，结果分别为 $Lx$ = 487.41718030204123 和 $Ly$ = 487. 4144693682443。 $L$ 是这两者中的最小值： $L = min (Lx, Ly)$ ，在这种情况下对应于 $Lx$ 。
现在，设 $P'$ = 2500，我们可以使用方程 (4) 和 (8) 找到 $x'$ 和 $y'$ ：
答案是 $x$ = 0.85 ETH 和 $y$ = 6572.89 USDC。可以从这个结果计算出无常损失。
我们也可以使用白皮书提供的 delta 计算，即方程 6.14 和 6.16，来计算这个答案：
变动值为： $∆x$ = −1.15 ETH 和 $∆y$ = +2572.89 USDC。
3.3 包含刻度数学和价格转换的示例
刻度数学是解释 Uniswap v3 API 提供的值和 Uniswap v3 子图中索引的数据所必需的。首先，让我们简要回顾一下白皮书中的刻度数学。Uniswap v3 将所有可能价格的连续空间映射到由刻度索引的离散子集。刻度与价格之间的唯一关系由刻度基本参数定义，在 Uniswap 中等于 1.0001。第 i 个刻度对应的价格是：
这表明：
在使用价格的平方根进行计算时：
3.3.2 将价格转换为人类可读的形式
在内部，价格只是 $y$ 和 $x$ 之间的关系。然而，为了在用户界面中显示它，必须考虑资产 $X$ 和 $Y$ 的小数位数。让我们分析一个 Uniswap v3 NFT 来证明这一点，以 ID 为 600006 的 NFT 为例。如图 1 所示，这个 NFT 的「Min Tick」为 200240， “Max Tick”等于200700。将这些刻度数转换为价格，我们得到以下价格范围：
问题在于 USDC（第一个资产 $X$ ）只有小数位数 decimalsx = 6，而 wrapped ETH（第二个资产 $Y$ ）有小数位数 decimals = 18。由于价格是 $y/x$ ，不带小数的价格等于：
在示例池中：
这些表示了以 ETH 为单位的 USDC 的价格。要以 USDC 为单位获取 ETH，需要反转价格：
请注意，最后一个范围是反转的，反转价格在顶部刻度处低于底部刻度处的反转价格。
3.3.3 当前刻度范围内资产的数量
为了了解价格范围内的资产数量，首先我们必须得到池的流动性。一个池变量跟踪特定的刻度范围内的流动性。Uniswap v2 只要不向池中添加或删除资产，池的流动性就保持不变，而与 v2 不同的是， Uniswap v3 中的池流动性可能会在价格跨越价格变动范围边界时发生变化。特别是在跨越具有非零流动性净值的刻度边界时会发生变化。
在本文中，该查询返回的 $L$ 为 22402462192838616433，刻度为 195574，对应价格为 3211.84 USDC/ETH。
池的价格变动间隔由其费率决定。对于 0.3% 的池，刻度间距等于 60。这意味着 只能初始化可被 60 整除的刻度。最接近具有此属性的返回值的刻度是：
因此，价格范围为：
使用方程 (11) 和 (12) 来计算可以从当前价格范围内的流动性中获取的 USDC 或 ETH 的最大值：
在调整各代币的小数后：
大约有 165 万 USDC 和 671 ETH 被锁定在当前的价格范围内。这意味着购买 165 万 USDC 的人将当前价格移动到当前价格范围的一端。相反，如果有人购买 671 ETH，那将把价格移动到范围的另一端。
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-2,150,492,415,715,726,800 | Online JudgeProblem SetAuthorsOnline ContestsUser
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Rooted Trees Isomorphism
Time Limit: 1000MSMemory Limit: 10000K
Total Submissions: 651Accepted: 175
Description
Isomorphism is the problem of testing whether two graphs are really the same. Suppose we are given a collection of graphs and must perform some operation on each of them. If we can identify which of the graphs are duplicate, they can be discarded so as to avoid redundant work.
First we have to explain what is meant when we say two graphs are the same. Two labeled graphs G1 = (V1,E2) and G2 = (V2,E2) are identical when we can find a mapping f of the vertices of G1 to the vertices of G2 such that (x, y) is an edge of G1 if and only if (f(x), f(y)) is an edge of G2. Such a mapping is called an isomorphism.
No efficient algorithm is known for the general graph isomorphism problem, but the problem is easier to determine whether two trees are isomorphic to each other. In Figure 7, it is not hard to verify that tree T1 is isomorphic to tree T3, but T1 is not isomorphic to T2.
You are given a collection of k trees C = {T1, T2, ... , Tk} such that each Ti has exactly n nodes. The objective of the problem is to partition these trees into isomorphic (equivalent) classes such that any two trees within the same isomorphic class are isomorphic to each other.
A naive method of enumerating all possible mapping functions would require generating all possible n! different mappings. What resulted is a very time-consuming O(n!) time algorithm just to test two trees. You need to figure out a somehow clever way for solving the problem.
Input
A collection of k (n-node) trees C = {T1, T2, ... , Tk}. The inputs are just a list of integers. The first 2 integers (in a single line) represent the number of trees, k, and the size of each tree, n. Note that k can be as large as 150 and n can be as large as 100. After the two integers, there will be k lines representing the edge sets for each tree Ti; each line contains exactly n-1 pairs of integers, representing the n-1 (directed) edges of each tree. Thus, there are totally 2n-2 integers for each tree, and the total input will be 2k(n-1) integers except the first two parameters. Each tree is indexed by their appearance ordering; that is, the first line represents the tree T1, the second line is T2, ... , etc, and the last (kth) line is just Tk.
Output
For the given collection of trees, partition these trees into isomorphic (equivalent) classes such that any two trees within the same isomorphic class are isomorphic to each other. For each isomorphic class, output the indices of these isomorphic trees in a line. Suppose that there are m isomorphic classes, you need to print out m lines. For example, a line
t1 = t2 = ... = tp;
represents an isomorphic class of size p such that two trees Tti and Ttj , 1<=i, j <=p, are isomorphic to each other. For each line, output indices of those isomorphic trees in increasing order; that is, t1 < t2 < ... < tp. Further, print out these m isomorphic classes by their increasing lexical ordering; that is, by the ordering of their first indices. For example, suppose that there are 4 isomorphic classes {4, 2, 7}, {5, 1, 3}, {8, 9}, {6}. The output shall be
1 = 3 = 5 ;
2 = 4 = 7 ;
6 ;
8 = 9 ;
Sample Input
3 7
7 2 7 1 7 6 2 3 1 4 6 5
7 2 7 1 2 3 1 4 1 5 5 6
4 3 3 2 4 1 1 7 5 6 4 5
Sample Output
1 = 3 ;
2 ;
Source
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-1,046,098,056,656,233,100 | Related Posts
This Post Has 10 Comments
1. angle BOC
Explanation:
As you can see, they share the same vertex and together they form a right angle so they are definitely complements.
Hope it helps!
2. c
step-by-step explanation:
it would be see because your formula is f(x)=6x, so 6 times 1 is 6 and you go on and the best answer is c.
[tex]49 points me! it's a multiple choice, but still show your work so i know you're right. you so so[/tex]
3. My imaginary angle "in mah mind" is complimentary to COD, hence 90 degree since it's "COMPLIMENTARY"
Step-by-step explanation:
Hehehehe
4. ∠BOC and ∠EOF because ∠BOD is already a right angle and ∠BOC ≅ ∠EOF (because they are opposite angles)
5. Then since ∠BOD = 90°, we know that ∠BOC + ∠COD = 90°. In other words, ∠BOC and ∠COD are complementary. Thus, the answer is angle BOC.
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-6,063,874,223,035,738,000 | Nita B. GCSE Maths tutor, A Level Maths tutor
Nita B.
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Degree: BSc Accounting and Finance (Bachelors) - Bristol University
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Hi! I'm Nita, second year Accounting and Finance student at the University of Bristol. I am keen on sharing my experiences and help out in Maths at GCSE level.
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Questions Nita has answered
Surd Calculations?
Surds are numbers left in 'square root form' (or 'cube root form' etc). They are therefore irrational numbers. Surds are the root of numbers and not whole numbers. Multiplication Example: √2 × √6 = √12 (= 2× 6) = √4 × √3 = 2√3 Addition and Subtraction 4√3 - 2√3 = 2√3 5√5 + 8√5 = 13...
Surds are numbers left in 'square root form' (or 'cube root form' etc). They are therefore irrational numbers. Surds are the root of numbers and not whole numbers.
Multiplication
square root of ab = square root of a x square root of b
square root of a x square root of a = a
Example:
√2 × √6 = √12 (= 2× 6)
= √4 × √3
= 2√3
Addition and Subtraction
4√3 - 2√3 = 2√3
5√5 + 8√5 = 13√5
Note: 5√7 + 3√3 cannot be manipulated because the surds are different (one is √7 and one is √3)
Example:
Simplify √12 + √27
12 = 3 × 4. So √12 = √(3 × 4) = √3 × √4 = 2 × √3.
Similarly, √27 = 3√3.
Hence √12 + √27 = 2√3 + 3√3 = 5√3
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5,029,176,812,725,151,000 | Transum Software
Exam-Style Question on Normal Distribution
A mathematics exam-style question with a worked solution that can be revealed gradually
List Of Questions Exam-Style Question More Normal Questions More on this Topic
Question id: 103. This question is similar to one that appeared on an IB Studies paper in 2014. The use of a calculator is allowed.
A group of students sat a Biology examination and a Computer Studies examination. The students' marks in the Biology examination are normally distributed with a mean of 70 and a standard deviation of 9.
(a) Draw a diagram that shows this information.
(b) Find the probability that a randomly chosen student who sat the Biology examination scored at most 70 marks.
Eric scored 82 marks in the Biology examination.
(c) Find the probability that a randomly chosen student who sat the Biology examination scored more than Eric.
The students' marks in the Computer Studies examination are normally distributed with a mean of 68 and a standard deviation of 11. Eric also scored 82 marks in the Computer Studies examination.
(d) Find the probability that a randomly chosen candidate who sat the Computer Studies examination scored less than Eric.
(e) Determine whether Eric's Computer Studies mark, compared to the other students, is better than his mark in Biology. Give a reason for your answer.
To obtain a grade A a student must be in the top 12% of the students who sat the Computer Studies examination.
(f) Find the minimum possible mark to obtain a grade A. Give your answer correct to the nearest integer.
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Introduction to Washer Method: How to Calculate volume by using this method
Introduction to Washer Method and How to Calculate volume by using this method
Introduction to Washer Method and How to Calculate volume by using this method
In Mathematics, there are different ways to calculate the volume of various kinds of solid shapes. With the help of integration in calculus, you can find the volume of irregular closed shapes like a round shape with a hole in the center and this shape is known as a washer or disk.
The result is represented by the rectangular section with R representing the higher function and r representing the lower function. This explains why there is a huge hole in the middle of solids generated by the space between two functions.
In this article, the basic definition of its formula and methods to calculate volume by using washer methods with detailed explanations with the help of examples will be discussed.
What is the washer method?
The washer method can be used to determine an object’s volume of revolution. To accommodate items with holes, it modifies the disc approach for solid objects. The “washer technique” got its name from the way the cross sections resemble washers.
Formula:
Vy = π dc [F(y)]2 – [G(y)]2 dy
Two functions are rotated around the x-axis or the y-axis to create this shape. Create slices of the shape to determine its volume, then after determining its overall volume, take the middle space out. Calculus refers to this approach as the washing method.
How to derive the formula?
Calculating the washer’s area is as easy as subtracting the area of the inner circle from the area of the outer circle.
A washer = A large – A small = πR2 – πr2 = π [R2 – r2]
By integrating the area function across the range x = 0 to x = h, you can determine the total number of washers:
h0 π [R2 – r2] dx
Now integrating with respect to x
π [R2 – r2] ∫h0 1 dx = π [R2 – r2] [x |h0] = π [R2 – r2] [(h) – (0)] = π [R2 – r2] [h] = πh [R2 – r2]
From this method, it is a derived form of a hollow cylinder: πh [R2 – r2]
How to Calculate Washer Method Problems?
Follow the below examples to learn how to calculate washer method problems.
Example 1
Calculate the volume of the enclosed region and you have a solid obtained by rotating the regions of parabola x = y2 and the other function is x = √y around the y-axis.
Solution:
Step 1: Draw the curves on the coordinate plane. Determine the axis of rotation.
The function x = y2 and x = √y, as well as the region bounded by both curves.
The limits of integration given below for revolving both functions around the y-axis
Step 2: Equating both functions to calculate the points of intersection
y2 = √y
Taking square on both sides it becomes
(y2)2 = (√y)2
y4 = y
y4 – y = 0
Taking common y
y (y3 – 1) = 0
Equating both equations
y = 0 , y3 – 1 = 0
y = 0 , y3 = 1
y = 0 , (y3)1/3 = (1)1/3
y = 0 , y = 1
Therefore, roots are c = 0 and d = 1
Step 3: Now applying the formula
F(y) = y2, and G(y) = √y
Vy = π ∫dc [F(y)]2 – [G(y)]2 dy
Vy = π ∫10 [√y]2 – [(y)2]2 dy
Vy = π ∫10 (y – y4) dy
Vy = π ∫10 (y) dy 10 (y4) dy
Vy = π [(y2 / 2)10 – (y5 / 5)10]
Vy = π [(1 / 2) – (1 / 5)]
Vy = π [(5-2 / 10)]
Vy = π [(5-2 / 10)]
Vy = π [(3 / 10)]
Vy = 3π / 10 cubic units
To avoid such a larger calculations, you can take help from a washer method calculator to find the step-by-step solution to the given problems.
Example 2
Calculate the volume of the enclosed region and you have a solid obtained by rotating the regions of parabola x = – y2 + 4 and the other function is x = – y + 2 around the y-axis.
Solution:
Step 1: Draw the curves on the coordinate plane. Determine the axis of rotation.
The function x = – y2 + 4 and x = – y + 2, as well as the region bounded by both curves.
The limits of integration given below for revolving both functions around the y-axis
Step 2: Equating both functions to calculate the points of intersection
– y2 + 4 = – y + 2
Making it standard form
y2 – y – 2 = 0
Using quadratic formula
y = (- b ± √ (b2 – 4ac)) / 2a
y = {-(-1) ± √ ((-1)2 – 4(1) (-2))} / 2(1)
y = (1 ± √ (1 +8)) / 2
y = (1 ± √9) / 2
y = (1 ± 3) / 2
y = (1+3) / 2, y = (1 -3) / 2
y = 4 / 2, y = -2 / 2
y = 2, y = -1
So, the roots c = -1 and d = 2
Step 3: Now applying the formula
F(y) = – y2 + 4, G(y) = – y + 2
Vy = π ∫dc [F(y)]2 – [G(y)]2 dy
Vy = π ∫2-1 [- y2 + 4]2 – [- y + 2]2 dy
Vy = π ∫2-1 [y4– 8y2+16 – (y2-4y+4)] dy
Vy = π ∫2-1 [y4– 8y2+16 – y2+4y-4)] dy
Vy = π ∫2-1 [y4– 9y2+4y+12)] dy
Vy = π ∫2-1 (y4) dy – 9 ∫2-1 (y2) dy + 4∫2-1 (y)dy + 12∫2-1 dy
Vy = π (y5 / 5) |2-1 -9 (y3/3) |2-1 + 4 (y2 / 2) |2-1 + 12(y) |2-1
Vy = π [(2)5/ 5 – (-1)5 / 5) – 9 {(2)3/ 3 – (-1)3 / 3)} + 4 {(2)2/ 2 – (-1)2 / 2)} +12{2 – (-1)}]
Vy = π [29.6]
It is a required answer.
Summary
In this article, the help of definition its formula, and derivation with detailed explanations and examples are discussed. With the help of simple integration, volume is calculated by using the washer method formula.
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I am currently trying to understand the linear regression fit by least squares for my machine learning homework, where I implement it and have to plot the result:
I have given two data sets, containing each a matrix $X$ and a vector $y$.
The goal is to predict the output $\hat Y$ for arbitrary data, isn't it? For this
In order to compute optimal parameters $\beta$ which will be used as coefficients:
$$\hat Y = \beta_0 + \sum_{j = 1}^p (X_j \cdot \beta_j)$$
For an easier computation I add the constant variable $1$ into each row of the matrix $X$, this way the prediction vector $\hat Y$can be calculated by:
$\hat Y = X^T \beta$
This way I compute $\beta = (X^T X)^{-1} X^T y$
Have I understand this correctly, $y$ is the training data?
For this exercise I have to generate a test data along a grid and collect it in a matrix $Z$. Then I should compute the prediction vector by $\hat y = Z \beta$. This works fine, but now I should plot it.
Here is my question, what, respectively how is this data plotted?
(I have 1-dim test data $\rightarrow X^{100 \times 1}$ and 2-dim test data $\rightarrow X^{100 \times 2}$)
My guesses so far:
1. The function I received through this method $f(x) = \beta_0 + x \cdot \beta_1$
2. The generated test data $Z$ and the prediction vector, but how is this plotted? How do you plot 1-dim data? What are exactly the $(x,y)$ points I should plot?
I tried for the coordinates, $x$ from the test data matrix $Z$ and $y$ from the prediction vector. This way all points will be put along the function $f(x)$ I don't know whether this is right, I would have thought that the points are scattered around the line.
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Some examples: here, here and here. – dtldarek Apr 22 '12 at 11:12
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The DP problems belonging to this category, in its simplest form, looks like below or some kind of variations of it:
• Given a target find minimum (maximum) cost / path / sum to reach the target.
The solution for this kind of problems, in a very generalized form, would often look like below:
• Choose optimal (minimal or maximal, as the case may be) path among all possible paths that lead to the current state, and then add value for the current state.
• routes[curr] = min(routes[curr - 1], routes[curr - 2], ... , routes[curr - k]) + cost[i]
where current target can be reached only from (curr - 1), (curr - 2), ... (curr - k).
• Overall the solution would look like this :
for (int curr = 1; curr <= target; curr++) {
for (int k = 0; k < waysToReachCurrentTarget.size(); k++) {
dp[i] = min(dp[curr], dp[waysToReachCurrentTarget[k]] + cost / path ) ;
}
}
return dp[target];
The below problem and its solution beautifully demonstrate this approach:
Problem Statement:
Given a m x n grid filled with non-negative numbers, find a path from top left to bottom right which minimizes the sum of all numbers along its path.
Note: You can only move either down or right at any point in time.
Example:
Input:
[
[1,3,1],
[1,5,1],
[4,2,1]
]
Output: 7
Explanation: Because the path 1→3→1→1→1 minimizes the sum.
Naive Solution:
This is a Premium Content.
Please subscribe to Algorithms course to access the solution.
Optimized Solution:
This is a Premium Content.
Please subscribe to Algorithms course to access the solution.
Problem Solving:
Instructor:
If you have any feedback, please use this form: https://thealgorists.com/Feedback.
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2,886,822,669,290,198,000 | If x^2+ kx - 6 = (x - 2)(x + 3), then k = ?
Expert Answers
An illustration of the letter 'A' in a speech bubbles
We are given that x^2+ kx - 6 = (x - 2)(x + 3)
Now x^2+ kx - 6 = (x - 2)(x + 3)
=> x^2 + kx - 6 = x^2 -2x + 3x - 6
=> x^2 + kx - 6 = x^2 + x - 6
canceling the common terms
=> kx = x
=> k = 1
Therefore k = 1
Approved by eNotes Editorial Team
An illustration of the letter 'A' in a speech bubbles
Given that x^2 + kx - 6 = (x-2)(x+3)
We need to find k.
First we will need to open the brackets on the left side and then compare the terms.
==> x^2 + kx -6 = x^2 -2x + 3x - 6
==> x^2 + kx -6 = x^2 + x -6
Now we will add 6 to both sides.
==> x^2 + kx = x^2 + x
Now we will subtract x^2
==> kx = x
Now we will subtract x from both sides.
==> kx -x = 0
Now we will factor x.
==> x (k-1) = 0
x can not be zero.
Then k-1 = 0 ==> k= 1
Then the value of k is 1.
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7,868,596,128,554,151,000 | RD Sharma Solutions Class 8 Rational Numbers
RD Sharma Solutions Class 8 Chapter 1
A rational number is a number which can be expressed in pq form or as a ratio. Rational numbers can also be expressed as a fraction in which “p” is the numerator and “q” is the denominator. The value of “q” should always be a whole number (non-zero denominators). Also, every integer is a rational number.
Examples:
1. 4 is a rational number since it can be expressed as 41.
2. 53 is a rational number since it can be expressed as a fraction.
3. A big number like 6633598,52594/25698 is also a rational number since it can be expressed as a fraction.
Here the RD Sharma solutions for class 8 maths provides the exercise problems of the chapter “Rational Numbers”. With these detailed solutions, the students will be able to clear all their confusion and learn about rational numbers and their use in everyday life. Here you can learn these topics in depth with step-wise and easily understandable solved questions from RD Sharma class 8 maths book.
Practise This Question
Is it possible to construct a unique quadrilateral ABCD with AB = 5cm, BC = 6cm, CD= 7cm , ABC = 45 and BCD = 55? | {
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-7,003,314,187,919,112,000 | Friday
October 21, 2016
Homework Help: ALGEBRA
Posted by Katie on Tuesday, April 15, 2008 at 7:51pm.
How do you solve the rational equation:
(n/n-2)+(n/n+2)=(n/n^2-4)
I know that the LCD of the equation is (n+2)(n-2) because that is what you get when you factor out the n^2-4 in the last part of the question. I thought I did it right until I reached the end where I don't know how to solve it.
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-2,042,565,849,093,205,500 | Calculadora de Subtração de Matriz
Subtrair matrizes passo a passo
A calculadora encontrará a diferença de duas matrizes (se possível), com as etapas mostradas. Subtrai matrizes de qualquer tamanho até 10x10 (2x2, 3x3, 4x4, etc.).
$$$\times$$$
$$$\times$$$
Se a calculadora não calculou algo ou você identificou um erro, ou tem uma sugestão/comentário, escreva nos comentários abaixo.
Sua entrada
Calcule $$$\left[\begin{array}{ccc}1 & 2 & -3\\2 & -3 & -5\\1 & 7 & 1\end{array}\right] - \left[\begin{array}{ccc}2 & -3 & 0\\1 & 1 & 5\\1 & 0 & -1\end{array}\right].$$$
Solução
$$$\left[\begin{array}{ccc}{\color{Purple}1} & {\color{DarkCyan}2} & {\color{Blue}-3}\\{\color{Chocolate}2} & {\color{OrangeRed}-3} & {\color{Crimson}-5}\\{\color{SaddleBrown}1} & {\color{GoldenRod}7} & {\color{Fuchsia}1}\end{array}\right] - \left[\begin{array}{ccc}{\color{Purple}2} & {\color{DarkCyan}-3} & {\color{Blue}0}\\{\color{Chocolate}1} & {\color{OrangeRed}1} & {\color{Crimson}5}\\{\color{SaddleBrown}1} & {\color{GoldenRod}0} & {\color{Fuchsia}-1}\end{array}\right] = \left[\begin{array}{ccc}{\color{Purple}\left(1\right)} - {\color{Purple}\left(2\right)} & {\color{DarkCyan}\left(2\right)} - {\color{DarkCyan}\left(-3\right)} & {\color{Blue}\left(-3\right)} - {\color{Blue}\left(0\right)}\\{\color{Chocolate}\left(2\right)} - {\color{Chocolate}\left(1\right)} & {\color{OrangeRed}\left(-3\right)} - {\color{OrangeRed}\left(1\right)} & {\color{Crimson}\left(-5\right)} - {\color{Crimson}\left(5\right)}\\{\color{SaddleBrown}\left(1\right)} - {\color{SaddleBrown}\left(1\right)} & {\color{GoldenRod}\left(7\right)} - {\color{GoldenRod}\left(0\right)} & {\color{Fuchsia}\left(1\right)} - {\color{Fuchsia}\left(-1\right)}\end{array}\right] = \left[\begin{array}{ccc}-1 & 5 & -3\\1 & -4 & -10\\0 & 7 & 2\end{array}\right]$$$
Responder
$$$\left[\begin{array}{ccc}1 & 2 & -3\\2 & -3 & -5\\1 & 7 & 1\end{array}\right] - \left[\begin{array}{ccc}2 & -3 & 0\\1 & 1 & 5\\1 & 0 & -1\end{array}\right] = \left[\begin{array}{ccc}-1 & 5 & -3\\1 & -4 & -10\\0 & 7 & 2\end{array}\right]$$$A | {
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Results 1 to 6 of 6
Math Help - Newton raphson method
1. #1
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Newton raphson method
i need to solve this equation with newton raphson method
x^3-3x+1
thank you
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2. #2
MHF Contributor FernandoRevilla's Avatar
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By Bolzano's Theorem there is a solution in (0,1) . Choose x_0=0 and construct the corresponding sequence.
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3. #3
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Surely you are not expecting people to do this problem for you. Show what you have done. Of course, Newton-Raphson is an iterative method- for each approximation, x_n, to the solution of f(x)= 0, x_{n+1}= x_n- \frac{f(x_n)}{f'(x_n)} should be a better approximation and, hopefully, the sequence will converge to a solution.
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4. #4
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Quote Originally Posted by HallsofIvy View Post
Surely you are not expecting people to do this problem for you. Show what you have done. Of course, Newton-Raphson is an iterative method- for each approximation, x_n, to the solution of f(x)= 0, x_{n+1}= x_n- \frac{f(x_n)}{f'(x_n)} should be a better approximation and, hopefully, the sequence will converge to a solution.
maybe i should find the location of the root by the method of tabulation ?
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5. #5
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Look at FernandoRevilla's post!
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6. #6
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for an equation like this ... x^3 -8x-4
newton raphson iterative scheme is given by
x_n_+_1 = x_n - f(x_n)/f'(x_n)
for the given equation f(x)= x^3-8x-4
first we find the location of the root by the method of tabulation . .
the table for f(x) is
x 0 1 2 3 4
f(x) -4 -13 -12 -1 28
the positive root is near x = 3
we take x_0 = 3 in newton raphson iterative scheme
x_n_+_1 =x_n - x_n^3-8x_n-4/3x_n^2-8
we get , x_1 = 3 - 27-24-4/27-8 = 3.0526
similarly , x_2 = 3.05138
and
x_3= 3.05138
thus , the positive root is 3.0514 , correct to five significant digits
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5,672,268,804,652,566,000 | Te Kete Ipurangi Navigation:
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Level Three > Geometry and Measurement
Belonging
Purpose:
This is an activity based on the picture book Belonging
Achievement Objectives:
Achievement Objective: GM3-1: Use linear scales and whole numbers of metric units for length, area, volume and capacity, weight (mass), angle, temperature, and time.
AO elaboration and other teaching resources
Specific Learning Outcomes:
1. Students will be able to illustrate units of time using number lines and accurate linear scales.
2. Students will be able to use units of time to calculate the duration of events and make comparative statements.
Description of mathematics:
1. The units for measuring time can be visualised using numberlines in which the linear scale is used to illustrate the passage of time.
2. Units of time can be converted to other units using the multiplicative relationship between the two.
Required Resource Materials:
Belonging by Jeannie Baker
strips of paper
metre rulers
tape
Activity:
Life Span Timeline
This activity is based on the picture book: Belonging
Author: Jeannie Baker
Illustrator: Jeannie Baker
Publisher: Walker (2004)
ISBN: 0-7445-9227-5
Summary:
This is an almost wordless book of scenes from a girl’s bedroom window over the course of 24 years. Each double page spread depicts the changes to the community outside the window and changes in the young girl’s life. There are clues about her age and the milestones she reaches as a young person. But, there is also a strong environmental message about the community and the need to nurture and care for our environment.
Lesson Sequence:
1. Prior to reading, warm up with some of the vocabulary associated with time and the units. Ask the class to create a series of units from the smallest to the largest they know.
Are there any gaps that need to be researched later? (such as 1million years or 1 thousandth of a second?)
What is the average life span in years of NZers?
If we measured it in weeks or months what would we have to multiply by?
What animals would have life spans measured in units other than years?
What do we mean that 1 dog year is equivalent to 7 human years?
2. Share the book with your students. As you share each page, discuss the passage of time since the previous page and note changes that have occurred in the main character and in her environment.
Which kinds of things can change dramatically in 2 years? Which kinds of things seem to change more slowly?
Keep track of how much time passes from the beginning of the book to the end (don’t miss the illustrations on the dedication page and on the page after the author’s notes!)
3. Discuss how a timeline could be used to represent the time span of the events. Ask students to construct a timeline scale to represent the time that passes in the story. They will need to make the line long enough to have space to record key events. There will be some variance in the timelines depending on how they decide to construct their scales (1 cm = 1 year or 10 cm= 1year or 1cm = 1 month etc). On a strip of paper make the timeline and mark the year units and record some of the key events from the story.
4. Now using the same scale create a century timeline from strips of paper the same width. Mark the decades. Line the story timeline up beside the century timeline and create fractional statements about the two lines.
For example: The story covers about 25 years, which is ¼ a century long. If the main character lived to 75 this would represent 1/3 of her life so far. A person is a child for about 18/76 of their life- that is less than ¼ of their life.
5. Measure the length of the century timeline.
How long would a millennium timeline be using that scale?
Mark this out on a netball court or in a corridor.
How far we would have to go to create a time line from 1AD to present day?
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-791,882,951,872,058,000 | Map projection
From Wikipedia, the free encyclopedia
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A medieval depiction of the Ecumene (1482, Johannes Schnitzer, engraver), constructed after the coordinates in Ptolemy's Geography and using his second map projection
In cartography, a map projection is a way to flatten a globe's surface into a plane in order to make a map. This requires a systematic transformation of the latitudes and longitudes of locations from the surface of the globe into locations on a plane.[1] All projections of a sphere on a plane necessarily distort the surface in some way and to some extent. Depending on the purpose of the map, some distortions are acceptable and others are not; therefore, different map projections exist in order to preserve some properties of the sphere-like body at the expense of other properties. The study of map projections is the characterization of the distortions. There is no limit to the number of possible map projections.[2]:1 Projections are a subject of several pure mathematical fields, including differential geometry, projective geometry, and manifolds. However, "map projection" refers specifically to a cartographic projection.
Despite the name's literal meaning, projection is not limited to perspective projections, such as those resulting from casting a shadow on a screen, or the rectilinear image produced by a pinhole camera on a flat film plate. Rather, any mathematical function that transforms coordinates from the curved surface distinctly and smoothly to the plane is a projection. Few projections in practical use are perspective.[citation needed]
Most of this article assumes that the surface to be mapped is that of a sphere. The Earth and other large celestial bodies are generally better modeled as oblate spheroids, whereas small objects such as asteroids often have irregular shapes. The surfaces of planetary bodies can be mapped even if they are too irregular to be modeled well with a sphere or ellipsoid.[3] Therefore, more generally, a map projection is any method of flattening a continuous curved surface onto a plane.[citation needed]
A model globe does not distort surface relationships the way maps do, but maps can be more useful in many situations: they are more compact and easier to store; they readily accommodate an enormous range of scales; they are viewed easily on computer displays; they can be measured to find properties of the region being mapped; they can show larger portions of the Earth's surface at once; and they are cheaper to produce and transport. These useful traits of maps motivate the development of map projections.
The best known map projection is the Mercator projection.[2]:45 Despite its important conformal properties, it has been criticized throughout the twentieth century for enlarging area further from the equator.[2]:156–157 Equal area map projections such as the Sinusoidal projection and the Gall–Peters projection show the correct sizes of countries relative to each other, but distort angles. The National Geographic Society and most atlases favor map projections that compromise between area and angular distortion, such as the Robinson projection or the Winkel tripel projection[2][4]
Metric properties of maps[edit]
An Albers projection shows areas accurately, but distorts shapes.
Many properties can be measured on the Earth's surface independent of its geography:
Map projections can be constructed to preserve some of these properties at the expense of others. Because the curved Earth's surface is not isometric to a plane, preservation of shapes inevitably leads to a variable scale and, consequently, non-proportional presentation of areas. Vice versa, an area-preserving projection can not be conformal, resulting in shapes and bearings distorted in most places of the map. Each projection preserves, compromises, or approximates basic metric properties in different ways. The purpose of the map determines which projection should form the base for the map. Because many purposes exist for maps, a diversity of projections have been created to suit those purposes.
Another consideration in the configuration of a projection is its compatibility with data sets to be used on the map. Data sets are geographic information; their collection depends on the chosen datum (model) of the Earth. Different datums assign slightly different coordinates to the same location, so in large scale maps, such as those from national mapping systems, it is important to match the datum to the projection. The slight differences in coordinate assignation between different datums is not a concern for world maps or other vast territories, where such differences get shrunk to imperceptibility.
Distortion[edit]
Carl Friedrich Gauss's Theorema Egregium proved that a sphere's surface cannot be represented on a plane without distortion. The same applies to other reference surfaces used as models for the Earth, such as oblate spheroids, ellipsoids and geoids. Since any map projection is a representation of one of those surfaces on a plane, all map projections distort.
Tissot's Indicatrices on the Mercator projection
The classical way of showing the distortion inherent in a projection is to use Tissot's indicatrix. For a given point, using the scale factor h along the meridian, the scale factor k along the parallel, and the angle θ′ between them, Nicolas Tissot described how to construct an ellipse that characterizes the amount and orientation of the components of distortion.[2]:147–149[5] By spacing the ellipses regularly along the meridians and parallels, the network of indicatrices shows how distortion varies across the map.
Other distortion metrics[edit]
Many other ways have been described for characterizing distortion in projections.[6][7] Like Tissot's indicatrix, the Goldberg-Gott indicatrix is based on infinitesimals, and depicts flexion and skewness (bending and lopsidedness) distortions.[8]
Rather than the original (enlarged) infinitesimal circle as in Tissot's indicatrix, some visual methods project finite shapes that span a part of the map. For example, a small circle of fixed radius (e.g., 15-degrees angular radius).[9] Sometimes spherical triangles are used.[citation needed] In the first half of the 20th century, projecting a human head onto different projections was common to show how distortion varies across one projection as compared to another.[10] In dynamic media, shapes of familiar coastlines and boundaries can be dragged across an interactive map to show how the projection distorts sizes and shapes according to position on the map.[11]
Another way to visualize local distortion is through grayscale or color gradations whose shade represents the magnitude of the angular deformation or areal inflation. Sometimes both are shown simultaneously by blending two colors to create a bivariate map.[12]
The problem of characterizing distortion globally across areas instead of at just a single point is that it necessarily involves choosing priorities to reach a compromise. Some schemes use distance distortion as a proxy for the combination of angular deformation and areal inflation; such methods arbitrarily choose what paths to measure and how to weight them in order to yield a single result. Many have been described.[8][13][14][15][16]
Design and construction[edit]
The creation of a map projection involves two steps:
1. Selection of a model for the shape of the Earth or planetary body (usually choosing between a sphere or ellipsoid). Because the Earth's actual shape is irregular, information is lost in this step.
2. Transformation of geographic coordinates (longitude and latitude) to Cartesian (x,y) or polar plane coordinates. In large-scale maps, Cartesian coordinates normally have a simple relation to eastings and northings defined as a grid superimposed on the projection. In small-scale maps, eastings and northings are not meaningful, and grids are not superimposed.
Some of the simplest map projections are literal projections, as obtained by placing a light source at some definite point relative to the globe and projecting its features onto a specified surface. Although most projections are not defined in this way, picturing the light source-globe model can be helpful in understanding the basic concept of a map projection.
Choosing a projection surface[edit]
A Miller cylindrical projection maps the globe onto a cylinder.
A surface that can be unfolded or unrolled into a plane or sheet without stretching, tearing or shrinking is called a developable surface. The cylinder, cone and the plane are all developable surfaces. The sphere and ellipsoid do not have developable surfaces, so any projection of them onto a plane will have to distort the image. (To compare, one cannot flatten an orange peel without tearing and warping it.)
One way of describing a projection is first to project from the Earth's surface to a developable surface such as a cylinder or cone, and then to unroll the surface into a plane. While the first step inevitably distorts some properties of the globe, the developable surface can then be unfolded without further distortion.
Aspect of the projection[edit]
This transverse Mercator projection is mathematically the same as a standard Mercator, but oriented around a different axis.
Once a choice is made between projecting onto a cylinder, cone, or plane, the aspect of the shape must be specified. The aspect describes how the developable surface is placed relative to the globe: it may be normal (such that the surface's axis of symmetry coincides with the Earth's axis), transverse (at right angles to the Earth's axis) or oblique (any angle in between).
Notable lines[edit]
Comparison of tangent and secant cylindrical, conic and azimuthal map projections with standard parallels shown in red
The developable surface may also be either tangent or secant to the sphere or ellipsoid. Tangent means the surface touches but does not slice through the globe; secant means the surface does slice through the globe. Moving the developable surface away from contact with the globe never preserves or optimizes metric properties, so that possibility is not discussed further here.
Tangent and secant lines (standard lines) are represented undistorted. If these lines are a parallel of latitude, as in conical projections, it is called a standard parallel. The central meridian is the meridian to which the globe is rotated before projecting. The central meridian (usually written λ0) and a parallel of origin (usually written φ0) are often used to define the origin of the map projection.[17][18]
Scale[edit]
A globe is the only way to represent the Earth with constant scale throughout the entire map in all directions. A map cannot achieve that property for any area, no matter how small. It can, however, achieve constant scale along specific lines.
Some possible properties are:
• The scale depends on location, but not on direction. This is equivalent to preservation of angles, the defining characteristic of a conformal map.
• Scale is constant along any parallel in the direction of the parallel. This applies for any cylindrical or pseudocylindrical projection in normal aspect.
• Combination of the above: the scale depends on latitude only, not on longitude or direction. This applies for the Mercator projection in normal aspect.
• Scale is constant along all straight lines radiating from a particular geographic location. This is the defining characteristic of an equidistant projection such as the Azimuthal equidistant projection. There are also projections (Maurer's Two-point equidistant projection, Close) where true distances from two points are preserved.[2]:234
Choosing a model for the shape of the body[edit]
Projection construction is also affected by how the shape of the Earth or planetary body is approximated. In the following section on projection categories, the earth is taken as a sphere in order to simplify the discussion. However, the Earth's actual shape is closer to an oblate ellipsoid. Whether spherical or ellipsoidal, the principles discussed hold without loss of generality.
Selecting a model for a shape of the Earth involves choosing between the advantages and disadvantages of a sphere versus an ellipsoid. Spherical models are useful for small-scale maps such as world atlases and globes, since the error at that scale is not usually noticeable or important enough to justify using the more complicated ellipsoid. The ellipsoidal model is commonly used to construct topographic maps and for other large- and medium-scale maps that need to accurately depict the land surface. Auxiliary latitudes are often employed in projecting the ellipsoid.
A third model is the geoid, a more complex and accurate representation of Earth's shape coincident with what mean sea level would be if there were no winds, tides, or land. Compared to the best fitting ellipsoid, a geoidal model would change the characterization of important properties such as distance, conformality and equivalence. Therefore, in geoidal projections that preserve such properties, the mapped graticule would deviate from a mapped ellipsoid's graticule. Normally the geoid is not used as an Earth model for projections, however, because Earth's shape is very regular, with the undulation of the geoid amounting to less than 100 m from the ellipsoidal model out of the 6.3 million m Earth radius. For irregular planetary bodies such as asteroids, however, sometimes models analogous to the geoid are used to project maps from.[19][20][21][22][23] Other regular solids are sometimes used as generalizations for smaller bodies' geoidal equivalent. For example, Io is better modeled by triaxial ellipsoid or prolated spheroid with small eccentricities. Haumea's shape is a Jacobi ellipsoid, with its major axis twice as long as its minor and with its middle axis one and half times as long as its minor.
Classification[edit]
A fundamental projection classification is based on the type of projection surface onto which the globe is conceptually projected. The projections are described in terms of placing a gigantic surface in contact with the Earth, followed by an implied scaling operation. These surfaces are cylindrical (e.g. Mercator), conic (e.g. Albers), and plane (e.g. stereographic). Many mathematical projections, however, do not neatly fit into any of these three conceptual projection methods. Hence other peer categories have been described in the literature, such as pseudoconic, pseudocylindrical, pseudoazimuthal, retroazimuthal, and polyconic.
Another way to classify projections is according to properties of the model they preserve. Some of the more common categories are:
• Preserving direction (azimuthal or zenithal), a trait possible only from one or two points to every other point[24]
• Preserving shape locally (conformal or orthomorphic)
• Preserving area (equal-area or equiareal or equivalent or authalic)
• Preserving distance (equidistant), a trait possible only between one or two points and every other point
• Preserving shortest route, a trait preserved only by the gnomonic projection
Because the sphere is not a developable surface, it is impossible to construct a map projection that is both equal-area and conformal.
Projections by surface[edit]
The three developable surfaces (plane, cylinder, cone) provide useful models for understanding, describing, and developing map projections. However, these models are limited in two fundamental ways. For one thing, most world projections in use do not fall into any of those categories. For another thing, even most projections that do fall into those categories are not naturally attainable through physical projection. As L.P. Lee notes,
No reference has been made in the above definitions to cylinders, cones or planes. The projections are termed cylindric or conic because they can be regarded as developed on a cylinder or a cone, as the case may be, but it is as well to dispense with picturing cylinders and cones, since they have given rise to much misunderstanding. Particularly is this so with regard to the conic projections with two standard parallels: they may be regarded as developed on cones, but they are cones which bear no simple relationship to the sphere. In reality, cylinders and cones provide us with convenient descriptive terms, but little else.[25]
Lee's objection refers to the way the terms cylindrical, conic, and planar (azimuthal) have been abstracted in the field of map projections. If maps were projected as in light shining through a globe onto a developable surface, then the spacing of parallels would follow a very limited set of possibilities. Such a cylindrical projection (for example) is one which:
1. Is rectangular;
2. Has straight vertical meridians, spaced evenly;
3. Has straight parallels symmetrically placed about the equator;
4. Has parallels constrained to where they fall when light shines through the globe onto the cylinder, with the light source someplace along the line formed by the intersection of the prime meridian with the equator, and the center of the sphere.
(If you rotate the globe before projecting then the parallels and meridians will not necessarily still be straight lines. Rotations are normally ignored for the purpose of classification.)
Where the light source emanates along the line described in this last constraint is what yields the differences between the various "natural" cylindrical projections. But the term cylindrical as used in the field of map projections relaxes the last constraint entirely. Instead the parallels can be placed according to any algorithm the designer has decided suits the needs of the map. The famous Mercator projection is one in which the placement of parallels does not arise by projection; instead parallels are placed how they need to be in order to satisfy the property that a course of constant bearing is always plotted as a straight line.
Cylindrical[edit]
The Mercator projection shows rhumbs as straight lines. A rhumb is a course of constant bearing. Bearing is the compass direction of movement.
A normal cylindrical projection is any projection in which meridians are mapped to equally spaced vertical lines and circles of latitude (parallels) are mapped to horizontal lines.
The mapping of meridians to vertical lines can be visualized by imagining a cylinder whose axis coincides with the Earth's axis of rotation. This cylinder is wrapped around the Earth, projected onto, and then unrolled.
By the geometry of their construction, cylindrical projections stretch distances east-west. The amount of stretch is the same at any chosen latitude on all cylindrical projections, and is given by the secant of the latitude as a multiple of the equator's scale. The various cylindrical projections are distinguished from each other solely by their north-south stretching (where latitude is given by φ):
• North-south stretching equals east-west stretching (sec φ): The east-west scale matches the north-south scale: conformal cylindrical or Mercator; this distorts areas excessively in high latitudes (see also transverse Mercator).
• North-south stretching grows with latitude faster than east-west stretching (sec2 φ): The cylindric perspective (or central cylindrical) projection; unsuitable because distortion is even worse than in the Mercator projection.
• North-south stretching grows with latitude, but less quickly than the east-west stretching: such as the Miller cylindrical projection (sec 4/5φ).
• North-south distances neither stretched nor compressed (1): equirectangular projection or "plate carrée".
• North-south compression equals the cosine of the latitude (the reciprocal of east-west stretching): equal-area cylindrical. This projection has many named specializations differing only in the scaling constant, such as the Gall–Peters or Gall orthographic (undistorted at the 45° parallels), Behrmann (undistorted at the 30° parallels), and Lambert cylindrical equal-area (undistorted at the equator). Since this projection scales north-south distances by the reciprocal of east-west stretching, it preserves area at the expense of shapes.
In the first case (Mercator), the east-west scale always equals the north-south scale. In the second case (central cylindrical), the north-south scale exceeds the east-west scale everywhere away from the equator. Each remaining case has a pair of secant lines—a pair of identical latitudes of opposite sign (or else the equator) at which the east-west scale matches the north-south-scale.
Normal cylindrical projections map the whole Earth as a finite rectangle, except in the first two cases, where the rectangle stretches infinitely tall while retaining constant width.
Pseudocylindrical[edit]
A sinusoidal projection shows relative sizes accurately, but grossly distorts shapes. Distortion can be reduced by "interrupting" the map.
Pseudocylindrical projections represent the central meridian as a straight line segment. Other meridians are longer than the central meridian and bow outward, away from the central meridian. Pseudocylindrical projections map parallels as straight lines. Along parallels, each point from the surface is mapped at a distance from the central meridian that is proportional to its difference in longitude from the central meridian. Therefore, meridians are equally spaced along a given parallel. On a pseudocylindrical map, any point further from the equator than some other point has a higher latitude than the other point, preserving north-south relationships. This trait is useful when illustrating phenomena that depend on latitude, such as climate. Examples of pseudocylindrical projections include:
• Sinusoidal, which was the first pseudocylindrical projection developed. On the map, as in reality, the length of each parallel is proportional to the cosine of the latitude.[26] The area of any region is true.
• Collignon projection, which in its most common forms represents each meridian as two straight line segments, one from each pole to the equator.
Tobler hyperelliptical projection SW.jpg
Mollweide projection SW.jpg
Goode homolosine projection SW.jpg
Ecker IV projection SW.jpg
Ecker VI projection SW.jpg
Kavraiskiy VII projection SW.jpg
Hybrid[edit]
The HEALPix projection combines an equal-area cylindrical projection in equatorial regions with the Collignon projection in polar areas.
Conic[edit]
Albers conic.
The term "conic projection" is used to refer to any projection in which meridians are mapped to equally spaced lines radiating out from the apex and circles of latitude (parallels) are mapped to circular arcs centered on the apex.[27]
When making a conic map, the map maker arbitrarily picks two standard parallels. Those standard parallels may be visualized as secant lines where the cone intersects the globe—or, if the map maker chooses the same parallel twice, as the tangent line where the cone is tangent to the globe. The resulting conic map has low distortion in scale, shape, and area near those standard parallels. Distances along the parallels to the north of both standard parallels or to the south of both standard parallels are stretched; distances along parallels between the standard parallels are compressed. When a single standard parallel is used, distances along all other parallels are stretched.
Conic projections that are commonly used are:
• Equidistant conic, which keeps parallels evenly spaced along the meridians to preserve a constant distance scale along each meridian, typically the same or similar scale as along the standard parallels.
• Albers conic, which adjusts the north-south distance between non-standard parallels to compensate for the east-west stretching or compression, giving an equal-area map.
• Lambert conformal conic, which adjusts the north-south distance between non-standard parallels to equal the east-west stretching, giving a conformal map.
Pseudoconic[edit]
• Bonne, an equal-area projection on which most meridians and parallels appear as curved lines. It has a configurable standard parallel along which there is no distortion.
• Werner cordiform, upon which distances are correct from one pole, as well as along all parallels.
• American polyconic and other projections in the polyconic projection class.
Azimuthal (projections onto a plane)[edit]
An azimuthal equidistant projection shows distances and directions accurately from the center point, but distorts shapes and sizes elsewhere.
Azimuthal projections have the property that directions from a central point are preserved and therefore great circles through the central point are represented by straight lines on the map. These projections also have radial symmetry in the scales and hence in the distortions: map distances from the central point are computed by a function r(d) of the true distance d, independent of the angle; correspondingly, circles with the central point as center are mapped into circles which have as center the central point on the map.
The mapping of radial lines can be visualized by imagining a plane tangent to the Earth, with the central point as tangent point.
The radial scale is r′(d) and the transverse scale r(d)/(R sin d/R) where R is the radius of the Earth.
Some azimuthal projections are true perspective projections; that is, they can be constructed mechanically, projecting the surface of the Earth by extending lines from a point of perspective (along an infinite line through the tangent point and the tangent point's antipode) onto the plane:
• The gnomonic projection displays great circles as straight lines. Can be constructed by using a point of perspective at the center of the Earth. r(d) = c tan d/R; so that even just a hemisphere is already infinite in extent.[28][29]
• The orthographic projection maps each point on the Earth to the closest point on the plane. Can be constructed from a point of perspective an infinite distance from the tangent point; r(d) = c sin d/R.[30] Can display up to a hemisphere on a finite circle. Photographs of Earth from far enough away, such as the Moon, approximate this perspective.
• Near-sided perspective projection, which simulates the view from space at a finite distance and therefore shows less than a full hemisphere, such as used in The Blue Marble 2012).[31]
• The General Perspective projection can be constructed by using a point of perspective outside the Earth. Photographs of Earth (such as those from the International Space Station) give this perspective. It is a generalization of near-sided perspective projection, allowing tilt.
• The stereographic projection, which is conformal, can be constructed by using the tangent point's antipode as the point of perspective. r(d) = c tan d/2R; the scale is c/(2R cos2 d/2R).[32] Can display nearly the entire sphere's surface on a finite circle. The sphere's full surface requires an infinite map.
Other azimuthal projections are not true perspective projections:
• Azimuthal equidistant: r(d) = cd; it is used by amateur radio operators to know the direction to point their antennas toward a point and see the distance to it. Distance from the tangent point on the map is proportional to surface distance on the Earth (;[33] for the case where the tangent point is the North Pole, see the flag of the United Nations)
• Lambert azimuthal equal-area. Distance from the tangent point on the map is proportional to straight-line distance through the Earth: r(d) = c sin d/2R[34]
• Logarithmic azimuthal is constructed so that each point's distance from the center of the map is the logarithm of its distance from the tangent point on the Earth. r(d) = c ln d/d0); locations closer than at a distance equal to the constant d0 are not shown.[35][36]
Comparison of some azimuthal projections centred on 90° N at the same scale, ordered by projection altitude in Earth radii. (click for detail)
Projections by preservation of a metric property[edit]
A stereographic projection is conformal and perspective but not equal area or equidistant.
Conformal[edit]
Conformal, or orthomorphic, map projections preserve angles locally, implying that they map infinitesimal circles of constant size anywhere on the Earth to infinitesimal circles of varying sizes on the map. In contrast, mappings that are not conformal distort most such small circles into ellipses of distortion. An important consequence of conformality is that relative angles at each point of the map are correct, and the local scale (although varying throughout the map) in every direction around any one point is constant. These are some conformal projections:
Equal-area[edit]
The equal-area Mollweide projection
Equal-area maps preserve area measure, generally distorting shapes in order to do that. Equal-area maps are also called equivalent or authalic. These are some projections that preserve area:
Equidistant[edit]
If the length of the line segment connecting two projected points on the plane is proportional to the geodesic (shortest surface) distance between the two unprojected points on the globe, then we say that distance has been preserved between those two points. An equidistant projection preserves distances from one or two special points to all other points. The special point or points may get stretched into a line or curve segment when projected. In that case, the point on the line or curve segment closest to the point being measured to must be used to measure the distance.
Gnomonic[edit]
The Gnomonic projection is thought to be the oldest map projection, developed by Thales in the 6th century BC
Great circles are displayed as straight lines:
Retroazimuthal[edit]
Direction to a fixed location B (the bearing at the starting location A of the shortest route) corresponds to the direction on the map from A to B:
Compromise projections[edit]
The Robinson projection was adopted by National Geographic magazine in 1988 but abandoned by them in about 1997 for the Winkel tripel.
Compromise projections give up the idea of perfectly preserving metric properties, seeking instead to strike a balance between distortions, or to simply make things look right. Most of these types of projections distort shape in the polar regions more than at the equator. These are some compromise projections:
Which projection is best?[edit]
The mathematics of projection do not permit any particular map projection to be best for everything.[37] Something will always be distorted. Thus, many projections exist to serve the many uses of maps and their vast range of scales.
Modern national mapping systems typically employ a transverse Mercator or close variant for large-scale maps in order to preserve conformality and low variation in scale over small areas. For smaller-scale maps, such as those spanning continents or the entire world, many projections are in common use according to their fitness for the purpose, such as Winkel tripel, Robinson and Mollweide.[38] Reference maps of the world often appear on compromise projections. Due to distortions inherent in any map of the world, the choice of projection becomes largely one of aesthetics.
Thematic maps normally require an equal area projection so that phenomena per unit area are shown in correct proportion.[39] However, representing area ratios correctly necessarily distorts shapes more than many maps that are not equal-area.
The Mercator projection, developed for navigational purposes, has often been used in world maps where other projections would have been more appropriate.[40][41][42][43] This problem has long been recognized even outside professional circles. For example, a 1943 New York Times editorial states:
The time has come to discard [the Mercator] for something that represents the continents and directions less deceptively ... Although its usage ... has diminished ... it is still highly popular as a wall map apparently in part because, as a rectangular map, it fills a rectangular wall space with more map, and clearly because its familiarity breeds more popularity.[2]:166
A controversy in the 1980s over the Peters map motivated the American Cartographic Association (now Cartography and Geographic Information Society) to produce a series of booklets (including Which Map Is Best[44]) designed to educate the public about map projections and distortion in maps. In 1989 and 1990, after some internal debate, seven North American geographic organizations adopted a resolution recommending against using any rectangular projection (including Mercator and Gall–Peters) for reference maps of the world.[45][46]
See also[edit]
References[edit]
Citations[edit]
1. ^ Snyder, J.P. (1989). Album of Map Projections, United States Geological Survey Professional Paper. United States Government Printing Office. 1453.
2. ^ a b c d e f g Snyder, John P. (1993). Flattening the earth: two thousand years of map projections. University of Chicago Press. ISBN 0-226-76746-9.
3. ^ Hargitai, Henrik; Wang, Jue; Stooke, Philip J.; Karachevtseva, Irina; Kereszturi, Akos; Gede, Mátyás (2017), "Map Projections in Planetary Cartography", Lecture Notes in Geoinformation and Cartography, Springer International Publishing, pp. 177–202, doi:10.1007/978-3-319-51835-0_7, ISBN 978-3-319-51834-3
4. ^ "Which is the best map projection?". Geoawesomeness. 25 April 2017.
5. ^ Snyder. Working Manual, p. 24.
6. ^ Karen A. Mulcahy and Keith C. Clarke (2001) "Symbolization of Map Projection Distortion: A Review", Cartography and Geographic Information Science, 101.28, No.3, pp.167-181
7. ^ Frank Canters (2002), Small-Scale Map Projection Design, CRC Press
8. ^ a b Goldberg, David M.; Gott III, J. Richard (2007). "Flexion and Skewness in Map Projections of the Earth" (PDF). Cartographica. 42 (4): 297–318. arXiv:astro-ph/0608501. doi:10.3138/carto.42.4.297. S2CID 11359702. Retrieved 2011-11-14.
9. ^ "Real-time projection visualisation with Indicatrix Mapper QGIS Plugin" (PDF).
10. ^ "Strange Maps: This is your brain on maps". 18 September 2013.
11. ^ "Mercator Puzzle Redux". Retrieved 2018-01-24.
12. ^ "A cornucopia of map projections".
13. ^ Peters, A. B. (1978). "Uber Weltkartenverzerrunngen und Weltkartenmittelpunkte". Kartographische Nachrichten [de]: 106–113.
14. ^ Gott, III, J. Richard; Mugnolo, Charles; Colley, Wesley N. (2006). "Map projections for minimizing distance errors". arXiv:astro-ph/0608500v1.
15. ^ Laskowski, P. (1997). "Distortion-spectrum fundamentals: A new tool for analyzing and visualizing map distortions". Cartographica. 34 (3). doi:10.3138/Y51X-1590-PV21-136G.
16. ^ Airy, G.B. (1861). "Explanation of a projection by balance of errors for maps applying to a very large extent of the Earth's surface; and comparison of this projection with other projections". London, Edinburgh, and Dublin Philosophical Magazine. 4. 22 (149): 409–421. doi:10.1080/14786446108643179.
17. ^ "Projection parameters".
18. ^ "Map projections".
19. ^ Cheng, Y.; Lorre, J. J. (2000). "Equal Area Map Projection for Irregularly Shaped Objects". Cartography and Geographic Information Science. 27 (2): 91. doi:10.1559/152304000783547957. S2CID 128490229.
20. ^ Stooke, P. J. (1998). "Mapping Worlds with Irregular Shapes". The Canadian Geographer. 42: 61. doi:10.1111/j.1541-0064.1998.tb01553.x.
21. ^ Shingareva, K.B.; Bugaevsky, L.M.; Nyrtsov, M. (2000). "Mathematical Basis for Non-spherical Celestial Bodies Maps" (PDF). Journal of Geospatial Engineering. 2 (2): 45–50.
22. ^ Nyrtsov, M.V. (August 2003). "The Classification of Projections of Irregularly-shaped Celestial Bodies" (PDF). Proceedings of the 21st International Cartographic Conference (ICC): 1158–1164.
23. ^ Clark, P. E.; Clark, C. S. (2013). "CSNB Mapping Applied to Irregular Bodies". Constant-Scale Natural Boundary Mapping to Reveal Global and Cosmic Processes. SpringerBriefs in Astronomy. p. 71. doi:10.1007/978-1-4614-7762-4_6. ISBN 978-1-4614-7761-7.
24. ^ Snyder, John Parr (1987). Map Projections – a Working Manual. U.S. Government Printing Office. p. 192. ISBN 9780318235622.
25. ^ Lee, L.P. (1944). "The nomenclature and classification of map projections". Empire Survey Review. VII (51): 190–200. doi:10.1179/sre.1944.7.51.190. p. 193
26. ^ Weisstein, Eric W. "Sinusoidal Projection". MathWorld.
27. ^ Carlos A. Furuti. "Conic Projections"
28. ^ Weisstein, Eric W. "Gnomonic Projection". MathWorld.
29. ^ "The Gnomonic Projection". Retrieved November 18, 2005.
30. ^ Weisstein, Eric W. "Orthographic Projection". MathWorld.
31. ^ "Near-sided perspective". PROJ 7.1.1 documentation. 2020-09-17. Retrieved 2020-10-05.
32. ^ Weisstein, Eric W. "Stereographic Projection". MathWorld.
33. ^ Weisstein, Eric W. "Azimuthal Equidistant Projection". MathWorld.
34. ^ Weisstein, Eric W. "Lambert Azimuthal Equal-Area Projection". MathWorld.
35. ^ Snyder, John P. "Enlarging the Heart of a Map". Archived from the original on July 2, 2010. Retrieved April 14, 2016.
36. ^ Snyder, John P. "Enlarging the Heart of a Map (accompanying figures)". Archived from the original on April 10, 2011. Retrieved November 18, 2005. (see figure 6-5)
37. ^ Choosing a map projection. Lapaine, Miljenko,, Usery, E. Lynn (Eddy Lynn), 1951-. Cham, Switzerland. 2017-04-04. ISBN 978-3-319-51835-0. OCLC 981765011.CS1 maint: others (link)
38. ^ Choosing a World Map. Falls Church, Virginia: American Congress on Surveying and Mapping. 1988. p. 1. ISBN 0-9613459-2-6.
39. ^ Slocum, Terry A.; Robert B. McMaster; Fritz C. Kessler; Hugh H. Howard (2005). Thematic Cartography and Geographic Visualization (2nd ed.). Upper Saddle River, NJ: Pearson Prentice Hall. p. 166. ISBN 0-13-035123-7.
40. ^ Bauer, H.A. (1942). "Globes, Maps, and Skyways (Air Education Series)". New York. p. 28
41. ^ Miller, Osborn Maitland (1942). "Notes on Cylindrical World Map Projections". Geographical Review. 32 (3): 424–430. doi:10.2307/210384. JSTOR 210384.
42. ^ Raisz, Erwin Josephus. (1938). General Cartography. New York: McGraw–Hill. 2d ed., 1948. p. 87.
43. ^ Robinson, Arthur Howard. (1960). Elements of Cartography, second edition. New York: John Wiley and Sons. p. 82.
44. ^ American Cartographic Association's Committee on Map Projections, 1986. Which Map is Best p. 12. Falls Church: American Congress on Surveying and Mapping.
45. ^ Robinson, Arthur (1990). "Rectangular World Maps—No!". Professional Geographer. 42 (1): 101–104. doi:10.1111/j.0033-0124.1990.00101.x.
46. ^ "Geographers and Cartographers Urge End to Popular Use of Rectangular Maps". American Cartographer. 16: 222–223. 1989. doi:10.1559/152304089783814089.
Sources[edit]
• Fran Evanisko, American River College, lectures for Geography 20: "Cartographic Design for GIS", Fall 2002
• Map Projections—PDF versions of numerous projections, created and released into the Public Domain by Paul B. Anderson ... member of the International Cartographic Association's Commission on Map Projections
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Stochastic process
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In probability theory, a stochastic process, or sometimes random process, is the counterpart to a deterministic process (or deterministic system). Instead of dealing with only one possible reality of how the process might evolve under time (as is the case, for example, for solutions of an ordinary differential equation), in a stochastic or random process there is some indeterminacy in its future evolution described by probability distributions. This means that even if the initial condition (or starting point) is known, there are many possibilities the process might go to, but some paths may be more probable and others less.
In the simplest possible case (discrete time), a stochastic process amounts to a sequence of random variables known as a time series (for example, see Markov chain). Another basic type of a stochastic process is a random field, whose domain is a region of space, in other words, a random function whose arguments are drawn from a range of continuously changing values. One approach to stochastic processes treats them as functions of one or several deterministic arguments (inputs, in most cases regarded as time) whose values (outputs) are random variables: non-deterministic (single) quantities which have certain probability distributions. Random variables corresponding to various times (or points, in the case of random fields) may be completely different. The main requirement is that these different random quantities all have the same type.[1] Although the random values of a stochastic process at different times may be independent random variables, in most commonly considered situations they exhibit complicated statistical correlations.
Familiar examples of processes modeled as stochastic time series include signals such as speech, audio and video, medical data such as a patient's EKG, EEG, blood pressure or temperature. Examples of random fields include static images,
Formal definition and basic properties Edit
Definition Edit
Given a probability space (\Omega, \mathcal{F}, P), a stochastic process (or random process) with state space X is a collection of X-valued random variables indexed by a set T ("time"). That is, a stochastic process F is a collection
\{ F_t : t \in T \}
where each F_t is an X-valued random variable.
A modification G of the process F is a stochastic process on the same state space, with the same parameter set T such that
P ( F_t = G_t) =1 \qquad \forall t \in T.
A modification is indistinguishable if
P ( \forall t \in T \, F_t = G_t) =1 .
Finite-dimensional distributions Edit
Let F be an X-valued stochastic process. For every finite subset T' \subseteq T, we may write T'=\{ t_1, \ldots, t_k \}, where k=\left|T'\right| and the restriction F|_{T'}=(F_{t_1}, F_{t_2},\ldots, F_{t_k}) is a random variable taking values in X^k. The distribution \mathbb{P}_{T'}= \mathbb{P} (F|_{T'})^{-1} of this random variable is a probability measure on X^k. Such random variables are called the finite-dimensional distributions of F.
Under suitable topological restrictions, a suitably "consistent" collection of finite-dimensional distributions can be used to define a stochastic process (see Kolmogorov extension in the next section).
Construction Edit
In the ordinary axiomatization of probability theory by means of measure theory, the problem is to construct a sigma-algebra of measurable subsets of the space of all functions, and then put a finite measure on it. For this purpose one traditionally uses a method called Kolmogorov extension.
There is at least one alternative axiomatization of probability theory by means of expectations on C-star algebras of random variables. In this case the method goes by the name of Gelfand-Naimark-Segal construction.
This is analogous to the two approaches to measure and integration, where one has the choice to construct measures of sets first and define integrals later, or construct integrals first and define set measures as integrals of characteristic functions.
Kolmogorov extension Edit
The Kolmogorov extension proceeds along the following lines: assuming that a probability measure on the space of all functions f: X \to Y exists, then it can be used to specify the joint probability distribution of finite-dimensional random variables f(x_1),\dots,f(x_n). Now, from this n-dimensional probability distribution we can deduce an (n − 1)-dimensional marginal probability distribution for f(x_1),\dots,f(x_{n-1}). Note that the obvious compatibility condition, namely, that this marginal probability distribution be in the same class as the one derived from the full-blown stochastic process, is not a requirement. Such a condition only holds, for example, if the stochastic process is a Wiener process (in which case the marginals are all gaussian distributions of the exponential class) but not in general for all stochastic processes. When this condition is expressed in terms of probability densities, the result is called the Chapman–Kolmogorov equation.
The Kolmogorov extension theorem guarantees the existence of a stochastic process with a given family of finite-dimensional probability distributions satisfying the Chapman-Kolmogorov compatibility condition.
Separability, or what the Kolmogorov extension does not provide Edit
Recall that in the Kolmogorov axiomatization, measurable sets are the sets which have a probability or, in other words, the sets corresponding to yes/no questions that have a probabilistic answer.
The Kolmogorov extension starts by declaring to be measurable all sets of functions where finitely many coordinates [f(x_1), \dots , f(x_n)] are restricted to lie in measurable subsets of Y_n. In other words, if a yes/no question about f can be answered by looking at the values of at most finitely many coordinates, then it has a probabilistic answer.
In measure theory, if we have a countably infinite collection of measurable sets, then the union and intersection of all of them is a measurable set. For our purposes, this means that yes/no questions that depend on countably many coordinates have a probabilistic answer.
The good news is that the Kolmogorov extension makes it possible to construct stochastic processes with fairly arbitrary finite-dimensional distributions. Also, every question that one could ask about a sequence has a probabilistic answer when asked of a random sequence. The bad news is that certain questions about functions on a continuous domain don't have a probabilistic answer. One might hope that the questions that depend on uncountably many values of a function be of little interest, but the really bad news is that virtually all concepts of calculus are of this sort. For example:
1. boundedness
2. continuity
3. differentiability
all require knowledge of uncountably many values of the function.
One solution to this problem is to require that the stochastic process be separable. In other words, that there be some countable set of coordinates \{f(x_i)\} whose values determine the whole random function f.
The Kolmogorov continuity theorem guarantees that processes that satisfy certain constraints on the moments of their increments are continuous.
Examples and special cases Edit
Time Edit
A notable special case is where the time is a discrete set, for example the nonnegative integers {0, 1, 2, 3, ...}. Another important special case is T = \mathbb{R}.
Stochastic processes may be defined in higher dimensions by attaching a multivariate random variable to each point in the index set, which is equivalent to using a multidimensional index set. Indeed a multivariate random variable can itself be viewed as a stochastic process with index set T = {1, ..., n}.
Examples Edit
The paradigm of continuous stochastic process is that of the Wiener process. In its original form the problem was concerned with a particle floating on a liquid surface, receiving "kicks" from the molecules of the liquid. The particle is then viewed as being subject to a random force which, since the molecules are very small and very close together, is treated as being continuous and, since the particle is constrained to the surface of the liquid by surface tension, is at each point in time a vector parallel to the surface. Thus the random force is described by a two component stochastic process; two real-valued random variables are associated to each point in the index set, time, (note that since the liquid is viewed as being homogeneous the force is independent of the spatial coordinates) with the domain of the two random variables being R, giving the x and y components of the force. A treatment of Brownian motion generally also includes the effect of viscosity, resulting in an equation of motion known as the Langevin equation.
If the index set of the process is N (the natural numbers), and the range is R (the real numbers), there are some natural questions to ask about the sample sequences of a process {Xi}iN, where a sample sequence is {X(ω)i}iN.
1. What is the probability that each sample sequence is bounded?
2. What is the probability that each sample sequence is monotonic?
3. What is the probability that each sample sequence has a limit as the index approaches ∞?
4. What is the probability that the series obtained from a sample sequence from f(i) converges?
5. What is the probability distribution of the sum?
Similarly, if the index space I is a finite or infinite interval, we can ask about the sample paths {X(ω)t}t I
1. What is the probability that it is bounded/integrable/continuous/differentiable...?
2. What is the probability that it has a limit at ∞
3. What is the probability distribution of the integral?
See alsoEdit
Notes Edit
1. Mathematically speaking, the type refers to the codomain of the function.
ReferencesEdit
1. Papoulis, Athanasios & Pillai, S. Unnikrishna (2001). Probability, Random Variables and Stochastic Processes, McGraw-Hill Science/Engineering/Math.
2. Boris Tsirelson. Lecture notes in Advanced probability theory.
3. J. L. Doob (1953). Stochastic Processes, Wiley.
4. (2006). An Exploration of Random Processes for Engineers. Free e-book.
External linksEdit
This page uses Creative Commons Licensed content from Wikipedia (view authors).
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-5,899,995,266,388,964,000 | Basic Mathematics MCQs Preparation For Commissions Test page6
Basic Mathematics MCQs Preparation For Commissions Test page6
98. The number, whose 5% is 10, is
(a) 100′ (b) 200
(c) 300 (d) 400
99. The difference between simple and
compound interest on Rs. 1625 for 3
years at 4% per annum in rupees is
(a) 7.95 (b) 7.90′
(c) 7.70 , (d) 7.75
1. 100. The number, whose 5% is 55,,is
(a) 1100′
(b) 2100
(c) 1200 (d) 1300
2. 101. We covered a 150 km distance of
Islamabad road. The total distance is
370 km. What percentage of distance
have we covered?
(a) 30%‘ (b) 40.54%
(c) 67% (d) 41%
102. Which number will come next?
1,0,3,2,5,6,______
(a) 9 (b) 8
(c) 7′ (d) 10
103. lnser|the missing number:
4 5 7 11 19
(a) 22 (b) 32
(c) 35^ (d) 24
104.Insert the missing number:
8 7 9 13 21
(a) 40 . (b) 38
(c) 37^ (d) 39
105.Insert the missing number:
64 48 36 34
(a) 33 (b)32
(c) 38 (d) 30
106. Insert the missing number:
718 (26) 582 474 (…)226
(a) 16
(b) 14
(c) 18 (d) 12
107.Insert the missing number:
11 12 14 ? 26 42
(a) 28 (b) 24
(c) 18 (d) 32
108. Which one number will complete
following number series?
8,9,8,7,10,9,6,11,10_________ 12
(a) 11 (b) 7
(c) 7 (d) 5′
109. Which number will come next?
Series: 1, 3, 7,15, 31, 63,______
(a) 123 (b) 125
(c) 127′ (d) 129
110. By selling a fan for Rs. 475, a person j loses 5%. To get a gain of 5%, he should 1 seU the fan for:
(a) Rs. 500 (b) Rs. 525′
(c) Rs. 535 (d) Rs. 575
111. The enrollment in a certain secondary school was 450 in 1979. By 1980 the enrollment had increased by 16%, what was the enrollment in 1980?
(a) Rs.512 (b) Rs.518
(c) Rs.522^ (d) Rs.526
113. Which number will come next?
5,3,6,2,7,1————
(a) 0 (b) 2
(c) 8′ (d) 4
114. All gained 510 marks in matriculation examination. What percentage of marks did he gain?
(a) 60%‘ (b) 57%
(c) 71% (d) 73%
115. Akhtar scored 178 marks out of 300 marks in a certain test. What percentage of marks did he score?
(a) 48% (b) 52.41%
(c) 61.41% (d) 59.33%
116. The price of a book increases from Rs.120 to Rs.150. What is the percentage increases?
(a) 35% (b) 25%
(c) 115% (d) 25.25.1
117. .A cyclist covers 660 feet in 66 seconds. How many yards will he cover in the same time?
(a) 220′ (b) 600
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Basic Mathematics MCQs Preparation For Commissions Test page6
One thought on “Basic Mathematics MCQs Preparation For Commissions Test page6
1. HELLO DEAR,GOOD EFFORT,but the 1st answer is incorrect by all means,plz rectify it to avoid any misguidense to the students
Many thanks
Leave a Reply | {
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5,884,231,388,167,262,000 | data-ad-format="horizontal">
Right Triangle
A triangle where one of its interior angles is a right angle (90 degrees).
Try this Drag the orange dots on each vertex to reshape the triangle. Notice it always remains a right triangle, because the angle ∠ABC is always 90 degrees.
Right triangles figure prominently in various branches of mathematics. For example, trigonometry concerns itself almost exclusively with the properties of right triangles, and the famous Pythagoras Theorem defines the relationship between the three sides of a right triangle:
a2 + b2 = h2
where h is the length of the hypotenuse
a,b are the lengths of the the other two sides
Attributes
Hypotenuse The side opposite the right angle. This will always be the longest side of a right triangle.
Sides The two sides that are not the hypotenuse. They are the two sides making up the right angle itself.
Properties
• A right triangle can also be isosceles if the two sides that include the right angle are equal in length (AB and BC in the figure above)
• A right triangle can never be equilateral, since the hypotenuse (the side opposite the right angle) is always longer than either of the other two sides.
Constructions
You can construct right triangles with compass and straightedge given various combinations of sides and angles. For a complete list see Constructions - Table of Contents.
Other triangle topics
General
Perimeter / Area
Triangle types
Triangle centers
Congruence and Similarity
Solving triangles
Triangle quizzes and exercises | {
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7,457,835,323,158,785,000 | Decimals on a number line
3 videos
3 skills
Let's think about where decimals are on a number line. It will help us understand what decimals represent in general!
Decimals on a number line
VIDEO 1:39 minutes
Let's compare decimals on a number line, shall we?
Decimals on the number line 1
PRACTICE PROBLEMS
Practice finding decimal numbers on the number line. Decimals are limited to tenths in these problems.
Decimals and fractions on a number line
VIDEO 1:58 minutes
We're mixing it up by placing both fractions and decimals on the same number line. Great practice because you need to move effortlessly between the two.
Decimals on the number line 2
PRACTICE PROBLEMS
Practice finding decimal numbers on the number line. Decimals are limited to hundredths in these problems.
Placing positive and negative decimals on a number line
VIDEO 1:15 minutes
These example exercises require us to interpret a number line in order to locate where our positive and negative decimals should be placed.
Decimals on the number line 3
PRACTICE PROBLEMS
Practice placing positive and negative decimal numbers on the number line.
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-2,513,873,182,498,224,600 | 213214 (number)
213,214 (two hundred thirteen thousand two hundred fourteen) is an even six-digits composite number following 213213 and preceding 213215. In scientific notation, it is written as 2.13214 × 105. The sum of its digits is 13. It has a total of 3 prime factors and 8 positive divisors. There are 100,320 positive integers (up to 213214) that are relatively prime to 213214.
Basic properties
Name
Short name 213 thousand 214
Full name two hundred thirteen thousand two hundred fourteen
Notation
Scientific notation 2.13214 × 105
Engineering notation 213.214 × 103
Prime Factorization of 213214
Prime Factorization 2 × 17 × 6271
Composite number
ω(n) Distinct Factors 3
Total number of distinct prime factors
Ω(n) Total Factors 3
Total number of prime factors
rad(n) Radical 213214
Product of the distinct prime numbers
λ(n) Liouville Lambda -1
Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n)
μ(n) Mobius Mu -1
Returns:
• 1, if n has an even number of prime factors (and is square free)
• −1, if n has an odd number of prime factors (and is square free)
• 0, if n has a squared prime factor
Λ(n) Mangoldt function 0
Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 213,214 is 2 × 17 × 6271. Since it has a total of 3 prime factors, 213,214 is a composite number.
Divisors of 213214
8 divisors
Even divisors 4
Odd divisors 4
4k+1 divisors 2
4k+3 divisors 2
τ(n) Total Divisors 8
Total number of the positive divisors of n
σ(n) Sum of Divisors 338688
Sum of all the positive divisors of n
s(n) Aliquot Sum 125474
Sum of the proper positive divisors of n
A(n) Arithmetic Mean 42336
Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n))
G(n) Geometric Mean 461.7510151586
Returns the nth root of the product of n divisors
H(n) Harmonic Mean 5.0362339380197
Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 213,214 can be divided by 8 positive divisors (out of which 4 are even, and 4 are odd). The sum of these divisors (counting 213,214) is 338,688, the average is 42,336.
Other Arithmetic Functions (n = 213214)
1 φ(n) n
φ(n) Euler Totient 100320
Total number of positive integers not greater than n that are coprime to n
λ(n) Carmichael Lambda 50160
Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n
π(n) Prime Pi ≈ 19012
Total number of primes less than or equal to n
r2(n) Sum of 2 squares 0
The number of ways n can be represented as the sum of 2 squares
There are 100,320 positive integers (less than 213,214) that are coprime with 213,214. And there are approximately 19,012 prime numbers less than or equal to 213,214.
Divisibility of 213214
m 2 3 4 5 6 7 8 9
n mod m 0 1 2 4 4 1 6 4
The number 213,214 is divisible by 2.
Classification of 213214
By Arithmetic functions
• Arithmetic
• Deficient
Expressible via specific sums
• Polite
By Powers
• Square Free
Other numbers
• Sphenic
Base conversion (213214)
Base System Value
2 Binary 110100000011011110
3 Ternary 101211110211
4 Quaternary 310003132
5 Quinary 23310324
6 Senary 4323034
8 Octal 640336
10 Decimal 213214
12 Duodecimal a347a
16 Hexadecimal 340de
20 Vigesimal 16d0e
36 Base36 4kim
Basic calculations (n = 213214)
Multiplication
n×y
n×2 426428
n×3 639642
n×4 852856
n×5 1066070
Division
n÷y
n÷2 106607.000
n÷3 71071.333
n÷4 53303.500
n÷5 42642.800
Exponentiation
ny
n2 45460209796
n3 9692753171444344
n4 2066630674696334361616
n5 440634592674704234577593824
Nth Root
y√n
2√n 461.7510151586
3√n 59.740919939267
4√n 21.488392568049
5√n 11.634913090104
213214 as geometric shapes
Circle
Radius = n
Diameter 426428
Circumference 1339663.072085
Area 142817461125.76
Sphere
Radius = n
Volume 4.0600909541958E+16
Surface area 571269844503.06
Circumference 1339663.072085
Square
Length = n
Perimeter 852856
Area 45460209796
Diagonal 301530.13048782
Cube
Length = n
Surface area 272761258776
Volume 9692753171444344
Space diagonal 369297.48088499
Equilateral Triangle
Length = n
Perimeter 639642
Area 19684848272.353
Altitude 184648.7404425
Triangular Pyramid
Length = n
Surface area 78739393089.412
Volume 1.1423019159826E+15
Height 174088.50200592
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-7,390,179,276,083,944,000 | sQuare
Please wait...
--}}
PERMUTATION GAME TYPE
PERMUTATION GAME PLAY
• PERM-1: (ANY OF THE SELECTED MUST BE THE FIRST NUMBER TO WIN)
A player places bet in favour of any number from a selected range of numbers he has chosen. E.g. amongst 45,48,80 on the possibility that any of the numbers selected will match the Very First ball of the five winning numbers to be selected in the draw entered for. The odds of perm 1 is N40/N1 staked. Note: If any number from these selected numbers is the first ball in the winning numbers, then the player is a winner.
• PERM-2: (ANY TWO MUST COME TO WIN)
To increase a players chances of winning, a player can place bet in favour of any two numbers from a selected set of numbers. This means the system will generate all possible 2 number combinations from the selected numbers without repetitions. if any 2 or more of the selected numbers are in the winning numbers of the draw, the players has won. e.g. if a player plays perm 2 amongst 45,48,78,80, this will result in (45-48, 45-78, 45-80, 48-78, 48-80, 78-80) 6 ticket plays.
If the numbers 78,45,80,17,90 are the winning numbers, then the players has won with (45-78, 45-80, 78-80). This means 3 of his 6 tickets won. The odds for perm 2 is N280/N1 staked. E.g. for every N1 staked you win N280 only on the Set Lotto Nigeria platform.
Same principles applies to Perm 3(any three number among the selected must come to win) Perm 4 (any four numbers among the selected must come to win) and Perm 5 (any five must come to win). However please note: in permutation, the numbers in a set selected by a player is expect to be at least 1 greater than the expected permutation.
For example:
Perm 1-- minimum set selection is 2 and max is 25
Perm 2-- minimum set selection is 3 and max is 25
Perm 3-- minimum set selection is 4 and max is 20
Perm 4-- minimum set selection is 5 and max is 15
Perm 5-- minimum set selection is 6 and max is 10
Winning odds for Perm 3, Perm 4 and Perm 5 remains N2,100, N6,000 and N44,000 respectively.
Play Now
Recent Winners: Jamnocase - ₦3,600. Elegbede768 - ₦2,400. Elegbede768 - ₦1,200. frank76 - ₦2,400. Elegbede768 - ₦3,600. Elegbede768 - ₦1,200. Elegbede768 - ₦1,200. Elegbede768 - ₦1,200. Elegbede768 - ₦1,200. Blessed-Grace - ₦3,600. | {
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"rps_doc_frac_chars_top_3gram": 0.05523255839943886,
"rps_doc_frac_chars_top_4gram": 0.06104651093482971,
"rps_doc_books_importance": -292.3184814453125,
"rps_doc_books_importance_length_correction": -292.3184814453125,
"rps_doc_openwebtext_importance": -159.92283630371094,
"rps_doc_openwebtext_importance_length_correction": -159.92283630371094,
"rps_doc_wikipedia_importance": -180.67721557617188,
"rps_doc_wikipedia_importance_length_correction": -180.67721557617188
},
"fasttext": {
"dclm": 0.023361269384622574,
"english": 0.8964030146598816,
"fineweb_edu_approx": 1.0431972742080688,
"eai_general_math": 0.6831412315368652,
"eai_open_web_math": 0.3937099575996399,
"eai_web_code": 0.15469276905059814
}
} | {
"free_decimal_correspondence": {
"primary": {
"code": "368.01",
"labels": {
"level_1": "Social sciences",
"level_2": "Social service and Societies",
"level_3": "Insurance"
}
},
"secondary": {
"code": "519.2",
"labels": {
"level_1": "Science and Natural history",
"level_2": "Mathematics",
"level_3": "Probabilities; or, Mathematical statistics"
}
}
},
"bloom_cognitive_process": {
"primary": {
"code": "2",
"label": "Understand"
},
"secondary": {
"code": "3",
"label": "Apply"
}
},
"bloom_knowledge_domain": {
"primary": {
"code": "1",
"label": "Factual"
},
"secondary": {
"code": "3",
"label": "Procedural"
}
},
"document_type_v1": {
"primary": {
"code": "3",
"label": "Reference/Encyclopedic/Educational"
},
"secondary": {
"code": "6",
"label": "Promotional/Advertisement"
}
},
"extraction_artifacts": {
"primary": {
"code": "0",
"label": "No Artifacts"
},
"secondary": {
"code": "3",
"label": "Irrelevant Content"
}
},
"missing_content": {
"primary": {
"code": "0",
"label": "No missing content"
},
"secondary": {
"code": "2",
"label": "Click Here References"
}
},
"document_type_v2": {
"primary": {
"code": "17",
"label": "Product Page"
},
"secondary": {
"code": "10",
"label": "Knowledge Article"
}
},
"reasoning_depth": {
"primary": {
"code": "2",
"label": "Basic Reasoning"
},
"secondary": {
"code": "1",
"label": "No Reasoning"
}
},
"technical_correctness": {
"primary": {
"code": "4",
"label": "Highly Correct"
},
"secondary": {
"code": "3",
"label": "Mostly Correct"
}
},
"education_level": {
"primary": {
"code": "1",
"label": "General Audience"
},
"secondary": {
"code": "2",
"label": "High School Level"
}
}
} | a246fc342e934853762e5c7f05dc3a09 |
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