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https://glasp.co/youtube/p/special-relativity-lecture-2
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# Special Relativity | Lecture 2 | Summary and Q&A
515.7K views
May 9, 2012
by
Stanford
Special Relativity | Lecture 2
## Summary
This video discusses basic Lorentz transformations in frames of reference and the concept of proper time or interval. It explains how the coordinates of one observer relate to the coordinates of another observer and introduces the notion of four-vectors, including four-velocity.
### Q: What is the main concept discussed in this video?
The main concept discussed in this video is Lorentz transformations and the notion of proper time or interval.
### Q: How are Lorentz transformations related to frames of reference?
Lorentz transformations relate the coordinates of one observer to the coordinates of another observer in different frames of reference. They account for the relative motion of observers and ensure that all reference frames see the speed of light the same way.
### Q: What are the Lorentz transformations based on?
The Lorentz transformations are based on the hypothesis of Einstein that all reference frames see the speed of light exactly the same.
### Q: How are the coordinates of one observer related to the coordinates of another observer?
The coordinates of one observer (stationary observer) can be related to the coordinates of another observer (moving observer) through Lorentz transformations. The coordinates depend on the relative velocity between the two observers.
### Q: What if the motion of observers is not only along one axis?
In that case, Lorentz transformations can still be used to relate the coordinates of the stationary observer to the coordinates of the moving observer in multiple dimensions. The transformations for the additional axes are simpler and do not involve changes in velocity.
### Q: What is the proper time or interval between two points in space-time?
The proper time or interval between two points in space-time is an invariant quantity that is the same in all reference frames. It is given by T^2 - X^2, where T is the time coordinate and X is the spatial coordinate.
### Q: How are Lorentz transformations represented mathematically?
Lorentz transformations can be represented mathematically using matrices. For example, for the transformation in the x-direction, the matrix would be [[1, -V], [-V, 1]].
### Q: What is a four-vector?
A four-vector is a four-dimensional object that includes both the space and time components. It is represented by coordinates X^(mu), where mu runs from 0 to 3 and corresponds to T, X, Y, and Z, respectively.
### Q: How does four velocity differ from ordinary velocity?
Four velocity, represented by U^(mu), is a four-dimensional quantity that incorporates both space and time components. It is obtained by dividing the change in coordinates (Delta X^(mu)) by the invariant distance (Delta Tau) between two points in space-time.
### Q: How is four velocity related to ordinary velocity?
Four velocity is related to ordinary velocity by dividing the space component of four velocity by the time component. This gives the velocity in three-dimensional space.
## Takeaways
In this video, we learned about Lorentz transformations and their application in relating coordinates between different frames of reference. We also explored the concept of proper time or interval, which is an invariant quantity in all reference frames. Additionally, we discussed four-vectors and four velocity, which extend the notion of velocity to incorporate the space and time components in a four-dimensional framework. These concepts are crucial in understanding the motion and dynamics of particles in the context of relativity theory.
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# Fraction Games for 9-Year-Olds
Fractions can often be a difficult concept for 9-year-olds to grasp because fractions aren't whole numbers. Below, you will find games that your child will enjoy playing while perfecting his or her understanding of fractions.
## How to Improve Your Child's Fraction Skills
Your 9-year-old child will need to understand equivalent fractions, how to correctly order fractions and also how to build fractions using mathematical operations. It is also important for your fourth grader to understand how decimal amounts correspond to fractional amounts. These interactive games will allow you the opportunity to gauge your child's understanding of fractions and to work on any problem areas.
### Family Fractions
To begin, have your child draw a family picture. It would be a good idea to include grandparents, cousins, aunts and uncles. Ask your child to create fractions based on family members' gender, hair color or other characteristics. For instance, if five of the nine family members have brown hair, your child would create the fraction 5/9. As an extension, your 9-year-old could also make a drawing to represent each of the family fractions he or she created.
### Fraction Eggs
You can play this fraction game with your child with just an empty egg carton and individual candies, such as jellybeans. Fill in some sections of the egg carton with jellybeans. Ask your child to identify the fraction being modeled. For example, if you filled seven of the sections with jellybeans, the fraction represented would be 7/12. When possible, ask your child to reduce the created fractions.
### Crafting a Pizza
Gather your favorite pizza toppings, step into the kitchen and help your child master fractions! This activity will have you and your child working together to create one large, personalized pizza for your family.
For example, let's say you divide the pizza into eight slices. Dad, whose favorite is pepperoni, wants three slices of pizza. Mom and sister prefer hamburger pizza; they will each have two slices. Brother asked for Canadian bacon for his one slice. Have your child determine the fraction represented by each pizza topping. In this situation, the pizza would be 3/8 pepperoni, 1/2 hamburger and 1/8 Canadian bacon. Of course, have your child reduce fractions wherever possible.
Did you find this useful? If so, please let others know!
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# Lazarus
## Free Pascal => Beginners => Topic started by: JLWest on July 04, 2020, 07:29:01 pm
Title: Roundto - Can't get it to work
Post by: JLWest on July 04, 2020, 07:29:01 pm
Var MPMin, MPHS, Wrk : Single;
begin
MPMin := (MPHs / 60); { = 8.71666622}
WrK := MPMin;
MPMin := Roundto(Wrk); {= 0}
Tried this:
MPMin := RoundTo( 8.71666622 , 2); {= 0}
Title: Re: Roundto - Can't get it to work
Post by: wp on July 04, 2020, 07:40:38 pm
The function RoundTo is a bit counter-intuitive because everybody calling RoundTo(x, 2) thinks that it rounds x to 2 decimal places. No, it rounds to multiples of 10 to the power of the second argument. In other words: 10^2 is 100, and thus RoundTo(x, 2) rounds to full hundreds, e.g. RoundTo(871.6666, 2) becomes 900, but RoundTo(8.71666, 2) becomes 0 because it is less than 100.
In order to round to two decimal places you round to multiples of 0.01 = 10^-2. therefore, you must call RoundTo(x, -2).
Title: Re: Roundto - Can't get it to work
Post by: winni on July 04, 2020, 07:41:59 pm
Hi!
Yes - that is correct.
roundTo (8.7,2) means: round to 100 - so it is zero
roundTo (8.7,1) means: round to 10 - so it is 10
roundTo (8.7,0) means: round to 1 - so it is 9
Everything as it should be.
Winni
Title: Re: Roundto - Can't get it to work
Post by: JLWest on July 04, 2020, 08:11:22 pm
A "bit counter-intuitive"! I felt bad about posting the question. Now I feel better.
Thank you gentlemen.
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# Write the Coordinates of the Centre of the Circle Inscribed in the Square Formed by the Lines X = 2, X = 6, Y = 5 and Y = 9. - Mathematics
Write the coordinates of the centre of the circle inscribed in the square formed by the lines x = 2, x = 6, y = 5 and y = 9.
#### Solution
From the figure, we can see that the vertices of the square are (2, 5), (6, 5), (2, 9) and (6, 9).
The vertices of the diagonals are (2, 9), (6, 5) and (2, 5),(6, 9).
∴ Coordinates of the centre = $\left( \frac{6 + 2}{2}, \frac{5 + 9}{2} \right) = \left( 4, 7 \right)$
Is there an error in this question or solution?
#### APPEARS IN
RD Sharma Class 11 Mathematics Textbook
Chapter 24 The circle
Q 9 | Page 38
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Thread: Two quick questions
1. Two quick questions
These are stumping me for some odd reason...
1. Find the remainder of $\displaystyle 2009^{2008}$ upon division by 9.
2. Determine with explanation, whether there exists an integer n such that $\displaystyle n^{10} + 1$ is divisible by 151. (Note that 151 is prime)
2. Hello, Aryth!
1. Find the remainder of $\displaystyle 2009^{2008}$ upon division by 9.
Since $\displaystyle 2009 \:=\:223(9) + 2$, then: .$\displaystyle 2009 \div 9$ has a remainder of 2.
Then: .$\displaystyle 2009^{2008} \div 9$ has a remainder of $\displaystyle 2^{2008}$
We find that: . $\displaystyle 2^6 \,=\,64$ . . . and $\displaystyle 64 \div 9$ has remainder 1.
$\displaystyle \text{Then: }\;2^{2008} \;=\;2^{334(6) + 4} \;=\;2^{334(6)}\cdot 2^4 \;=\;\underbrace{\left(2^6\right)^{334}}_{\text{re m. 1}}\cdot16$
Hence: .$\displaystyle 16 \div 9 \quad\to\quad \text{remainder } 7$
3. Would anyone be able to help with the second one?
4. Originally Posted by Aryth
Would anyone be able to help with the second one?
well, it's quite easy: since $\displaystyle \left(\frac{-1}{151} \right)=-1,$ the equation $\displaystyle x^2 \equiv -1 \mod 151$ has no solution and thus, obviously, $\displaystyle x^{10} \equiv -1 \mod 151$ cannot have any solution either.
5. Originally Posted by NonCommAlg
well, it's quite easy: since $\displaystyle \left(\frac{-1}{151} \right)=-1,$ the equation $\displaystyle x^2 \equiv -1 \mod 151$ has no solution and thus, obviously, $\displaystyle x^{10} \equiv -1 \mod 151$ cannot have any solution either.
another way: suppose $\displaystyle n^{10} \equiv -1 \mod 151$ has a solution. then $\displaystyle n^{150}=(n^{10})^{15} \equiv -1 \mod 151.$ but, since $\displaystyle \gcd(n,151)=1,$ by Fermat's little theorem $\displaystyle n^{150} \equiv 1 \mod 151.$ contradiction!
6. I seriously appreciate the help.
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### Home > CCA2 > Chapter 9 > Lesson 9.1.2 > Problem9-28
9-28.
Convert the following angle measurements from degrees to radians.
1. $30^\circ$
This is a special angle. It is one that is worth memorizing!
$\frac{\pi}{6}$
1. $15^\circ$
Half of part (a).
1. $-75^\circ$
Negative $(30^\circ + 30^\circ + 15^\circ)$
1. $630^\circ$
$360^\circ + 270^\circ$
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0
# Find the definite integral?
Find ∫ 1/(sqrt(4-x^2)) dx from 1 to 2.
Answer: (1/3)pi
### Comments
Note that this is an improper integral since the integrand is undefined at 2. The integral is defined as
limb→2- ( ∫1b [4 - x2] dx )
### 2 Answers by Expert Tutors
Tutors, sign in to answer this question.
Jim S. | Physics (and math) are fun, reallyPhysics (and math) are fun, really
4.7 4.7 (197 lesson ratings) (197)
0
the integral is sin-1(x/2) evaluated between 1 and 2 which is an angle whose sine is 1-an angle whose sine is 1/2 which is π/2-π/6=π/3
Regards
Jim
William S. | Experienced scientist, mathematician and instructor - WilliamExperienced scientist, mathematician and...
4.4 4.4 (10 lesson ratings) (10)
0
∫[4 - x2] dx = sin-1 (x/2) evaluated between 1 and 2
sin-1(2) - sin-1(1/2) = (1/3)pi
### Comments
William S., how did you get sin^-1 (x/2)?
The way you get arcsin(x/2)from ∫dx/√(4-x2)=∫dx/(2√(1-(x/2)2)=∫d(x/2)/(√(1-(x/2)2)
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## Precalculus: Mathematics for Calculus, 7th Edition
$a)$ $4$ $b)$ $\dfrac{3}{2}$ $c)$ $\dfrac{8}{27}$
$a)$ $32^{2/5}$ Rewrite the expression as a radical expression and evaluate: $32^{2/5}=\sqrt[5]{32^{2}}=\sqrt[5]{1024}=4$ $b)$ $\Big(\dfrac{4}{9}\Big)^{-1/2}$ Switch the numerator and the denominator and change the sign of the exponent: $\Big(\dfrac{4}{9}\Big)^{-1/2}=\Big(\dfrac{9}{4}\Big)^{1/2}=...$ Rewrite as a radical expression and evaluate: $...=\sqrt{\dfrac{9}{4}}=\dfrac{3}{2}$ $c)$ $\Big(\dfrac{16}{81}\Big)^{3/4}$ Rewrite this as a radical expression and evaluate: $\Big(\dfrac{16}{81}\Big)^{3/4}=\sqrt[4]{\Big(\dfrac{16}{81}\Big)^{3}}=\sqrt[4]{\dfrac{4096}{531441}}=\dfrac{8}{27}$
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Search a number
219044 = 2277823
BaseRepresentation
bin110101011110100100
3102010110202
4311132210
524002134
64410032
71601420
oct653644
9363422
10219044
1113a631
12a6918
1378917
1459b80
1544d7e
hex357a4
219044 has 12 divisors (see below), whose sum is σ = 438144. Its totient is φ = 93864.
The previous prime is 219041. The next prime is 219053. The reversal of 219044 is 440912.
219044 = T230 + T231 + ... + T237.
It is not an unprimeable number, because it can be changed into a prime (219041) by changing a digit.
It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 3884 + ... + 3939.
It is an arithmetic number, because the mean of its divisors is an integer number (36512).
2219044 is an apocalyptic number.
It is an amenable number.
219044 is a primitive abundant number, since it is smaller than the sum of its proper divisors, none of which is abundant.
It is a pseudoperfect number, because it is the sum of a subset of its proper divisors.
219044 is a wasteful number, since it uses less digits than its factorization.
219044 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 7834 (or 7832 counting only the distinct ones).
The product of its (nonzero) digits is 288, while the sum is 20.
The square root of 219044 is about 468.0213670336. The cubic root of 219044 is about 60.2805381136.
Subtracting from 219044 its sum of digits (20), we obtain a square (219024 = 4682).
Adding to 219044 its reverse (440912), we get a palindrome (659956).
It can be divided in two parts, 21904 and 4, that multiplied together give a square (87616 = 2962).
The spelling of 219044 in words is "two hundred nineteen thousand, forty-four".
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# Defining variable in projectile motion
I'm trying to understand drag on projectile motion but I don't know what this variable $$b$$ in Eq.1 below on p. 24 in this document is.
We define the drag force to be $$\mathbf{F}_D$$ and the gravitational force is $$\mathbf{F_g}$$. We have $$m\mathbf{a}=\mathbf{F}=\mathbf{F_g}+\mathbf{F_D}=mg\mathbf{\hat y}-b(\mathbf{\hat x}+\mathbf{\hat y}),\tag{1}$$ and letting $$k=b/m$$, we can separate the above equation into $$x-$$ and $$y-$$equations. We have $$x''(t)=-kx'(t), \quad y''(t)=-h-ky'(t).\tag{2}$$ Next we will solve the above differential equations using the initial conditions: $$x(0)=0; \quad y(0)=h;\tag{3.1}$$ $$x'(0)=v\cos\theta; \quad y'(0)=v\sin\theta;\tag{3.2}$$ where $$v$$ is the initial velocity of the projectile. Using separation of variables to solve the $$x-$$equation, we obtain $$x''(t)=-kx'(t),\tag{4.1}$$ $$x'(t)=Ce^{-kt}=v\cos\theta e^{-kt}\tag{4.2}$$ $$x(t)=-\frac{v\cos\theta}{k}e^{-kt}+C=\frac{v\cos\theta}{k}\left(1-e^{-kt}\right).\tag{4.3}$$ Similarly, for the $$y-$$equation, we have $$y''(t)=-h-ky'(t)\tag{5.1}$$ $$dy'=(-g-ky')dt\tag{5.2}$$ $$\frac{dy'}{-g-ky'}=dt\tag{5.3}$$ $$\frac{1}{k}\ln(g+ky')=-t+C\tag{5.4}$$
References:
1. Nina Henelsmith, Projectile Motion:Finding the Optimal Launch Angle, Whitman College, 2016; p. 24.
• Hi WaterRocket123, it's against our rules to post images of text you want to quote. Please type it out instead so it can be indexed by search engines. For formulas, use MathJax. Commented Mar 28, 2020 at 5:51
$$b$$ is the drag constant for a linear drag force. OP's question might have been spurred by the fact that the first eq. on p. 24 has a typo: The factor $$\hat{\bf x}+\hat{\bf y}$$ should have been the velocity $$\vec{\bf v}=x^{\prime}(t)\hat{\bf x}+y^{\prime}(t)\hat{\bf y}$$.
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### Algebra Puzzle
```
Date: 04/09/2001 at 18:26:47
From: Johann Matias H. L.
Subject: Hard equation
Okay, here you go... if this is true:
x + y + z = 1
x*x + y*y + z*z = 2
x*x*x + y*y*y + z*z*z = 3
find (without solving the equation):
x*x*x*x + y*y*y*y + z*z*z*z = ?
```
```
Date: 04/16/2001 at 12:06:35
From: Doctor Peterson
Subject: Re: Hard equation
Hello, Johann.
This question stuck in my mind until I had the time to solve it, and I
want to give you a hint you can try.
I will write "x^2" for x squared, and so on. I will also generalize
the problem a little. Suppose we know that:
x + y + z = A
x^2 + y^2 + z^2 = B
x^3 + y^3 + z^3 = C
Now we want to find D:
x^4 + y^4 + z^4 = D
I tried combining the original equations in various ways to get
something that includes D and other symmetrical expressions; here is
one that I found useful:
AC = (x + y + z)(x^3 + y^3 + z^3)
= (x^4 + y^4 + z^4) + (xy^3 + xz^3 + yx^3 + yz^3 + zx^3 + zy^3)
= (x^4 + y^4 + z^4) + (xy*x^2 + xy*y^2 + xy*z^2)
+ (yz*x^2 + yz*y^2 + yz*z^2) + (zx*x^2 + zx*y^2 + zx*z^2)
- (xy*z^2 + yz*x^2 + zx*y^2)
= (x^4 + y^4 + z^4) + (xy + yz + zx)*(x^2 + y^2 + z^2)
- xyz*(x + y + z)
= D + PB - QA
where
P = xy + yz + zx
Q = xyz
The fact that P and Q are symmetrical with respect to the three
variables suggests that they might appear in other expressions
involving A, B, and C; if we can find P and Q in terms of A, B, and C,
we will be able to find D.
To do this, I used:
A^2 = (x + y + z)(x + y + z)
= (x^2 + y^2 + z^2) + 2(xy + yz + zx)
= B + 2P
and
AB = (x + y + z)(x^2 + y^2 + z^2)
= (x^3 + y^3 + z^3) + (xy^2 + xz^2 + yx^2 + yz^2 + zx^2 + zy^2)
= (x^3 + y^3 + z^3) + (xy*x + xy*y + xy*z - xyz)
+ (yz*x + yz*y + yz*z - xyz) + (zx*x + zx*y + zx*z - xyz)
= (x^3 + y^3 + z^3) + xy*(x + y + z) + yz*(x + y + z)
+ zx*(x + y + z) - 3xyz
= (x^3 + y^3 + z^3) + (xy + yz + zx)*(x + y + z) - 3xyz
= C + PA - 3Q
You'll be able to solve this for D. What I'm not sure of is whether
there are any actual solutions in x, y, and z at all; but if there
are, this will tell you what x^4 + y^4 + z^4 has to be. (I think
actually solving for x, y, and z requires solving a sixth-degree
polynomial, and they may be complex.) I'd feel a lot surer of my
solution if I could check it!
- Doctor Peterson, The Math Forum
http://mathforum.org/dr.math/
```
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## Sue de Coq Help Needed
If you invented that new way to solve these little puzzles, tell us about it
nj3h
Gold Member
Posts: 111
Joined: Mon Jul 10, 2006 12:58 pm
Location: Virginia / USA
### Sue de Coq Help Needed
All,
I am hoping someone can help me in understanding the criteria for the subject technique. Please use the Sudopedia example for this technique in any replies.
The following is a quote from the Sudopedia: "The Sue de Coq technique uses two intersecting sets A and B, where A is a set of N cells with N candidates in a line, B is a set of N cells with N candidates in a box, and the sets A - B and B - A have no common candidates. Candidates can be eliminated from the cells in the line that are not in A, and the cells in the box that are not in B."
My questions are:
a. Please list the members of set A and B. I think they are:
A={1, 2, 4, 5, 8, 9} and B={1, 2, 5, 6, 8, 9}
b. I presume N equals 6 in this example. I presume since N is used in the above quote for both line and box, that N has to be the same for both the line and box. In other words, do the unsolved number of cells in the line and in the box have to be the same?
c. What is A-B and B-A? I think the answer is 4 and 6 respectively. I am puzzled how you could have common candidates.
d. The last sentence totally puzzles me. In the line, 1, 2, and 9 are eliminated, but aren't they in Set A? Likewise, 1 is eliminated in the box, but 1 is in B.
Anyway, any help, using the example in the Suopedia and/or additional examples, will be most appreciated. I guess I must have missed the chapter on sets or just totally have forgotten the concept.
Once I have an explanation I can grasp, then the next trick will be how to easily identify this technique.
Regards,
George
Ron Moore
Posts: 72
Joined: Sun Aug 13, 2006 3:34 am
Location: New Mexico
### Sue de Coq Hypotheses
George,
I commented on some problems with the Sue de Coq page in the Sudopedia in my post in this thread, although I didn't complete my thoughts on how everything should read and how the technique can be explained. So I will attempt to do so here (you don't need to consult that thread).
First of all, just a comment on the notation S - T, where S and T are two sets of cells (or widgets, or whatever). This denotes the set of cells (or whatever) which are in S but not in T.
As you've noticed, Sue de Coq in its simplest form is sometimes referred to as the "Two Sector Disjoint Subsets" technique. The usage of the term "disjoint" here is probably behind some of your confusion. In the presentation in the Sudopedia, A and B are sets of cells. Now A - B and B - A are the sectors here, but we always have (tautologically) that A - B and B - A are disjoint, so that's not what is really meant here. What is meant is that the set of candidate digits in A - B is disjoint from the set of candidate digits in B - A, or alternatively, A - B and B - A have no common candidates.
It's not absolutely necessary, but I think things will be clearer if some simple notation is used to avoid this confusion. Here I'll adopt this notation:
• If A is a set of cells, then D(A) denotes the set of all candidate digits in A.
Now the Sue de Coq position and eliminations can be described as follows. Things would be neater if the standard symbols for the union and intersection of two sets were available, but I can't figure out how to make those work. In the abstract, this may seem a bit cumbersome but things should clear up with the subsequent discussion of the example in the Sudopedia:
*************************************************************
Suppose that A is a set of cells in a box, and B is a set of cells in a line which intersects that box, and that (A intersect B) contains 2 or 3 cells from the line box/intersection; furthermore, suppose that D(A - B) is disjoint from D(B - A), that both D(A - B) and D(B - A) are subsets of D(A intersect B), and that the number of cells in (A union B) equals the number of digits in D(A union B). Then for any digit d in D(A union B), d can be removed as a candidate from any cell outside of (A union B) which sees all cells with candidate d in (A union B). This means that:
For any digit d in D(A - B), d can be removed as a candidate from any cell in the box which is not in A.
Similarly, for any digit d in D(B - A), d can be removed as a candidate from any cell in the line which is not in B.
Finally, for any digit d which appears in (A intersect B) only, that is, any d in D(A intersect B) - D(A - B) - D(B - A), d may be removed as a candidate from any cell in the line or the box which is not in (A union B). Note that in many Sue de Coq positions, this set of digits will be empty, in which case this conclusion is not applicable.
*************************************************************
Below is the relevant portion of the grid in the Sudopedia example, with sets A and B identified by corresponding prefixes to the cells. Just a note on a notational convention -- I'm listing all eliminated digits after the "-" sign.
Code: Select all
`````` .------------------.---------------------.-----------------------.
row 1 | 1 29 7 | 259 6 28 | A58 4 3 |
row 2 | 5 8 4 | 129 3 12 | 7 69-1 126-1 |
row 3 | 3 B29 6 | 45-129 48-29 7 | AB158 AB159 AB125 |
.------------------.---------------------.-----------------------.``````
Set A is r1c7, r3c789.
Set B is r3c2, r3c789.
Set A - B is r1c7, with D(A - B) = {5,8}
Set B - A is r3c2, with D(B - A) = {2,9}. Note that D(B - A) and D(A - B) are disjoint.
Set (A intersect B) is r3c789, with D(A intersect B) = {1,2,5,8,9}. Note that D(B - A) and D(A - B) are subsets of this set.
Set (A union B) contains 5 cells and D(A union B) contains 5 digits, {1,2,5,8,9}.
Then the first conclusion tells us that 5 and 8 (the digits in D(A - B)) can be removed from any cell in box 3 which is not in A. In this case, the first conclusion yields nothing.
The second conclusion tells us that 2 and 9 (the digits in D(B - A)) can be removed from any cell in row 3 which is not in B -- thus the eliminations of 2 and 9 shown in r3c45.
The third conclusion does apply in this case. Digit 1 appears in the intersection only; that is, "1" is in D(A intersect B), but in neither of D(A - B) nor D(B - A). So the conclusion tells us that 1 can be removed from all cells in the box and the line which are not in (A union B) -- thus, the eliminations of 1 shown in r2c89 and r3c4.
It's possible to understand all of this from the reasoning given in the Sudopedia, but that's not really looking at the position from the Sue de Coq perspective. By hypothesis we have the same number of cells in the entire structure (A union B) as the number of candidates in the entire structure -- D(A union B). It follows from our hypotheses that no digit d can be placed twice in the structure. Why is this true? Well, digit d certainly cannot appear twice in box 3, nor twice in row 3; thus, the only conceivable way to place d twice in the entire structure would be for d to appear once in A - B and once in B - A. However, this would be contrary to the hypothesis that D(A - B) and D(B - A) are disjoint.
Thus, in order to fill all 5 cells of the entire structure (A union B), each digit d in D(A union B) -- which contains 5 digits -- must appear exactly once in the structure. Thus, if any cell outside of (A union B) sees all d candidates in (A union B), it has to see digit d somewhere and thus cannot itself contain d. From this the three conclusions above follow, although you can simply work from this principle alone if you prefer.
To answer one of your direct questions, A and B do not have to be the same size.
Another observation: If (A intersect B) contains 2 cells, then the remaining cell in the line/box intersection may be the target for both of the eliminations in the box and the eliminations in the line (the first and second conclusions).
The hypotheses for the Sue de Coq eliminations are a bit restrictive, but they do produce a nice symmetry between the eliminations in the line and the box. There may be some "almost" Sue de Coq positions from which eliminations of multiple digits can be made using similar counting arguments, but the symmetry between the line and box eliminations is lost. If you like, I can continue on with an example, but for the moment I'll stop here and let you absorb the strictly Sue de Coq positions.
mhparker
Grandmaster
Posts: 345
Joined: Sat Jan 20, 2007 10:47 pm
Location: Germany
Nevertheless, it has to be said that the Sudopedia definition is just plain wrong! Somebody should change it. I also didn't understand the Sudopedia definition, and only started to understand the technique after consulting other sources, which defeats the whole purpose of a centralized knowledge base, which is what Sudopedia is aspiring to be.
I also don't like the fact the Sudopedia definition ends with the sentence "Candidates can be eliminated from the cells...", without being explicit about which candidates can be eliminated. This information should be part of any formal definition.
Also confusing is the fact that the explanation of the example does not seem to bear any resemblance to the formal definition. This encourages people to ignore the confusing definition, and latch onto the concrete example instead. Unfortunately, there is only one example provided, which also happens to correspond to the same pattern used in the original sources and most other solving guides (i.e., three common cells, and a single bivalue cell in each of A - B and B - A). This encourages people to think that once they understand this one pattern, they have completely understood the principle, which is clearly not the case.
Cheers,
Mike
Ron Moore
Posts: 72
Joined: Sun Aug 13, 2006 3:34 am
Location: New Mexico
### Alternate Formulation of Sue de Coq Technique
George and Mike,
I formulated my first post in terms of two intersecting sets of cells A and B, as I was trying to maintain some similarity with the presentation in the Sudopedia. However, as Mike points out, there is really little which can be salvaged from that page, so I wish I hadn't limited myself that way. It's a bit cleaner (especially with my inability to use some special characters) to present this in terms of three disjoint sets of cells, as was done on the Sudoku Players Forum, in the thread in which the "Sue de Coq" pattern was first presented (here). So George, I'm sorry if you've been struggling to understand my first post, but I'm going to shift gears and present things slightly differently. I think it will be easier in the long run. None of this is original to me -- it's just my understanding of the thread in the Sudoku Players Forum. However, I don't really see any one easily identifiable place in that thread where the Sue de Coq constraints and eliminations are summarized. This may be why there seems to be some confusion over exactly what "Sue de Coq" is.
********************************************************************************
Suppose, for some box and some line intersecting that box, that C is a set of two or three cells in the line/box intersection, that A is a set of cells in the box, but disjoint from the line/box intersection, and that B is a set of cells in the line, but disjoint from the line/box intersection, such that:
• A and B have no common candidate digits.
All candidate digits in A, and all candidate digits in B, are candidates in C.
The total number of cells in A, B, and C equals the number of distinct candidate digits in C.
Then:
• All candidate digits in A can be eliminated from any cell in the box which is not in either of A or C.
All candidate digits in B can be eliminated from any cell in the line which is not in either of B or C.
If C contains some candidate d which does not appear as a candidate in either of A or B, then d can be eliminated from any cell in the line or the box which is not in C.
********************************************************************************
Of course, the counting argument that was used in the first formulation is still applicable here. From the hypothesis that A and B have no common candidates, we conclude that each of the candidate digits in C must placed (exactly once) somewhere in the entire structure, A union B union C, and from this the three conclusions follow.
As a second example, consider the position below, which is reached after initial basic techniques are used in the 19 September 2006 Daily Nightmare. By way of explanation, when I first worked on this puzzle, I was trying to learn the ALS XZ technique, and in my eagerness to use this, I overlooked a simpler way to proceed. Please ignore any simpler approaches you may see.
Code: Select all
``````.---------------------.----------------------.---------------------.
| 2 1 3 | 489 49 #48 | 57 57 6 |
| 7 5 9 | 16 16 2 | 8 34 34 |
| 6 8 4 | 7 3 5 | 9 2 1 |
:---------------------+----------------------+---------------------:
| 3 9 1 | 456 46 7 | 2 456 8 |
| 8 4 2 | 13569 169 136 | 567 1567 57 |
| 5 6 7 | 2 8 #14 | 3 149 49 |
:---------------------+----------------------+---------------------:
| 9 237 56 | 1368 27 #1368 | 4 35678 2357 |
| 4 237 8 | *36 5 9 | 1 367 237 |
| 1 237 56 | 3468 27 #3468 | 567 356789 23579 |
'---------------------'----------------------'---------------------'``````
Actually, something good came from my oversight. For those readers familiar with Almost Locked Sets (ALS's), I observed that the sets
• r8c4 -- 1 cell with 2 candidates, {3,6}
and
• r1679c6 -- 4 cells with 5 candidates, {1,3,4,6,8}
are two ALS's which share 2 restricted common digits, 3 and 6. At the time I noted that there were eliminations in two different houses (box 8 and column 6) which could be made in this position, but it was some time later before I really understood what was going on to make this work. Eventually I came up with the idea I presented in my post in this thread. The eliminations are the same as those which we'll derive from the Sue de Coq technique.
In the position (repeated below), sets A, B, and C in the new Sue de Coq formulation are identified by corresponding prefixes to the cells.
Code: Select all
``````.---------------------.----------------------.---------------------.
| 2 1 3 | 489 49 B48 | 57 57 6 |
| 7 5 9 | 16 16 2 | 8 34 34 |
| 6 8 4 | 7 3 5 | 9 2 1 |
:---------------------+----------------------+---------------------:
| 3 9 1 | 456 46 7 | 2 456 8 |
| 8 4 2 | 13569 169 36-1 | 567 1567 57 |
| 5 6 7 | 2 8 B14 | 3 149 49 |
:---------------------+----------------------+---------------------:
| 9 237 56 | 18-36 27 C1368 | 4 35678 2357 |
| 4 237 8 | A36 5 9 | 1 367 237 |
| 1 237 56 | 48-36 27 C3468 | 567 356789 23579 |
'---------------------'----------------------'---------------------'``````
• set A = r8c4, with candidates {3,6}
set B = r16c6, with candidates {1,4,8}
set C = r79c6, with candidates {1,3,4,6,8}
We quickly confirm that:
• A and B have no common candidates.
All candidates in A, and all candidates in B, are candidates in C.
Together, sets A, B, and C contain 5 cells, and 5 distinct candidate digits.
Then the Sue de Coq eliminations are:
• All candidate digits in A can be removed from any cell in the box which is not in either of A or C -- thus the eliminations of 3 and 6 in r79c4.
All candidate digits in B can be removed from any cell in the line which is not in either of B or C -- thus the elimination of 1 shown in r5c6.
In this case, there are no candidates which appear only in C.
As an aside, we can also view this as an Almost Locked Candidates (ALC) position, yielding the same eliminations. ALC was the topic of discussion in a thread I recently initiated (here). Again, a big part of the problem here was determining exactly what "Almost Locked Candidates" means. From the ALC viewpoint, the argument would go something like this (position repeated for convenience):
Code: Select all
``````.---------------------.----------------------.---------------------.
| 2 1 3 | 489 49 48 | 57 57 6 |
| 7 5 9 | 16 16 2 | 8 34 34 |
| 6 8 4 | 7 3 5 | 9 2 1 |
:---------------------+----------------------+---------------------:
| 3 9 1 | 456 46 7 | 2 456 8 |
| 8 4 2 | 13569 169 *36-1 | 567 1567 57 |
| 5 6 7 | 2 8 14 | 3 149 49 |
:---------------------+----------------------+---------------------:
| 9 237 56 | 18-36 27 *1368 | 4 35678 2357 |
| 4 237 8 | *36 5 9 | 1 367 237 |
| 1 237 56 | 48-36 27 *3468 | 567 356789 23579 |
'---------------------'----------------------'---------------------'``````
In column 6, the digits 3 and 6 appear only in r5c6, and in the line/box intersection, r79c6. Digits 3 and 6 cannot both be placed in the line/box intersection, as that would conflict with the bivalued "36" cell in r8c4. In column 6, digits 3 and 6 cannot both lie outside of the line box intersection, since there is only one cell available to hold them outside of the line/box intersection, namely r5c6. Therefore, one of 3, 6 must appear in r5c6 (so that r5c6 can't be 1) and the other must appear in one of r79c6. Whichever it is will pair up with r8c4 to eliminate 3's and 6's in other cells of the box.
If you've followed this then you've probably realized that the reasoning presented in the Sudopedia example is really using the ALC viewpoint rather than the Sue de Coq viewpoint.
Anyway, we've seen that there are many ways to skin this cat. I'd be interested to see a real life Sue de Coq position which can't be considered as a "2 ALS's with two restricted common digits" position. (I don't mean to suggest that Sue de Coq is not useful. Players not familiar with ALS's can still use Sue de Coq to advantage.)
mhparker
Grandmaster
Posts: 345
Joined: Sat Jan 20, 2007 10:47 pm
Location: Germany
Ron,
Many thanks for another impressive post, and for taking the time to explain this so thoroughly. I certainly find your reasoning significantly easier to follow than the original Sue de coq thread!
One point I didn't realize up to now is the fact that the intersection of the two sets (C in your example) must consist of at least 2 cells. But this has now become clear. To avoid being a naked subset, the disjoint sets A and B must each have at least one excess candidate (compared to the number of cells of which they consist). Because of the condition that (A + B + C) must consist of N candidates distributed across N cells, these excess candidates from A and B must be "absorbed" by the intersection C, which must therefore consist of at least 2 cells.
"All candidate digits in A, and all candidate digits in B, are candidates in C."
By similar logic to that applied above, I can see that this must indeed be the case in all the examples I've seen so far, where A and B both consist of a single cell. Otherwise, if C were to lose a candidate, either (A + C) or (B + C) would reduce to a naked subset of (N - 1) candidates distributed across (N - 1) cells.
However, in a more complicated example, where either A or B (or both) consist of more than one cell, this restriction would (according to my current understanding) no longer apply. I assume that, in general, C can afford to lose up to 2 candidates less than the combined number of cells in (A + B).
Ron, am I on the right track here?
Cheers,
Mike
Ron Moore
Posts: 72
Joined: Sun Aug 13, 2006 3:34 am
Location: New Mexico
Mike,
I don't have time for a response with examples, but I did want to acknowledge your comments. I believe you're definitely on the right track, maybe a bit ahead of me. I included the condition that you question because, as I remember, that was the way the originator (our "boy named Sue," if you've read the "Discussion" tab in the Sudopedia topic) described the pattern in the original Sudoku Player's Forum thread. I don't have time at the moment to confirm that, because, as you suggest, it takes a while to plod through all that's in that thread.
You made some good observations. I agree with what you seem to be suggesting, but it's late for me so you should check me on this.
We can delete the condition in question (the second one) if the third is reworded slightly, something like:
The total number of cells in A, B, and C equals the number of distinct candidate digits in A, B, and C. (Formerly, this was the same as the number of distinct candidate digits in C.)
Then the conclusions can be something like this:
Every digit (in the combined structure) which is not a candidate in B can be removed from any cell in the box which is in neither A nor C.
Similarly, every digit (in the combined structure) which is not a candidate in A can be removed from any cell in the line which is in neither B nor C.
Now we don't have to mention specifically the third conclusion as it's covered by these two. So it looks like this more general formulation is actually easier to express.
mhparker
Grandmaster
Posts: 345
Joined: Sat Jan 20, 2007 10:47 pm
Location: Germany
Ron,
Sorry it took so long for me to get back to you on this one.
I just want to say that the new generalized description looks fine to me now. Maybe it can serve as the basis for a new formal definition in Sudopedia.
Thanks ever so much for your elaborate and very helpful answers. I look forward to "meeting" you again on another topic in this forum soon!
Cheers,
Mike
Ron Moore
Posts: 72
Joined: Sun Aug 13, 2006 3:34 am
Location: New Mexico
### Generalized 2 Sector Sue de Coq
(Edited 18 Apr 2007 to improve wording of hypotheses and add another example.)
Mike,
Please don't feel you need to excuse yourself for any delays in responding. I'm usually not timely myself even in more limited discussions. Anyway, I have given all this some more thought in the interim.
After reviewing everything to date, I find that I still didn't quite state things in the most general way, and that some more examples would be useful. So here I will attempt to consolidate and make some minor adjustments on everything stated so far, and also provide several hypothetical examples.
*************************************************************************************
Generalized Two Sector Sue de Coq
Suppose, for some box and some line intersecting that box, that A is some set of unsolved cells in the box, B is some set of unsolved cells in the line, and C is a set of two or three unsolved cells in the line/box intersection, such that:
• A, B, and C are pairwise disjoint (no two share a common cell).
A and B have no common candidate digits.
The total number of cells in A, B, and C equals the number of distinct candidate digits in A, B, and C.
Then, letting T denote the set of all cells in the pattern, i.e., T = A union B union C:
• All candidate digits in T which are not candidates in B can be eliminated from any cell in the box which is in neither A nor C.
All candidate digits in T which are not candidates in A can be eliminated from any cell in the line which is in neither B nor C.
Note: For a "classic" Sue de Coq pattern, an additional constraint is added -- that all candidate digits in A, and all candidate digits in B, are also candidates in C. However, this constraint is not necessary and the same eliminations follow without it. I suspect that in practice that the vast majority of generalized patterns will in fact be classic patterns. However, I'm wondering if any solver programs have been coded to recognize the generalized pattern described here.
*************************************************************************************
The following diagram is intended to facilitate discussion, not to be physically suggestive of the Sue de Coq pattern.
Code: Select all
``````.--------.--------.
| | |
| A | A' |
| | |
'--------'--------+--------. A, A', C are all in the box
| |
| C |
| |
'--------'
| |
| B' | B, B', C are all in the line
| |
'--------'
| |
| B | T is the set A union B union C
| |
'--------'``````
A' is the set of all unsolved cells in the box which are in neither A nor C, and similarly B' is the set of all unsolved cells in the line which are in neither B nor C. In an actual Sudoku grid, the cells in any of these sets do not need to be physically contiguous or connected. Also, if there are three unsolved cells in the line/box intersection and C contains only two of them, then the remaining cell of the line/box intersection will be in both A' and B'.
In terms of the diagram, the conclusions are that all digits appearing in T (A union B union C) which are not candidates in B can be removed from A', and that all digits in T which are not candidates in A can be removed from B'.
My assumption has been that the argument from the Sue de Coq perspective is something like this:
By hypothesis, if there are N total cells in T, then in total there are N distinct candidate digits in T. None of these digits can be placed twice in T: certainly no digit can be placed twice in the box, and similarly no digit can be placed twice in the line; and, since by hypothesis A and B have no common candidates, no digit can appear in both A and B. Thus, in order for all N cells of T to be filled, each of the N digits must appear (exactly once) in it.
Each candidate digit of T which is not a candidate in B must reside somewhere in A or C, and thus can be eliminated from all cells in A'. Likewise, each candidate digit of T which is not a candidate in A must reside somewhere in B or C, and thus can be eliminated from all cells in B'.
However, upon a careful rereading of the original Sue de Coq thread in the Sudoku Players Forum, (here), I see that the argument used there is not along the line of reasoning presented here, but in fact resembles the Sudopedia argument. I perhaps should back off from my claim earlier in this thread that the Sudopedia presentation is an Almost Locked Candidates argument, although it does have much of that flavor.
I believe that it was after I read the statement in the Sudopedia that the Sue de Coq technique is a special case of subset counting, that (after a bit of thought) I jumped to the conclusion that the entire pattern was the subset being counted. (This view was recently reinforced when I read the second external reference link given on the Sudopedia page (here). The author, Andries E. Brouwer, explains the Sue de Coq eliminations with an argument similar to the one given above -- a subset counting argument based on the set of cells in the entire pattern. That reference also states some general subset counting principles which are useful.) What we see in the Sudopedia, and in the Sudoku Player's Forum thread, is really a subset counting argument for the set A union C (and similarly for B union C). I don't feel that the argument from this viewpoint is as easily understood as the reasoning presented above, nor is it clearly expressed in general terms in either source. At the risk of being long-winded, I will attempt to elaborate the argument from that viewpoint, since some interesting insights can be gained along the way. (Some readers may prefer to skip to the "Examples" section below.)
To this end, some symbology and definitions (some of which have appeared earlier in this thread) will be useful:
Definitions
If S is any set of cells (or widgets, etc.), then |S| denotes the number of elements in S.
If S denotes a set of cells, then D(S) denotes the set of all candidate digits in S.
If a set of N cells in some house contains N + k candidate digits, then the set will be said to be an Almost ... Almost Locked Set (A...ALS) with k degrees of freedom.
So k is just the candidate digit count minus the cell count.
If S is an A...ALS with k degrees of freedom, then
• |D(S)| = |S| + k
For k = 0, we have a locked (or naked) set; for k = 1, we have an Almost Locked Set; and for k = 2, we have an Almost Almost Locked Set, etc.
If we agree that two sets of cells "see" each other if the two sets are disjoint, and all cells of the two sets reside in some common house, then we have the following self-evident principle:
• If A is an A...ALS with k degrees of freedom, and S is any set of cells which sees A, then S can contain at most k digits of D(A).
*************************************************************************************
Now suppose that A, B, and C are sets of cells satisfying the generalized two sector hypotheses, and again let T denote the combined set of cells: T = A union B union C.
First we note that from the hypothesis that sets A, B, and C are pairwise disjoint, the number of cells in T is |A| + |B| + |C|.
Now let kA and kB denote the number of degrees of freedom of sets A and B, respectively.
Since A and B have no common candidates, the total set of candidate digits in T can be broken down into three mutually exclusive sets:
• Those digits which appear as candidates in A (i.e., those in D(A)). There are |A| + kA such digits.
Those digits which appear as candidates in B. There are |B| + kB such digits.
Those digits which appear as candidates only in C (in neither of A or B). I'll call these digits "C only" candidates. This set may be empty. Let nCO denote the number of such digits.
From the hypothesis that the total number of cells in T equals the total number of candidate digits appearing in T, we have the relation
• |A| + |B| + |C| = |A| + kA + |B| + kB + nCO
which reduces to
• |C| = kA + kB + nCO
**********************************************************************************
Aside: As Mike pointed out earlier in the thread, to make things interesting, we can assume that both kA and kB are at least one. (If either of A or B is a naked subset, there are corresponding eliminations in its house, independent of the other sets in the pattern.)
Now since |C| is 2 or 3, and for "interesting" patterns we have kA, kB >= 1, the above relation shows that there's not much leeway in the configuration:
• If |C| = 2, we must have kA = kB = 1, nCO = 0.
If nCO > 0, we must have |C| = 3, kA = kB = nCO = 1.
So nCO is 0 or 1 -- there is at most one "C only" digit.
We also note that at least one of the sets A, B must be an Almost Locked Set. The other can be an Almost Almost Locked Set if |C| = 3 and nCO = 0.
So it's not surprising that many Sue de Coq configurations can be entirely understood in terms of Almost Locked Set arguments.
*********************************************************************************
Diagram repeated for convenience:
Code: Select all
``````.--------.--------.
| | |
| A | A' |
| | |
'--------'--------+--------. A, A', C are all in the box
| |
| C |
| |
'--------'
| |
| B' | B, B', C are all in the line
| |
'--------'
| |
| B | T is the set A union B union C
| |
'--------'``````
Now consider the set C. Again we can break down the candidate digits in C into three mutually exclusive sets -- those in D(A), those in D(B), and the "C only" digit, if any. Since C sees set A (in the box), which has kA degrees of freedom, C can contain at most kA digits of D(A). Likewise, C can contain at most kB digits of D(B). Finally, C can contain at most nCO of the "C only" candidate digits (because that's all there are).
So C can contain at most
• kA + kB + nCO digits
but we've shown above that C has exactly that many cells.
So, in order for all cells of C to be filled, C must contain (some set of) kA digits of D(A), (some set of) kB digits of D(B), and the "C only" digit, if any.
In the box, the |A| cells of set A will combine with the kA + nCO cells of C which contain the kA digits of D(A), and the "C only" digit (if any), to form a locked set -- |A| + kA + nCO cells with |A| + kA + nCO digits. So, these digits can be removed as candidates from all cells in A'. Note that this set of digits contains exactly those digits in T which are not in B.
Of course, everything in this discussion is symmetrical with respect to the box and the line so we similarly derive the conclusion that those digits in T which are not in A can be removed from all cells in B'.
***************************************************************************
Examples
Note: In the following diagrams, all eliminated digits are shown after the "-" sign.
In this classic 4-cell Sue de Coq pattern, |C| = 2, and the remaining cell in the line box intersection benefits from both the eliminations in the box and in the line.
Code: Select all
`````` .----------------------.-----------------.------------------.
row 1 | A12 *-12 *-12 | . . . | . . . |
row 2 | *-12 *-12 *-12 | . . . | . . . |
row 3 | *-1234 C1234 C1234 | B34 *-34 *-34 | *-34 *-34 *-34 |
.----------------------.-----------------.------------------.
Set C = r3c23, 2 cells with digits 1234
Set A = r1c1, 1 cell with digits 12
Set B = r3c4, 1 cell with digits 34``````
This 5-cell example is not a classic Sue de Coq pattern, since digit 5 is in A but not in C. Again we have |C| = 2, and the remaining cell of the line/box intersection benefits from both the eliminations in the box and in the line.
Code: Select all
``````
.-----------------------.-----------------.------------------.
row 1 | A125 *-125 *-125 | . . . | . . . |
row 2 | *-125 A125 *-125 | . . . | . . . |
row 3 | *-12345 C1234 C1234 | B34 *-34 *-34 | *-34 *-34 *-34 |
.-----------------------.-----------------.------------------.
Set C = r3c23, 2 cells with digits 1234
Set A = r1c1|r2c2 2 cells with digits 125
Set B = r3c4, 1 cell with digits 34``````
This 5-cell pattern shows that if |C| = 2, it is possible that one of the sets A or B can contain the remaining cell in the line/box intersection:
Code: Select all
`````` .---------------------.-----------------.------------------.
row 1 | *-125 *-125 *-125 | . . . | . . . |
row 2 | *-125 A125 *-125 | . . . | . . . |
row 3 | A125 C1234 C1234 | B34 *-34 *-34 | *-34 *-34 *-34 |
.---------------------.-----------------.------------------.
Set C = r3c23, 2 cells with digits 1234
Set A = r2c2|r3c1, 2 cells with digits 125
Set B = r3c4, 1 cell with digits 34``````
In this 6-cell pattern, |C| = 2 and each of sets A and B contains a digit not in C.
Code: Select all
`````` .------------------------.-------------------.---------------------.
row 1 | A125 *-125 *-125 | . . . | . . . |
row 2 | *-125 A125 *-125 | . . . | . . . |
row 3 | *-123456 C1234 C1234 | B346 *-346 *-346| B346 *-346 *-346 |
.------------------------.-------------------.---------------------.
Set C = r3c23, 2 cells with digits 1234
Set A = r1c1|r2c2, 2 cells with digits 125
Set B = r3c47, 2 cells with digits 346``````
For completeness, this is a 5-cell classic Sue de Coq pattern with a "C only" digit (namely, 5):
Code: Select all
`````` .------------------------.-------------------.---------------------.
row 1 | A12 *-125 *-125 | . . . | . . . |
row 2 | *-125 *-125 *-125 | . . . | . . . |
row 3 | C12345 C12345 C12345 | B34 *-345 *-345 | *-345 *-345 *-345 |
.------------------------.-------------------.---------------------.
Set C = r3c123, 3 cells with digits 12345
Set A = r1c1, 1 cell with digits 12
Set B = r3c4, 1 cell with digits 34``````
Clearly, this last example could be extended to a 6-cell non-classic pattern if B consisted of two cells in the line, say r3c47, containing digits 3,4, and 6, for example.
This is a 5-cell pattern with set A an Almost Locked Set and set B an Almost Almost Locked Set.
Code: Select all
`````` .------------------------.--------------------.---------------------.
row 1 | A12 *-12 *-12 | . . . | . . . |
row 2 | *-12 *-12 *-12 | . . . | . . . |
row 3 | C12345 C12345 C12345 | B345 *-345 *-345 | *-345 *-345 *-345 |
.------------------------.--------------------.---------------------.
Set C = r3c123, 3 cells with digits 12345
Set A = r1c1, 1 cell with digits 12
Set B = r3c4, 1 cell with digits 345``````
Last edited by Ron Moore on Wed Apr 18, 2007 9:05 pm, edited 1 time in total.
mhparker
Grandmaster
Posts: 345
Joined: Sat Jan 20, 2007 10:47 pm
Location: Germany
Ron,
All of what you say really appeals to me. Indeed, I can't really fault any of it, and particularly liked your formalization based on degrees of freedom. I can honestly say - and I'm not just saying this to be polite - it's the best description of Sue de Coq I've ever read.
The only thing I would like to point out is that maybe it would be a good idea to generalize even more (if and when the definitions need to be migrated to another source such as Sudopedia)! In particular, I'm thinking about jigsaws here, where a box/line intersection may consist of more than 3 cells. For example, instead of saying:
C is a set of two or three unsolved cells in the line/box intersection
it might instead be better to say:
"C is a set of two or more unsolved cells in the line/box intersection".
I would certainly be very interested in seeing a generalized Sue de Coq implementation, especially when properly adapted and applied to jigsaws, where I have high hopes for it. Now, with Ruud's Daily Jigsaws (in particular, the "Ultra Hard" ones at the weekend), we have - possibly for the first time on the web (?!) - a consistent source of high-quality jigsaws that require advanced vanilla techniques to solve. I'm hoping that these jigsaws will start attracting a new "hard-core" audience because of this.
Cheers,
Mike
Ron Moore
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# Homework Help: Evaluate the definite integral for the area of the surface.
1. Jun 2, 2010
### lude1
1. The problem statement, all variables and given/known data
Evaluate the definite integral for the area of the surface generated by revolving the curve about the x-axis.
y=(x3/6) + (1/2x), [1,2]
2. Relevant equations
2π∫[r(x)](1+[f'(x)2])
3. The attempt at a solution
First I found the derivative.
f'(x)= (x2/2) + (1/2x2)dx
And since y is a function of x, r(x) is
r(x)= (x3/6) + (1/2x)
Then I plug everything in and get
2π∫ [(x3/6) + (1/2x)] * {1 + [(x2/2) + (1/2x2)]2}1/2}dx
And then I'm stuck. The book tells me that I am suppose to get
2π∫ [(x3/6) + (1/2x)] * [(x2/2) + (1/2x2)]dx
But I have no idea how they got that. Specifically, I don't know how they got rid of the radical...
2. Jun 2, 2010
### mmmboh
For starters you derived it wrong, the derivative of f(x) is f'(x)=x2/2-1/2x2. Maybe this is the problem. Unless you just wrote your f(x) wrong.
Edit: Yes that is the problem, use the correct f'(x), and then expand f'(x)2 and put everything over a common denominator, and then you can make an equation which is being squared that is equal to that, and then the root cancels out the squares.
Last edited: Jun 2, 2010
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## McGraw Hill My Math Grade 5 Chapter 10 Lesson 8 Answer Key Multiplication as Scaling
All the solutions provided in McGraw Hill Math Grade 5 Answer Key PDF Chapter 10 Lesson 8 Multiplication as Scaling will give you a clear idea of the concepts.
## McGraw-Hill My Math Grade 5 Answer Key Chapter 10 Lesson 8 Multiplication as Scaling
Scaling is the process of resizing a number when you multiply by a fraction that is greater than or less than 1.
Draw It Multiply the number 2 by three fractions greater than 1.
Multiply the number 2 by three fractions less than 1.
Talk About it
Question 1. Predict whether the product of 3 and $$\frac{4}{5}$$ is greater than, less than, or equal to 3. Explain. Answer: The above-given equation: product of 3 and 4/5. This can be written as: 3 x 4/5 = 12/5 = 2 2/5. 12/5 = 2.4 which is less than 3 Therefore, the product is less than 3.
Question 2. Mathematical PRACTICE 3 Draw a Conclusion Predict whether the product of 2 and 2$$\frac{1}{5}$$ is greater than, less than, or equal to 2. Explain. Answer: The above-given equation: product of 2 and 2 1/5 2 x 2 1/5 2 1/5 is a mixed fraction. Convert it into an improper fraction. 2 1/5 = 11/5 Now multiply: 2 x 11/5 = 22/5 = 4.4 Therefore, the product is greater than 2.
Practice It
Without multiplying, circle whether each product is greater than, less than, or equal to the whole number.
Algebra Without multiplying, circle whether the unknown in each equation is greater than, less than, or equal to the whole number.
For Exercises 13-15, analyze each product in the table.
Question 13. Why is the first product less than $$\frac{3}{4}$$? Answer: The above-given: The first product and its factors are given in the table. 1/2 x 3/4 = 3/8 3/8 is smaller than 3/4 3/8 = 0.375 3/4 = 0.75 3/8 is 50% smaller than 3/4. Question 14. Why is the product of 1 and $$\frac{3}{4}$$ equal to $$\frac{3}{4}$$? Answer: The above-given: 1 x 3/4 = 3/4 if we multiply anything by 1 we get the same product So, they are equal.
Write About It
Question 18. How can I use scaling to help predict the product of a number and a fraction? Answer: I can predict the size of the product based on the size of the factors. For example fraction × fraction = smaller fraction, fraction × whole number = smaller number, whole number × mixed number = larger than the original whole number.
## McGraw Hill My Math Grade 5 Chapter 10 Lesson 8 My Homework Answer Key
Problem Solving
Vocabulary Check
Question 7. Fill in the blank with the correct term or number to complete the sentence. ____ is the process of resizing a number when you multiply by a fraction that is greater than or less than 1. Answer: Scaling – Scaling is the factor which is used to represent the object size. The size of the object can be shown by increasing or decreasing its original size. In general, the represented size of the object increased for the small object whereas it decreased for the bigger object. Scaling is used for better viewing of an object. The ratio by which an object’s size increases or decreases is called a scale factor. – Scale factor is the ratio between the corresponding measurement of an object and the representation of that object. A scale factor is a whole number or greater than 1 to make a larger copy. The scale factor is a fraction or less than 1 to make a smaller copy. Therefore, the scale factor can be expressed by any number or fraction. Also, the scale factor can be expressed using the colon (:), such as Original sized object: Representation-sized object where, Representation sized object = Any number x Original sized object (in case of larger copy) or, Representation sized object = 1/ any number x Original sized object (in case of smaller copy)
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## Free Printable Properties of Multiplication Worksheets for 8th Grade
Properties of Multiplication: Discover a collection of free printable worksheets for Grade 8 math teachers, designed to help students understand and master the essential multiplication properties.
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## Explore printable Properties of Multiplication worksheets for 8th Grade
Properties of Multiplication worksheets for Grade 8 are an essential resource for teachers looking to help their students master the fundamental concepts of multiplication in math. These worksheets provide a variety of exercises and problems that focus on the different properties of multiplication, such as the commutative, associative, and distributive properties. By incorporating these worksheets into their lesson plans, teachers can ensure that their Grade 8 students have a solid understanding of multiplication and its properties, which will serve as a strong foundation for more advanced mathematical concepts in the future. Additionally, these worksheets can be used for in-class practice, homework assignments, or even as assessment tools to gauge students' progress in their understanding of multiplication. Properties of Multiplication worksheets for Grade 8 are a valuable tool for any math teacher looking to enhance their students' learning experience.
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## Lesson Plan
Associative property of multiplication, view aligned standards, learning objectives.
Students will be able to apply the associative property to multiply single-digit factors.
## Introduction
• Ask students what the word associate means. Use it in a sentence. For example, "I associate with Matthew during recess," or "We often associate the color blue with sadness." Give students a moment to discuss the word with peers.
• Call on a few students to give a definition for the word associate and then develop a meaning with the class (i.e. joined or connected)
• On the board, draw a quick picture to illustrate the word (i.e. draw two people holding hands)
• Explain: Today we are going to explore the associative property of multiplication .
• Write the name of the property on the board and underline the word associative .
• Have students share their ideas with a supportive partner or a peer with the same home language (L1), if possible.
• Allow students to share their ideas in their L1.
Intermediate
• Ask students to restate the definition of associate in their own words to a partner and then to the whole group.
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The Identity Property of Multiplication states that the product of any factor and 1 equals the factor. McGraw Hill My Math Grade 5 Chapter 6 Lesson 8 My Homework Answer Key. Practice 3. p. 427-428 Name Numbers and Operations in Base Ten 5.NBT.5, 5.NBT.7 Lesson 8 Multiplication Properties eHelp Homework Helper Need help? connectED.mcgraw-hill.com Use properties of multiplication to find (1.7 × 5) × 2. (1.7 × 5) × 2 = 1.7 × (5 × 2) = 1.7 × = 17 10 Associative Property Multiply. ... Lesson 6: The ... 4. Multiplication Properties Students learn the following multiplication properties: the commutative property of multiplication, which states that a x b = b x a, the associative property of multiplication, which states that (a x b) x c = a x (b x c), the identity property of multiplication, which states that a x 1 = a, and the zero property of multiplication, which states that a x 0 = 0. 5. Module 4 Lesson 8 Math properties Flashcards Module 4 Lesson 8 Math properties. 2.8 (5 reviews) Flashcards; Learn; Test; Match; Get a hint. when multiplying a number by one, the number will not change and keeps its identity Example: 5 x 1 = 5. 6. McGraw Hill My Math Grade 5 Answer Key Pdf Lesson 4 Multiplication Patterns; Lesson 5 Problem-Solving Investigation: Make a Table; Lesson 6 Use Partial Products and the Distributive Property; Lesson 7 The Distributive Property; Lesson 8 Estimate Products; Lesson 9 Multiply by One-Digit Numbers; Lesson 10 Multiply by Two-Digit Numbers; McGraw Hill My Math Grade 5 Chapter 2 Check My Progress 7. Lesson 5, Multiplication properties and division rules Students will learn how to solve math exercises using multiplication properties and division rules. My Math book. ISBN 978--07-905763-1 ,McGraw-Hill 8. IXL skill plan Lesson 2: Hands On: Divide Using Base-Ten Blocks. Divide by 2-digit numbers using models. Lesson 3: Divide by a Two-Digit Divisor. Divide 2-digit and 3-digit numbers by 2-digit numbers. Divide 2-digit and 3-digit numbers by 2-digit numbers: word problems. Lesson 3: Divide by a Two-Digit Divisor. 9. IXL Improve your math knowledge with free questions in "Properties of addition and multiplication" and thousands of other math skills. 10. McGraw Hill My Math Grade 5 Chapter 10 Lesson 8 Answer Key All the solutions provided in McGraw Hill Math Grade 5 Answer Key PDF Chapter 10 Lesson 8 Multiplication as Scaling will give you a clear idea of the concepts.. McGraw-Hill My Math Grade 5 Answer Key Chapter 10 Lesson 8 Multiplication as Scaling. Scaling is the process of resizing a number when you multiply by a fraction that is greater than or less than 1. 11. 8th Grade Math Other. 8th grade 7 units · 121 skills. Unit 1 Numbers and operations. Unit 2 Solving equations with one unknown. Unit 3 Linear equations and functions. Unit 4 Systems of equations. Unit 5 Geometry. Unit 6 Geometric transformations. Unit 7 Data and modeling. 12. 50+ Properties of Multiplication worksheets for 8th Grade on Quizizz Additionally, these worksheets can be used for in-class practice, homework assignments, or even as assessment tools to gauge students' progress in their understanding of multiplication. Properties of Multiplication worksheets for Grade 8 are a valuable tool for any math teacher looking to enhance their students' learning experience. 13. Associative Property of Multiplication Explain: Today we are going to explore the associative property of multiplication. Write the name of the property on the board and underline the word associative. Beginning. Have students share their ideas with a supportive partner or a peer with the same home language (L1), if possible. Allow students to share their ideas in their L1. 14. PDF 0479 0480 Gr3 S C08L8HW 115022 Lesson 8 Multiply by 11 and 12 Practice Write an addition sentence and a multiplication sentence for each. 1. 5 rows of 11 counters + + + + = × = 2. 3 rows of 12 counters + + = × = Homework Helper Felisa can put 6 photos on each page of her scrapbook. How many photos can she place altogether on 11 pages? Find 6 × 11. 15. PDF Homework Helper eHelpeHelp Multiplication Lesson 6 My Homework 171 ... Commutative Property of Multiplication • 8 × 9 = 9 × 8 • (4 × 7) × 2 = 4 × (7 × 2) Draw a line to match each property with the equation that represents it. Vocab 172 Need more practice? Download Extra Practice at connectED.mcgraw-hill.com 16. PDF Lesson 8 Homework 5 Lesson 8: Use the area model and multiplication to show the equivalence of two fractions. 4Lesson 8 Homework 5 3. Draw area models to prove that the following number sentences are true. a. 1 3 = 2 6 b. 2 5 = 4 10 c. 5 ; = 10 14 d. 3 6 = 9 18 4. Use multiplication to create an equivalent fraction for each fraction below. a. 2 3 b. 5 6 c. 6 5 d ... 17. My Homework Lesson 8 Hands On Multiplication As Scaling Answer Key connectED.mcgraw-hill.com Lesson 8 Hands On: Distributive Property and Partial Quotients Homework Helper Find 375 ÷ 5. You can use the Distributive Property and an area model to divide. Model 375 as (300 + 70 + 5). Divide each section by 5. Add the partial quotients. 60 + 14 + 1 = 75 375 ÷ 5 = 75 So, 375 ÷ 5 = 75. 18. Multiplication Properties Rationale for teaching this lesson: I feel that it is important for the students to know the multiplication properties because they will help the students with computation. If the students know that 7 x 3 = 21, then the students will automatically know that 3 x 7 = 21 because of the commutative property. 19. 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## Steve on Image Processing with MATLABImage processing concepts, algorithms, and MATLAB
Note
Steve on Image Processing with MATLAB has been archived and will not be updated.
# Learning Lessons from a One-Liner
My colleague Greg Wilson, creator of the course Software Carpentry, proposed a couple of programming problems as prerequisites for the course. The idea is that if you can solve the problems using any language at all, then you're ready for the course.
Here's Problem 1:
"Write a function that takes a list or array of numbers as input and return the largest number that is adjacent to a zero. For example, if the input is:
[1, 5, 3, 0, 2, 7, 0, 8, 9, 1 0]
the output is 8."
When I saw this problem I immediately fired off an e-mail to Greg with a MATLAB one-liner:
max(a(imdilate(a == 0, [1 1 1])))
a = [1, 5, 3, 0, 2, 7, 0, 8, 9, 1 0];
max(a(imdilate(a == 0, [1 1 1])))
ans =
8
Of course, one common problem with one-liners tossed off by "experts" is that they are so often wrong. And my answer above is no exception! If all the nonzero elements adjacent to 0s in the input vector are negative, the above solution would incorrectly return 0.
b = [5 4 -1 0 -2 0 -5 8];
max(b(imdilate(b == 0, [1 1 1])))
ans =
0
Greg raised two questions in response to my e-mail:
• How many MATLAB users would be able to understand that solution?
• How many MATLAB users would be able to come up with it on their own?
Well, I hope by writing this blog to increase the number of MATLAB and Image Processing Toolbox users who could come up with it on their own. But let me tackle Greg's first question - how many users would understand the solution?
I think the answer is "not very many." That's because my code does not clearly express my intent. This is a common problem with one-liners.
I suspect that most MATLAB programmers go through a phase of being overly enamored of the one-liner. (Some never leave that phase!) Now that I've been in the software development business for quite a while, I've (mostly) lost my fondness for the one-liner. I'll still write code that way if I'm experimenting at the MATLAB command prompt, but in code that's going to ship I'm much more likely to break that expression into parts and give each part a meaningful name. I might write it like this (including the bug fix):
zero_mask = (a == 0);
max_value_adjacent_to_zero = max(a(adjacent_to_zero_mask));
zero_mask = (a == 0)
zero_mask =
0 0 0 1 0 0 1 0 0 0 1
adjacent_to_zero_mask = imdilate(zero_mask, [1 0 1])
adjacent_to_zero_mask =
0 0 1 0 1 1 0 1 0 1 0
max_value_adjacent_to_zero = max(a(adjacent_to_zero_mask))
max_value_adjacent_to_zero =
8
Is this solution understandable? Well, it's more understandable than the first version, but it does assume that the code reader is familiar with a few key concepts:
• MATLAB logical indexing. If you do image processing in MATLAB and you're not familiar with logical indexing, stop right now and go read my short tutorial on the subject. We'll wait here until you get back.
• The use of the term "mask" to mean a logical matrix indicating which elements in another matrix satisfy some property.
• The use of binary image dilation (or binary image erosion) to find pixels that are adjacent to a specified set of pixels.
And maybe you need to be comfortable with the idea of applying image processing operators to things that don't appear to be images, such 1-by-11 vectors. :-)
I have used logical indexing and binary image dilation and erosion in countless ways to do useful things in MATLAB, sometimes with images and sometimes not. I strongly recommend that you add these ideas to your bag of tricks.
Published with MATLAB® 7.8
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Debate: Lorentz invariance of certain zero angles
Discussion in 'Formal debates' started by Pete, Nov 25, 2011.
Not open for further replies.
1. TachBannedBanned
Messages:
5,265
Please stop telling me what I am doing. It is very simple, really: t is NOT a function of $\theta$.
$t'=\gamma(t-vx/c^2)$ implies:
$\frac{dt'}{d \theta}=\gamma( \frac{\partial t}{\partial \theta}-v/c^2 \frac {\partial x}{\partial \theta})=-\gamma vr/c^2 sin (\omega t +\theta)$
Nonsense, see above. Please look up the definition of total derivative.
I am not "simply declaring", they are independent variables by virtue of how the rotating wheel is described mathematically. I explained that several times already.
No, I cannot, since there is no definition of how $\theta'$ transforms between frames. Determining this transform is equivalent to the problem that we are trying to solve in this debate.
Last edited: Feb 8, 2012
3. PeteIt's not rocket surgeryRegistered Senior Member
Messages:
10,167
What general rule or definition are you using to conclude that t and $\theta$ are independent?
Would you object if I opened a thread in Physics and Maths about partial differentiation and the independence of variables in this situation?
Please stop ignoring what I point out.
Only if you leave t' out of the equation.
If you include t', then t can potentially be expressed as a function of t' and $\theta$.
Similarly for t'.
In this equation:
$t'=\gamma(t-vr\cos(\omega t + \theta)/c^2)$
t' is dependent on t and $\theta$.
In this equation:
$x' = \gamma r\cos(\omega\gamma t' + \omega\gamma vx'/c^2 + \theta) - v\gamma^2t' - v^2\gamma^2x'/c^2$
t' and $\theta$ are independent.
For any value of $\theta$, t' can have any value, and vice versa.
Wrong.
You have t' as a function of variables t and x, where (it is implied) x is a function of t and $\theta$.
You can't get the total derivative $\frac{dt'}{d \theta}$, unless you define t as a function of $\theta$.
If t and $\theta$ are independent, then the expression $\frac{dt'}{d\theta}$ just doesn't make sense.
You can get the partial derivative by treating t as a constant:
\begin{align} \frac{\partial t'}{\partial \theta} &= -\gamma v/c^2 \frac {\partial x}{\partial \theta} \\ &= -\gamma vr/c^2 sin (\omega t +\theta) \end{align}
Note that $\frac{dt}{d\theta}$ and $\frac{\partial t}{\partial \theta}$ are undefined, unless t is considered constant or a function of $\theta$.
As I said before, if a and b are independent, then $da/db$ is undefined (it is not zero).
You've said it several times, but you haven't explained your logic.
What is it about the wheel equations that make $\theta$ and t independent?
The angle we're chasing at the core of the debate is the angle between the tangent vector and the velocity vector.
But that's different.
$\theta$ is simply a continuous parameter that distinguishes different points around the wheel.
In S, it corresponds to the angle between the x-axis and a point on the wheel at t=0, but that particular definition is incidental to its purpose of acting as a tangent-defining parameter - all that really matters is that a particular value of $\theta$ corresponds to a unique physical wheel element, and that the smaller the difference in value of $\theta$ between two wheel elements, the closer together those wheel elements are.
In S', it can serve the same purpose - a small change in $\theta$ still corresponds to a small change along the wheel, so it still works for the purpose of defining the tangent.
5. TachBannedBanned
Messages:
5,265
Nonsense, t' is a function of t and x whereby x IS a function of $\theta$ and t is NOT a function of $\theta$. Read the example, is self-explanatory.
Err, wrong. $\theta$ happens to be the angle between the positional vector and the x axis. Surely you should have recognized the equation of a circle in polar coordinates by now.
Last edited: Feb 8, 2012
7. PeteIt's not rocket surgeryRegistered Senior Member
Messages:
10,167
What general rule or definition are you using to conclude that t and $\theta$ are independent?
What is it about the wheel equations that make $\theta$ and t independent?
Would you object if I opened a thread in Physics and Maths about partial differentiation and the independence of variables in this situation?
Sorry, I can't make out your intended meaning.
Yes, we have t' as a function of t and x.
Yes, we have x as a function of $\theta$ (and t).
Yes, in this context t is independent of $\theta$.
That means you can't take a total derivative $dt'/d\theta$, because $dt/d\theta$ is undefined:
\begin{align} \frac{dt'}{d\theta} &= \frac{\partial t'}{\partial x}\frac{dx}{d\theta} + \frac{\partial t'}{\partial t}\frac{dt}{d\theta} \\ &= \frac{\partial t'}{\partial x}(\frac{\partial x}{\partial \theta} + \frac{\partial x}{\partial t}\frac{dt}{d\theta}) + \frac{\partial t'}{\partial t}\frac{dt}{d\theta} \end{align}
What example are you referring to?
At t=0 only, like I said in the next sentence. Or if you prefer, it's the angle between the positional vector of a point on the circle and the positional vector of the point which was at (r,0) at t=0.
But that particular definition is incidental to its purpose of acting as a tangent-defining parameter - all that really matters is that a particular value of corresponds to a unique physical wheel element, and that the smaller the difference in value of between two wheel elements, the closer together those wheel elements are.
8. TachBannedBanned
Messages:
5,265
Look, we need to break state, this is not constructive and it is leading nowhere.
I answered this several times, let's see if this explanation registers.
The stationary wheel is described by:
$x=r cos (\theta)$
$y=r sin (\theta)$
When the wheel starts rotating with angular speed $\omega$, each angle changes from $\theta$ to $\omega t +\theta$. There is nothing connecting the angle $\theta$ t the time $t$, hence the two variables are independent.
As an alternative, you can start with the equation of the stationary circle. When the circle rotates wrt frame S, the rotation is expressed as :
$x_s=x cos (\omega t)+ y sin (\omega t)= r cos (-\omega t + \theta)$
$y_s=-x sin (\omega t)+ y cos (\omega t)= r sin (-\omega t + \theta)$
Now, for the calculus part that you seem not to get either:
$f=f(t,x(\theta))$
Therefore:
$\frac{df}{d \theta}=\frac{\partial f}{\partial t} \frac{dt}{d \theta}+\frac{\partial f}{\partial x} \frac{dx}{d \theta}=\frac{\partial f}{\partial t} 0+\frac{\partial f}{\partial x} \frac{dx}{d \theta}$
Contrary to what you might think, $\frac{dt}{d \theta}=0$
Now, having said that, I do not plan to waste any more time on this basic stuff, it is irrelevant any way, since what is really needed, is not $\frac{df}{d \theta}$ but $\frac{df}{d \theta'}$ and I have just figured the way $\theta'$ transforms between frames.
If you want to do something constructive, we can talk about that.
Feel free to do so but count me out, I am not interested in wasting my time. You can start with the example that I have just given you above.
Last edited: Feb 8, 2012
9. PeteIt's not rocket surgeryRegistered Senior Member
Messages:
10,167
Is there any more rigorous way of determining whether two variables are independent?
It seems that you're working on intuition rather than rigor.
So far, the same as my post.
Whoops, wrong.
$\frac{dt}{d \theta}=0$
implies that t is a constant.
If t and $\theta$ are independent, then $\frac{dt}{d \theta}$ is undefined.
Great! Let's see it.
10. TachBannedBanned
Messages:
5,265
I just gave you another proof. Should be rigorous enough.
Nonsense, it implies that t is not a function of $\theta$
I don't know where you get this thing but it is wrong.
11. PeteIt's not rocket surgeryRegistered Senior Member
Messages:
10,167
You said "There is nothing connecting the angle $\theta$ to the time t, hence the two variables are independent."
I don't call that rigorous, but let's explore the concept anyway.
What if we wrote an equation that added something to connect $\theta$ to t?
In that context, they would be dependent.
And what if we then derived a new equation that took the connection away?
In that context, they would be independent.
Well, that's what we did with t'.
It is t that connects t' to $\theta$.
With that connection removed, t' and $\theta$ are independent.
Look at this equation:
$x' = \gamma r\cos(\omega\gamma t' + \omega\gamma vx'/c^2 + \theta) - v\gamma^2t' - v^2\gamma^2x'/c^2$
In that equation, there is nothing connecting the angle $\theta$ to the time t', hence the two variables are independent.
Really?
What happens if you integrate it?
\begin{align} \frac{dt}{d\theta} &= 0 \\ \int \frac{dt}{d\theta} d\theta &= \int 0 d\theta \\ t + C_1 &= C_2 \\ t &= C \end{align}
I really don't understand how you conclude that it should be zero.
Is it something you read in a text?
It doesn't work out from the definition of the derivative - the definition implicitly requires one variable to be dependent on the other.
12. TachBannedBanned
Messages:
5,265
What in the concept of a 2D rotation transformation applied to a circle (expressed in polar coordinates) don't you understand?
Really. I'll give it on more shot:
$f=f(t,x(\theta))$
Then, by definition, $\frac{df}{d \theta}=\frac{\partial f}{\partial x} \frac{dx}{d \theta}$
This is the last time I answer this basic stuff for you, I see that you have a whole thread going. PM me when you are ready to discuss the $\theta$ transformation.
13. PeteIt's not rocket surgeryRegistered Senior Member
Messages:
10,167
One, you're being rude.
Two, I understand it fine.
You don't seem to understand what I'm getting at.
Yes, I agree that it is intuitively obvious that t and $\theta$ are independent.
But I also think it's intuitively obvious that t' and $\theta$ are independent.
So, I'm trying to explore whether we can rigorously say whether two variables are independent.
By what definition?
And why did you ignore this proof:
\begin{align} \frac{dt}{d\theta} &= 0 \\ \int \frac{dt}{d\theta} d\theta &= \int 0 d\theta \\ t + C_1 &= C_2 \\ t &= C \end{align}
Again, rude.
We're supposed to be engaged in mutual exploration, but you don't seem interested in exploring... please, don't let your pride hold you back.
Last edited: Feb 9, 2012
14. TachBannedBanned
Messages:
5,265
...because it is incorrect, you are trying to generalize incorrectly a conclusion from univariate functions to multivariate ones. The correct generalization is:
\begin{align} \frac{dt}{d\theta} &= 0 \\ t=t(u,v,....) \\ \int \frac{dt}{d\theta} d\theta &= \int 0 d\theta \\ t + C_1(u,v,....) &= C_2 \\ t &= C(u,v,...) \end{align}
This might remove your confusion on the subject of total derivatives. It is a rigorous treatment of total derivatives from base principles. I apologize if I came across as rude but so did you, every explanation that I gave you was met with "wrong" , though it was correct.
Last edited: Feb 9, 2012
15. PeteIt's not rocket surgeryRegistered Senior Member
Messages:
10,167
My understanding is that the expression $\frac{dt}{d\theta}$ only makes sense in a context where t depends only on $\theta$ (directly or indirectly).
It seems to me that you are trying to derive a total derivative in a context where you can only derive a partial derivative.
I think this step is incorrect:
This would be correct:
It seems to me that you are at least as confused as I on this topic.
Case 2 starts with t and $\theta$ as independent variables, then treats t as a function of $\theta$, such that:
$t(\theta) = t(\theta + \Delta \theta)$
I don't think that is a valid step. In effect, you are treating t as a constant, not a variable.
Can you show me a similar step in a maths text?
The thing is that you presented essentially the same explanation in several different ways, without exploring my objections. Even now, you're still assuming that your explanations were correct and I just misunderstood them.
You're still using the same assumption that I objected to many posts ago, that if two variables are independent, then one can be treated as a constant function of the other.
You haven't tried to support that assumption, you've just presented it again in a different way.
Last edited: Feb 9, 2012
16. TachBannedBanned
Messages:
5,265
You managed to get this just as wrong as the rest. Case 2 says clearly that t is INDEPENDENT of $\theta$. That means that one can use the general formalism developed for Case 1 and use the fact that $t(\theta+\delta \theta)=t(\theta)=t$ (i.e, that t does NOT depend on $\theta$).
This is precisely why I don't want to waste my time trying to correct your misconceptions. I prepared a rigorous presentation, you did not understand it , yet you keep claiming "wrong".
Last edited: Feb 10, 2012
17. PeteIt's not rocket surgeryRegistered Senior Member
Messages:
10,167
That's what I'm talking about, Tach. Not on.
As I said.
No, that makes no sense.
In Case 2, t is not a function of $\theta$. $t(\theta)$ is meaningless.
I suggest that your time would be more productively spent on testing your own understanding.
I'm tested my own in the [thread=112351]Independent variables and partial differentiation[/thread] thread, and I'm growing more and more suspicious that you just don't know what you're talking about.
I suggest for your consideration the possibility that you may, in fact, be wrong.
-------
This discussion hasn't progressed for a while.
Perhaps we could change tack and examine the low velocity case, i.e. consider $\frac{\partial y'}{partial \theta}\frac{\partial \theta}{\partial x'}$ and $\frac{\partial y'}{partial t'}\frac{\partial t'}{\partial x'}$ under the transforms:
\begin{align}x' &= x - vt \\ y' &= y \\ t' &= t \\ \theta' &= \theta \end{align}
I haven't worked it through yet, but I think we should be able to do it while avoiding the problems that we a bogged down in now.
18. TachBannedBanned
Messages:
5,265
...which is precisely what is expressed in the basic math that you seem totally unable to follow.
I can see that people tried to set you straight and you still don't get it. So, I made an even simpler version , from base principles. If you manage to misunderstand this, we are done, there is no point in continuing.
This is basic stuff, Pete, if you don't get this, then there is no point in continuing.
Not if you don't understand how multivariate functions are differentiated.
$\theta' \ne \theta$, I already told you that.
19. PeteIt's not rocket surgeryRegistered Senior Member
Messages:
10,167
You seem to be agreeing that $t(\theta)$ is meaningless.
Is that what you meant?
James had some of the same thoughts as you, but wasn't definite about it. Trippy made a contribution, but again didn't pretend to be authoritative.
I'm inclined to give most weight to temur, who I believe has the best maths background, but I might be biased because what he says matches what I've been saying to you.
This:
...is actually using the definition of the partial derivative:
(use $x_1 = \theta, x_2 = t$)
In other words, you're setting t constant and varying $\theta$ - that's a partial derivative.
I wouldn't call multivariate calculus basic. There's no shame in learning something new.
Look, you obviously are beyond considering that I could be right... but have you tried checking your understanding against a textbook, or with a professor, or a tutor, or any of the maths people on sciforums?
Or are you so convinced that you're right that you don't feel the need to double check?
We agree on enough for the simpler case, because there is no problem with whether t' is independent - we can use t in both cases.
In the low velocity limit, $\theta'=\theta$.
Look:
\begin{align} x'&=x-vt \\ y'&=y \\ t'&=t \\ \\ r_x &= r \cos (\omega t + \theta) \\ r_y &= r \sin (\omega t + \theta) \\ \\ r_x' &= r \cos (\omega t + \theta) - vt \\ r_y' &= r \sin (\omega t + \theta) \end{align}
In both frames, $\theta$ is a true angle around the wheel axis.
Last edited: Feb 10, 2012
20. TachBannedBanned
Messages:
5,265
...which is the same exact thing as the total derivative in the case when t does not depend of $\theta$. Since you seem to refuse this simple explanation, I gave you the slightly more complicated one, where I took the total derivative while observing that $t(\theta+\Delta \theta)=t(\theta)=t$. You rejected that explanation as well, so I give up, there is no way of convincing you that you are wrong.
Maybe this will work:
$f=f(u(\theta),v,w,...)$
where $v,w,...$ are not functions of $\theta$.
Then, the total derivative of $f$ wrt $\theta$ is:
$\frac{df}{d \theta}=\frac{\partial f}{\partial u} \frac{du}{d \theta}$
It makes no sense to talk about the partial derivative of $f$ wrt $\theta$ (as you've been trying) since $f$ does not depend explicitly of $\theta$.
Now, if $f=f(u(\theta),v (\theta),w,...)$
then:
$\frac{df}{d \theta}=\frac{\partial f}{\partial u} \frac{du}{d \theta}+\frac{\partial f}{\partial v} \frac{dv}{d \theta}$
and so on.
By contrast, if: $f=f(\theta, u(\theta),v,w,...)$
then:
The total derivative wrt $\theta$is:
$\frac{df}{d \theta}=\frac{\partial f}{\partial \theta}+\frac{\partial f}{\partial u} \frac{du}{d \theta}$
while the partial derivative wrt $\theta$ is:
$\frac{\partial f}{\partial \theta}$
Last edited: Feb 10, 2012
21. PeteIt's not rocket surgeryRegistered Senior Member
Messages:
10,167
It seems we are at an impasse in this section.
I suggest we put it on hold for now, and instead consider (your choice):
• A differential approach to the low velocity case, as suggested in post 196, or
• Rindler's proof of angle invariance, last addressed in post 135, or
• My approach to my method, which (as agreed) uses the rod T1 to define the tangent to the wheel, or
22. TachBannedBanned
Messages:
5,265
We are at an impasse because you repeatedly rejected the textbook definition of the Differentiation of Composite Functions. Why don't we wait and see what you have learned on the subject?
Once this is cleared, we can proceed with the correct way of doing the differentiation , not only at low speeds but also at any speed. There is no need to use galilean transforms.
23. PeteIt's not rocket surgeryRegistered Senior Member
Messages:
10,167
Here's the post that began this sidetrack:
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# Order (group theory)
In mathematics, the order of a finite group is the number of its elements. If a group is not finite, one says that its order is infinite. The order of an element of a group (also called period length or period) is the order of the subgroup generated by the element. If the group operation is denoted as a multiplication, the order of an element a of a group, is thus the smallest positive integer m such that am = e, where e denotes the identity element of the group, and am denotes the product of m copies of a. If no such m exists, the order of a is infinite.
Lagrange's theorem states that for any subgroup H of a finite group G, the order of the subgroup divides the order of the group; that is, |H| is a divisor of |G|. In particular, the order |a| of any element is a divisor of |G|.
This group has six elements, so ord(S3) = 6. By definition, the order of the identity, e, is one, since e 1 = e. Each of s, t, and w squares to e, so these group elements have order two: |s| = |t| = |w| = 2. Finally, u and v have order 3, since u3 = vu = e, and v3 = uv = e.
The order of a group G and the orders of its elements give much information about the structure of the group. Roughly speaking, the more complicated the factorization of |G|, the more complicated the structure of G.
The relationship between the two concepts of order is the following: if we write
In general, the order of any subgroup of G divides the order of G. More precisely: if H is a subgroup of G, then
As an immediate consequence of the above, we see that the order of every element of a group divides the order of the group. For example, in the symmetric group shown above, where ord(S3) = 6, the possible orders of the elements are 1, 2, 3 or 6.
The following partial converse is true for finite groups: if d divides the order of a group G and d is a prime number, then there exists an element of order d in G (this is sometimes called Cauchy's theorem). The statement does not hold for composite orders, e.g. the Klein four-group does not have an element of order four). This can be shown by inductive proof.[1] The consequences of the theorem include: the order of a group G is a power of a prime p if and only if ord(a) is some power of p for every a in G.[2]
If a has infinite order, then all non-zero powers of a have infinite order as well. If a has finite order, we have the following formula for the order of the powers of a:
for every integer k. In particular, a and its inverse a−1 have the same order.
Suppose G is a finite group of order n, and d is a divisor of n. The number of order-d-elements in G is a multiple of φ(d) (possibly zero), where φ is Euler's totient function, giving the number of positive integers no larger than d and coprime to it. For example, in the case of S3, φ(3) = 2, and we have exactly two elements of order 3. The theorem provides no useful information about elements of order 2, because φ(2) = 1, and is only of limited utility for composite d such as d=6, since φ(6)=2, and there are zero elements of order 6 in S3.
Group homomorphisms tend to reduce the orders of elements: if fG → H is a homomorphism, and a is an element of G of finite order, then ord(f(a)) divides ord(a). If f is injective, then ord(f(a)) = ord(a). This can often be used to prove that there are no (injective) homomorphisms between two concretely given groups. (For example, there can be no nontrivial homomorphism h: S3 → Z5, because every number except zero in Z5 has order 5, which does not divide the orders 1, 2, and 3 of elements in S3.) A further consequence is that conjugate elements have the same order.
An important result about orders is the class equation; it relates the order of a finite group G to the order of its center Z(G) and the sizes of its non-trivial conjugacy classes:
where the di are the sizes of the non-trivial conjugacy classes; these are proper divisors of |G| bigger than one, and they are also equal to the indices of the centralizers in G of the representatives of the non-trivial conjugacy classes. For example, the center of S3 is just the trivial group with the single element e, and the equation reads |S3| = 1+2+3.
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## DEV Community is a community of 637,088 amazing developers
We're a place where coders share, stay up-to-date and grow their careers.
# Discussion on: Project Euler #2 - Even Fibonacci numbers
## Replies for: Math to the rescue again! To compute the n-th Fibonacci number, you can use the Binet formula: where φ is the Golden Ratio, (1 + √5)/2. Also, it...
Kushan Joshi • Edited
Ah nice to see maths, I came up with the same thing however with a different approach of using generator functions.
Let us forget about fibonacci and think of the even fibonacci as a new series. (I probably think there would be a mathematical way of deducing it, but I just used brute force to find it out)
``````series = 2, 8, 34, 144, 610, 2584...
34 = 8*4 + 2,
144 = 34*4 + 8,
610 = 144*4 + 34;
Number = 4 * (Previous Number) + Previous Previous Number
``````
So the recurrence relation for this series would be
Fn= 4Fn-1 + Fn-2.
To find out the Sum of n elements in this series (let's call this Sn), we can rewrite the recurrence relation like this:
4Fn-1 = Fn - Fn-2
and we can move all n's by 1:
4Fn = Fn+1 - Fn-1
Now we can use the same logic to finder other n's:
4Fn = Fn+1 - Fn-1
4Fn-1 = Fn - Fn-2
4Fn-2 = Fn-1 - Fn-3
4Fn-3 = Fn-2 - Fn-3
...
...
4F4 = F5 - F3
4F3 = F4 - F2
4F2 = F3 - F1
4F1 = 4F1
For folks not familiar with this technique, we are simply doing to cancel out similar terms when we add all equations. (Note all items in pair are canceled out since `x - x = 0`).
4Fn = Fn+1 - Fn-1
4Fn-1 = Fn - Fn-2
4Fn-2 = Fn-1 - Fn-3
4Fn-3 = Fn-2 - Fn-3
...
...
4F4 = F5 - F3
4F3 = F4 - F2
4F2 = F3 - F1
4F1 = 4F1
4Fn + 4Fn-1 + 4Fn-2 + ... + 4F2 + 4F1 = Fn+1 + Fn - F2 + 3F1
If you notice the left hand side is essentially equivalent to
4Sn
Now we can finally write the relation between sum and series as:
Sn= ( Fn+1 + Fn - F2 + 3F1 ) / 4
The point is this avoids the for loop for summing all the values. Go ahead and try substituting values and you will find this sums it up. You will have to use `F(1) = 2` and `F(2) = 8`.
Now that we have removed one `for loop`, how do we get a direct formula for getting the Fn.
This involves a bit more complicated math(for adventurous folks visit austinrochford.com/posts/2013-11-0...).
In short the closed form for this recurrence relation is
And if you compute the roots and follow along the url I posted above, you will end up with this
This gives us a function to compute Fn .
Now we can combine our summing and this formula to get a simple O(1) arithmetic function to get the sum. Side note: Running a for loop will take much less brain time :P
Massimo Artizzu
I applaud your solution, fellow math lover! 👏
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https://homebrew.stackexchange.com/questions/7426/how-quickly-can-you-pull-beer-through-a-jockey-box-with-a-70-coil
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How quickly can you pull beer through a jockey box with a 70' coil?
I'm looking at building a jockey box with a 70', 5/16" (21.3m, 8mm) stainless steel coil (installed in a drum-style cooler.) Seems to be a good compromise between size, weight and cost.
How quickly can I pull cold beer out of it without it foaming?
My back-of-envelope calculation says a 2133.6cm by 0.8cm coil holds π × 0.42 × 2133.6 = 10723 or just over a litre of beer. The Mythbusters say you can cool a six pack of beer in five minutes with salted ice water, so I'd interpolate that one beer per minute would be achievable.
Can anyone confirm my theory with experimental evidence?
• the updated maths were incorrect - you'd used 0.3125cm - it should have been 0.3125in or ~0.8cm (0.793) for the diameter, and then you only need half that squared for computing the area.
– mdma
Aug 10, 2012 at 18:30
1 Answer
The cooling effectiveness of the coil is a function of several things:
• Contact surface area
• Contact time
• The difference in temperature between the warm beer and cool water.
The MoreBeer Draft Box has 50 feet of 3/8" tubing, providing about 700 square inches of contact surface versus your proposed 69 square inches. So I would say you need to run the beer approximately 10x slower than the MoreBeer model to get similar cooling, which may be prohibitively slow.
Also, Mythbusters is a fun show but I wouldn't trust it for any real measurements or proof of anything. They say that a 6-pack of beer cooled in 5 minutes, but how warm was the beer to start? How much cold water did they use? Was there any circulation? Assuming they had enough water, it probably doesn't matter if you're cooling 1 beer or 6, since all the cans have the same volume and surface area, so I don't think you could cool a single can in only 1 minute.
• Looks like there's a mistake - the surface area of 70 feet x 5/16" is 0.3125 x 3.141 x 70 x 12 = 824 sq inches, so larger than the morebeer. I think you forgot to multiply by the 70 ft by 12.
– mdma
Aug 10, 2012 at 14:48
• @mdma: The original poster proposed a 70 inch coil, not 70 feet. Unless that was a typo on Robert's part.
– Hank
Aug 10, 2012 at 16:07
• Ah yes, hehe, or my mistake... I read it as 70 feet, but the OP does write 70 inches, seems very short.
– mdma
Aug 10, 2012 at 16:15
• You're right, it's 70 feet, not 70 inches. I've adjusted the maths. Join us in the 21st century in order to avoid this confusion in the future won't you please, America? :-P Aug 10, 2012 at 17:42
• Sure - I'm from Europe, I would have happily done it in metric if you wrote all lengths in metric in the original post. PS: It still says 70 inches, I'll edit it.
– mdma
Aug 10, 2012 at 18:22
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# (solved)Question 8 SSC-CGL 2020 March 4 Shift 1
A shopkeeper gives two successive discounts of 25% and 20% respectively and still earns 20% profit. If no discount was given by the shopkeeper, then the profit% would have been?
(Rev. 18-Jun-2024)
## Categories | About Hoven's Blog
,
### Question 8SSC-CGL 2020 Mar 4 Shift 1
A shopkeeper gives two successive discounts of 25% and 20% respectively and still earns 20% profit. If no discount was given by the shopkeeper, then the profit% would have been?
### Solution in Detail
Let CP = $\displaystyle 100$
After discounts profit is $\displaystyle 20\%$
$\displaystyle \therefore$ SP = $\displaystyle 100 + 20\% = 120$
Let MP = x
There are two successive discounts
$\displaystyle 25\% \equiv \frac 14, 20\%\equiv \frac 15$
SP after these discounts
$\displaystyle 120 = x (1 - 1/4) ( 1 - 1/5)$
$\displaystyle \therefore x = 200$
We have to calculate his profit if he had sold at this marked price, i.e., without giving any discounts. So take his $\displaystyle SP$ as $\displaystyle = MP = 200$
Profit $\displaystyle = \frac{SP - CP}{CP} \times 100$
$\displaystyle = \frac{200 - 100}{100} \times 100$
$\displaystyle = 100\%\:\underline {Ans}$
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https://www.physicsforums.com/threads/verifying-the-accuracy-of-a-gun-aimed-at-a-falling-target.272347/
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# Verifying the Accuracy of a Gun Aimed at a Falling Target
• TITO09
Newton's First Law states that objects in motion will stay in motion unless acted upon by an external force. Since the bullet and the can are both moving in the y-direction, they will continue to move in that direction unless an external force, such as gravity, acts upon them. Therefore, the statement is verified that the bullet will not miss the can as long as it is fired with enough initial speed to reach the falling target before it hits the ground. In summary, the conversation discusses the idea of a gun being fired directly at a falling can, and whether the bullet will miss or not. By applying Newton's First Law, it is determined that as long as the bullet is fired with enough initial speed, it will reach the falling target
TITO09
## Homework Statement
A gun is aimed directly at a can, which is simultaneously released when the gun is fired. This gun won't miss as long as the initial speed of the bullet is sufficient to reach the falling target before the target hits the floor. Verify this statement.
## Homework Equations
Yo=(x)(tan(angle))
## The Attempt at a Solution
I have no idea how to do this.
TITO09 said:
## Homework Statement
A gun is aimed directly at a can, which is simultaneously released when the gun is fired. This gun won't miss as long as the initial speed of the bullet is sufficient to reach the falling target before the target hits the floor. Verify this statement.
## Homework Equations
Yo=(x)(tan(angle))
## The Attempt at a Solution
I have no idea how to do this.
You already know that the bullet is fired fast enough to get to the can. So, what do you know about the motion of the objects in the y-direction?
the movement in the y-direction is a projectile movement
TITO09 said:
the movement in the y-direction is a projectile movement
Apply Newton's First Law to each of the objects and tell me what you decide.
## 1. How can I verify the accuracy of a gun aimed at a falling target?
To verify the accuracy of a gun aimed at a falling target, you can use a ballistic pendulum or a shooting chronograph. A ballistic pendulum measures the momentum of a projectile by measuring the swing of a pendulum after impact. A shooting chronograph measures the velocity of the projectile as it passes through two sensors. By using either of these methods, you can determine the accuracy of your gun by comparing the actual results with the predicted results based on the gun's specifications.
## 2. What factors can affect the accuracy of a gun aimed at a falling target?
The accuracy of a gun aimed at a falling target can be affected by several factors. Some of these factors include the type and quality of the ammunition used, the condition of the gun, the shooter's skill and technique, environmental conditions such as wind and temperature, and the distance between the shooter and the target.
## 3. Can I use a laser sight to verify the accuracy of my gun aimed at a falling target?
While a laser sight can help improve the accuracy of your gun, it is not an effective method for verifying accuracy. This is because a laser sight only indicates the direction in which the gun is pointed, not the actual impact of the projectile. To verify accuracy, you need to measure the actual impact on the target.
## 4. How many shots should I fire to verify the accuracy of my gun aimed at a falling target?
The number of shots you should fire to verify accuracy depends on the method you are using. If you are using a ballistic pendulum, a minimum of three shots is recommended to obtain an accurate average. If you are using a shooting chronograph, it is recommended to fire at least five shots to account for any outliers and obtain a more precise average.
## 5. Are there any safety precautions I should take when verifying the accuracy of a gun aimed at a falling target?
Yes, there are several safety precautions you should take when verifying the accuracy of a gun aimed at a falling target. First, always ensure that the area is clear of any people or animals. Use a designated shooting range or a safe and controlled environment. Additionally, always follow the proper handling and storage procedures for firearms, and wear protective gear such as eye and ear protection. Finally, make sure to follow all local and state laws and regulations regarding the use of firearms.
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Multiplication of a Matrix Questions and Answers : Quiz 10
MCQ 46: Two matrices A and B are multiplied to get AB if
1. both are rectangular
2. both have same order
3. no of columns of A is equal to columns of B
4. no of rows of A is equal to no of columns of B
MCQ 47: Sin(A+45°)sin(A-45°) =
1. −1/2cos(2A)
2. −1/2sin(2A)
3. 1/2cos(2A)
4. None of Above
MCQ 48: The identity (x+3)(x+4) = x² + 7x + 12 is true for
1. two values of x
2. one value of x
3. all value of x
4. None of Above
MCQ 49: If a < 0, then the function ƒ(x) = a x² + bx + c has
1. maximum value
2. minimum value
3. constant value
4. positive value
MCQ 50: {X | x ε N Ʌ x < 1} is the
1. singleton set
2. The set with two points
3. a set with three points
4. Empty set
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Causation doesn’t imply Correlation either
You may have misread the title as the old correlation does not imply causation mantra, but the opposite is also true! If you don’t believe me, read on…
First I want to provide you with some intuition on what correlation is really all about! For many people (and many of my students for sure) the implications of the following formula for the correlation coefficient of two variables and are not immediately clear:
In fact the most interesting part is this: . We see a product of two differences. The differences consist of the data points minus the respective means (average values): in effect this leads to the origin being moved to the means of both variables (as if you moved the crosshair right into the centre of all data points).
There are now four possible quadrants for every data point: top or bottom, left or right. Top right means that both differences are positive, so the result will be positive too. The same is true for the bottom left quadrant because minus times minus equals plus (it often boils down to simple school maths)! The other two quadrants give negative results because minus times plus and plus times minus equals minus.
After that we sum over all products and normalize them by dividing by the respective standard deviations (how much the data are spread out), so that we will only get values between and .
Let us see this in action with an example. First we define a helper function for visualizing this intuition:
cor.plot <- function(data) {
x_mean <- mean(data[ , 1])
y_mean <- mean(data[ , 2])
plot(data, type = "n") # plot invisibly
limits = par()$usr # limits of plot # plot correlation quadrants rect(x_mean, y_mean, limits[2], limits[4], col = "lightgreen") rect(x_mean, y_mean, limits[1], limits[4], col = "orangered") rect(x_mean, y_mean, limits[1], limits[3], col = "lightgreen") rect(x_mean, y_mean, limits[2], limits[3], col = "orangered") points(data, pch = 16) # plot scatterplot on top colnames(data) <- c("x", "y") # rename cols instead of dynamic variable names in lm abline(lm(y ~ x, data), lwd = 2) # add regression line title(paste("cor =", round(cor(data[1], data[2]), 2))) # add cor as title } Now for the actual example (in fact the same example we had in this post: Learning Data Science: Modelling Basics): age <- c(21, 46, 55, 35, 28) income <- c(1850, 2500, 2560, 2230, 1800) data <- data.frame(age, income) plot(data, pch = 16) cor.plot(data) The correlation is very high because most of the data points are in the positive (green) quadrants and the data is close to its regression line (linear regression and correlation are closely related mathematically). Now, let us get to the actual topic of this post: Causation doesn’t imply Correlation either. What could be “more causal” than a parabolic shot? When you shoot a projectile without air resistance the trajectory will form a perfect parabola! This is in fact rocket science! Let us simulate such a shot and calculate the correlation between time and altitude, two variables that are perfectly causally dependent: t <- c(-30:30) x <- -t^2 data <- data.frame(t, x) plot(data, pch = 16) cor.plot(data) The correlation is exactly zero, zip, nada! And it is clear why: the data points in the positive and in the negative quadrants cancel each other out completely because of the perfect symmetry! This leads us to the following very important insight: Correlation is a measure of linear dependance (and linear dependance only!). Even a strong causal relationship can be overlooked by correlation because of its non-linear nature (as in this case with the quadratic relationship). The following example conveys the same idea in a somewhat more humorous manner – it is the by now infamous datasaurus: library(datasauRus) # on CRAN dino <- datasaurus_dozen[datasaurus_dozen$dataset == "dino", 2:3]
plot(dino, pch = 16, cex = 2)
cor.plot(dino)
As with the above example we can clearly see why the correlation is so low, although there is a whole dinosaur hiding in your data…
The learning is that you should never just blindly trust statistical measures on their own, always visualize your data when possible: there might be some real beauties hiding inside your data, waiting to be discovered…
Understanding AdaBoost – or how to turn Weakness into Strength
Many of you might have heard of the concept “Wisdom of the Crowd”: when many people independently guess some quantity, e.g. the number of marbles in a jar glass, the average of their guesses is often pretty accurate – even though many of the guesses are totally off.
The same principle is at work in so called ensemble methods, like bagging and boosting. If you want to know more about boosting and how to turn pseudocode of a scientific paper into valid R code read on…
We start from an original paper of one of the authors of the first practical boosting algorithm, i.e. AdaBoost: Robert E. Schapire: Explaining AdaBoost. The first sentence of the introduction gives the big idea:
Boosting is an approach to machine learning based on the idea of creating a highly accurate prediction rule by combining many relatively weak and inaccurate rules.
The second page gives the pseudocode of Adaboost…:
Given: where .
Initialize: for .
For :
• Train weak learner using distribution .
• Get weak hypothesis : .
• Aim: select with low weighted error:
• Choose .
• Update, for :
where is a normalization factor (chosen so that will be a distribution).
Output the final hypothesis:
… with some explanation:
[…] we are given labeled training examples where the are in some domain , and the labels . On each round , a distribution is computed as in the figure over the training examples, and a given weak learner or weak learning algorithm is applied to find a weak hypothesis : , where the aim of the weak learner is to find a weak hypothesis with low weighted error relative to . The final or combined hypothesis computes the sign of a weighted combination of weak hypotheses
This is equivalent to saying that is computed as a weighted majority vote of the weak hypotheses where each is assigned weight . ([…] we use the terms “hypothesis” and “classifier” interchangeably.)
So, AdaBoost is adaptive in the sense that subsequent weak learners are tweaked in favor of those instances misclassified by previous ones. But to really understand what is going on my approach has always been that you haven’t really understood something before you didn’t build it yourself…
Perhaps you might want to try to translate the pseudocode into R code before reading on… (to increase your motivation I frankly admit that I also had some errors in my first implementation… which provides a good example of how strong the R community is because I posted it on stackoverflow and got a perfect answer two hours later: What is wrong with my implementation of AdaBoost?
Anyway, here is my implementation (the data can be found here: http://freakonometrics.free.fr/myocarde.csv):
library(rpart)
library(OneR)
maxdepth <- 1
T <- 100 # number of rounds
# Given: (x_1, y_1),...,(x_m, y_m) where x_i element of X, y_i element of {-1, +1}
y <- (myocarde[ , "PRONO"] == "SURVIE") * 2 - 1
x <- myocarde[ , 1:7]
m <- nrow(x)
data <- data.frame(x, y)
# Initialize: D_1(i) = 1/m for i = 1,...,m
D <- rep(1/m, m)
H <- replicate(T, list())
a <- vector(mode = "numeric", T)
set.seed(123)
# For t = 1,...,T
for(t in 1:T) {
# Train weak learner using distribution D_t
# Get weak hypothesis h_t: X -> {-1, +1}
H[[t]] <- rpart(y ~., data = data, weights = D, maxdepth = maxdepth, method = "class")
# Aim: select h_t with low weighted error: e_t = Pr_i~D_t[h_t(x_i) != y_i]
h <- predict(H[[t]], x, type = "class")
e <- sum((h!=y) * D)
# Choose a_t = 0.5 * log((1-e) / e)
a[t] <- 0.5 * log((1-e) / e)
# Update for i = 1,...,m: D_t+1(i) = (D_t(i) * exp(-a_t * y_i * h_t(x_i))) / Z_t
# where Z_t is a normalization factor (chosen so that Dt+1 will be a distribution)
D <- D * exp(-a[t] * y * as.numeric(as.character(h)))
D <- D / sum(D)
}
# Output the final hypothesis: H(x) = sign(sum of a_t * h_t(x) for t=1 to T)
newdata <- x
H_x <- sapply(H, function(x) as.numeric(as.character(predict(x, newdata = newdata, type = "class"))))
H_x <- t(a * t(H_x))
pred <- sign(rowSums(H_x))
eval_model(pred, y)
##
## Confusion matrix (absolute):
## Actual
## Prediction -1 1 Sum
## -1 29 0 29
## 1 0 42 42
## Sum 29 42 71
##
## Confusion matrix (relative):
## Actual
## Prediction -1 1 Sum
## -1 0.41 0.00 0.41
## 1 0.00 0.59 0.59
## Sum 0.41 0.59 1.00
##
## Accuracy:
## 1 (71/71)
##
## Error rate:
## 0 (0/71)
##
## Error rate reduction (vs. base rate):
## 1 (p-value < 2.2e-16)
Let’s compare this with the result from the package JOUSBoost (on CRAN):
library(JOUSBoost)
## JOUSBoost 2.1.0
boost <- adaboost(as.matrix(x), y, tree_depth = maxdepth, n_rounds = T)
pred <- predict(boost, x)
eval_model(pred, y)
##
## Confusion matrix (absolute):
## Actual
## Prediction -1 1 Sum
## -1 29 0 29
## 1 0 42 42
## Sum 29 42 71
##
## Confusion matrix (relative):
## Actual
## Prediction -1 1 Sum
## -1 0.41 0.00 0.41
## 1 0.00 0.59 0.59
## Sum 0.41 0.59 1.00
##
## Accuracy:
## 1 (71/71)
##
## Error rate:
## 0 (0/71)
##
## Error rate reduction (vs. base rate):
## 1 (p-value < 2.2e-16)
As you can see: zero errors as with my implementation. Two additional remarks are in order:
An accuracy of 100% hints at one of the problems of boosting: it is prone to overfitting (see also Learning Data Science: Modelling Basics).
The second problem is the lack of interpretability: whereas decision trees are normally well interpretable ensembles of them are not. This is also known under the name Accuracy-Interpretability Trade-Off (another often used ensemble method is random forests, see also here: Learning Data Science: Predicting Income Brackets).
Hope that this post was helpful for you to understand the widely used boosting methodology better and to see how you can get from pseudocode to valid R code. If you have any questions or feedback please let me know in the comments – Thank you and stay tuned!
From Coin Tosses to p-Hacking: Make Statistics Significant Again!
One of the most notoriously difficult subjects in statistics is the concept of statistical tests. We will explain the ideas behind it step by step to give you some intuition on how to use (and misuse) it, so read on…
Let us begin with some coin tosses and the question how to find out whether a coin is fair, i.e. shows heads and tails with fifty-fifty probability. If you had all the time of the world you could throw it indefinitely often to see where the probabilities stabilize. Yet, in reality we only have some sample of tosses and have to infer results based on this. Because of this the area that deals with those problems is called inferential statistics (also inductive statistics).
So, to keep things simple let us throw our coin only ten times. Where is the point where you would suspect that something is not quite right with this coin? As soon as you don’t get five times heads and five times tails? Or only when you get ten times one side only? Possibly somewhere in between? But where exactly?
The first thing to understand is that this is a somewhat arbitrary decision… but one that should be set in advance and which should adhere to some general standard. The idea is to set some fixed probability with which you compare the probability of the result you see in your experiment (i.e. the number of heads and tails in our example). When you obtain a result that is as improbable or even more so than that pre-fixed probability you conclude that the result happened not just by chance but that something significant happened (e.g. that your coin is unfair). The generally accepted probability below which one speaks of a significant result is 5% (or 0.05), it is therefore called significance level.
After fixing the significance level at 0.05 we have to compare it with the probability of our observed result (or a result that is even more extreme). Let us say that we got nine times one side of the coin and only one time the other. What is the probability of getting this or an even more extreme result? The general probability formula is:
For the numerator we get 22 possibilities for observing the above mentioned or an even more extreme result: two times ten possibilities that each of the ten coins could be the odd one out and two more extreme cases showing only heads or only tails. For the denominator we get possibilities for throwing coins:
22 / 2^10
## [1] 0.02148438
So, the probability is about 2% which is below 5%, so we would conclude that our coin is not fair!
Another possibility is to use the binomial coefficient which gives us the number of possibilities to choose k items out of n items. Conveniently enough the R function is called choose(n, k):
(choose(10, 0) + choose(10, 1) + choose(10, 9) + choose(10, 10))/2^10
## [1] 0.02148438
Or, because the situation is symmetric:
2 * (choose(10, 0) + choose(10, 1))/2^10
## [1] 0.02148438
To make things even simpler we can use the binomial distribution for our calculation, where basically all of the needed binomial coefficients are summed up automatically:
prob <- 0.5 # fair coin
n <- 10 # number of coin tosses
k <- 1 # number of "successes", i.e. how many coins show a different side than the rest
2 * pbinom(k, n, prob)
## [1] 0.02148438
As you can see, we get the same probability value over and over again. If you have understood the line of thought so far you are ready for the last step: performing the statistical test with the binom.test() function:
binom.test(k, n)
##
## Exact binomial test
##
## data: k and n
## number of successes = 1, number of trials = 10, p-value = 0.02148
## alternative hypothesis: true probability of success is not equal to 0.5
## 95 percent confidence interval:
## 0.002528579 0.445016117
## sample estimates:
## probability of success
## 0.1
As you can see, we get the same number again: it is named p-value in the output:
The p-value is the probability of getting the observed – or an even more extreme – result under the condition that the null hypothesis is true.
The only thing that is new here is the null hypothesis. The null hypothesis is just a fancy name for the assumption that in general things are unremarkable, examples could be: our coin is fair, a new medical drug doesn’t work (its effect is just random noise) or one group of students is not better than the other.
So, after having a look at the output we would formally say that we have a significant result and therefore reject the null hypothesis that the coin is fair.
In most real world scenarios the situation is not that clear cut as it is with our coins. Therefore we won’t normally use a binomial test but the workhorse of statistical inference: the famous t-test! On the positive side you can use the t-test in many situations where you don’t have all the information on the underlying population distribution, e.g. its variance. It is based on comparatively mild assumptions, e.g. that the population distribution is normal (which is itself effectively based on the Central Limit Theorem (CLT)) – and even this can be relaxed as long as its variance is finite and when you have big sample sizes. If it is normal though small sample sizes suffice. On the negative side the t-test often only gives approximations of exact tests. Yet the bigger the sample sizes the better the approximation. Enough of the theory, let us perform a t-test with our data. We use the t.test() function for that:
data <- c(rep(0, n-k), rep(1, k))
t.test(data, mu = prob)
##
## One Sample t-test
##
## data: data
## t = -4, df = 9, p-value = 0.00311
## alternative hypothesis: true mean is not equal to 0.5
## 95 percent confidence interval:
## -0.1262157 0.3262157
## sample estimates:
## mean of x
## 0.1
As you can see, we get a different p-value this time but the test is still significant and leads to the right conclusion. As said before, the bigger the sample size the better the approximation.
Let us try two more examples. First we use the inbuilt mtcars dataset to test whether petrol consumption of manual and automatic transmissions differ significantly:
head(mtcars)
## mpg cyl disp hp drat wt qsec vs am gear carb
## Mazda RX4 21.0 6 160 110 3.90 2.620 16.46 0 1 4 4
## Mazda RX4 Wag 21.0 6 160 110 3.90 2.875 17.02 0 1 4 4
## Datsun 710 22.8 4 108 93 3.85 2.320 18.61 1 1 4 1
## Hornet 4 Drive 21.4 6 258 110 3.08 3.215 19.44 1 0 3 1
## Hornet Sportabout 18.7 8 360 175 3.15 3.440 17.02 0 0 3 2
## Valiant 18.1 6 225 105 2.76 3.460 20.22 1 0 3 1
# make new column with better labels
mtcars$trans <- ifelse(mtcars$am == 0, "aut", "man")
mtcars$trans <- factor(mtcars$trans)
mtcars$mpg ## [1] 21.0 21.0 22.8 21.4 18.7 18.1 14.3 24.4 22.8 19.2 17.8 16.4 17.3 15.2 ## [15] 10.4 10.4 14.7 32.4 30.4 33.9 21.5 15.5 15.2 13.3 19.2 27.3 26.0 30.4 ## [29] 15.8 19.7 15.0 21.4 mtcars$trans
## [1] man man man aut aut aut aut aut aut aut aut aut aut aut aut aut aut
## [18] man man man aut aut aut aut aut man man man man man man man
## Levels: aut man
by(mtcars$mpg, mtcars$trans, summary)
## mtcars$trans: aut ## Min. 1st Qu. Median Mean 3rd Qu. Max. ## 10.40 14.95 17.30 17.15 19.20 24.40 ## -------------------------------------------------------- ## mtcars$trans: man
## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 15.00 21.00 22.80 24.39 30.40 33.90
boxplot(mpg ~ trans, data = mtcars)
As we can see the consumption is obviously different – but is it also significantly different at the 5% significance level? Let us find out:
M <- mtcars$trans == "man" #manual mpg_man <- mtcars[M, ]$mpg #consumption of manual transmission
mpg_man
## [1] 21.0 21.0 22.8 32.4 30.4 33.9 27.3 26.0 30.4 15.8 19.7 15.0 21.4
mpg_aut <- mtcars[!M, ]$mpg #consumption of automatic transmission mpg_aut ## [1] 21.4 18.7 18.1 14.3 24.4 22.8 19.2 17.8 16.4 17.3 15.2 10.4 10.4 14.7 ## [15] 21.5 15.5 15.2 13.3 19.2 t.test(mpg_man, mpg_aut) ## ## Welch Two Sample t-test ## ## data: mpg_man and mpg_aut ## t = 3.7671, df = 18.332, p-value = 0.001374 ## alternative hypothesis: true difference in means is not equal to 0 ## 95 percent confidence interval: ## 3.209684 11.280194 ## sample estimates: ## mean of x mean of y ## 24.39231 17.14737 Yes, it is significantly different! And now for our final example, a classic use of the t-test to find out whether a medical drug has a significant effect (we use the inbuilt sleep dataset): sleep ## extra group ID ## 1 0.7 1 1 ## 2 -1.6 1 2 ## 3 -0.2 1 3 ## 4 -1.2 1 4 ## 5 -0.1 1 5 ## 6 3.4 1 6 ## 7 3.7 1 7 ## 8 0.8 1 8 ## 9 0.0 1 9 ## 10 2.0 1 10 ## 11 1.9 2 1 ## 12 0.8 2 2 ## 13 1.1 2 3 ## 14 0.1 2 4 ## 15 -0.1 2 5 ## 16 4.4 2 6 ## 17 5.5 2 7 ## 18 1.6 2 8 ## 19 4.6 2 9 ## 20 3.4 2 10 by(sleep$extra, sleep$group, summary) ## sleep$group: 1
## Min. 1st Qu. Median Mean 3rd Qu. Max.
## -1.600 -0.175 0.350 0.750 1.700 3.700
## --------------------------------------------------------
## sleep$group: 2 ## Min. 1st Qu. Median Mean 3rd Qu. Max. ## -0.100 0.875 1.750 2.330 4.150 5.500 boxplot(extra ~ group, sleep) Again, there is obviously a difference… but is it significant? t.test(extra ~ group, sleep) ## ## Welch Two Sample t-test ## ## data: extra by group ## t = -1.8608, df = 17.776, p-value = 0.07939 ## alternative hypothesis: true difference in means is not equal to 0 ## 95 percent confidence interval: ## -3.3654832 0.2054832 ## sample estimates: ## mean in group 1 mean in group 2 ## 0.75 2.33 The p-value is above 0.05 which means that this time it cannot be ruled out that the difference is random! As you can imagine whenever there is money to be made people get creative on how to game the system… unfortunately statistical tests are no exception. Especially in the medical field huge amounts of money are at stake, so one should be well aware of a manipulation technique called p-hacking (also known under the names data snooping or data dredging)! To summarize our journey into statistical testing so far we have seen that you set a significance level (normally at 0.05) and draw a random sample. You calculate how probable the drawing of this sample (or an even more extreme sample) is and compare it to the significance level. If the probability is below the significance level you say that the test shows a significant result, otherwise you say that one cannot rule out that the difference (if there is one) is just due to chance. Now, if you don’t like the result… why not draw a sample again? And again? And again? This is p-hacking! Just draw a lot of samples and take the one that fits your agenda! Of course this is a stark misuse of the idea of statistical tests. But the fact remains that sometimes samples will be significant just by chance! In the following example we “prove” that homeopathy works after all (as we all know it just doesn’t work beyond the so-called placebo effect as numerous high quality studies over many decades have shown). For that end we just conduct one hundred trials and take one out that fits our bill: p_hack <- function() { homeopathy <- rnorm(20, 1.2, 5) t.test(homeopathy, mu = 1.2)$p.value
}
set.seed(123)
trials <- replicate(100, p_hack())
trials[6] # homeopathy works!
## [1] 0.033461758
# wait... no it doesn't!
trials
## [1] 0.522741341 0.785376466 0.624588751 0.587983281 0.057318309
## [6] 0.033461758 0.876112179 0.461733176 0.495169397 0.519897894
## [11] 0.800638084 0.836667733 0.915272904 0.340430844 0.087656634
## [16] 0.879783049 0.824428157 0.784957654 0.898447738 0.494603038
## [21] 0.279707663 0.438005042 0.886043148 0.158778621 0.167712182
## [26] 0.135880072 0.185375902 0.662883154 0.840370636 0.331157973
## [31] 0.283332330 0.423746570 0.849937345 0.503256673 0.546014504
## [36] 0.725568387 0.727886195 0.167482514 0.586386335 0.493916303
## [41] 0.937060320 0.271651762 0.236448565 0.191925375 0.983841382
## [46] 0.373146285 0.520463358 0.323242616 0.193315250 0.757835487
## [51] 0.429311063 0.284986685 0.868272041 0.844042454 0.885548528
## [56] 0.996021648 0.978855283 0.588192375 0.495521737 0.610192393
## [61] 0.242524547 0.404220265 0.136245145 0.004912541 0.408530447
## [66] 0.458030143 0.784011725 0.357773131 0.818207705 0.698330582
## [71] 0.451268449 0.740943226 0.266786179 0.120894921 0.050307044
## [76] 0.387838555 0.232600995 0.834739682 0.669270240 0.516910985
## [81] 0.273077802 0.291004360 0.369842912 0.132130995 0.454371585
## [86] 0.567029545 0.219163798 0.524362414 0.737950629 0.855945701
## [91] 0.959611992 0.901298484 0.931203165 0.204489683 0.843491761
## [96] 0.567982673 0.329414884 0.107350495 0.911279358 0.696661191
Lo and behold: trial no. 6 shows a significant result – although by design both means are the same (1.2)! So we take this study, publish it and sell lots and lots of useless homeopathy drugs!
How many significant results will we get just by chance on average? Well, exactly the amount of our significance level!
trials <- replicate(1e5, p_hack())
round(length(trials[trials < 0.05]) / length(trials), 2)
## [1] 0.05
As always our friends over at xkcd summarize the situation brilliantly:
Learning R: Permutations and Combinations with base R
The area of combinatorics, the art of systematic counting, is dreaded territory for many people, so let us bring some light into the matter: in this post we will explain the difference between permutations and combinations, with and without repetitions, will calculate the number of possibilities and present efficient R code to enumerate all of them, so read on…
The cases we cover in this post:
• Permutations with repetitions
• Combinations without repetitions
• Permutations (of all elements) without repetitions
We won’t cover permutations without repetition of only a subset nor combinations with repetition here because they are more complicated and would be beyond the scope of this post. We will perhaps cover those in a later post.
In all cases, you can imagine somebody drawing elements from a set and the different ways to do so. Permutation implies that the order does matter, with combinations it does not (e.g. in a lottery it normally does not matter in which order the numbers are drawn). Without repetition simply means that when one has drawn an element it cannot be drawn again, so with repetition implies that it is replaced and can be drawn again.
Let us start with permutations with repetitions: as an example take a combination lock (should be permutation lock really!) where you have three positions with the numbers zero to nine each. We use the expand.grid() function for enumerating all possibilities:
P_wi <- expand.grid(rep(list(0:9), 3))
## Var1 Var2 Var3
## 1 0 0 0
## 2 1 0 0
## 3 2 0 0
## 4 3 0 0
## 5 4 0 0
## 6 5 0 0
tail(P_wi)
## Var1 Var2 Var3
## 995 4 9 9
## 996 5 9 9
## 997 6 9 9
## 998 7 9 9
## 999 8 9 9
## 1000 9 9 9
The formula for calculating the number of permutations is simple for obvious reasons ( is the number of elements to choose from, is the number of actually chosen elements):
In R:
10^3
## [1] 1000
nrow(P_wi)
## [1] 1000
The next is combinations without repetitions: the classic example is a lottery where six out of 49 balls are chosen. We use the combn() function for finding all possibilities:
C_wo <- combn(1:49, 6)
C_wo[ , 1:10]
## [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
## [1,] 1 1 1 1 1 1 1 1 1 1
## [2,] 2 2 2 2 2 2 2 2 2 2
## [3,] 3 3 3 3 3 3 3 3 3 3
## [4,] 4 4 4 4 4 4 4 4 4 4
## [5,] 5 5 5 5 5 5 5 5 5 5
## [6,] 6 7 8 9 10 11 12 13 14 15
To calculate the number of combinations the binomial coefficient is used:
To give you some intuition consider the above example: you have possibilities for choosing the first ball, for the second, for the third and so on up to the sixth ball. So that gives . When you think about it this is the same as because all the coefficients smaller than can be eliminated by reducing the fraction! Now, there are possible positions for the first ball that is drawn, for the second… and so on and because the order doesn’t matter we have to divide by , which gives the binomial coefficient. In R we use the choose() function to calculate it:
choose(49, 6)
## [1] 13983816
ncol(C_wo)
## [1] 13983816
So, you see that the probability of winning the lottery are about the same, no matter whether you play it… or not 😉
Our last case is permutations (of all elements) without repetitions which is also the most demanding one because there is no readily available function in base R. So, we have to write our own:
perm <- function(v) {
n <- length(v)
if (n == 1) v
else {
X <- NULL
for (i in 1:n) X <- rbind(X, cbind(v[i], perm(v[-i])))
X
}
}
As you can see it is a recursive function, to understand recursion read my post: To understand Recursion you have to understand Recursion…. What makes matters a little bit more complicated is that the recursive call is within a for loop.
The idea is to fix one element after the other [for (i in 1:n) and cbind(v[i], ...)] and permute the remaining elements [perm(v[-i])] down to the base case when only one element remains [if (n == 1) v], which cannot be permuted any further. Collect all sets on the respective higher level [X <- (rbind(X, ...)] and return the whole matrix X.
Let us see this in action, as an example we’ll see how many different ways there are of four runners reaching the finishing line:
P_wo <- perm(1:4)
P_wo
## [,1] [,2] [,3] [,4]
## [1,] 1 2 3 4
## [2,] 1 2 4 3
## [3,] 1 3 2 4
## [4,] 1 3 4 2
## [5,] 1 4 2 3
## [6,] 1 4 3 2
## [7,] 2 1 3 4
## [8,] 2 1 4 3
## [9,] 2 3 1 4
## [10,] 2 3 4 1
## [11,] 2 4 1 3
## [12,] 2 4 3 1
## [13,] 3 1 2 4
## [14,] 3 1 4 2
## [15,] 3 2 1 4
## [16,] 3 2 4 1
## [17,] 3 4 1 2
## [18,] 3 4 2 1
## [19,] 4 1 2 3
## [20,] 4 1 3 2
## [21,] 4 2 1 3
## [22,] 4 2 3 1
## [23,] 4 3 1 2
## [24,] 4 3 2 1
After this rather complicated function the calculation of the number of ways is simple, it is just the factorial function (it should again be obvious why):
In R we have the factorial() function:
factorial(4)
## [1] 24
nrow(P_wo)
## [1] 24
As you will see when solving real world problems with R the above functions often come in handy, so you should add them to your ever growing tool set – have fun and stay tuned!
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# SSC CPO : Quantitative Aptitude Quiz | 13- 09 - 18
Mahendra Guru
In SSC exam, quantitative Aptitude section is more scoring and easy, if you know the shorts tricks and formulas of all the topics. So, it is important to know the basic concepts of all the topics so you can apply the short tricks and solve the question with the new concepts in lesser time while giving the quiz. It will help you to score more marks from this section in less time period. Quantitative Aptitude section basically measures your mathematical and calculation approach of solving the question. SSC Quiz of quantitative Aptitude section helps you to analyse your preparation level for upcoming SSC examination. Mahendra Guru provides you Quantitative Aptitude Quiz for SSC examination based on the latest pattern so that you can practice on regular basis. It will definitely help you to score good marks in the exam. It is the most important section for all the govt exams like Insurance, SSC-MTS, SSC CPO, CGL, CHSL, State Level, and other Competitive exams.
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Q.1 – ABCD is a cyclic quadrilateral. A tangent PQ is drawn on the point B of the circle. If ∠DBP = 650 then find ∠BCD-
ABCD एक चक्रीय चतुर्भुज है। एक स्पर्श रेखा PQ वृत्त के बिंदु B पर खींची गई है। यदि ∠ DBP = 6500तो ∠ BCD ज्ञात कीजिये-
(A) 1050
(B) 1150
(C) 1250
(D) 950
Q.2 – Two chords AB and CD of a circle intersect at E such that AE = 2.4 cm, BE = 3.2 cm, and CE = 1.6 cm. The length of DE is-
एक वृत्त की दो जीवा AB और CD, E पर इस प्रकार प्रतिच्छेदित करती है कि AE = 2.4 सेमी, BE = 3.2 सेमी और CE = 1.6 सेमी. DE की लम्बाई क्या है?
(A) 1.6 cm/सेमी.
(B) 3.2 cm/सेमी
(C) 4.8 cm/सेमी
(D) 6.4 cm/सेमी
Q.3 – The areas of two similar triangle are 12 cm2 and 48 cm2 . If the height of the smaller one is 2.1 cm, then the corresponding height of the bigger one is-
दो समरूप त्रिभुजों के क्षेत्रफल 12 सेमी2 और 48 सेमी2 है यदि छोटे त्रिभुज की ऊँचाई 2.1 सेमी है तो बड़े त्रिभुज की संगत ऊँचाई क्या है?
(A) 4.8 cm/सेमी
(B) 8.4 cm/सेमी
(C) 4.2 cm/सेमी
(D) 0.525 cm/सेमी
Q.4 – 10 years ago Ram was 5 times as old as Shyam but 20 year later from now he will be only twice as old as Shyam. How many years old is Shyam?
10 वर्ष पहले राम, श्याम से आयु में 5 गुना था, लेकिन अब से 20 वर्ष बाद वह, श्याम की आयु का केवल दोगुना होगा। श्याम की आयु क्या है?
(A) 20 years/वर्ष
(B) 30 years/वर्ष
(C) 40 years/वर्ष
(D) 50 years/वर्ष
Q.5 – When x40 + 2 is divided by x4 +1, what is the remainder?
जब x40 + 2 को x4 +1 से भग दिया जाता है तो शेषफल क्या है?
(A) 1
(B) 2
(C) 3
(D) 4
Q-6 If x = 2015, y = 2014 and z = 2013, then value of x2+y2 +z2 – xy – yz – zx is
यदि x = 2015, y = 2014 और z = 2013 हो तो x2+y2 +z2 – xy – yz – zx का मान है-
(A) 7.5
(B) 5
(C) 4.5
(D) 3
Q-7 If a+b+c = 0, then the value of (a+b – c)2 + (b+c – a)2 + (c+a – b)2 is -
यदि a+b+c = 0 हो तो (a+b – c)2 + (b+c – a)2 + (c+a – b)2 का मान है-
(A) (a2 + b2 + c2)
(B) 2 (a2 + b2 + c2)
(C) 8 (a2 + b2 + c2)
(D) 4 (a2 + b2 + c2)
Q-8 DE is a tangent to the circum-circle of ΔABC at the vertex A such that DE||BC. If AB = 17 cm., then the length of AC is equal to -
ΔABC के परिवृत्त में DE रेखा A शीर्ष बिन्दु पर इस प्रकार स्पर्श करती है कि DE||BC यदि AB = 17 सेमी. है तो AC की लम्बाई बराबर है-
(A) 20
(B) 18
(C) 15
(D) 17
Q-9 The distance between the centres of two circles with radii 9 cm. and 16 cm. is 25 cm. Then find the length of the common transverse tangent.
दो 9 सेमी. और 16 सेमी. त्रिज्या वाले वृत्तों के केन्द्रों के बीच की दूरी 25 सेमी. है तो उनके उभयनिष्ठ अनुस्पर्श रेखा खण्ड की लम्बाई ज्ञात कीजिए।
(A) 27
(B) 24
(C) 25
(D) 29
Q-10 The sum of two numbers is 232 and their HCF is 29. What is the numbers of such pairs of numbers satisfied the above condition.
दो संख्याओं का योग 232 और उनका म.स.प. 29 है। ऐसी संख्याओं के कितने जोड़े संभव है जो उपरोक्त शर्त को संतुष्ट करेंगे।
(A) 0
(B) 1
(C) 2
(D) 3
Q1. (B)
∠DBQ = 180o – ∠DBP = 180o – 65o = 115o
∠DBP = ∠DCB = 115o (alternate angle/एकान्तर कोण)
∠BCD = 115o
Q2. (C)
AE × BE = CE × DE
2.4 × 3.2 = 1.6 × x
x = = 4.8 cm. /सेमी
Q3. (C)
Q4. (A)
Let the present age of Ram and Shyam be x and y years./माना राम और श्याम की वर्तमान आयु x और y वर्ष हैं |
(x – 10) = 5 (y – 10)
x – 5y = – 40 ________(I)
and/और (x+20) = 2(y+20)
x – 2y = 20 _________(II)
On solving/हल करने पर,
y = 20 years/वर्ष
Q5. (C)
Let/माना f(x) = x40 + 2
Put/रखिये x4 = –1,
f(x) = (–1)10 + 2 = 3
Q6. (D)
Q7. (D)
a + b + c = 0
a + b = - c, b + c = - a, c + a = - b
(a + b - c)2 + (b + c - a)2 + (c + a - b)2
= 4c2 + 4a2 + 4b2 = 4 (a2 + b2 + c2 )
Q8. (D)
OA = OB = OC
AB = BC = AC
AC = 17 cm
Q9. (A) Length of tangent /स्पर्श रेखा की लम्बाई
Q10. (C) Let two number is 29a and 29b
माना दो संख्याएं 29a तथा 29b हैं
29a + 29b = 232
a + b = 232/29 = 8
(a,b)= (1,7) (3,5)
The pair is / युग्म हैं (87, 145) (29,203)
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# Solving for the trajectory of the center of mass
• I
• nDever
In summary, the conversation discusses the physics of a square with mass M that is free to translate and rotate in the xy plane. An angular impulse is generated when a linear impulse is applied at a point above the center of mass, causing both rotational and translational motion. The overall motion of the system can be computed using momentum and angular momentum conservation.
nDever
I'm working on the physics engine component of a game engine I'm building, and I need some guidance with this particular situation.
Consider a square with mass M that is free to translate in the xy plane and free to rotate about any axis perpendicular to the page (Fig. 1)
If a linear impulse J is applied at a point above the center of mass (CM) as shown below, I know there must be some angular impulse (momentary torque) generated since there is a component of J that is perpendicular to the displacement vector from CM. I imagine this angular impulse will tend to rotate the square clockwise.
However, I can also imagine that the CM will also undergo translation since the square is not constrained. How would I go about computing the overall rotational + translational motion of this system?
Delta2
The linear motion is the easiest since overall momentum conservation gives
$$\frac{d\vec p}{dt} = m\dot{\vec v}_{com} = \vec F_{tot}$$
The rotational part can be slightly trickier due to the overall acceleration of the system. You should be able to do it by overall angular momentum, preferably relative to the CoM in the comoving frame.
vanhees71
## 1. What is the center of mass?
The center of mass is the point in an object or system where the mass is evenly distributed in all directions. It is also known as the center of gravity.
## 2. Why is it important to solve for the trajectory of the center of mass?
Solving for the trajectory of the center of mass allows us to predict the motion and behavior of an object or system. This information is crucial in understanding and analyzing physical phenomena.
## 3. How is the trajectory of the center of mass calculated?
The trajectory of the center of mass is calculated using the principles of Newtonian mechanics, specifically the laws of motion and the concept of the center of mass. It involves determining the position, velocity, and acceleration of the center of mass at different points in time.
## 4. What factors affect the trajectory of the center of mass?
The trajectory of the center of mass is affected by various factors such as the mass distribution of the object or system, external forces acting on the object, and the initial conditions of the object's motion.
## 5. Can the trajectory of the center of mass be calculated for any object or system?
Yes, the trajectory of the center of mass can be calculated for any object or system as long as the principles of Newtonian mechanics apply. However, for complex systems, the calculations may become more challenging and may require the use of advanced mathematical techniques.
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How Do You Calculate The Magnification Of A Telescope? (Question)
It is equal to the product of the focal length of the telescope divided by the focal length of the eyepiece. As a matter of thumb, the maximum usable magnification of a telescope is 50 times the aperture in inches of the telescope (or twice its aperture in millimeters).
What is the equation for magnifying a telescope’s magnification power?
• When calculating the magnification of a telescope, the most generally used formula in amateur astronomy is as follows: magnification = focal length of telescope / focal length of eyepiece. Using a 10mm eyepiece in a 1000mm focal length telescope results in a magnification of 100x (1000 / 10 = 100), as an example.
What is the formula for magnification of a telescope?
where M is the magnification of the image and fe is the fraction of the image. The focal length of the objective is denoted by the letter fo (sometimes referred to the telescope focal length).
What is the magnification of a telescope *?
In reality, the magnification provided by a telescope is a connection between two distinct optical systems: the telescope itself and the eyepiece you are using. If you want to know how much power your telescope has, divide the focal length of the telescope (in mm) by the focal length of the eyepiece (in mm).
You might be interested: How Was Galileo'S Telescope Better Tha The Oone First Invented By The Dutch? (Solution)
How is magnification used on a telescope?
Simple division of the focal length of the eyepiece by the focal length of the telescope yields the formula for viewing distance. As an example, dividing a 1000mm telescope by a 10mm eyepiece will result in a 100x magnification result. 1000 divided by ten equals one hundred. This is due to the fact that 10 multiplies by 1000 100 times.
How do you calculate Barlow magnification?
A Barlow telescope works by essentially extending the focal length of the telescope and, as a result, the magnification of the telescope when used with a certain eyepiece. For example, if you use the Ultrascopic 30mm eyepiece in conjunction with a 1,200mm focal length telescope, the combined magnification is 40X (1,200/30=40).
What does 50x magnification mean?
With a telescope, the magnification power is roughly equivalent to the ratio of the size of an item visible inside its eyepiece compared to the size of the same object when examined with the naked eye. For example, while seeing Mars with a magnification of 50x, the red planet will appear 50 times larger than it would appear if you were simply looking at it with your eyes.
What magnification do I need to see the rings of Saturn?
If you use even the tiniest telescope at 25x [25 times the magnification], you should be able to see Saturn’s rings. A decent 3-inch scope at 50x [50 times magnification] can reveal them as a distinct structure that is completely isolated from the orb of the planet on all sides.
What magnification telescope do I need to see planets?
Planetary watchers with years of experience employ 20x to 30x magnification per inch of aperture to view the most planetary detail. Double-star observers can magnify objects up to 50 times per inch (which corresponds to an exit pupil of 12 mm). Beyond that, the vision is hampered by the magnifying power of the telescope and the limits of the human eye.
You might be interested: How Does A Camera Lense Telescope? (Solution)
How good is a 70mm telescope?
It is quite easy to observe every planet in the Solar System using a telescope of 70mm aperture. On the Moon, you will be able to get a close look at the surface and easily discern the majority of its distinguishable features and craters. Mars is going to look fantastic.
What can you see with a 100mm telescope?
To What Can You Look Forward When Using 100mm Telescopes? (With Illustrations)
• When using a 100mm telescope, the greatest magnitude achieved is 13.6. As a point of comparison, the Moon has a magnitude of -12.74 while Mars has a magnitude of -2.6. The Moon is a celestial body. The Moon appears spectacularly in these telescopes, as do Mars, Venus, Jupiter, Saturn, Neptune, Pluto, and the Dwarf Planets.
• Mercury is also visible with these telescopes.
What is the difference between a 10mm and 20mm telescope lens?
The focal length of an eyepiece is the most crucial feature to consider. The result is that a smaller number on an eyepiece corresponds to a greater magnifying power. A 10mm eyepiece would offer two times the magnification of a 20mm eyepiece, and vice versa. Moreover, it implies that the same eyepiece provides variable magnifications when used with different scopes.
What is telescope Barlow?
When used with a telescope, the Barlow is an auxiliary lens system contained within a tube that is mounted in front of the focal point, between the telescope and the eyepiece. It does this by extending the effective focal length of the telescope, which boosts the power of any eyepiece used with the telescope. (
You might be interested: Who Improved The Telescope? (Best solution)
Is a 2x or 3x Barlow lens better?
Simply explained, Barlow lenses are a cost-effective solution to enhance the magnification of your eyepieces without increasing the size of your eyepieces. Their effect is to magnify any eyepiece that is used in conjunction with them by a factor of 2 or 3, depending on the model. As you might assume, a 2x Barlow increases the magnification of your eyepiece by a factor of two, while a 3x Barlow increases it by a factor of three.
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Dictionary
1decimal
mathematics : based on the number 10
Full Definition of DECIMAL
: numbered or proceeding by tens:
a : based on the number 10; especially : expressed in or utilizing a decimal system especially with a decimal point
b : subdivided into 10th or 100th units <decimal coinage>
Examples of DECIMAL
1. In the number 8.901, the 9 is in the first decimal place.
Origin of DECIMAL
French décimal, from Medieval Latin decimalis of a tithe, from Latin decima tithe — more at dime
First Known Use: 1608
Other Mathematics and Statistics Terms
abscissa, denominator, divisor, equilateral, exponent, hypotenuse, logarithm, oblique, radii, rhomb
2decimal
noun
mathematics : a number that is written with a dot between the part of the number that is equal to 1 or more and the part of the number that is less than 1
Full Definition of DECIMAL
: any real number expressed in base 10; especially : decimal fraction
Examples of DECIMAL
1. The number 67.398 is a decimal. It is equal to the whole number 67 plus the decimal.398.
2. Seven-tenths written as a decimal is.7. Seven-tenths written as a fraction is {frac7.
3. The decimal.2 is equal to the fraction {frac2.
1651
Other Number-Related Terms
DECIMALLY Defined for Kids
Definition of DECIMAL for Kids
1
: based on the number 10 : numbered or counting by tens <We use a decimal system of writing numbers.>
2
: expressed in or including a decimal <The decimal form of 1⁄4 is .25.>
noun
Definition of DECIMAL for Kids
: a proper fraction in which the denominator is 10 or 10 multiplied one or more times by itself and is indicated by a point placed at the left of the numerator <the decimal .2=2⁄10, the decimal .25=25⁄100, the decimal .025=25⁄1000>
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## 2440.75 – Around the Reservoir
Amy and Felicia stand at opposite each other across a circular reservoir. At a signal, they start racing around the reservoir in opposite directions. The meet for the first time after Felicia has traveled 100 yards and they meet again 60 yards before Amy has completed her first lap. What is the circumference of the circular path around the reservoir?
Solution
Let $c$ be the circumference of the reservoir; $t_1$ be the time of the first meeting; and $t_2$ be the time of the second meeting. Let $r_A$ and $r_F$ be the speeds of the two women. Then because time equals distance over rate,
$t_1 = \frac{\frac{c}{2}-100}{r_A} = \frac{100}{r_F},$
$t_2 = \frac{c+60}{r_A} = \frac{\frac{c}{2}+60}{r_F}.$
$\frac{r_F}{r_A} = \frac{100}{\frac{c}{2}-100} = \frac{\frac{c}{2}+60}{c-60}.$
$\left( \frac{c}{2}-100 \right) \left(\frac{c}{2}+60\right) = 100 (c-60).$
$c^2 = 480 c,$
so that $c$ must be 480 yards.
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https://www.physicsforums.com/threads/boat-wavelength-oscillation-problem.290493/
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# Boat wavelength oscillation problem
1. Feb 7, 2009
### phys1618
1. The problem statement, all variables and given/known data
a father and son went on a cruise thats 20 ft long. the dad observed the cruise swayed thru 22 complete oscillations in 38 s, and one wave crest pass per/oscillation. the son noticed that each crest required 13s to reach land and the wavelenght is twice the length of the cruise. how far were they from land?
2. Relevant equations
velocity=wavelengh/T
velocity=wavelengh *f
3. The attempt at a solution
for the wavelenghth I have 13.33m because i converted ft into meters. then since they nentioned that the wavelenth is twice the size of the boat.
I wanted to divide the number of complete ossiclation 22 by the 38s, but when it added that it takes 13 seconds for each crest to reach land. I got a little confused..
should i continue to do that or should i figure how many crest had hiut land already??
Last edited: Feb 7, 2009
2. Feb 7, 2009
### LowlyPion
Re: plz help asap!!
How frequently were the wave lengths completing an oscillation?
That gives you the T for 1 oscillation.
Each wavelength is twice the length of the boat which is 20'.
So that tells you the wave velocity doesn't it?
Knowing the velocity and the time to shore, how far again is the shore?
3. Feb 7, 2009
### phys1618
Re: plz help asap!!
yes, that's what i was doing until they told me that it takes 13 second to shore. I didnt know where that time to shore belongs. So i did all that and my answer came to be 299.39m. is that correct?
ThankYou LowlyPion. You have always been a great help and quick too.
thank you thank you. I appreciate it
4. Feb 7, 2009
### LowlyPion
Re: plz help asap!!
Once you have your wave speed, then surely you know that D = v*t
Preserving more precision and noting that the dimensions used were kept in feet and not meters I get slightly more.
13*40/(38/22) = ...
5. Feb 7, 2009
### phys1618
Re: plz help asap!!
yes, keeping in ft, i got 301 .. thank you so much for your help
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# How To Calculate Concrete
Enlarge
## How To Calculate Concrete
This video makes calculating concrete simple. It informs you of the measurements you will need to know to work out the cubic amount of concrete needed.
In this video, I am going to demonstrate how to work out the area of concrete. For this video, I am going to demonstrate it based on building a concrete footing for a brick wall. The same principle would apply if you were going to concrete a driveway or a path.
So, down here, I have my wall, you need to measure past the length of the wall, add two marks on the floor, one here and the brick wall ends here, and then you need to go wider than the wall, so the brick wall is a hundred mil and you need to go past it as is demonstrated by the length of this brick. The reason we go past it is because the footing needs to hold the wall and to support the wall, for this demonstration, I have got the length and then the width and you need to know how deep you are going. With these three measurements, you times them together which gives you the area and I am going to show you how to do that now.
Right, now, we have our three figures, we times the length by the depth and the width, we times these together and it gives us our cubic meters. For instance, if I was going to concrete an area of 0.054 cubic meters, I would need approximately one ton bag of ballast and approximately five 25-kilogram bags of cement.
That is how you calculate concrete. .
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https://testbook.com/question-answer/if-12-315-0-4319-find-the-value-of-10-0002--5c2dba9dfdb8bb5d5d6eb727
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# If (1/2.315) = 0.4319, find the value of (1/0.0002315).
This question was previously asked in
SSC CHSL Previous Paper 59 (Held On: 16 March 2018 Shift 2)
View all SSC CHSL Papers >
1. 4319
2. 2315
3. 431.9
4. 231.5
Option 1 : 4319
Free
SSC CHSL: Full Mock Test
233549
100 Questions 200 Marks 60 Mins
## Detailed Solution
Given,
⇒ (1/2.315) = 0.4319
Now,
∴ (1/0.0002315) = (1/2.315) × 10000 = 0.4319 × 10000 = 4319
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# 12 Aug, 2021
Courses βΈ± Algorithms
You can also access the quiz on Itempool here if you want to practice, or see the notes below for more context.
#### In-class Quiz Questions
π
We continue exploring examples and non-examples of matroids.
1. Let be a undirected and connected graph. Let be the ground set. A subset is independent if the complementary subgraph of is connected. Consider the example shown in the figure below. Do the highlighted edges form an independent set?
1. Yes
2. No
β
Yes
There are no cycles among the highlighted edges.
1. Let be a undirected and connected graph. Let be the ground set. A subset is independent if the complementary subgraph of is connected. Consider the example where the graph G is a star, as shown below. What is the family of independent sets?
1. Empty set.
2. All singleton edges; i.e, the size of the family is 7 for the example above.
3. All subsets of edges belong to the family; i.e, the size of the family is 128 for the example above
4. None of the above
β
Empty set.
There is no edge whose removal keeps connected.
1. Suppose is subset of edges such that is connected. Further, let be a subset of What can you say about ? (Hint: think about in terms of . Is it a subgraph? A supergraph?)
1. Also connected.
2. May or may not be connected.
3. Will not be connected.
4. None of the above.
β
Also connected
is a supergraph of which was already connected.
1. Suppose is a subset of edges such that is connected. Suppose is a subset of edges such that is connected. Further, suppose . What can we say about the number of edges in ? Pick the strongest claim you can make. Hint: since is connected, it must have at least how many edges? Now put two and two together given that .
1. at least
2. at least
3. at least
β
at least
A connected graph with exactly one cycle will have at least edges.
1. Let be an undirected and connected graph. Let be the ground set. A subset is independent if the complementary subgraph of is connected. Does this collection of independent sets form a matroid? Hint: for the exchange axiom, consider that a graph with at least n edges has a cycle.
1. Yes
2. No, it violates the first axiom (empty set membership)
3. No, it violates the second axiom (heredity)
4. No, it violates the third axiom (exchange axiom)
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# Parameter
Parameter from Ancient Greek παρά also “para” meaning “beside, subsidiary” and μέτρον also “metron” meaning “measure”, can be interpreted in mathematics, logic, linguistics, environmental science[1] and other disciplines.
In its common meaning, the term is used to identify a characteristic, a feature, a measurable factor that can help in defining a particular system. It is an important element to take into consideration for the evaluation or for the comprehension of an event, a project or any situation.
## Mathematical functions
In mathematics, statistics, and the mathematical sciences, a parameter (from Ancient Greek παρά “beside” and μέτρον “measure”) is a quantity that serves to relate functions and variables using a common variable when such a relationship would be difficult to explicate with an equation.
Mathematical functions have one or more arguments that are designated in the definition by variables, while their definition can also contain parameters. The variables are mentioned in the list of arguments that the function takes, but the parameters are not. When parameters are present, the definition actually defines a whole family of functions, one for every valid set of values of the parameters. For instance, one could define a general quadratic function by defining
f(x) = ax2 + bx + c;
here, the variable x designates the function argument, but a, b, and c are parameters that determine which quadratic function one is considering. The parameter could be incorporated into the function name to indicate its dependence on the parameter. For instance, one may define the base a logarithm by
$\log_a(x)=\frac{\log(x)}{\log(a)}$
where a is a parameter that indicates which logarithmic function is being used. It is not an argument of the function, and will, for instance, be a constant when considering the derivative $\textstyle\log_a'(x)$.
In some informal situations it is a matter of convention (or historical accident) whether some or all of the symbols in a function definition are called parameters. However, changing the status of symbols between parameter and variable changes the function as a mathematical object. For instance, the notation for the falling factorial power
$n^{\underline k}=n(n-1)(n-2)\cdots(n-k+1)$,
defines a polynomial function of n (when k is considered a parameter), but is not a polynomial function of k (when n is considered a parameter). Indeed, in the latter case, it is only defined for non-negative integer arguments.
In the special case of parametric equations, the independent variables are called the parameters.
### Examples
• In a section on frequently misused words in his book The Writer's Art, James J. Kilpatrick quoted a letter from a correspondent, giving examples to illustrate the correct use of the word parameter:
“ W.M. Woods ... a mathematician ... writes ... "... a variable is one of the many things a parameter is not." ... The dependent variable, the speed of the car, depends on the independent variable, the position of the gas pedal. ”
“ [Kilpatrick quoting Woods] "Now ... the engineers ... change the lever arms of the linkage ... the speed of the car ... will still depend on the pedal position ... but in a ... different manner. You have changed a parameter" ”
• A parametric equaliser is an audio filter that allows the frequency of maximum cut or boost to be set by one control, and the size of the cut or boost by another. These settings, the frequency level of the peak or trough, are two of the parameters of a frequency response curve, and in a two-control equaliser they completely describe the curve. More elaborate parametric equalisers may allow other parameters to be varied, such as skew. These parameters each describe some aspect of the response curve seen as a whole, over all frequencies. A graphic equaliser provides individual level controls for various frequency bands, each of which acts only on that particular frequency band.
• If asked to imagine the graph of the relationship y = ax2, one typically visualizes a range of values of x, but only one value of a. Of course a different value of a can be used, generating a different relation between x and y. Thus a is considered to be a parameter: it is less variable than the variable x or y, but it is not an explicit constant like the exponent 2. More precisely, changing the parameter a gives a different (though related) problem, whereas the variations of the variables x and y (and their interrelation) are part of the problem itself.
• In calculating income based on wage and hours worked (income equals wage multiplied by hours worked), it is typically assumed that the number of hours worked is easily changed, but the wage is more static. This makes 'wage' a parameter, 'hours worked' an independent variable, and 'income' a dependent variable.
### Mathematical models
In the context of a mathematical model, such as a probability distribution, the distinction between variables and parameters was described by Bard as follows:
We refer to the relations which supposedly describe a certain physical situation, as a model. Typically, a model consists of one or more equations. The quantities appearing in the equations we classify into variables and parameters. The distinction between these is not always clear cut, and it frequently depends on the context in which the variables appear. Usually a model is designed to explain the relationships that exist among quantities which can be measured independently in an experiment; these are the variables of the model. To formulate these relationships, however, one frequently introduces "constants" which stand for inherent properties of nature (or of the materials and equipment used in a given experiment). These are the parameters.[2]
### Analytic geometry
In analytic geometry, curves are often given as the image of some function. The argument of the function is invariably called "the parameter". A circle of radius 1 centered at the origin can be specified in more than one form:
• implicit form
x2 + y2 = 1
• parametric form
$(x,y)=(\cos \; t,\sin \; t)$
where t is the parameter.
A somewhat more detailed description can be found at parametric equation.
### Mathematical analysis
In mathematical analysis, integrals dependent on a parameter are often considered. These are of the form
$F(t)=\int_{x_0(t)}^{x_1(t)}f(x;t)\,dx.$
In this formula, t is the argument of the function F, and on the right-hand side the parameter on which the integral depends. When evaluating the integral, t is held constant, and so it is considered a parameter. If we are interested in the value of F for different values of t, we now consider it to be a variable. The quantity x is a dummy variable or variable of integration (confusingly, also sometimes called a parameter of integration).
### Statistics and econometrics
In statistics and econometrics, the probability framework above still holds, but attention shifts to estimating the parameters of a distribution based on observed data, or testing hypotheses about them. In classical estimation these parameters are considered "fixed but unknown", but in Bayesian estimation they are treated as random variables, and their uncertainty is described as a distribution.[citation needed]
It is possible to make statistical inferences without assuming a particular parametric family of probability distributions. In that case, one speaks of non-parametric statistics as opposed to the parametric statistics described in the previous paragraph. For example, Spearman is a non-parametric test as it is computed from the order of the data regardless of the actual values, whereas Pearson is a parametric test as it is computed directly from the data and can be used to derive a mathematical relationship.
Statistics are mathematical characteristics of samples which can be used as estimates of parameters, mathematical characteristics of the populations from which the samples are drawn. For example, the sample mean ($\overline X$) can be used as an estimate of the mean parameter (μ) of the population from which the sample was drawn.
In short, parameter can be defined as the numerical summary of a population.
### Probability theory
These traces all represent Poisson distributions, but with different values for the parameter λ
In probability theory, one may describe the distribution of a random variable as belonging to a family of probability distributions, distinguished from each other by the values of a finite number of parameters. For example, one talks about "a Poisson distribution with mean value λ". The function defining the distribution (the probability mass function) is:
$f(k;\lambda)=\frac{e^{-\lambda} \lambda^k}{k!}.$
This example nicely illustrates the distinction between constants, parameters, and variables. e is Euler's Number, a fundamental mathematical constant. The parameter λ is the mean number of observations of some phenomenon in question, a property characteristic of the system. k is a variable, in this case the number of occurrences of the phenomenon actually observed from a particular sample. If we want to know the probability of observing k1 occurrences, we plug it into the function to get f(k1;λ). Without altering the system, we can take multiple samples, which will have a range of values of k, but the system will always be characterized by the same λ.
For instance, suppose we have a radioactive sample that emits, on average, five particles every ten minutes. We take measurements of how many particles the sample emits over ten-minute periods. The measurements will exhibit different values of k, and if the sample behaves according to Poisson statistics, then each value of k will come up in a proportion given by the probability mass function above. From measurement to measurement, however, λ remains constant at 5. If we do not alter the system, then the parameter λ is unchanged from measurement to measurement; if, on the other hand, we modulate the system by replacing the sample with a more radioactive one, then the parameter λ would increase.
Another common distribution is the normal distribution, which has as parameters the mean μ and the variance σ².
In these above examples, the distributions of the random variables are completely specified by the typpe of distribution, i.e. Poisson or normal, and the parameter values, i.e. mean and variance. In such a case, we have a parameterized distribution.a
It is possible to use the sequence of moments (mean, mean square, ...) or cumulants (mean, variance, ...) as parameters for a probability distribution: see Statistical parameter.
## Computer science
When the terms formal parameter and actual parameter are used, they generally correspond with the definitions used in computer science. In the definition of a function such as
f(x) = x + 2,
x is a formal parameter. When the function is used as in
y = f(3) + 5 or just the value of f(3),
3 is the actual parameter value that is substituted for the formal parameter in the function definition. These concepts are discussed in a more precise way in functional programming and its foundational disciplines, lambda calculus and combinatory logic.
In computing, parameters are often called arguments, and the two words are used interchangeably. However, some computer languages such as C define argument to mean actual parameter (i.e., the value), and parameter to mean formal parameter.
## Engineering
In engineering (especially involving data acquisition) the term parameter sometimes loosely refers to an individual measured item. This usage isn't consistent, as sometimes the term channel refers to an individual measured item, with parameter referring to the setup information about that channel.
"Speaking generally, properties are those physical quantities which directly describe the physical attributes of the system; parameters are those combinations of the properties which suffice to determine the response of the system. Properties can have all sorts of dimensions, depending upon the system being considered; parameters are dimensionless, or have the dimension of time or its reciprocal."[3]
The term can also be used in engineering contexts, however, as it is typically used in the physical sciences.
## Environmental science
In environmental science and particularly in chemistry and microbiology, a parameter is used to describe a discrete chemical or microbiological entity which can be assigned a value which is commonly a concentration. The value may also be a logical entity (present or absent), a statistical result such as a 95%ile value or in some cases a subjective value
## Linguistics
Within linguistics, the word "parameter" is almost exclusively used to denote a binary switch in a Universal Grammar within a Principles and Parameters framework.
## Logic
In logic, the parameters passed to (or operated on by) an open predicate are called parameters by some authors (e.g., Prawitz, "Natural Deduction"; Paulson, "Designing a theorem prover"). Parameters locally defined within the predicate are called variables. This extra distinction pays off when defining substitution (without this distinction special provision has to be made to avoid variable capture). Others (maybe most) just call parameters passed to (or operated on by) an open predicate variables, and when defining substitution have to distinguish between free variables and bound variables.
## References
1. ^ Parameter from the Free Dictionary
2. ^ Bard, Yonathan (1974). Nonlinear Parameter Estimation. New York: Academic Press. p. 11. ISBN 0120782502.
3. ^ Trimmer, John D. (1950). Response of Physical Systems. New York: Wiley. p. 13.
Wikimedia Foundation. 2010.
### Look at other dictionaries:
• Parameter — Parameter, im allgemeinen jede Größe, die in einer Funktion neben den. Veränderlichen enthalten ist. Bei einem Kegelschnitt (s.d.) ist der Parameter die Länge der im Brennpunkt auf der Hauptachse senkrechten Sehne. Der Halbparameter ist der… … Lexikon der gesamten Technik
• parameter — In technical use, a parameter is a measurable factor that contributes to determining a system or event. In the 20c it developed rapidly into the kind of word that Fowler (1926) described as a ‘popularized technicality’, many instances of which he … Modern English usage
• Parameter — Pa*ram e*ter, n. [Pref. para + meter: cf. F. param[ e]tre.] 1. A constant number which is part of a theory, function, or calculation, whose value is not determined by the form of the theory or equation itself, and may in some cases be arbitrary… … The Collaborative International Dictionary of English
• parameter — (n.) 1650s, from Mod.L. parameter (1630s), from Gk. para beside, subsidiary + metron measure (see METER (Cf. meter) (n.2)). A geometry term until 1920s when it yielded sense of measurable factor which helps to define a particular system (1927).… … Etymology dictionary
• Paramĕter — (gr.), jede Constante, welche in der Gleichung einer geraden Linie vorkommt. P. einer Parabel, die durch den Brennpunkt der Parabel gelegte, auf der Achse senkrecht stehende Sehne. Ist also y2 = px die Scheitelgleichung der Parabel für… … Pierer's Universal-Lexikon
• Parameter — (griech., Nebenmaß) einer Funktion, Gleichung, Kurve oder Fläche heißt jede unbestimmte Konstante, von der die Funktion etc. abhängt und durch deren verschiedene Wahl sich die Gestalt der Funktion etc. ändert. Über P. bei den Kegelschnitten s. d … Meyers Großes Konversations-Lexikon
• Parameter — Paramĕter (grch.), jede in der Gleichung einer Kurve vorkommende Konstante; bei einem Kegelschnitt die durch den Brennpunkt gehende, auf der Hauptachse senkrechte Sehne … Kleines Konversations-Lexikon
• Parameter — Parameter, in der Geometrie doppelte Ordinate, die durch den Brennpunkt eines Kegelschnitts geht. Vergl. Ellipse, Hyperbel, Parabel … Herders Conversations-Lexikon
• parameter — index guideline Burton s Legal Thesaurus. William C. Burton. 2006 … Law dictionary
• Parameter — (matematik), når man i analytisk geometri lader nogle af konstanterne i ligningen for en kurve eller flade efterhånden antager alle mulige værider, vil ligningen komme til at fremstille uendelig mange kurver eller flader med visse fælles… … Danske encyklopædi
• parameter — [n] limit constant, criterion, framework, guideline, limitation, restriction, specification; concept 688 … New thesaurus
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# Resources tagged with: Mathematical reasoning & proof
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### Cows and Sheep
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Use your logical reasoning to work out how many cows and how many sheep there are in each field.
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If you can copy a network without lifting your pen off the paper and without drawing any line twice, then it is traversable. Decide which of these diagrams are traversable.
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You have been given nine weights, one of which is slightly heavier than the rest. Can you work out which weight is heavier in just two weighings of the balance?
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Look at what happens when you take a number, square it and subtract your answer. What kind of number do you get? Can you prove it?
### Online
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A game for 2 players that can be played online. Players take it in turns to select a word from the 9 words given. The aim is to select all the occurrences of the same letter.
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Can you cross each of the seven bridges that join the north and south of the river to the two islands, once and once only, without retracing your steps?
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From a group of any 4 students in a class of 30, each has exchanged Christmas cards with the other three. Show that some students have exchanged cards with all the other students in the class. How. . . .
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In how many distinct ways can six islands be joined by bridges so that each island can be reached from every other island...
### Tri-colour
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Six points are arranged in space so that no three are collinear. How many line segments can be formed by joining the points in pairs?
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Is it possible to rearrange the numbers 1,2......12 around a clock face in such a way that every two numbers in adjacent positions differ by any of 3, 4 or 5 hours?
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Toni Beardon has chosen this article introducing a rich area for practical exploration and discovery in 3D geometry
### Sprouts Explained
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This article invites you to get familiar with a strategic game called "sprouts". The game is simple enough for younger children to understand, and has also provided experienced mathematicians with. . . .
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Eight children enter the autumn cross-country race at school. How many possible ways could they come in at first, second and third places?
### Sticky Numbers
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### What Numbers Can We Make?
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Imagine we have four bags containing a large number of 1s, 4s, 7s and 10s. What numbers can we make?
### Advent Calendar 2011 - Secondary
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Advent Calendar 2011 - a mathematical activity for each day during the run-up to Christmas.
### Three Neighbours
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Look at three 'next door neighbours' amongst the counting numbers. Add them together. What do you notice?
### What Numbers Can We Make Now?
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Imagine we have four bags containing numbers from a sequence. What numbers can we make now?
### Reasoning: the Journey from Novice to Expert (article)
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This article for primary teachers suggests ways in which we can help learners move from being novice reasoners to expert reasoners.
### Always, Sometimes or Never?
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Are these statements relating to odd and even numbers always true, sometimes true or never true?
### Reasoning: Identifying Opportunities (article)
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In this article for primary teachers we consider in depth when we might reason which helps us understand what reasoning 'looks like'.
### Always, Sometimes or Never? Number
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### Cycle It
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Carry out cyclic permutations of nine digit numbers containing the digits from 1 to 9 (until you get back to the first number). Prove that whatever number you choose, they will add to the same total.
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Some puzzles requiring no knowledge of knot theory, just a careful inspection of the patterns. A glimpse of the classification of knots and a little about prime knots, crossing numbers and. . . .
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Pick the number of times a week that you eat chocolate. This number must be more than one but less than ten. Multiply this number by 2. Add 5 (for Sunday). Multiply by 50... Can you explain why it. . . .
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This addition sum uses all ten digits 0, 1, 2...9 exactly once. Find the sum and show that the one you give is the only possibility.
### Always the Same
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Arrange the numbers 1 to 16 into a 4 by 4 array. Choose a number. Cross out the numbers on the same row and column. Repeat this process. Add up you four numbers. Why do they always add up to 34?
### Flight of the Flibbins
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Blue Flibbins are so jealous of their red partners that they will not leave them on their own with any other bue Flibbin. What is the quickest way of getting the five pairs of Flibbins safely to. . . .
### Geometry and Gravity 2
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This is the second of two articles and discusses problems relating to the curvature of space, shortest distances on surfaces, triangulations of surfaces and representation by graphs.
### Dicing with Numbers
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In how many ways can you arrange three dice side by side on a surface so that the sum of the numbers on each of the four faces (top, bottom, front and back) is equal?
### More Mathematical Mysteries
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Write down a three-digit number Change the order of the digits to get a different number Find the difference between the two three digit numbers Follow the rest of the instructions then try. . . .
### What Do You Need?
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Four of these clues are needed to find the chosen number on this grid and four are true but do nothing to help in finding the number. Can you sort out the clues and find the number?
### Tower of Hanoi
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The Tower of Hanoi is an ancient mathematical challenge. Working on the building blocks may help you to explain the patterns you notice.
### Aba
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In the following sum the letters A, B, C, D, E and F stand for six distinct digits. Find all the ways of replacing the letters with digits so that the arithmetic is correct.
### Eleven
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Replace each letter with a digit to make this addition correct.
### Seven Squares - Group-worthy Task
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Choose a couple of the sequences. Try to picture how to make the next, and the next, and the next... Can you describe your reasoning?
### Not Necessarily in That Order
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Baker, Cooper, Jones and Smith are four people whose occupations are teacher, welder, mechanic and programmer, but not necessarily in that order. What is each person’s occupation?
### Even So
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Find some triples of whole numbers a, b and c such that a^2 + b^2 + c^2 is a multiple of 4. Is it necessarily the case that a, b and c must all be even? If so, can you explain why?
### Convex Polygons
##### Age 11 to 14 Challenge Level:
Show that among the interior angles of a convex polygon there cannot be more than three acute angles.
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| 4.3125 | 4 |
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| 0.885323 |
https://edurev.in/p/349641/RD-Sharma-Solutions-Polynomials
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RD Sharma Solutions: Polynomials
# Polynomials RD Sharma Solutions | Mathematics (Maths) Class 10 PDF Download
``` Page 1
Exercise 2.1
1. Find the zeroes of each of the following quadratic polynomials and verify the relationship
between the zeroes and their co efficient:
(i) f(x) = ?? 2
- 2?? - 8
(ii) g(s) = 4?? 2
- 4?? + 1
(iii) h(t) = ?? 2
- 15
(iv) p(x) = ?? 2
+ 2v2?? + 6
(v) q(x) = v3?? 2
+ 10?? + 7v3
(vi) f(x) = ?? 2
- ( v3 + 1) ?? + v3
(vii) g(x) = ?? ( ?? 2
+ 1)- ?? ( ?? 2
+ 1)
(viii) 6?? 2
- 3 - 7?? Sol:
(i) f(x) = ?? 2
- 2?? - 8
?? ( ?? ) = ?? 2
- 2?? - 8 = ?? 2
- 4?? + 2?? - 8
= ?? ( ?? - 4)+ 2( ?? - 4)
= ( ?? + 2) ( ?? - 4)
Zeroes of the polynomials are -2 and 4
Sum of the zeroes =
- ???? ?????????????????? ???? ?? ???? ?????????????????? ???? ??
-2 + 4 =
-( -2)
1
2 = 2
Product of the zeroes =
???????????????? ???????? ???? ?????????????????? ???? ?? 2
= 24 =
-8
1
- 8 = -8
? Hence the relationship verified
(ii) 9( 5) = 45 - 45 + 1 = 45
2
- 25 - 25 + 1 = 25( 25 - 1)- 1( 25 - 1)
= ( 25 - 1) ( 25 - 1)
Zeroes of the polynomials are
1
2
?????? 1
2
Sum of zeroes =
- ???? ?????????????????? ???? ?? ???? ?????????????????? ???? ?? 2
1
2
+
1
2
=
-( -4)
4
1 = 1
Product of the zeroes =
???????????????? ???????? ???? ?????????????????? ???? ?? 2
1
2
×
1
2
=
1
4
?
1
4
=
1
4
? Hence the relationship verified.
(iii) h(t) = ?? 2
- 15 = ( ?? 2
)- ( v15)
2
= ( ?? + v15) ( ?? - v15)
zeroes of the polynomials are -v15 ?????? v15
sum of zeroes = 0
-v15 + v15 = 0
0 = 0
Page 2
Exercise 2.1
1. Find the zeroes of each of the following quadratic polynomials and verify the relationship
between the zeroes and their co efficient:
(i) f(x) = ?? 2
- 2?? - 8
(ii) g(s) = 4?? 2
- 4?? + 1
(iii) h(t) = ?? 2
- 15
(iv) p(x) = ?? 2
+ 2v2?? + 6
(v) q(x) = v3?? 2
+ 10?? + 7v3
(vi) f(x) = ?? 2
- ( v3 + 1) ?? + v3
(vii) g(x) = ?? ( ?? 2
+ 1)- ?? ( ?? 2
+ 1)
(viii) 6?? 2
- 3 - 7?? Sol:
(i) f(x) = ?? 2
- 2?? - 8
?? ( ?? ) = ?? 2
- 2?? - 8 = ?? 2
- 4?? + 2?? - 8
= ?? ( ?? - 4)+ 2( ?? - 4)
= ( ?? + 2) ( ?? - 4)
Zeroes of the polynomials are -2 and 4
Sum of the zeroes =
- ???? ?????????????????? ???? ?? ???? ?????????????????? ???? ??
-2 + 4 =
-( -2)
1
2 = 2
Product of the zeroes =
???????????????? ???????? ???? ?????????????????? ???? ?? 2
= 24 =
-8
1
- 8 = -8
? Hence the relationship verified
(ii) 9( 5) = 45 - 45 + 1 = 45
2
- 25 - 25 + 1 = 25( 25 - 1)- 1( 25 - 1)
= ( 25 - 1) ( 25 - 1)
Zeroes of the polynomials are
1
2
?????? 1
2
Sum of zeroes =
- ???? ?????????????????? ???? ?? ???? ?????????????????? ???? ?? 2
1
2
+
1
2
=
-( -4)
4
1 = 1
Product of the zeroes =
???????????????? ???????? ???? ?????????????????? ???? ?? 2
1
2
×
1
2
=
1
4
?
1
4
=
1
4
? Hence the relationship verified.
(iii) h(t) = ?? 2
- 15 = ( ?? 2
)- ( v15)
2
= ( ?? + v15) ( ?? - v15)
zeroes of the polynomials are -v15 ?????? v15
sum of zeroes = 0
-v15 + v15 = 0
0 = 0
Product of zeroes =
-15
1
-v15 × v15 = -15
-15 = -15
? Hence the relationship verified.
(iv) p(x) = ?? 2
+ 2v2?? - 6 = ?? 2
+ 3v2?? + v2 × 3v2
= ?? ( ?? + 3v2)- v2( 2 + 3v2) = ( ?? - v2) ( ?? + 3v2)
Zeroes of the polynomial are 3v2 and -3v2
Sum of the zeroes =
-3v2
1
v2 - 3v2 = -2v2
-2v2 = -2v2
?????????????? ???? ???????????? ? v2 × -3v2 = -
6
1
-6 = -6
?????????? ?? h?? ???????????????? h???? ????????????????
(v) 2(x) = v3?? 2
+ 10?? + 7v3 = v3?? 2
+ 7?? + 3?? + 7v3
= v3?? ( ?? + v3)+ 7( ?? + v3)
= ( v3?? + 7) ( ?? + v3)
Zeroes of the polynomials are -v3,
-7
v3
Sum of zeroes =
-10
v3
? -v3 -
7
v3
=
-10
v3
?
-10
v3
=
-10
v3
Product of zeroes =
7v3
3
?
v3?? -7
v30
= 7
? 7 = 7
Hence, relationship verified.
(vi) f(x) = ?? 2
- ( v3 + 1) ?? + v3 = ?? 2
- v3?? - ?? + v3
= x (x - v3) – 1 (x - v3)
= (x – 1) (x - v3)
Zeroes of the polynomials are 1 and v3
Sum of zeroes =
-{?????????????????????? ???? ?? }
???? ?????????????????? ???? ?? 2
=
-[-v3-1]
1
1 + v3 = v3 + 1
Product of zeroes =
???????????????? ???????? ???? ?????????????????? ???? ?? 2
=
v3
1
1 × v3 = v3 = v3 = v3
? Hence, relationship verified
(vii) g(x) = ?? [( ?? 2
+ 1)- ?? ( ?? 2
+ 1) ]
2
= ?? ?? 2
+ ?? - ?? 2
?? - ??
= ?? ?? 2
- [( ?? 2
+ 1)- ?? ] + 0 = ?? ?? 2
- ?? 2
?? - ?? + ??
Page 3
Exercise 2.1
1. Find the zeroes of each of the following quadratic polynomials and verify the relationship
between the zeroes and their co efficient:
(i) f(x) = ?? 2
- 2?? - 8
(ii) g(s) = 4?? 2
- 4?? + 1
(iii) h(t) = ?? 2
- 15
(iv) p(x) = ?? 2
+ 2v2?? + 6
(v) q(x) = v3?? 2
+ 10?? + 7v3
(vi) f(x) = ?? 2
- ( v3 + 1) ?? + v3
(vii) g(x) = ?? ( ?? 2
+ 1)- ?? ( ?? 2
+ 1)
(viii) 6?? 2
- 3 - 7?? Sol:
(i) f(x) = ?? 2
- 2?? - 8
?? ( ?? ) = ?? 2
- 2?? - 8 = ?? 2
- 4?? + 2?? - 8
= ?? ( ?? - 4)+ 2( ?? - 4)
= ( ?? + 2) ( ?? - 4)
Zeroes of the polynomials are -2 and 4
Sum of the zeroes =
- ???? ?????????????????? ???? ?? ???? ?????????????????? ???? ??
-2 + 4 =
-( -2)
1
2 = 2
Product of the zeroes =
???????????????? ???????? ???? ?????????????????? ???? ?? 2
= 24 =
-8
1
- 8 = -8
? Hence the relationship verified
(ii) 9( 5) = 45 - 45 + 1 = 45
2
- 25 - 25 + 1 = 25( 25 - 1)- 1( 25 - 1)
= ( 25 - 1) ( 25 - 1)
Zeroes of the polynomials are
1
2
?????? 1
2
Sum of zeroes =
- ???? ?????????????????? ???? ?? ???? ?????????????????? ???? ?? 2
1
2
+
1
2
=
-( -4)
4
1 = 1
Product of the zeroes =
???????????????? ???????? ???? ?????????????????? ???? ?? 2
1
2
×
1
2
=
1
4
?
1
4
=
1
4
? Hence the relationship verified.
(iii) h(t) = ?? 2
- 15 = ( ?? 2
)- ( v15)
2
= ( ?? + v15) ( ?? - v15)
zeroes of the polynomials are -v15 ?????? v15
sum of zeroes = 0
-v15 + v15 = 0
0 = 0
Product of zeroes =
-15
1
-v15 × v15 = -15
-15 = -15
? Hence the relationship verified.
(iv) p(x) = ?? 2
+ 2v2?? - 6 = ?? 2
+ 3v2?? + v2 × 3v2
= ?? ( ?? + 3v2)- v2( 2 + 3v2) = ( ?? - v2) ( ?? + 3v2)
Zeroes of the polynomial are 3v2 and -3v2
Sum of the zeroes =
-3v2
1
v2 - 3v2 = -2v2
-2v2 = -2v2
?????????????? ???? ???????????? ? v2 × -3v2 = -
6
1
-6 = -6
?????????? ?? h?? ???????????????? h???? ????????????????
(v) 2(x) = v3?? 2
+ 10?? + 7v3 = v3?? 2
+ 7?? + 3?? + 7v3
= v3?? ( ?? + v3)+ 7( ?? + v3)
= ( v3?? + 7) ( ?? + v3)
Zeroes of the polynomials are -v3,
-7
v3
Sum of zeroes =
-10
v3
? -v3 -
7
v3
=
-10
v3
?
-10
v3
=
-10
v3
Product of zeroes =
7v3
3
?
v3?? -7
v30
= 7
? 7 = 7
Hence, relationship verified.
(vi) f(x) = ?? 2
- ( v3 + 1) ?? + v3 = ?? 2
- v3?? - ?? + v3
= x (x - v3) – 1 (x - v3)
= (x – 1) (x - v3)
Zeroes of the polynomials are 1 and v3
Sum of zeroes =
-{?????????????????????? ???? ?? }
???? ?????????????????? ???? ?? 2
=
-[-v3-1]
1
1 + v3 = v3 + 1
Product of zeroes =
???????????????? ???????? ???? ?????????????????? ???? ?? 2
=
v3
1
1 × v3 = v3 = v3 = v3
? Hence, relationship verified
(vii) g(x) = ?? [( ?? 2
+ 1)- ?? ( ?? 2
+ 1) ]
2
= ?? ?? 2
+ ?? - ?? 2
?? - ??
= ?? ?? 2
- [( ?? 2
+ 1)- ?? ] + 0 = ?? ?? 2
- ?? 2
?? - ?? + ??
= ???? ( ?? - ?? )- 1( ?? - ?? ) = ( ?? - ?? ) ( ???? - 1)
Zeroes of the polynomials =
1
?? ?????? ??
Sum of the zeroes =
-[-?? 2
-1]
??
?
1
?? + ?? =
?? 2
+1
?? ?
?? 2
+1
?? =
?? 2
+1
??
Product of zeroes =
?? ??
?
1
?? × ?? =
?? ?? ?
?? 2
+1
?? =
?? 2
+1
??
Product of zeroes =
?? ?? ? 1 = 1
Hence relationship verified
(viii) 6?? 2
- 3 - 7?? = 6?? 2
- 7?? - 3 = ( 3?? + 11) ( 2?? - 3)
Zeroes of polynomials are +
3
2
?????? -1
3
Sum of zeroes =
-1
3
+
3
2
=
7
6
=
-( -7)
6
=
-( ???? ?????????????????? ???? ?? )
???? ?????????????????? ???? ?? 2
Product of zeroes =
-1
3
×
3
2
=
-1
2
=
-3
6
=
???????????????? ???????? ???? ?????????????????? ???? ?? 2
? Hence, relationship verified.
2. If ?? and ?? are the zeros of the quadratic polynomial f(x) = ax
2
+ bx + c, then evaluate:
(i) ?? - ??
(ii)
1
?? -
1
??
(iii)
1
?? +
1
?? - 2?? ??
(iv) ?? 2
?? + ?? ?? 2
(v) ?? 4
+ ?? 4
(vi)
1
???? +?? +
1
???? +??
(vii)
?? ???? +?? +
?? ???? +??
(viii) ?? [
?? 2
?? +
?? 2
?? ] +
?? [
?? ?? +
?? ?? ]
Sol:
f(x) = ?? ?? 2
+ ???? + ??
?? + ?? =
-?? ??
???? =
?? ??
?????????? ?? + ?? ?????? ?? h?? ?????????? ( ???? ) ???????????? ???? ?? h?? ?????????? ??????????????????????
(i) ?? - ??
The two zeroes of the polynomials are
-?? +v?? 2
-4????
2?? - (?? -v?? 2
-4????
2?? ) = -?? +
v?? 2
-4???? +?? +v?? 2
-4????
2?? =
2v?? 2
-4????
2?? =
v?? 2
-4????
2??
(ii)
1
?? -
1
?? =
?? -?? ????
=
-( ?? - ?? )
????
… ( ?? )
From (i) we know that ?? - ?? =
v?? 2
-4????
2?? [???????? ( ?? ) ]???? =
?? ??
Putting the values in the (a) = - (
v?? 2
-4???? ×?? ?? ×?? ) =
-v?? 2
-4????
??
(iii)
1
?? +
1
?? - 2?? ??
Page 4
Exercise 2.1
1. Find the zeroes of each of the following quadratic polynomials and verify the relationship
between the zeroes and their co efficient:
(i) f(x) = ?? 2
- 2?? - 8
(ii) g(s) = 4?? 2
- 4?? + 1
(iii) h(t) = ?? 2
- 15
(iv) p(x) = ?? 2
+ 2v2?? + 6
(v) q(x) = v3?? 2
+ 10?? + 7v3
(vi) f(x) = ?? 2
- ( v3 + 1) ?? + v3
(vii) g(x) = ?? ( ?? 2
+ 1)- ?? ( ?? 2
+ 1)
(viii) 6?? 2
- 3 - 7?? Sol:
(i) f(x) = ?? 2
- 2?? - 8
?? ( ?? ) = ?? 2
- 2?? - 8 = ?? 2
- 4?? + 2?? - 8
= ?? ( ?? - 4)+ 2( ?? - 4)
= ( ?? + 2) ( ?? - 4)
Zeroes of the polynomials are -2 and 4
Sum of the zeroes =
- ???? ?????????????????? ???? ?? ???? ?????????????????? ???? ??
-2 + 4 =
-( -2)
1
2 = 2
Product of the zeroes =
???????????????? ???????? ???? ?????????????????? ???? ?? 2
= 24 =
-8
1
- 8 = -8
? Hence the relationship verified
(ii) 9( 5) = 45 - 45 + 1 = 45
2
- 25 - 25 + 1 = 25( 25 - 1)- 1( 25 - 1)
= ( 25 - 1) ( 25 - 1)
Zeroes of the polynomials are
1
2
?????? 1
2
Sum of zeroes =
- ???? ?????????????????? ???? ?? ???? ?????????????????? ???? ?? 2
1
2
+
1
2
=
-( -4)
4
1 = 1
Product of the zeroes =
???????????????? ???????? ???? ?????????????????? ???? ?? 2
1
2
×
1
2
=
1
4
?
1
4
=
1
4
? Hence the relationship verified.
(iii) h(t) = ?? 2
- 15 = ( ?? 2
)- ( v15)
2
= ( ?? + v15) ( ?? - v15)
zeroes of the polynomials are -v15 ?????? v15
sum of zeroes = 0
-v15 + v15 = 0
0 = 0
Product of zeroes =
-15
1
-v15 × v15 = -15
-15 = -15
? Hence the relationship verified.
(iv) p(x) = ?? 2
+ 2v2?? - 6 = ?? 2
+ 3v2?? + v2 × 3v2
= ?? ( ?? + 3v2)- v2( 2 + 3v2) = ( ?? - v2) ( ?? + 3v2)
Zeroes of the polynomial are 3v2 and -3v2
Sum of the zeroes =
-3v2
1
v2 - 3v2 = -2v2
-2v2 = -2v2
?????????????? ???? ???????????? ? v2 × -3v2 = -
6
1
-6 = -6
?????????? ?? h?? ???????????????? h???? ????????????????
(v) 2(x) = v3?? 2
+ 10?? + 7v3 = v3?? 2
+ 7?? + 3?? + 7v3
= v3?? ( ?? + v3)+ 7( ?? + v3)
= ( v3?? + 7) ( ?? + v3)
Zeroes of the polynomials are -v3,
-7
v3
Sum of zeroes =
-10
v3
? -v3 -
7
v3
=
-10
v3
?
-10
v3
=
-10
v3
Product of zeroes =
7v3
3
?
v3?? -7
v30
= 7
? 7 = 7
Hence, relationship verified.
(vi) f(x) = ?? 2
- ( v3 + 1) ?? + v3 = ?? 2
- v3?? - ?? + v3
= x (x - v3) – 1 (x - v3)
= (x – 1) (x - v3)
Zeroes of the polynomials are 1 and v3
Sum of zeroes =
-{?????????????????????? ???? ?? }
???? ?????????????????? ???? ?? 2
=
-[-v3-1]
1
1 + v3 = v3 + 1
Product of zeroes =
???????????????? ???????? ???? ?????????????????? ???? ?? 2
=
v3
1
1 × v3 = v3 = v3 = v3
? Hence, relationship verified
(vii) g(x) = ?? [( ?? 2
+ 1)- ?? ( ?? 2
+ 1) ]
2
= ?? ?? 2
+ ?? - ?? 2
?? - ??
= ?? ?? 2
- [( ?? 2
+ 1)- ?? ] + 0 = ?? ?? 2
- ?? 2
?? - ?? + ??
= ???? ( ?? - ?? )- 1( ?? - ?? ) = ( ?? - ?? ) ( ???? - 1)
Zeroes of the polynomials =
1
?? ?????? ??
Sum of the zeroes =
-[-?? 2
-1]
??
?
1
?? + ?? =
?? 2
+1
?? ?
?? 2
+1
?? =
?? 2
+1
??
Product of zeroes =
?? ??
?
1
?? × ?? =
?? ?? ?
?? 2
+1
?? =
?? 2
+1
??
Product of zeroes =
?? ?? ? 1 = 1
Hence relationship verified
(viii) 6?? 2
- 3 - 7?? = 6?? 2
- 7?? - 3 = ( 3?? + 11) ( 2?? - 3)
Zeroes of polynomials are +
3
2
?????? -1
3
Sum of zeroes =
-1
3
+
3
2
=
7
6
=
-( -7)
6
=
-( ???? ?????????????????? ???? ?? )
???? ?????????????????? ???? ?? 2
Product of zeroes =
-1
3
×
3
2
=
-1
2
=
-3
6
=
???????????????? ???????? ???? ?????????????????? ???? ?? 2
? Hence, relationship verified.
2. If ?? and ?? are the zeros of the quadratic polynomial f(x) = ax
2
+ bx + c, then evaluate:
(i) ?? - ??
(ii)
1
?? -
1
??
(iii)
1
?? +
1
?? - 2?? ??
(iv) ?? 2
?? + ?? ?? 2
(v) ?? 4
+ ?? 4
(vi)
1
???? +?? +
1
???? +??
(vii)
?? ???? +?? +
?? ???? +??
(viii) ?? [
?? 2
?? +
?? 2
?? ] +
?? [
?? ?? +
?? ?? ]
Sol:
f(x) = ?? ?? 2
+ ???? + ??
?? + ?? =
-?? ??
???? =
?? ??
?????????? ?? + ?? ?????? ?? h?? ?????????? ( ???? ) ???????????? ???? ?? h?? ?????????? ??????????????????????
(i) ?? - ??
The two zeroes of the polynomials are
-?? +v?? 2
-4????
2?? - (?? -v?? 2
-4????
2?? ) = -?? +
v?? 2
-4???? +?? +v?? 2
-4????
2?? =
2v?? 2
-4????
2?? =
v?? 2
-4????
2??
(ii)
1
?? -
1
?? =
?? -?? ????
=
-( ?? - ?? )
????
… ( ?? )
From (i) we know that ?? - ?? =
v?? 2
-4????
2?? [???????? ( ?? ) ]???? =
?? ??
Putting the values in the (a) = - (
v?? 2
-4???? ×?? ?? ×?? ) =
-v?? 2
-4????
??
(iii)
1
?? +
1
?? - 2?? ??
? [
?? + ?? ????
] - 2????
?
-?? ?? ×
?? ?? - 2
?? ?? = -2
?? ?? -
?? ?? =
-???? -2?? 2
????
- [
?? ?? +
2?? ?? ]
(iv) ?? 2
?? + ?? ?? 2
?? ?? ( ?? + ?? )
=
?? ?? (
-?? ?? )
=
-????
?? 2
(v) ?? 4
+ ?? 4
= ( ?? 2
+ ?? 2
)
2
- 2?? 2
+ ?? 2
= ( ( ?? + ?? )
2
- 2???? )
2
- 2( ???? )
2
= [(-
?? ?? )
2
- 2
?? ?? ]
2
- [2 (
?? ?? )
2
]
= [
?? 2
-2????
?? 2
]
2
-
2?? 2
?? 2
=
( ?? 2
2???? )
2
-2?? 2
?? 2
?? 4
(vi)
1
???? +?? +
1
???? +??
?
???? +?? +???? +?? ( 3?? +?? ) ( ???? +?? )
=
?? ( ?? + ?? ) +2?? ?? 2
???? +???? ?? +?????? +?? 2
=
?? ( ?? +?? ) +?? ?? 2
???? +???? ( ?? 2
?? ) +?? 2
=
?? ×
?? +2?? ?? ?? ×
?? ?? +
?????? ( -?? ) +?? 2
?? =
?? ???? -?? 2
+?? 2
=
?? ????
(vii)
?? ???? +?? +
?? ???? +??
=
?? ( ???? +?? ) +?? ( ???? +?? )
( ???? +?? ) ( ???? +?? )
=
?? ?? 2
+???? +?? ?? 2
+????
?? 2
???? +?????? +?????? +?? 2
=
?? ?? 2
+?? ?? 2
+?? ?? 2
+????
?? ×
?? ?? +???? ( ?? +?? ) +?? 2
=
?? [( ?? 2
+?? 2
) +?? ( ?? +?? ) ]
???? +???? +?? (
-?? ?? )+?? 2
=
?? [( ?? + ?? )
2
-2???? ]+???? -
?? ?? ????
=
?? [
?? 2
?? -
2?? ?? ]-
?? 2
?? ????
=
?? ×[
?? 2
-2?? ?? ]-?? 2
????
=
-2
??
(viii) ?? [
?? 2
?? +
?? 2
?? ] + ?? [
?? ?? +
?? ?? ]
= ?? [
?? 3
+?? 3
????
] + ?? (
?? 2
+?? 2
????
)
Page 5
Exercise 2.1
1. Find the zeroes of each of the following quadratic polynomials and verify the relationship
between the zeroes and their co efficient:
(i) f(x) = ?? 2
- 2?? - 8
(ii) g(s) = 4?? 2
- 4?? + 1
(iii) h(t) = ?? 2
- 15
(iv) p(x) = ?? 2
+ 2v2?? + 6
(v) q(x) = v3?? 2
+ 10?? + 7v3
(vi) f(x) = ?? 2
- ( v3 + 1) ?? + v3
(vii) g(x) = ?? ( ?? 2
+ 1)- ?? ( ?? 2
+ 1)
(viii) 6?? 2
- 3 - 7?? Sol:
(i) f(x) = ?? 2
- 2?? - 8
?? ( ?? ) = ?? 2
- 2?? - 8 = ?? 2
- 4?? + 2?? - 8
= ?? ( ?? - 4)+ 2( ?? - 4)
= ( ?? + 2) ( ?? - 4)
Zeroes of the polynomials are -2 and 4
Sum of the zeroes =
- ???? ?????????????????? ???? ?? ???? ?????????????????? ???? ??
-2 + 4 =
-( -2)
1
2 = 2
Product of the zeroes =
???????????????? ???????? ???? ?????????????????? ???? ?? 2
= 24 =
-8
1
- 8 = -8
? Hence the relationship verified
(ii) 9( 5) = 45 - 45 + 1 = 45
2
- 25 - 25 + 1 = 25( 25 - 1)- 1( 25 - 1)
= ( 25 - 1) ( 25 - 1)
Zeroes of the polynomials are
1
2
?????? 1
2
Sum of zeroes =
- ???? ?????????????????? ???? ?? ???? ?????????????????? ???? ?? 2
1
2
+
1
2
=
-( -4)
4
1 = 1
Product of the zeroes =
???????????????? ???????? ???? ?????????????????? ???? ?? 2
1
2
×
1
2
=
1
4
?
1
4
=
1
4
? Hence the relationship verified.
(iii) h(t) = ?? 2
- 15 = ( ?? 2
)- ( v15)
2
= ( ?? + v15) ( ?? - v15)
zeroes of the polynomials are -v15 ?????? v15
sum of zeroes = 0
-v15 + v15 = 0
0 = 0
Product of zeroes =
-15
1
-v15 × v15 = -15
-15 = -15
? Hence the relationship verified.
(iv) p(x) = ?? 2
+ 2v2?? - 6 = ?? 2
+ 3v2?? + v2 × 3v2
= ?? ( ?? + 3v2)- v2( 2 + 3v2) = ( ?? - v2) ( ?? + 3v2)
Zeroes of the polynomial are 3v2 and -3v2
Sum of the zeroes =
-3v2
1
v2 - 3v2 = -2v2
-2v2 = -2v2
?????????????? ???? ???????????? ? v2 × -3v2 = -
6
1
-6 = -6
?????????? ?? h?? ???????????????? h???? ????????????????
(v) 2(x) = v3?? 2
+ 10?? + 7v3 = v3?? 2
+ 7?? + 3?? + 7v3
= v3?? ( ?? + v3)+ 7( ?? + v3)
= ( v3?? + 7) ( ?? + v3)
Zeroes of the polynomials are -v3,
-7
v3
Sum of zeroes =
-10
v3
? -v3 -
7
v3
=
-10
v3
?
-10
v3
=
-10
v3
Product of zeroes =
7v3
3
?
v3?? -7
v30
= 7
? 7 = 7
Hence, relationship verified.
(vi) f(x) = ?? 2
- ( v3 + 1) ?? + v3 = ?? 2
- v3?? - ?? + v3
= x (x - v3) – 1 (x - v3)
= (x – 1) (x - v3)
Zeroes of the polynomials are 1 and v3
Sum of zeroes =
-{?????????????????????? ???? ?? }
???? ?????????????????? ???? ?? 2
=
-[-v3-1]
1
1 + v3 = v3 + 1
Product of zeroes =
???????????????? ???????? ???? ?????????????????? ???? ?? 2
=
v3
1
1 × v3 = v3 = v3 = v3
? Hence, relationship verified
(vii) g(x) = ?? [( ?? 2
+ 1)- ?? ( ?? 2
+ 1) ]
2
= ?? ?? 2
+ ?? - ?? 2
?? - ??
= ?? ?? 2
- [( ?? 2
+ 1)- ?? ] + 0 = ?? ?? 2
- ?? 2
?? - ?? + ??
= ???? ( ?? - ?? )- 1( ?? - ?? ) = ( ?? - ?? ) ( ???? - 1)
Zeroes of the polynomials =
1
?? ?????? ??
Sum of the zeroes =
-[-?? 2
-1]
??
?
1
?? + ?? =
?? 2
+1
?? ?
?? 2
+1
?? =
?? 2
+1
??
Product of zeroes =
?? ??
?
1
?? × ?? =
?? ?? ?
?? 2
+1
?? =
?? 2
+1
??
Product of zeroes =
?? ?? ? 1 = 1
Hence relationship verified
(viii) 6?? 2
- 3 - 7?? = 6?? 2
- 7?? - 3 = ( 3?? + 11) ( 2?? - 3)
Zeroes of polynomials are +
3
2
?????? -1
3
Sum of zeroes =
-1
3
+
3
2
=
7
6
=
-( -7)
6
=
-( ???? ?????????????????? ???? ?? )
???? ?????????????????? ???? ?? 2
Product of zeroes =
-1
3
×
3
2
=
-1
2
=
-3
6
=
???????????????? ???????? ???? ?????????????????? ???? ?? 2
? Hence, relationship verified.
2. If ?? and ?? are the zeros of the quadratic polynomial f(x) = ax
2
+ bx + c, then evaluate:
(i) ?? - ??
(ii)
1
?? -
1
??
(iii)
1
?? +
1
?? - 2?? ??
(iv) ?? 2
?? + ?? ?? 2
(v) ?? 4
+ ?? 4
(vi)
1
???? +?? +
1
???? +??
(vii)
?? ???? +?? +
?? ???? +??
(viii) ?? [
?? 2
?? +
?? 2
?? ] +
?? [
?? ?? +
?? ?? ]
Sol:
f(x) = ?? ?? 2
+ ???? + ??
?? + ?? =
-?? ??
???? =
?? ??
?????????? ?? + ?? ?????? ?? h?? ?????????? ( ???? ) ???????????? ???? ?? h?? ?????????? ??????????????????????
(i) ?? - ??
The two zeroes of the polynomials are
-?? +v?? 2
-4????
2?? - (?? -v?? 2
-4????
2?? ) = -?? +
v?? 2
-4???? +?? +v?? 2
-4????
2?? =
2v?? 2
-4????
2?? =
v?? 2
-4????
2??
(ii)
1
?? -
1
?? =
?? -?? ????
=
-( ?? - ?? )
????
… ( ?? )
From (i) we know that ?? - ?? =
v?? 2
-4????
2?? [???????? ( ?? ) ]???? =
?? ??
Putting the values in the (a) = - (
v?? 2
-4???? ×?? ?? ×?? ) =
-v?? 2
-4????
??
(iii)
1
?? +
1
?? - 2?? ??
? [
?? + ?? ????
] - 2????
?
-?? ?? ×
?? ?? - 2
?? ?? = -2
?? ?? -
?? ?? =
-???? -2?? 2
????
- [
?? ?? +
2?? ?? ]
(iv) ?? 2
?? + ?? ?? 2
?? ?? ( ?? + ?? )
=
?? ?? (
-?? ?? )
=
-????
?? 2
(v) ?? 4
+ ?? 4
= ( ?? 2
+ ?? 2
)
2
- 2?? 2
+ ?? 2
= ( ( ?? + ?? )
2
- 2???? )
2
- 2( ???? )
2
= [(-
?? ?? )
2
- 2
?? ?? ]
2
- [2 (
?? ?? )
2
]
= [
?? 2
-2????
?? 2
]
2
-
2?? 2
?? 2
=
( ?? 2
2???? )
2
-2?? 2
?? 2
?? 4
(vi)
1
???? +?? +
1
???? +??
?
???? +?? +???? +?? ( 3?? +?? ) ( ???? +?? )
=
?? ( ?? + ?? ) +2?? ?? 2
???? +???? ?? +?????? +?? 2
=
?? ( ?? +?? ) +?? ?? 2
???? +???? ( ?? 2
?? ) +?? 2
=
?? ×
?? +2?? ?? ?? ×
?? ?? +
?????? ( -?? ) +?? 2
?? =
?? ???? -?? 2
+?? 2
=
?? ????
(vii)
?? ???? +?? +
?? ???? +??
=
?? ( ???? +?? ) +?? ( ???? +?? )
( ???? +?? ) ( ???? +?? )
=
?? ?? 2
+???? +?? ?? 2
+????
?? 2
???? +?????? +?????? +?? 2
=
?? ?? 2
+?? ?? 2
+?? ?? 2
+????
?? ×
?? ?? +???? ( ?? +?? ) +?? 2
=
?? [( ?? 2
+?? 2
) +?? ( ?? +?? ) ]
???? +???? +?? (
-?? ?? )+?? 2
=
?? [( ?? + ?? )
2
-2???? ]+???? -
?? ?? ????
=
?? [
?? 2
?? -
2?? ?? ]-
?? 2
?? ????
=
?? ×[
?? 2
-2?? ?? ]-?? 2
????
=
-2
??
(viii) ?? [
?? 2
?? +
?? 2
?? ] + ?? [
?? ?? +
?? ?? ]
= ?? [
?? 3
+?? 3
????
] + ?? (
?? 2
+?? 2
????
)
=
?? [( ?? +?? )
3
-3???? ( ?? +?? ) ]
?? ?? + ?? ( ?? + ?? )
2
- 2?? ??
=
?? [(
-?? 3
?? 3
)+
3?? ?? .
?? ?? +?? (
?? 2
?? 2
-
2?? ?? )]
?? ??
=
?? 2
?? [
-?? 3
?? 3
+
3????
?? 2
+
?? 3
?? 2
-
2????
?? ]
=
-?? 2
?? 3
?? ?? 3
+
3?? 2
????
?? ?? 2
+
?? 3
?? 2
?? 2
?? -
2???? ?? 2
????
=
-?? 3
????
+ 3?? +
?? 3
????
- 2??
= b
3. If ?? and ?? are the zeros of the quadratic polynomial f(x) = 6x
2
+ x - 2, find the value of
?? ?? +
?? ??
Sol:
f(x) = 6?? 2
- ?? - 2
Since ?? and ?? are the zeroes of the given polynomial
? Sum of zeroes [?? + ?? ] =
-1
6
Product of zeroes (???? ) =
-1
3
=
?? ?? +
?? ?? =
?? 2
+?? 2
????
=
( ?? + ?? )
2
-2????
????
=
(
1
6
)
2
-2×(
-1
3
)
-
1
3
=
1
6
-
2
3
-1
3
=
1+24
36
-1
3
=
2
36
1
3
=
-25
12
4. If a and are the zeros of the quadratic polynomial f(x) = ?? 2
- ?? - 4, find the value of
1
?? +
1
?? - ????
Sol:
Since ?? + ?? are the zeroes of the polynomial: ?? 2
- ?? - 4
Sum of the roots (?? + ?? ) = 1
Product of the roots (???? ) = -4
1
?? +
1
?? - ???? =
?? +?? ????
- ????
=
1
-4
+ 4 =
-1
4
+ 4 =
-1+16
4
=
15
4
5. If ?? and ?? are the zeros of the quadratic polynomial p(x) = 4x
2
- 5x -1, find the value of
?? 2
?? + ?? ?? 2
.
Sol:
Since ?? ?????? ?? are the roots of the polynomial: 4?? 2
- 5?? - 1
```
## Mathematics (Maths) Class 10
116 videos|420 docs|77 tests
## Mathematics (Maths) Class 10
116 videos|420 docs|77 tests
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Found in: Page 415
### Fundamentals Of Differential Equations And Boundary Value Problems
Book edition 9th
Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider
Pages 616 pages
ISBN 9780321977069
# In Problems 3–10, determine the Laplace transform of the given function.${{\mathbit{t}}}^{{\mathbf{2}}}{{\mathbit{e}}}^{\mathbf{-}\mathbf{9}\mathbf{t}}$
$L\left\{{t}^{2}{e}^{-9t}\right\}\left(s\right)=\frac{2}{{\left(s+9\right)}^{3}}$
See the step by step solution
## Step 1:Given Information.
The given function is .${t}^{2}{e}^{-9t}$
## Step 2: Determining the Laplace transform:
Using the Laplace transform definition, $L\left\{{t}^{n}{e}^{at}\right\}=\frac{n!}{{\left(s-a\right)}^{n+1}}$we get
$\begin{array}{c}L\left\{{t}^{2}{e}^{-9t}\right\}=\frac{2!}{{\left(s+9\right)}^{3}}\\ =\frac{2}{{\left(s+9\right)}^{3}}\end{array}$
## Step 3: Determining the Result
Thus, the required Laplace transform is .$L\left\{{t}^{2}{e}^{-9t}\right\}\left(s\right)=\frac{2}{{\left(s+9\right)}^{3}}$
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CC-MAIN-2023-23
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https://jeopardyapp.com/play/exam-11
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Quadratics Absolute Values Applications Concept Checks
(7/4, -121/8)
What is the vertex of y=2x^2 -7x-9
no solution
What is the solution of [7-x]+6=3
69ft
What is the Maximum Height If a football is kicked straight up with an initial velocity of 64 ft/sec from a height of 5 ft, then its height above the earth is a function of time given by h(t) = -16t^2 + 64t + 5.
Down 5 units
Compared to the graph of f(x) = x2, the graph of g(x) = x^2 - 5 is shifted ___
x=-8, 5
What is the solution of (x+4)(x-1)=36
x= 5, 1
What is the solution of 2[x-3]=4
1995
The quadratic function f(x) = 0.041x^2-0.45x+30.06 models the median age at which men
in the United States were first married x years after 1990. In which year was this average age
at a minimum? (Round to the nearest year.)
9
To complete the square of the expression x^2 -6x, add
x= 9/2, -1
What are the zeros of y=2x^2 -7x-9
(-1/3, 1)
What is the solution of [1-3x] +4 less then or equal to 6
~4.1 sec
What is the hang time of the ball if a football is kicked straight up with an initial velocity of 64 ft/sec from a height of 5 ft, then its height above the earth is a function of time given by h(t) = -16t^2 + 64t + 5.
1 real solution
If the discriminant of a quadratic equation is zero, the nature of solution(s) is(are):
# Exam 2
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# Circuit to provide 0-5 V output proportional to 0-30 A from C.T?
I want to design a circuit to protect oveload current of 30A.I have a controller board(PLC) which accepts the voltage range of 0-5 as an input and C.T with 1:7000 turns which can maesure 0.5 -60 A(RMS) value with burden resistance of 47 ohm(recommended).My question is how should i design a circuit which porvides 0-5 V output proportional to 0-30 A(RMS).
I'll assume you want "average reading" equivalent to RMS for a sinusoidal input. If you want "true RMS" you'll have to add a more complex conversion circuit. This is the same principle used in non-"true-RMS" multimeters.
The transformer gives you (60A/7000) * 47 = 402.8mV RMS AC full scale (full scale of the transformer)
So you need to convert an AC input of 402.8mV/2.0 = 201.4mV RMS into 5VDC (30A in for 5.0V out)
Here is one of many of possible precision rectifier circuits:-
It shunts the input resistor a bit, so 47.22 ohms can be used as the burden resistor to give you 47 ohms equivalent as recommended.
You feed the output of that circuit into a low-pass filter. The sophistication depends on response time requirement and ripple. Look up a Sallen Key type, for example.
The input voltage was 201.4mV RMS, which is 284.8mV peak. The output voltage of our absolute value circuit will also be 284.8mV peak. The filter will find the average of the absolute value circuit output, which we know to be (since it's equivalent to the average value of half a sine wave) $V_{PK}\cdot 2\over\pi$ = 181.3mV (assuming a filter gain of 1.0).
You then simply need an amplifier with a gain of $5.0 \over 0.1813$ = 27.58 to give a 5V output for 30A.
Edit: The full wave rectifier circuit shown works as shown below:
The output voltage is $2V_B - V_A$
For positive inputs, $V_B$ =0 and $V_A$ = -Vin, so overall gain is +1
For negative inputs, $V_B$ = (2/3)Vin and $V_A$ = (1/3)Vin so overall gain is +1
• Thanks for your valueble suggestion.I'm not good in op-amp circuits.First I will try to understand how the circuit works.Will ask you if find any doubts :) Commented Apr 6, 2014 at 5:40
• Okay, it's really just a series arrangement of three building blocks after the burden resistor. Rectifier->Filter->output amplifier. Commented Apr 6, 2014 at 5:50
• @Pefhany, in your example,i didn't understand why o/p (from node 6) is given to inverting input through 1k ohm and diode. and filter part, could you please elaborate how it works for me Commented Apr 14, 2014 at 12:38
• @user3396484 1K and diode? Not sure what you're asking. Commented Apr 14, 2014 at 12:44
• @user3396484 see edit. Both amplifiers are required to make a full wave rectifier. Commented Apr 14, 2014 at 15:32
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Q:
# How many electrons make up one coulomb of charge?
A:
One coulomb of charge has the equivalent charge of 6.25x10^18 electrons. This is determined from the value of charge on one electron and the value of charge for 1 coulomb.
Know More
## Keep Learning
The algebraic solution is laid out by DronStudy as follows:
One electron has a charge equivalent of 1.6x10^-19 coulomb. This is written as 1.6x10^-19 C = 1 electron.
To get the value of 1 C, both sides of the equation have to be divided by 1.6x10^-19. This is written as 1 C = 1/1.6x10^-19 electrons. Dividing 1 by 1.6x10^-19 gives a value of 6.25x10^18 electrons per coulomb.
Given this value, the number of electrons passing through a circuit over a given time can be calculated. One ampere equals 1 coulomb per second. Therefore, a vacuum cleaner with a 3.5-amp motor uses 3.5 coulomb per second multiplied by 6.25x10^18 electrons/coulomb, or 21.875x10^18 electrons per second. That is 21,875,000,000,000,000,000 electrons every second!
This equivalency can also be used to determine the number of electrons required to do a given amount of work. Given that 1 volt equals 1 joule/coulomb, a circuit producing 18 joules of work off of a 9-volt power source would require 2 coulombs worth of charge, or 12.5x10^18 electrons.
Sources:
## Related Questions
• A:
Selenium contains four unpaired electrons in its outermost orbital. These electrons can form bonds with other elements and are called valence electrons. On the periodic table, selenium is element number 34; it is located in group 16 below oxygen and sulfur. This group of elements is sometimes called the chalcogens.
Filed Under:
• A:
Helium has two valence electrons. As an element, helium is very stable and contains a single s-orbital on its outer shell.
Filed Under:
• A:
In its ground, or lowest-energy, state, carbon has two unpaired electrons. However, there are four total outer, or valence, electrons, meaning carbon atoms have four possible bonding sites.
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## Roots of unity proof
I'm having trouble following one step in a proof I'm studying. I'm sure I'm missing something obvious, but I just can't get it to work out (it supposed to be "obvious" which is why they left out the details).
Anyway, it's part of a proof showing that if you have a monic polynomial with all integer coefficients and all the zeros have absolute value 1 then all the zeros are roots of unity.
So say we have our polynomial:
$f(t)=(t-\alpha_1)...(t-\alpha_k)$
A simple argument is then used to show that you can create new monic integer polynomials from this by raising the alpha's to powers. For example:
$f_m(t)=(t-\alpha^m_1)...(t-\alpha^m_k)$
Then it's easy to put a finite limit on the the absolute value of the coefficients, from which you get a finite set of inequalities which have a finite set of solutions, and then it is an easy leap to see that there must be two different of these "power polynomials" that are equal, for different m.
So far so good (rushing through the proof here). We then create a permutation function $\pi$ so we end up with for m not equal to g:
$\alpha^m_j = \alpha^g_{\pi (j)}$
And then comes the next step that I don't really see. I think you only need the above, and not all the background. But my book claims that a simple induction argument gives you from the above that:
$\alpha^{m^r}_j = \alpha^{g^r}_{\pi^r (j)}$
And I just can't see how to get this. The rest of the proof is also easy. It's just this one step that is bugging me. So I would really appreciate an explanation. Thanks.
PhysOrg.com science news on PhysOrg.com >> Hong Kong launches first electric taxis>> Morocco to harness the wind in energy hunt>> Galaxy's Ring of Fire
nevermind, it was trivial. I just had a blind spot.
Recognitions: Homework Help Lol i've had those before, It laster much longer and it was much more obvious though. It was proving the identity for cos(x-y) and I couldnt see why the angle between the y angle and x angle was x-y...took me 6 months to realize :D
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# array to binary tree
For instance, int nums[] = {7,5,9,1,7,3} to a binary tree (root-7, root's left child-5, root's right child-9...). Binary search trees don’t have to be scary! So, we represented a complete binary tree using an array and saw how to get the parent and children of any node. Array Representation of Incomplete Binary Tree. We'll need an array of size (2**H)-1. Write a function that given an array representation of a binary tree will convert it into a typical tree format. The Binary Tree is converted to Binary Search Tree. If I want to make a binary tree from an array in the following order: Example: for the following array: { -1, 17, -1, 3, -1, -1, -1, 55, -1, 4, -1, 15, 11, 2, 3 } the following tree is created: 55 15 3 2 4 * 17 3 11 * * * * The function is recursive and returns a Tree input: the array and it's size. ( here, I made a . Python Binary Search Tree: Exercise-5 with Solution. I wrote this code in C++ to convert an array to a binary tree. The following is a visual representation of expected input and output: Input: [7, 3, 9, 2, 4, 8, 10,11,12,13,14] Output: 7 / \ 3 9 /\ /\ 2 4 8 10. Let's discuss about doing the same for an incomplete binary tree. For this problem, a height-balanced binary tree is defined as a binary tree in which the depth of the two subtrees of every node never differ by more than 1. They can actually be fun once we get to know their characteristics. Time Complexity = O(n log(n)) Space Complexity = O(n), as we used an array to store the in-order traversal where n is the number of node in given Binary Tree. Given an array where elements are sorted in ascending order, convert it to a height balanced BST. Our strategy is to fix the maximum height of the tree (H), and make the array big enough to hold any binary tree of this height (or less). Stack Exchange Network. Because an array's length is fixed at compile time, if we use an array to implement a tree we have to set a limit on the number of nodes we will permit in the tree. Simultaneously traverse the array and Binary Tree in in-order form and replace the corresponding node’s value in Binary Tree with the value in inOrder array. Note: The selection sort improves on the bubble sort by making only one exchange for every … Write a Python program to convert a given array elements to a height balanced Binary Search Tree (BST). One way to build a tree is that we know that array is like a breadth first traversal .
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Todays Question
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# Todays Question - PowerPoint PPT Presentation
Today’s Question. Example: Dave gets a 50 on his Statistics midterm and an 50 on his Calculus midterm. Did he do equally well on these two exams? Big question: How can we compare a person’s score on different variables?. Example 1. In one case, Dave’s exam score is 10 points above the mean
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Today’s Question
• Example: Dave gets a 50 on his Statistics midterm and an 50 on his Calculus midterm. Did he do equally well on these two exams?
• Big question: How can we compare a person’s score on different variables?
Example 1
• In one case, Dave’s exam score is 10 points above the mean
• In the other case, Dave’s exam score is 10 points below the mean
• In an important sense, we must interpret Dave’s grade relative to the average performance of the class
Statistics
Calculus
Mean Calculus = 60
Mean Statistics = 40
Example 2
• Both distributions have the same mean (40), but different standard deviations (10 vs. 20)
• In one case, Dave is performing better than almost 95% of the class. In the other, he is performing better than approximately 68% of the class.
• Thus, how we evaluate Dave’s performance depends on how much variability there is in the exam scores
Statistics
Calculus
Standard Scores
• In short, we would like to be able to express a person’s score with respect to both (a) the mean of the group and (b) the variability of the scores
• how far a person is from the mean
• variability
Standard Scores
• In short, we would like to be able to express a person’s score with respect to both (a) the mean of the group and (b) the variability of the scores
• how far a person is from the mean = X - M
• variability = SD
Standard (Z) Scores
• In short, we would like to be able to express a person’s score with respect to both (a) the mean of the group and (b) the variability of the scores
• how far a person is from the mean = X - M
• variability = SD
Standard score or
** How far a person is from the mean, in the metric of standard deviation units **
Example 1
Dave in Statistics:
(50 - 40)/10 = 1
(one SD above the mean)
Dave in Calculus
(50 - 60)/10 = -1
(one SD below the mean)
Statistics
Calculus
Mean Statistics = 40
Mean Calculus = 60
Example 2
An example where the means are identical, but the two sets of scores have different spreads
Dave’s Stats Z-score
(50-40)/5 = 2
Dave’s Calc Z-score
(50-40)/20 = .5
Statistics
Calculus
Thee Properties of Standard Scores
• 1. The mean of a set of z-scores is always zero
Properties of Standard Scores
• Why?
• The mean has been subtracted from each score. Therefore, following the definition of the mean as a balancing point, the sum (and, accordingly, the average) of all the deviation scores must be zero.
Three Properties of Standard Scores
• 2. The SD of a set of standardized scores is always 1
Why is the SD of z-scores always equal to 1.0?
M = 50
SD = 10
if x = 60,
x
20
30
40
50
60
70
80
z
-3
-2
-1
0
1
2
3
Three Properties of Standard Scores
• 3. The distribution of a set of standardized scores has the same shape as the unstandardized scores
• beware of the “normalization” misinterpretation
1. We can use standard scores to find centile scores: the proportion of people with scores less than or equal to a particular score. Centile scores are intuitive ways of summarizing a person’s location in a larger set of scores.
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# Implementation Of The Intermediate Value Theorem
1. Let $$f : \mathbb R \rightarrow \mathbb R$$ be a continuous function, such that $$f(1) = 3$$ $$\lim_{x \to \infty} f(x) = -3$$ Prove that there exists at least one $$c\in \mathbb R$$ such that $$f(c) = c$$.
2. Let $$f(x)$$ be a continuous function on $$\mathbb R$$, that is bounded for $$x ≥ 0$$ and non-positive for $$x < 0$$. Prove that the equation $$f(x)+7x=17$$ has at least one solution.
I know that I need to use the IVT, but I am also looking for the intuition behind solving questions of this nature. I am able to solve general problems with IVT if I know the function.
I have found a solution for a. For b I am claiming that $$x < 0 \rightarrow g(x) < 0$$ $$\lim_{x \to \infty} g(x) = \infty> 0$$can I use the IVT on an interval $$[0,\infty)$$ , or it must be a closed interval?
Let $$\varphi(x)=f(x)-x$$ and we have $$\lim_{x\rightarrow\infty}\varphi(x)=-\infty$$ and $$\varphi(1)=3-1=2>0$$, so some $$c$$ is such that $$\varphi(c)=0$$.
In the first case, you can (almost) apply apply the theorem to the function $$g(x)=f(x)-x$$. You know that $$g(1)=2$$ and that $$\lim_{x\to\infty}g(x)=-\infty$$. So, applying the theorem to $$g$$, there is some $$c\in(1,\infty)$$ such that $$g(c)=0$$.
For the other problem, take $$g(x)=f(x)+7x$$ and do something similar.
Concerning your solution of the second problem, no, you should not apply the intermediate value theorem directly to infinite intervals. However, since $$\lim_{x\to\infty}g(x)=\infty$$, there is a $$M\in\mathbb R$$ such that $$g(M)>17$$. Now, apply the intermediate value theorem to the interval $$[0,M]$$.
• I am sure that you mean $g (1)=2.$ – Fred Dec 16 '19 at 19:02
• Sure you're sure! I've edited my answer. Thank you. – José Carlos Santos Dec 16 '19 at 19:16
• How should I use the bounded statement to my advantage ? – johnadams12 Dec 16 '19 at 19:40
• Since $f|_{[0,\infty)}$ is bounded and $\lim_{x\to\infty}7x=\infty$, $\lim_{x\to\infty}g(x)=\infty$. – José Carlos Santos Dec 16 '19 at 19:44
• I have edited with a proposed solution , can you make sure I am on the right track ? – johnadams12 Dec 16 '19 at 20:00
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#### 12th Standard Maths Two Dimensional Analytical Geometry-II English Medium Free Online Test One Mark Questions 2020 - 2021
12th Standard
Reg.No. :
•
•
•
•
•
•
Maths
Time : 00:10:00 Hrs
Total Marks : 10
10 x 1 = 10
1. The equation of the circle passing through(1,5) and (4,1) and touching y -axis is x2+y2−5x−6y+9+(4x+3y−19)=0 whereλ is equal to
(a)
$0,-\frac { 40 }{ 9 }$
(b)
0
(c)
$\frac { 40 }{ 9 }$
(d)
$\frac { -40 }{ 9 }$
2. The radius of the circle3x2+by2+4bx−6by+b2 =0 is
(a)
1
(b)
3
(c)
$\sqrt {10}$
(d)
$\sqrt {11}$
3. The equation of the normal to the circle x2+y2−2x−2y+1=0 which is parallel to the line
2x+4y=3 is
(a)
x+2y=3
(b)
x+2y+3= 0
(c)
2x+4y+3=0
(d)
x−2y+3= 0
4. If the normals of the parabola y2 = 4x drawn at the end points of its latus rectum are tangents to the circle (x−3)2+(y+2)2=r2 , then the value of r2 is
(a)
2
(b)
3
(c)
1
(d)
4
5. The equation of the circle passing through the foci of the ellipse $\frac { { x }^{ 2 } }{ 16 } +\frac { { y }^{ 2 } }{ 9 } =1$ 1having centre at
(0,3) is
(a)
x2+y2−6y−7=0
(b)
x2+y2−6y+7=0
(c)
x2+y2−6y−5=0
(d)
x2+y2−6y+5=0
6. If the two tangents drawn from a point P to the parabolay2 = 4x are at right angles then the locus of P is
(a)
2x+1=0
(b)
x = −1
(c)
2x−1=0
(d)
x =1
7. If (0, 4) and (0, 2) are the vertex and focus of a parabola then its equation is
(a)
x2 + 8y = 32
(b)
y2 + 8x = 32
(c)
x2 - 8y= 32
(d)
y2 - 8x = 32
8. Equation of tangent at (-4, -4) on x2 = -4y is
(a)
2x - y + 4 = 0
(b)
2x + y - 4 = 0
(c)
2x - y - 12 = 0
(d)
2x + y + 4 = 0
9. In an ellipse, the distance between its foci is 6 and its minor axis is 8, then e is
(a)
$\frac { 4 }{ 5 }$
(b)
$\frac { 1 }{ \sqrt { 52 } }$
(c)
$\frac { 3 }{ 5 }$
(d)
$\frac { 1 }{ 2 }$
10. The locus of the foot of perpendicular from the focus on any tangent to y2 = 4ax is
(a)
x2 + y2 = a2 - b2
(b)
x2 + y2 = a2
(c)
x2 + y2 = a2 - b2
(d)
x = 0
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Date:
Class:
Group Activity: Stoichiometry
Background
The study of chemistry is concerned with understanding how and why chemicals react with one another. A critical tool in building this knowledge is called stoichiometry. The word originates in Greek roots for elements and measurement. This is appropriate because stoichiometry is a mathematical approach to understanding the proportions in which elements (and compounds) react in balanced chemical equations. These proportions are based on the whole-number ratio of molecules (or formula units) which react with one another. For example, consider the combustion of butane in oxygen to make carbon dioxide and water:
2C4H10(g) + 13O2(g) → 8CO2(g) + 10H2O(g)
In this balanced chemical equation two molecules of butane (C4H10) are shown to require 13 molecules of oxygen (O2) to react to make carbon dioxide (CO2) and water (H2O). Based on this observation it’s straightforward to conclude that 4 molecules of butane would require 26 molecules of oxygen to react. And 8 molecules of butane require 52 molecules of oxygen and so on. This mathematical relationship continues to hold no matter how many molecules of butane you start with. Consider the case when 6.022 × 1023 molecules of butane react with oxygen. In that case,
``` 13 molecules O2
6.022 × 1023 molecules C4H10 × ---------------- = 3.914 × 1024 molecules O2
2 molecules C4H10
```
You may recall that 6.022 × 1023 is the number of items in one mole (1 mol). By this reasoning it is just as true to say that two moles of butane react with thirteen moles of oxygen as to say that two molecules of butane reacts with thirteen molecules of oxygen. So the equation states that two moles of butane react with thirteen moles of oxygen to make eight moles of carbon dioxide and ten moles of water.
In fact, expressing the mathematical proportions of compounds and elements in chemical equations in terms of moles is far more useful to chemists than expressing them in terms of molecules. A mole of a substance is a measurable amount of it: one mole of butane has a mass of 58.122 g. Remember, molar mass is calculated by adding up all the atomic mass in the chemical formula: C4H10 has a molar mass of 4 × 12.011 g/mol + 10 × 1.00794 g/mol = 58.122 g/mol. Single molecules of butane are far too small to measure or work with. One molecule has a mass of 9.65 × 10-23 grams.
Molar Ratios
Mathematical proportions from chemical equations are expressed as molar ratios. A molar ratio is the relative number of formula units of two chemicals in a balanced chemical equation. The full set of possible molar ratios from the chemical equation for the burning of butane is as follows:
2 mol C4H10 = 13 mol O2 2 mol C4H10 = 8 mol CO2 2 mol C4H10 = 10 mol H2O 13 mol O2 = 8 mol CO2 13 mol O2 = 10 mol H2O 8 mol CO2 = 10 mol H2O ```any ratio can be expressed as a fraction: 2 mol C4H10 13 mol O2 ----------- or ----------- 13 mol O2 2 mol C4H10 ```
Stoichiometry Problems
Using Moles
When the ratios are expressed as fractions (see above, at right) they can be used as conversion factors of a sort. If you remember your dimensional analysis this will be familiar. Molar ratios are really proportions that can be used to find out how many moles of any of the other chemicals correspond to a given amount of the first. For example, 2 mol of butane corresponds to 13 mol of oxygen; how many moles of oxygen correspond to 4 mol of butane? Twenty-six:
``` 13 mol O2
4 mol C4H10 × ---------- = 26 mol O2
2 mol C4H10
```
To the well-informed this means that to burn four moles of butane you need twenty-six moles of oxygen gas. How about if you have a number like 3.19 moles of butane? It works even then:
``` 8 mol CO2
3.19 mol C4H10 × ---------- = 12.8 mol CO2
2 mol C4H10
```
page break
In this calculation, as in all of these examples, you use the molar ratio to convert moles of one substance to moles of another substance. In the example immediately above we found that burning 3.19 mol C4H10 produces 12.8 mol CO2.
Now let’s look at a few examples in the way they might be phrased in a set of chemistry problems or on a quiz.
Example 1
How many moles of carbon dioxide result from the burning of 8.9 moles of butane?
``` 8 mol CO2
8.9 mol C4H10 × ---------- = 35.6 mol CO2 (because 8.9 × 8 ÷ 2 = 35.6)
2 mol C4H10
```
Example 2
If 42 mol of carbon dioxide were produced then how many moles of oxygen were used up in making it?
``` 13 mol O2
42 mol CO2 × ---------- = 68 mol O2 (because 42 × 13 ÷ 8 = 68.25)
8 mol CO2
```
An Analogy
It is often helpful to relate new ideas to familiar ones. In this case a useful analogy is to the construction of a simple cheese sandwich. To make it more appetizing you may imagine grilling the sandwich in butter but that part does not play a part in our analogy.
One simple way to make a cheese sandwich is with two pieces of bread and a single slice of cheese. To make the chemical analogy clear, this recipe can be written as a sort of equation:
The equation expresses a set of proportions: exactly two slices of bread are needed for each one slice of cheese. Another proportion implied in the recipe (or the equation) is that exactly one slice of cheese is needed for each sandwich. These proportions can be written as fractions:
``` 2(Bread) 1(Cheese) 2(Bread)
----------- and ------------ and -------------
1(Cheese) 1(Sandwich) 1(Sandwich)
```
This is a complicated way of talking about something simple, but since the complicated way of thinking is what you are meant to learn, this is helpful. For example, it is probably obvious that if you want to make 10 sandwiches you need 20 slices of bread. But this can be calculated, with a little labor, as follows:
``` 2(Bread)
1(Sandwich)
```
These ratios can scale up to any amount so that if instead of 10, you had 100 sandwiches to make then you would need 200 slices of bread. Similarly, if you have a mole of sandwiches to make (which is an insane amount to think about, but still) the ratios still work. One mole of sandwiches would require 2 moles of slices of bread. This way of thinking and calculating is exactly what stoichiometry is all about, with the added complication that you are calculating the number of molecules, which are too small to see or to count directly, rather than the number of slices of bread or cheese, which are easy to count for normal situations. But if molecules are too small to count directly, then how can stoichometry be used in any practical way?
Using Masses
To make stoichiometry practical it is necessary to be able to relate the mass of a reactant or product to the mass of the other reactants and products. It’s not possible to measure individual molecules in the lab or to measure moles directly. But moles are still useful because theycan be measured indirectly in the lab. Every chemical has a characteristic called its molar mass. The molar mass of a chemical is the sum of the atomic masses of the atoms in its chemical formula. You have probably already learned about molar mass and how to calculate it. As a reminder, here is an example calculation:
page break
Example 3
```Molar Mass of Fe2O3 (iron(III) oxide, aka rust)
Fe: 55.845 g/mol and O: 15.9994 g/mol
2 × 55.845 g/mol + 3 × 15.9994 g/mol = 159.69 g/mol 159.69 g
The molar mass of Fe2O3 is ----------
1 mol
```
So one mole of iron(III) oxide has a mass of 159.69 g. Butane has a molar mass of 58.122 g/mol. The molar mass of a chemical is the proportion between the mass of a chemical and the number of moles of that chemical. Since moles are just a way of counting things in a (very big) group, this is a way of counting atoms and molecules by weighing the chemical. It is far easier to measure a chemical by mass than by counting atoms and molecules.
Chemical equations express the proportions that exist between the reactants and products in a chemical reaction. These proportions are based on numbers of atoms and molecules and not on mass. Mass is a quantity easy to measure in a lab and once the mass is known, it is a simple matter to convert the mass to the number of moles, which is equivalent to the number of molecules. Since chemical equations relate the number of atoms and molecules it is critical to convert masses to moles before using a chemical equation to determine how much of a reactant is needed to react with a given amount of some chemical. The following example shows how to use an amount of a reactant given in grams (a mass) to determine the mass of a chemical product which would be produced by reacting all of that chemical. The question asks about grams of carbon dioxide produced by burning a certain number of grams of butane. Remember, the key step is the one in the middle where the number of moles of butane is related to the number of moles of carbon dioxide using the ratio between these two chemicals in the balanced chemical equation. First, the number of grams of the reactant is converted to moles using the molar mass of that chemical. Then the molar ratio from the balanced equation is used to calculate the number of moles of the product. Finally, the moles of the product are converted to the mass using the molar mass of the product chemical.
Example 4
How many grams of carbon dioxide result from the burning of 392 grams of butane?
A ``` 1 mol 39.2 g C4H10 × --------- = 0.674 mol C4H10 58.122 g ``` B ``` 8 mol CO2 0.674 mol C4H10 × ---------- = 2.70 mol CO2 2 mol C4H10 ``` C ``` 44.009 g 2.70 mol CO2 × --------- = 119 g CO2 1 mol ```
This could all be done in one calculation as shown below:
``` 1 mol 39.2 g C4H10 × --------- 58.122 g ``` ``` 8 mol CO2 × ---------- 2 mol C4H10 ``` ``` 44.009 g × --------- = 119 g CO2 1 mol ```
In step A the mass of butane in grams is changed to the number of moles of butane using the molar mass of butane. In step B the moles of butane is changed to moles of carbon dioxide using the molar ratio from the balanced chemical equation. In step C the moles of carbon dioxide is changed to the mass of carbon dioxide in grams. All of these calculation steps can be combined into one calculation, as shown above. This simplifies matters and can mean fewer entries into a calculator, which can help to avoid errors due to putting the wrong number into your calculator. Just remember to ‘multiply by the top and divide by the bottom’ as your learned doing dimensional analysis.
page break
Exercises
The following exercises were written in the order given to help you to develop the skills necessary to master the topic of stoichiometry. You will need a calculator and a periodic table to complete them. You may also use your ions reference sheet.
Remember Dimensional Analysis
Dimensional analysis is an important skill to have when you are learning about stoichiometry. These problems are designed to help you to remember and practice this skill. Show all work using conversion factors and cancelling units as appropriate.
1. How many miles could you cover in eight hours of driving at an average speed of 55 miles per hour?
2. If 0.3048 m equals 1 foot and there are 5,280 feet in one mile, then how many meters is equal to 2 miles?
1. Say that grapes cost \$2.99 per pound at the grocery store. If one pound equals 0.4536 kilograms, then how much would it cost to buy 3.0 kg of grapes?
2. Hydrogen is a gas with a very low density. One liter of the gas weighs only 0.082 g. How many liters would you get if you had just 10.0 g of hydrogen gas?
Basic Stoichiometry
5C + 2SO2 CS2 + 4CO
Solid carbon reacts with sulfur dioxide to make carbon disulfide and carbon monoxide.
1. How many moles of carbon disulfide form when 5.0 mol of carbon react?
2. How many moles of carbon are needed to react with 4.0 mol of sulfur dioxide?
3. How many moles of carbon monoxide form at the same time that 0.50 mol of carbon disulfide form?
1. How many moles of sulfur dioxide are required to make 5.0 mol of carbon disulfide?
2. Calculate the number of moles of carbon disulfide and the number of moles of carbon monoxide that form when 30 mol of carbon react with enough sulfur dioxide.
page break
More Basic Stoichiometry
Use molar ratios found in this chemical equation to answer the following questions. Show your work for all calculations.
3CaCl2(aq) + 2Na3PO4(aq) → Ca3(PO4)2(s) + 6NaCl(aq)
Aqueous calcium chloride reacts with aqueous sodium phosphate to precipitate solid calcium phosphate and to produce aqueous sodium chloride.
1. If you have 3 moles of CaCl2 then how many moles of Na3PO4 will you need if you react all of the CaCl2? Also, how many moles of the solid product, Ca3(PO4)2, will be formed?
2. Say that you find that 0.25 mol of Ca3(PO4)2 formed in an experiment. How many moles of CaCl2 were required to get this result? Also, how many moles of Na3PO4 were required?
1. Say that you need to generate 3.5 mol Ca3(PO4)2 using this chemical reaction. How many moles of each reactant do you need?
2. How many moles of both products would be made if you reacted 0.125 mol CaCl2 with enough of the other reactant to react all of the CaCl2?
Stoichiometry with Masses
2KClO3(l) 2KCl(s) + 3O2(g)
Potassium chlorate decomposes to form potassium chloride and oxygen gas.
N2(g) + 3H2(g) 2NH3(g)
Nitrogen and hydrogen gases react to form ammonia.
1. How many grams of oxygen can be obtained by the decomposition of 15 grams of potassium chlorate?
2. How many grams of potassium chloride are produced in the same reaction?
1. How many grams of ammonia would be produced by the reaction of 5.00 g of nitrogen with excess hydrogen?
2. How many grams of nitrogen must be used to produce 240 grams of ammonia? How many grams of hydrogen were used in the reaction? (There are two ways to calculate the answer to the second question: show both).
page break
More Stoichiometry with Masses
Tin(II) fluoride and hydrogen gas can be produced by the reaction of metallic tin with hydrogen fluoride gas.
Balanced Equation:
Sn(s) + 2HFSnF2(s) + H2(g)
Limestone (calcium carbonate) can be decomposed into carbon dioxide and calcium oxide by heating to high temperature.
Balanced Equation:
CaCO3(s) → CO2(g) + CaO(s)
1. How many grams of tin(II) fluoride can be produced by the reaction of 128 grams of hydrogen fluoride with an excess amount of tin?
2. How many grams of hydrogen would be produced by the same reaction?
3. Is the mass of tin(II) fluoride after the reaction more than, less than, or equal to the mass of tin before the reaction? Why?
4. What mass of tin(II) fluoride is produced by reacting 115 g of tin with excess hydrogen fluoride?
5. What mass in grams of tin is required to make 10.0 g of hydrogen gas?
1. As calcium carbonate is heated, does the mass of the solid material in the container increase, decrease or stay the same? Why?
2. How many grams of calcium oxide are produced by the complete decomposition of 1,000 grams of limestone?
3. How many grams of carbon dioxide are produced at the same time?
4. What is the sum of the amount of carbon dioxide and calcium oxide produced by the complete decomposition of 1,000 g of limestone? What should it be? Why?
5. How many grams of calcium carbonate are required to produce 1,000 g of calcium oxide?
Do these problems together as a class: Stoichiometry Start-up
Here is the homework for this activity.
A related activity about Limiting Reagent and Percent Yield and the homework to go with it.
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Chapter6
# Chapter6 - CHAPTER6 Unemployment APowerPoint Tutorial...
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1 Chapter Six CHAPTER 6 Unemployment ® A PowerPoint Tutorial To Accompany MACROECONOMICS, 7th. Edition N. Gregory Mankiw Tutorial written by: Mannig J. Simidian B.A. in Economics with Distinction, Duke University M.P.A., Harvard University Kennedy School of Government M.B.A., Massachusetts Institute of Technology (MIT) Sloan School of Management
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2 Chapter Six Using this notation, the rate of unemployment is U/L. Now, we’ll denote the rate of job separation as s . Let f denote the rate of job finding. Together these determine the rate of unemployment. The average rate of unemployment around which the economy fluctuates is called the natural rate of unemployment. The natural rate is the rate of unemployment toward which the economy gravitates in the long run. Let’s start with some fundamental equations that will build a model of labor-force dynamics that shows what determines the natural rate. L = E + U L = E + U Labor force Labor force is composed of is composed of Number of employed workers Number of employed workers Number of unemployed workers Number of unemployed workers
3 Chapter Six f f U U = = E E Number of people finding jobs Number of people finding jobs Number of people loosing jobs Number of people loosing jobs Steady-state unemployment rate From an earlier equation, we known that E = L – U, that is the number of employed equals the labor force minus the number of unemployed. If we substitute ( L-U ) for E in the steady-state condition, we find: f f U = U = (L – U) (L – U) Then, divide both sides by L and to obtain: f f U/L = U/L = (1-U/L) (1-U/L) Now solve for U/L for find : U/L = U/L = s / (s + f) s / (s + f)
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4 Chapter Six U/L = U/L = s / (s + f) s / (s + f)
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# Search by Topic
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### Man Food
##### Stage: 2 and 3 Challenge Level:
Sam displays cans in 3 triangular stacks. With the same number he could make one large triangular stack or stack them all in a square based pyramid. How many cans are there how were they arranged?
### Sam Again
##### Stage: 3 Challenge Level:
Here is a collection of puzzles about Sam's shop sent in by club members. Perhaps you can make up more puzzles, find formulas or find general methods.
### Handshakes
##### Stage: 3 Challenge Level:
Can you find an efficient method to work out how many handshakes there would be if hundreds of people met?
### Picturing Triangular Numbers
##### Stage: 3 Challenge Level:
Triangular numbers can be represented by a triangular array of squares. What do you notice about the sum of identical triangle numbers?
### Mystic Rose
##### Stage: 4 Challenge Level:
Use the animation to help you work out how many lines are needed to draw mystic roses of different sizes.
### Take Ten Sticks
##### Stage: 3 and 4 Challenge Level:
Take ten sticks in heaps any way you like. Make a new heap using one from each of the heaps. By repeating that process could the arrangement 7 - 1 - 1 - 1 ever turn up, except by starting with it?
### Reasonable Algebra
##### Stage: 4 Challenge Level:
Use algebra to reason why 16 and 32 are impossible to create as the sum of consecutive numbers.
### Route to Infinity
##### Stage: 3 Challenge Level:
Can you describe this route to infinity? Where will the arrows take you next?
### Triangle Numbers
##### Stage: 3 Challenge Level:
Take a look at the multiplication square. The first eleven triangle numbers have been identified. Can you see a pattern? Does the pattern continue?
### Satisfying Statements
##### Stage: 2 and 3 Challenge Level:
Can you find any two-digit numbers that satisfy all of these statements?
### Iff
##### Stage: 4 Challenge Level:
Prove that if n is a triangular number then 8n+1 is a square number. Prove, conversely, that if 8n+1 is a square number then n is a triangular number.
### Slick Summing
##### Stage: 4 Challenge Level:
Watch the video to see how Charlie works out the sum. Can you adapt his method?
### Factors and Multiples Puzzle
##### Stage: 3 Challenge Level:
Using your knowledge of the properties of numbers, can you fill all the squares on the board?
### Forgotten Number
##### Stage: 3 Challenge Level:
I have forgotten the number of the combination of the lock on my briefcase. I did have a method for remembering it...
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# How many Numbers can be formed using 0,1,2,3,5,7,8
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Intern
Joined: 03 Nov 2019
Posts: 34
How many Numbers can be formed using 0,1,2,3,5,7,8 [#permalink]
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26 Nov 2019, 11:58
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### HideShow timer Statistics
How many Numbers can be formed using 0,1,2,3,5,7,8
Such that 300< Number< 3000 and repetition is not allowed?
A) 360
B) 450
C) 340
D) 180
E) 570
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Re: How many Numbers can be formed using 0,1,2,3,5,7,8 [#permalink]
### Show Tags
26 Nov 2019, 12:33
3
2
for 3-digit numbers, the possible combinations are:
3xy , 5xy , 7xy , 8xy
so we can see that the possibilities of the 3-digts are 4*6*5 = 120
for 4-digit numbers, the possible combinations are:
1xyz , 2xyz
so we can see that the possibilities of 4-digits are 2*6*5*4 = 240
and the total = 360
##### General Discussion
Intern
Joined: 25 Nov 2019
Posts: 10
Re: How many Numbers can be formed using 0,1,2,3,5,7,8 [#permalink]
### Show Tags
26 Nov 2019, 12:14
I got A
but I also didn't score well so we'll see
XXX has to be greater than 300 so the 100 spot can be 3,5,7,8 (4) for the 10 spot all are open minus 1 so that is (6) and the 1's all are open less 2 so (5) 4*6*5 = 120
XXXX is the same 1000 can be 2 or 1 (2) 100's all minus 1 (6) 10's (5) 1's (4) = 240
120 + 240 = 360 or choice A
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Re: How many Numbers can be formed using 0,1,2,3,5,7,8 [#permalink]
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28 Nov 2019, 04:26
what is the source of this question
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Joined: 16 Nov 2013
Posts: 3
How many Numbers can be formed using 0,1,2,3,5,7,8 [#permalink]
### Show Tags
28 Nov 2019, 04:56
Possible 3-digit numbers => 4×6×5 = 120
Possible 4-digit numbers => 2×6×5×4 = 240
Total possible numbers => (120+240) = 360
How many Numbers can be formed using 0,1,2,3,5,7,8 [#permalink] 28 Nov 2019, 04:56
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