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You start with the understanding of this formula: $(A-\lambda I)\vec x=0$, which is equivalent to $\det(A-\lambda I)=0$ $$\begin{vmatrix}-3-\lambda&6&-43\\0&-1-\lambda&9\\0&0&2-\lambda\end{vmatrix}=(-3-\lambda)(-1-\lambda)(2-\lambda)=0$$ Therefore, $\lambda_1=-3, \ \lambda_2=-1, \ \lambda_3=2$.
Let's do one example for eigenvectors:
Plug in the value of $\lambda$ into the augmented form of the matrix:
With $\lambda_1=-3$, $$\left[\begin{array}{ccc|c}-3-(-3)&6&-43&0\\0&-1-(-3)&9&0\\0&0&2-(-3)&0\end{array}\right]=\left[\begin{array}{ccc|c}0&6&-43&0\\0&2&9&0\\0&0&5&0\end{array}\right]$$ Solve this matrix and get $v_1=\begin{bmatrix}1\\0\\0\end{bmatrix}$
Now you can use similar approach to find the eigenvectors of the next two eigenvalues.
• @Yusha, double check your calculations. At least for $\lambda=-1$, I did not get $(-3,0,0)$ – Ron Jul 14 '16 at 2:18
• I'm lost, how are you getting (1,0,0). You have in your work (0,0,0). – Yusha Jul 14 '16 at 2:43
• Both answers are correct, since one is a scalar multiple of the other. But it is customary to use numbers are small as possible, so $(1,0,0)$ would be prefered. – imranfat Jul 14 '16 at 2:47
• Does (1,0,0) @imranfat come from the 2nd column after its been put in RREF? – Yusha Jul 14 '16 at 2:50
• RREF has to give infinite solutions, I looked at Sophia's work, realizing that both vectors serve as the same eigenvector. Your RREF must have all zero's in bottom row – imranfat Jul 14 '16 at 2:55
Hint: In an upper triangular matrix the characteristic polynomial is $(x_1-\lambda_1)^{\alpha_1}(x_2-\lambda_2)^{\alpha_2}\ldots(x_n-\lambda_n)^{\alpha_n}$, where $x_i$ are diagonal entries with $\alpha$'s as their multiplicities.
In your case the characteristic polynomial is $(3+\lambda)(1+\lambda)(2-\lambda)$ | {
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In your case the characteristic polynomial is $(3+\lambda)(1+\lambda)(2-\lambda)$
• Here is one: sosmath.com/matrix/eigen2/eigen2.html – imranfat Jul 14 '16 at 2:27
• Khan academy usually rocks: khanacademy.org/math/linear-algebra/alternate-bases/… – imranfat Jul 14 '16 at 2:28
• OK, so I assume that you DO know how to find eigenvalues theoretically, but somehow your algebra is letting you down? It would be helpful to see your work in attempting finding the eigenvalue, because a lot of times it is a small algebraic mistake... – imranfat Jul 14 '16 at 2:33
• Here is a worked out example of a 2 by 2 matrix: calvin.edu/~scofield/courses/m256/materials/eigenstuff.pdf In my view you should first be fluent with 2by2's before going for 3by3's... – imranfat Jul 14 '16 at 2:34
• It sounds like your exchange converged. Anyway, this conversation has been moved to chat. If you need to revisit the full exchange of comments, go to that chatroom. I left the posts with links to resources also here. – Jyrki Lahtonen Jul 14 '16 at 6:26 | {
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# By using De Moivre's Theorem, show that $\cos5\theta = 16\cos^5 \theta - 20 \cos^3 \theta + 5 \cos \theta$ [closed]
First step is
$$\cos5\theta + i \sin 5\theta = (\cos \theta + i \sin \theta)^5$$
thanks
• Because it's easier than using using other trigonometric formulae. Note that the real parts of the expressions on each side of the equation must be equal. – user164587 Sep 26 '14 at 14:39
• @user164587 well i meant how to get it? – problematic Sep 26 '14 at 14:41
• It is de Moivre – Seub Sep 26 '14 at 14:41
• @vera sorry bad english – problematic Sep 26 '14 at 14:41
Hint: Expand $(\cos\theta + i\sin\theta)^5$ using binomial theorem. i.e. $(a+b)^n = \sum\limits_{k = 0}^n {n \choose k}a^k b^{n-k}$
Here, $a = \cos\theta, b=i\sin\theta$.
\begin{align}(\cos\theta+i\sin\theta)^5 &= \sum\limits_{k = 0}^5 {5\choose k}(\cos\theta)^k (i\sin\theta)^{5-k}\\ &= {5\choose0}(\cos\theta)^0(i\sin\theta)^5 + {5\choose1}(\cos\theta)^1(i\sin\theta)^4 + {5\choose2}(\cos\theta)^2(i\sin\theta)^3\\&\quad+ {5\choose3}(\cos\theta)^3(i\sin\theta)^2+{5\choose4}(\cos\theta)^4(i\sin\theta)^1+{5\choose5}(\cos\theta)^5(i\sin\theta)^0 \\&= (i\sin\theta)^5 + 5(\cos\theta)(i^4\sin^4\theta) + 10\cos^2\theta(i^3\sin^3\theta)\\&\quad+10\cos^3\theta(i^2\sin^2\theta)+5\cos^4\theta(i\sin\theta)+\cos^5\theta \\&=\cdots \end{align}
Then, for even powers of $\sin \theta$, use $\sin^2 \theta = 1-\cos^2 \theta$.
Finally, equate the real parts of $\cos(5\theta)+i\sin(5\theta) = (\cos\theta+i\sin\theta)^5$. | {
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Finally, equate the real parts of $\cos(5\theta)+i\sin(5\theta) = (\cos\theta+i\sin\theta)^5$.
• am i doing it right ? lol $(_0^5)$$(cos^0 \theta)$$(i sin^5 \theta)$ – problematic Sep 26 '14 at 15:03
• Yes, and there will be $5$ more terms. – taninamdar Sep 26 '14 at 15:05
• actually i don't know how binomial theoram works, can you give an example and show me what $(_0^5)$$(cos^0 \theta)$$(sin^5 \theta)$ goes to, i will do the rest my self – problematic Sep 26 '14 at 15:06
• See the edit, ask if you need more help. – taninamdar Sep 26 '14 at 15:13
• thank you, how to get the first number? i mean it looks like $(_1^5)$ : 5x1, $(_2^5)$ 5x2 , then $(_3^5)$ onward? – problematic Sep 26 '14 at 15:22
The point is that $$\cos 5 \theta$$ is the even part of $$\exp(5 i \theta) = [\exp(i \theta)]^5 = (\cos \theta + i \sin \theta)^5,$$ where the second equality follows from de Moivre's Theorem. Then, by expanding the right-hand side using the Binomial Theorem, we can immediately write $\cos 5 \theta$ as a sum of products of powers of $\sin \theta$ and $\cos \theta$. Then, we can eliminate even powers of $\sin \theta$ using the Pythagorean identity $$\sin^2 \theta + \cos^2 \theta = 1.$$
• You're welcome, I hope you found it helpful. – Travis Willse Sep 26 '14 at 15:37
• yes, you showed me exp(5iθ)=[exp(iθ)]5=(cosθ+isinθ)5, i needed this to know why we have the first step, i can't choose 2 accepted answers... – problematic Sep 26 '14 at 15:40
• No problem, that's just a part of the site mechanics, the important thing is that you understand the concepts you asked about better than you did before. – Travis Willse Sep 26 '14 at 15:51
De Moivre's Theorem states $$\cos n\theta + \mathbb i \sin n\theta = \left( \cos \theta + \mathbb i \sin \theta \right)^n$$ | {
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×
# Proof
Show that, for any positive integer $$n\geq 1$$ it is true that: $\displaystyle \sum_{k=1}^{n} \sqrt{k} \geq \sqrt{\displaystyle \sum_{k=1}^{n} k}.$
Note by Hjalmar Orellana Soto
10 months, 3 weeks ago
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for $$n\geq 2$$ it can be easily proven by applying triangle inequality recursively
assume a,b,c are three terms in the seuence
we have $$\sqrt{a}+\sqrt{b}\geq \sqrt{a+b}$$ as $$\sqrt{a},\sqrt{b},\sqrt{a+b}$$ form sides of a right triangle
now we have $$\sqrt{a}+\sqrt{b}+\sqrt{c}\geq \sqrt{a+b}+\sqrt{c}\geq\sqrt{a+b+c}$$ and so on....
sum of each of the outer sides of the spiral form the LHS of the equation,and RHS is given by the line to centre
also note that for n=1 the line to centre is same as side, the only point of equality
- 10 months, 1 week ago
very nice solution, Anirudh
- 10 months, 1 week ago
thank you :)
- 10 months, 1 week ago
$$LHS^2 = 1+2+\cdots+k + 2\sum_{1\le x<y\le n}\sqrt{xy}\ge 1+2+\cdots+n = RHS^2$$ and equallity occurs when $$n=1$$
- 10 months, 3 weeks ago | {
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- 10 months, 3 weeks ago
In addition to the problem.... Is this necessary information for saying that the limit below converges? $\lim_{x\to \infty} \dfrac{\displaystyle \sqrt{\sum_{n=1}^{x}n}}{\displaystyle \sum_{n=1}^{x}\sqrt{n}}$
- 10 months, 3 weeks ago
(Assuming I'm interpreting you correctly,) the essence of what you're asking here is:
If $$a_n, b_n$$ are sequences of (positive) reals, such that $$a_n \geq b_n$$, does $$\lim \frac{ b_n } { a_n }$$ exist?
If yes, why?
If no, what is a counter example? What other assumptions do we need to add to guarantee a true statement?
Staff - 10 months, 3 weeks ago
Exactly that is what I ask, I was thinking of asking for the value of $\displaystyle \sum_{n=1}^{\infty} (\frac{\sqrt{\sum_{k=1}^{n}n}}{\sum_{k=1}^{n}\sqrt{n}})$ but I really don't know how to approach it
- 10 months, 3 weeks ago
The answer to "does $$\lim \frac{ b_n } { a_n }$$ exist?" is no for general sequences. Do you see an obvious counter example?
For your recent comment, are you looking for the summation, or for the limit of the term (as in the prior comment)?
Staff - 10 months, 3 weeks ago
The answer to the limit question is yes for this case, using the same argument which is used for $$\lim_{x\to \infty} \frac{x}{e^x}$$.... and I'm looking for the summation
- 10 months, 3 weeks ago
See How to ask for help to understand how to provide proper context of what you want. E.g. You can see that people quickly established the inequality, but that doesn't help with your concern.
Staff - 10 months, 3 weeks ago
The summation is something else, something I supposed and now I'm trying to approach, and I think the inequality may help for knowing if the summation converges... and then I'll try to approach the value of the summation, I don't know if I'm explaining right...
- 10 months, 3 weeks ago | {
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- 10 months, 3 weeks ago
Asymptotically, $$\sqrt{1} + \sqrt{2} + \dots + \sqrt{n}$$ can be approximated by the integral $\int_0^n \sqrt{x} \ dx = \frac{2}{3} n^{3/2}.$ And of course, $\sqrt{1 + 2 + \dots + n} = \sqrt{\frac{n(n + 1)}{2}}.$ That should help answer your question.
- 10 months, 3 weeks ago
We want to prove that $\sqrt{1} + \sqrt{2} + \dots + \sqrt{n} \ge \sqrt{1 + 2 + \dots + n}.$ Squaring the left-hand side, we get a term of $$\sqrt{k} \cdot \sqrt{k} = k$$ for each $$1 \le k \le n$$. The inequality follows.
- 10 months, 3 weeks ago
× | {
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# On the number of roots of the polynomial $x^3+Ax^2+1=0$
I have the following cubic equation
$$x^{3}+Ax^{2}+1=0$$
where $$A$$ is an arbitrary (real) number.
I know that either:
• The 3 roots will be real.
• One root will be real and the other two will be complex conjugates of each other.
I would like to find out
• For what value/values of A the roots change from 3 real roots to one real and two complex roots.
• The signs of each of the real roots (both when they are all real and when there is only one real root)
Is there an analytical way of finding this as a function of $$A$$ or the only option is to solve the cubic numerically?
• Google Cardano's method. – hamam_Abdallah Jan 6 at 19:44
If we can show that the value of $$A$$ changes the amount of maxima or minima from 1 to 2 such that one extreme is at a positive $$y$$-location and the other at a zero or negative $$y$$-location, then we know that this value of $$A$$ is the border for all real solutions versus 1 real and 2 complex solutions.
Denote $$f(x) = x^3 + Ax^2 + 1.$$
$$f'(x) = 3x^2+2Ax=x(3x+2A) = 0.$$
We see that $$f(0)$$ is always either a local maximum or minimum and $$f(0) > 0.$$ The other zero is given by $$x = -2A/3 = \xi.$$
We need $$f(\xi) \leq 0.$$
$$f(\xi) = \frac{4}{27}A^3 + 1 \leq 0$$
solves for the values of $$A$$ where 3 real solutions are guaranteed.
If there are 3 real roots, then since $$x = 0$$ is always a local extreme, at least one has to be positive and another has to be negative. Also, if $$A \leq -3/\sqrt[3]{4}$$ and the location of the moving extreme is $$x=-2A/3 > 0,$$ then the remaining real root has to be positive.
If there's only 1 real root, then $$A > -3/\sqrt[3]{4}$$ and the moving extreme is located at $$x < \sqrt[3]{2}.$$ Thus the only real root must be negative. | {
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• How do we find the signs of the real roots? By actually finding them or is there a shortcut? – JennyToy Jan 6 at 20:21
• @JennyToy I should've reread the questions before submitting the first time, sorry! How's the edit? – AEngineer Jan 6 at 20:45
hint
Put $$A=-\frac 32B$$ and $$f(x)=x^3-\frac 32Bx^2+1.$$
$$f'(x)=3x(x-B).$$
$$f(0)=1$$ to have three real roots, we need
$$B>0 \text{ and } f(B)<0.$$
• To find when there are 3 real roots or one only use the sign of the discriminant $\Delta = -4A^{3}-27$. Thus if $A=-3/2^{2/3}$ there are 3 equal real roots. If $A<-3/2^{2/3}$ there are 3 real distinct roots. And if $A>-3/2^{2/3}$ there is one real root and two (complex conjugate) complex roots. Unfortunately to find the signs of the real roots we probably need to actually find the roots. – user2175783 Jan 6 at 20:13
• @user2175783 If $A=-\frac{3}{2^{2/3}}$, then there are a negative zero and a double zero on the positive axis. – Hanno Jan 12 at 22:17
The stationary points of $$f(x)$$ lie at $$x=0$$ and at $$x=-\frac{2A}{3}$$ and the sign of $$f(0)f\left(-\tfrac{2A}{3}\right) = 1+\tfrac{4}{27}A^3$$ decides if there are three real roots (<0), a simple real root (>0) or a double real root and a simple real root (=0). This happens since the sign of $$f(0)f\left(-\tfrac{2A}{3}\right)$$ only depends on the stationary values having the same sign or not.
Too late at the party. But it looks good, so let's enter anyway.
TL;DR $$\quad f(x)=x^3+Ax^2+1\,$$ always has a simple zero on the negative axis. The other two zeros are
• real, positive and distinct, if $$\,A<-1.88988\:=\:-\dfrac{3}{\sqrt[3]{4}}\:=\:A_1$$
• merging into a double zero at $$x=\sqrt[3]{2}$$, if $$\,A=A_1$$
• non-real (thus complex-conjugate), if $$\,A>A_1$$ | {
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The party-giver also asked if there is an analytical way of finding this as a function of $$A$$?
Let's take a step in this direction, concentrating on the case of "one real zero", and express the negative zero as a function of $$A$$:
Let $$\,d=\big(\frac A3\big)^3 + \big(\frac 12\big)^2$$, and assume $$d>0$$ which corresponds to $$\,A>A_1$$. Define $$r \:=\:\left(-\frac 12 -\sqrt d\right)^\frac 13 +\, \left(-\frac 12 +\sqrt d\right)^\frac 13\,,$$ $$r\,$$ is observed to be strictly negative. Then \begin{align} r^2\:= & \;\left(-\frac 12 -\sqrt d\right)^\frac 23 +\, \left(-\frac 12 +\sqrt d\right)^\frac 23 -\frac23 A \\[2ex] r^3\:= & \; -\frac23 Ar + \left(-\frac 12 -\sqrt d\right) +\, \left(-\frac 12 +\sqrt d\right) \\ & \quad -\frac13 A \left\{ \left(-\frac 12 -\sqrt d\right)^\frac 13 +\, \left(-\frac 12 +\sqrt d\right)^\frac 13\right\} \\ = & -Ar - 1 \end{align} Whence $$1+Ar+r^3=0\quad\Longleftrightarrow\quad\frac1{r^3}+A\frac1{r^2}+1 =0 \\[2ex] \Longrightarrow\quad x \:=\: \frac1r \:=\: -r^2-A \:=\: -\frac13 A -\left(\sqrt d + \frac 12\right)^\frac 23 -\, \left(\sqrt d -\frac12\right)^\frac 23$$ satisfies $$x^3+Ax^2+1=0$$.
Added in Edit: Driven by user2175783's comment as of below, I found Lecture 4 in Mathematical Omnibus: 30 Lectures on Classic Maths by D. Fuchs + S. Tabachnikov: It gives an extensive and readable account of how to explicitly solve the cubic and quartic equations.
Visual TL;DR
• Wouldnt it be possible to generalize the algebra for the negative root for $A<A_{1}$? – user2175783 Jan 13 at 0:14
• @user2175783 I did not look closer at that case. I'd say there's no ambiguity for the sign of $\sqrt d$ because $r$ is a symmetrical expression. But afterwards complex cube roots have to be considered which might involve some choice. – Hanno Jan 13 at 9:31 | {
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The discriminant of a cubic polynomial with real coefficients is $$0$$ when the polynomial has two (or more) equal roots, positive when there are three distinct real roots, and negative when there are one real and two non-real roots. In your case it is $$-4 A^3 - 27$$. | {
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# Find all triplets (x,y,z)
#### anemone
##### MHB POTW Director
Staff member
Find all triplets of positive integers $(x, y, z)$ such that $$\displaystyle \left( 1+\frac{1}{x} \right)\left( 1+\frac{1}{y} \right)\left( 1+\frac{1}{z} \right)=2$$.
Last edited:
##### Well-known member
Re: Find all triplets of x, y and z
Find all triplets of positive integers $(x, y, z)$ such that $$\displaystyle \left( 1+\frac{1}{x} \right)\left( 1+\frac{1}{y} \right)\left( 1+\frac{1}{z} \right)=2$$.
(x,y,z ) = (2, 4, 15) , ( 2, 5, 9) , (2,6,7), (3,3,8) , ( 3,4,5) or any permutation of them
solved as
Without loss of generality we can choose x <=y <=z and ans shall be a permutation of if
Now x < 4 as (5/4)^3 < 2 ( it is 125/64)
x cannot be 1 as (1+1/y)(1+ 1/z) = 1 has no solution
so x = 2 or 3
if x = 2 we get
3/2(y+1)(z+1) = 2yz
Or 3(y+1)(z+1) = 4 yz
Or yz – 3y – 3z = 3
(y-3)(z-3) = 12 by adding 9 on both sides
( y-3)(z-3) = 1 * 12 or 2 * 6 or 3 * 4
So (x,y,z ) = (2, 4, 15) , ( 2, 5, 9) , (2,6,7)
if x = 3 we get
4/3(y+1)(z+1) = 2yz
Or 2(y+1)(z+1) = 3 yz
Or yz – 2y – 2z = 2
(y-2)(z-2) = 6
( y-2)(z-2) = 1 * 6 or 2 * 3
So (x,y,z ) = (3,3,8) , ( 3,4,5)
#### anemone
##### MHB POTW Director
Staff member
Re: Find all triplets of x, y and z
(x,y,z ) = (2, 4, 15) , ( 2, 5, 9) , (2,6,7), (3,3,8) , ( 3,4,5) or any permutation of them
solved as
Without loss of generality we can choose x <=y <=z and ans shall be a permutation of if
Now x < 4 as (5/4)^3 < 2 ( it is 125/64)
x cannot be 1 as (1+1/y)(1+ 1/z) = 1 has no solution
so x = 2 or 3
if x = 2 we get
3/2(y+1)(z+1) = 2yz
Or 3(y+1)(z+1) = 4 yz
Or yz – 3y – 3z = 3
(y-3)(z-3) = 12 by adding 9 on both sides
( y-3)(z-3) = 1 * 12 or 2 * 6 or 3 * 4
So (x,y,z ) = (2, 4, 15) , ( 2, 5, 9) , (2,6,7)
if x = 3 we get
4/3(y+1)(z+1) = 2yz
Or 2(y+1)(z+1) = 3 yz
Or yz – 2y – 2z = 2
(y-2)(z-2) = 6
( y-2)(z-2) = 1 * 6 or 2 * 3
So (x,y,z ) = (3,3,8) , ( 3,4,5) | {
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Or yz – 2y – 2z = 2
(y-2)(z-2) = 6
( y-2)(z-2) = 1 * 6 or 2 * 3
So (x,y,z ) = (3,3,8) , ( 3,4,5)
Wow! I forgot to thank you explicitly for participating and also for your unique way of attacking the problem. What's worst is that I forgot completely how I approached it two months ago and I promise you to add my reply once I solved it because I could only tell at this point that I solved it differently than what you did. Sorry, kaliprasad! | {
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# [SOLVED]2 coins are thrown 20 times
#### mathmari
##### Well-known member
MHB Site Helper
Hey!!
2 coins are thrown 20 times. I want to calculate the probability
(a) to achieve exactly 5 times the Tail/Tail
(b) to achieve at least 2 times Tail/Tail
If we throw the 2 coins once the probability that we get Tail/Tail is equal to $\frac{1}{2}\cdot \frac{1}{2}=\frac{1}{4}$, or not?
Is the probability then at (a) equal to $\left (\frac{1}{4}\right )^{20}$ ?
Could you give me a hint for (b) ?
#### Klaas van Aarsen
##### MHB Seeker
Staff member
Hey!!
2 coins are thrown 20 times. I want to calculate the probability
(a) to achieve exactly 5 times the Tail/Tail
(b) to achieve at least 2 times Tail/Tail
If we throw the 2 coins once the probability that we get Tail/Tail is equal to $\frac{1}{2}\cdot \frac{1}{2}=\frac{1}{4}$, or not?
Hey mathmari !!
Yep.
Is the probability then at (a) equal to $\left (\frac{1}{4}\right )^{20}$ ?
There seems to be a $5$ missing.
It's a binomial distribution, so we should use the corresponding formula, which includes a spot to put the $5$.
Could you give me a hint for (b) ?
Take the complement?
It's $1$ minus the probability of getting either 0 Tail/Tail or 1 Tail/Tail.
#### mathmari
##### Well-known member
MHB Site Helper
There seems to be a $5$ missing.
It's a binomial distribution, so we should use the corresponding formula, which includes a spot to put the $5$.
Oh yes.. So do we have the following? $$\binom{20}{5}\cdot \left (\frac{1}{4}\right )^5\cdot \left (1-\frac{1}{4}\right )^{20-5}$$
Take the complement?
It's $1$ minus the probability of getting either 0 Tail/Tail or 1 Tail/Tail.
So, we have $$P(X\geq 2)=1-P(X<2)=1-P(X=1)-P(X=0)=1-\binom{20}{1}\cdot \left (\frac{1}{4}\right )^1\cdot \left (1-\frac{1}{4}\right )^{20-1}-\binom{20}{0}\cdot \left (\frac{1}{4}\right )^0\cdot \left (1-\frac{1}{4}\right )^{20-0}$$ or not?
#### Klaas van Aarsen
##### MHB Seeker
Staff member
Yep. All correct.
#### Wilmer | {
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#### Klaas van Aarsen
##### MHB Seeker
Staff member
Yep. All correct.
#### Wilmer
##### In Memoriam
So do we have the following? $$\binom{20}{5}\cdot \left (\frac{1}{4}\right )^5\cdot \left (1-\frac{1}{4}\right )^{20-5}$$
Looks good.
Ran a simulation: got 202276 out of 1 million.
MHB Site Helper
Thank you!! | {
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# Thread: Problem #8 - Number Theory
1. ## Problem #8 - Number Theory
I got this one from our archives as well.
Show that $n^4 + 4^n$ is not prime for $n \geq 2$. (n is a positive integer!)
This is a more of a Number Theory problem than algebraic. As my proof relies more on a College Algebra level rather than Number Theory I'm curious to see what you might come up with.
-Dan
2. ## Re: Problem #8 - Number Theory
all n values that are even will give a non-prime number, as $(evennumber)^4$ is always even and $4^n$ is always even. even+even gives even numbers bigger than 2, which are never prime. i can't continue from here though, dont know about the odd n values
3. ## Re: Problem #8 - Number Theory
Originally Posted by muddywaters
all n values that are even will give a non-prime number, as $(evennumber)^4$ is always even and $4^n$ is always even. even+even gives even numbers bigger than 2, which are never prime. i can't continue from here though, dont know about the odd n values
if n is an odd integer >= 3 and n is not a multiple of 5 then n^4 + 4^n is a multiple of 5 and therefore not a prime
Proof:
n is 1,2,3,or 4 (mod 5)
n^4 is 1 (mod 5)
4^n is -1 (mod 5)
So n^4 + 4^n is congruent to 0 (mod 5)
How do we handle the case n=5, 15, 25, 35, 45, ...... ?
4. ## Re: Problem #8 - Number Theory
If $n\equiv 0\pmod{2}$, then $n^4+4^n \equiv 0\pmod{2}$.
If $n\equiv 1\pmod{2}$, then $n+1\equiv 0\pmod{2}$, so $2^{n+1}$ is a perfect square, and $n^4+4^n$ factors as
$(n^2+n\sqrt{2^{n+1}}+2^n)(n^2-n\sqrt{2^{n+1}} + 2^n)$
So, all that is left is to show that for odd $n>2$, both terms are greater than 1. Since $n^2+2^n>1$, the first term is guaranteed to be greater than 1. So, we just need to show the second term is greater than one.
Since $(n^2-1)^2+2^{2n} \ge 2^{2n} > 2^{n+1}$ (for all $n>1$), we have | {
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Since $(n^2-1)^2+2^{2n} \ge 2^{2n} > 2^{n+1}$ (for all $n>1$), we have
\begin{align*}n^4-2n^2+1+4^n-2^{n+1} > 0 & \Rightarrow n^4-2n^2+1+4^n-2^{n+1}+n^2 2^{n+1} > n^2 2^{n+1} \\ & \Rightarrow (n^2+2^n-1)^2 > n^2 2^{n+1} \\ & \Rightarrow n^2 + 2^n - 1 > n\sqrt{2^{n+1}} \\ & \Rightarrow n^2-n\sqrt{2^{n+1}}+2^n > 1\end{align*}
5. ## Re: Problem #8 - Number Theory
If n even, obvious.
If n odd, 2n+1 is a perfect square and the factorization are really neat observations. Thanks.
But what's the point of the congruences?
6. ## Re: Problem #8 - Number Theory
Originally Posted by Hartlw
But what's the point of the congruences?
$n\equiv 0\pmod{2}$ is the same as saying $n$ is even.
$n\equiv 1\pmod{2}$ is the same as saying $n$ is odd.
I have been working with modular arithmetic long enough that it feels more natural for me to use congruences rather than the words even and odd. The meaning is the same either way.
7. ## Re: Problem #8 - Number Theory
You have shown n^4+4^n=rs where r and s are integers and r positive. Obviously s is positive.
8. ## Re: Problem #8 - Number Theory
Originally Posted by Hartlw
You have shown n^4+4^n=rs where r and s are integers and r positive. Obviously s is positive.
If $n=1$, then the factorization I gave is $1^4+4^1 = (5)(1)$. However, 5 is a prime number. The problem asks to prove that for $n\ge 2$, $n^4+4^n$ is NOT prime. So, to do that, it is not enough to show the factors are positive. You must show they are both greater than 1.
9. ## Re: Problem #8 - Number Theory
Hey, lots of action on this one! Thanks to all.
SlipEternal gets some extra kudos for checking the "trivial" factorization problem. (That one of the factors might be 1.) I did eventually find the proof online (which was pretty much identical to Slip's and my own) but didn't address the trivial factorization problem.
-Dan | {
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What is the probability of getting three pairs that makes up seven when six dice are rolled?
Stumbled upon this question in a game of dice where a point is dealt for each pair that makes seven. In the game there were a total of six dice that were to be thrown up to three times. Each time a pair of dice made seven 1+6, 2+5 or 4+3), the pair(s) was taken out from the rest, a point was dealt and you got to roll the remaining dice. You can not use dice from the excluded pairs to make up new ones. You win the game by getting the most points/pairs in 3 rounds.
Example:
round one [1] [1] [1] [3] [5] [6] - one pair make 7 (1+6)
round two [1] [3] [4] [2] - one pair make 7 (3+4)
round one [2] [6] - no pairs make 7
points awarded: 2
Out of curiosity I have been trying to calculate the probabilities of getting three pairs of sevens in the first round?
• We have six dice and we throw them all together in the first round. Let's say the faces are $3,4,5,2,6,6$. Because there are two pairs that make $7$, we remove the pairs and continue with two dice. Round 2 and 3 continues this way. What is the definition of winning? Is it the situation that we're left with no dice at the end? – gunes Nov 10 '19 at 18:18
• and how is it possible to roll three dice following those rules? – carlo Nov 10 '19 at 18:39
• Your question is unclear. Your title refers to three dice while your body text refers to six. You describe a game where the number of dice change, altering the probabilities. At which point in that game are you calculating the chance of getting two dice that add to 7? – Glen_b Nov 10 '19 at 21:02
• Sorry if the original edit was confusing, I have tried to clarify. The game description is just for context. The question is in the title and reformulated in the last paragraph. – joho Nov 13 '19 at 14:28 | {
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Consider a sequence of $$2k$$ dice each with the possible values $$1, 2, \ldots, 2n.$$ (The question concerns $$k=n=3.$$) The possible pairs are $$\{1,n\},$$ $$\{2,2n-1\},$$ and so on, through $$\{n,n+1\}.$$ Denote such a pair by its smallest value $$i$$ and for each $$i$$ let $$k_i\gt 0$$ be the number of such pairs that can be located in the sequence. We need to count the number of equally probable sequences for which $$k_1+k_2+\cdots + k_n = k.$$
Such a sequence is determined by (a) which pairs occur in it and (b) where in the sequence of $$2k$$ values each such pair occurs. The permutation group on $$2k$$ elements acts on the set of such sequences. For all $$i,$$ the stabilizer of such a sequence permutes the $$k_i$$ values of $$i$$ among themselves and the $$k_i$$ values of $$2n+1-i$$ among themselves. Thus, applying the method described at https://stats.stackexchange.com/a/415878/919, the number of ways of producing a sequence designated by $$\mathrm{k}=(k_1,k_2,\ldots,k_n)$$ is
$$p(\mathrm{k}) = \frac{(2k)!}{(k_1!)^2(k_2!)^2\cdots(k_n!)^2}.$$
Thus, the chance is obtained by summing $$p(\mathrm{k})$$ over all possible $$\mathrm{k}$$ whose components sum to $$k$$ and dividing by the total number of sequences, $$(2n)^{2k}.$$
These possibilities correspond to the weak compositions of $$k$$ into $$n$$ parts, which number $$\binom{k+n-1}{n-1}.$$ However, the amount of calculation is smaller than this, because $$p(\mathrm{k})$$ does not depend on the order of the $$k_i.$$ We may therefore do the calculation for all $$k_1\ge k_2\ge \cdots \ge k_n \ge 0$$ (giving a partition of $$k$$), multiplying each by the number of distinct re-orderings of $$\mathrm k.$$ Such sequences correspond to the Ferrers diagrams for $$n$$ having at most $$k$$ rows. They are relatively easy to enumerate.
With $$k=n=3,$$ we have $$\binom{3+3-1}{3-1}=10$$ possibilities for $$\mathrm k,$$ but they fall into just three groups corresponding to the partitions $$3 = 2+1 = 1+1+1:$$ | {
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$$p(3,0,0)=p(0,3,0)=p(0,0,3) = \frac{6!}{3!^2} = 20;$$ $$p(2,1,0)=p(2,0,1)=p(1,2,0)=p(1,0,2)=p(0,2,1)=p(0,1,2) = \frac{6!}{2!^21!^2}= 180;$$ $$p(1,1,1) = \frac{6!}{1!^21!^21!^2} = 720.$$
$$\Pr(\text{three pairs}) = \frac{3\times 20 + 6\times 180 + 1\times 720}{6^6} = \frac{(5)(31)}{(2^4)(3^5)} \approx 3.9866\%.$$
• @Ron I haven't thoroughly checked my general answer, but I did check the specific answer with a brute force enumeration of all possibilities. At least one mistake in your post is that many sequences without three pairs will sum to 21, such as 5,5,5,4,1,1, which has no pairs at all. Thus you substantially overcount. Here is my implementation: n <- 3; k <- 3; X <- expand.grid(rep(list(seq_len(2*n)), 2*k)); p <- function(x, d=n) { x <- sort(x); sum((x + rev(x))[seq_len(d)] == 2*d+1) }; table(apply(X, 1, p)) – whuber Nov 13 '19 at 19:04
• @Ron If you would like to explore the situation in more detail, this code (which follows the creation of data frame X in my previous comment) breaks the population down into all possible pair patterns: pattern <- function(x, d=n) { x <- sort(x); paste0(x[which((x + rev(x))[seq_len(d)] == 2*d + 1)], collapse="") }; table(apply(X, 1, pattern)) – whuber Nov 13 '19 at 19:19 | {
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# Choosing a linear map $(\mathbb{Z}/2\mathbb{Z})^n \rightarrow \mathbb{Z}/2\mathbb{Z}$ which is nonzero on half of a sequence of vectors
Let $v_1,\ldots,v_m \in (\mathbb{Z}/2\mathbb{Z})^n$ be nonzero vectors. Is it always possible to choose a linear map $f : (\mathbb{Z}/2\mathbb{Z})^n \rightarrow \mathbb{Z}/2\mathbb{Z}$ such that $f$ is nonzero on at least half of the $v_i$, i.e. such that
$$|\{\text{i | 1 \leq i \leq m and f(v_i) \neq 0}\}| \geq \frac{1}{2}m?$$
My guess is that the answer is yes; at the very least, it is true for $m=1$ and when the $v_i$ enumerate all of the nonzero vectors.
-
Why don't you write in words what you want in the body of the question? Desciphering an equation when a perfectly good English sentence might be considerably easier to understand is silly! – Mariano Suárez-Alvarez May 10 '13 at 3:45
@MarianoSuárez-Alvarez : Sure thing. But isn't that a little pedantic? The equation isn't all that complicated... – Monica May 10 '13 at 3:47
I don't disagree with @Mariano's comment, but let me just add that this is a very nice question. Where did it come from? – Pete L. Clark May 10 '13 at 13:33
And I am someone who generally finds contest type problems quite boring (which is not to say that I know how to solve them). Looking back at the comments, Monica got a less warm welcome than she should have: $\mathbb{Z}/2$ could only mean one thing (unlike $\mathbb{Z}_2$!) and upon reflection I do disagree with Mariano's comment. I'm glad it turned out well in the end. – Pete L. Clark May 10 '13 at 17:10
@PeteL.Clark : Thanks! This experience definitely provides me another data point concerning using gendered names on the internet (I usually don't). It also appears that one can order the Babai book from U of C; see cs.uchicago.edu/research/publications/combinatorics – Monica May 10 '13 at 17:25
The statement is true and let us prove it by induction on $n$. | {
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The statement is true and let us prove it by induction on $n$.
Denote $\mathbb{Z}/2\mathbb{Z}:=\mathbb{Z}_2:=\{0,1\}$. When $n=1$, the statement is clearly true. Now assume that when $n\le n_0$, the statement is true for every $m\ge 1$, and let us show the statement is true for $n:=n_0+1$ and every $m\ge 1$.
Consider $(\mathbb{Z}_2)^n$ as $(\mathbb{Z}_2)^{n-1}\times\mathbb{Z}_2$ and denote by $P$ and $Q$ the projections of $(\mathbb{Z}_2)^n$ to its first $n-1$ coordinates and its last coordinate respectively, i.e. $$P: (\mathbb{Z}_2)^n\mapsto (\mathbb{Z}_2)^{n-1},\quad(a_1,\dots,a_{n-1},a_n)\mapsto (a_1,\dots,a_{n-1}),$$ and $$Q: (\mathbb{Z}_2)^n\mapsto \mathbb{Z}_2,\quad(a_1,\dots,a_n)\mapsto a_n.$$
Up to a rearrangement of $v_1,\dots,v_m$, we may assume that there exists $0\le m'\le m$, such that $Qv_i\ne 0$ if and only if $m'<i\le m$. If $m'=0$, simply choose $f=Q$ and we are done, so let us assume that $m'\ge 1$. By definition, $Pv_1,\dots,Pv_{m'}\ne 0$. By the induction assumption on $n$, there exists a linear function $g:(\mathbb{Z}_2)^{n-1}\to \mathbb{Z}_2$, such that $$|\{ 1\le i\le m' \mid g(P v_i) \neq 0\}| \ge \frac{1}{2}m'.\tag{1}$$ For $j=0,1$, define $$f_j:(\mathbb{Z}_2)^n\to \mathbb{Z}_2,\quad v\mapsto g(Pv)+j\cdot Qv.\tag{2}$$ By definition, for $j=0,1$, $f_j$ is linear and $f_j(v_i)=g(P v_i)$ when $1\le m\le m'$. Then by $(1)$, for both $j=0$ and $j=1$, $$|\{ 1\le i\le m' \mid f_j(v_i) \neq 0\}| \ge \frac{1}{2}m'.\tag{3}$$ If $m'=m$, then we are done. Otherwise, note that by $(2)$ and the choice of $m'$, for $j=0,1$, $$f_j(v_i)=0 \iff g(Pv_i)=1-j\iff f_{1-j}(v_i)=1,\quad \forall\, m'<i\le m.$$ It follows that $$|\{ m'< i\le m \mid f_0(v_i) \neq 0\}|+|\{ m'< i\le m \mid f_1(v_i) \neq 0\}|=m-m'.\tag{4}$$ Combining $(3)$ and $(4)$, we can conclude that the statement is true for either $f=f_0$ or $f=f_1$, which completes the proof. | {
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Remark: Thanks to PeteL.Clark's nice question in the comment, I realized that the conclusion can be generalized to arbitrary finite field with the same argument as above.
Claim: Let $\mathbb{F}_q$ be the finite field of $q$ elements and let $v_1,\dots,v_m$ be nonzero vectors in $(\mathbb{F}_q)^n$. Then there exists a linear function $f:(\mathbb{F}_q)^n\to\mathbb{F}_q$, such that $$|\{1\le i\le m\mid f(v_i)\ne 0\}|\ge\frac{(q-1)m}{q}.\tag{5}$$
Sketch of Proof: Define projections $P:(\mathbb{F}_q)^n\to (\mathbb{F}_q)^{n-1}$, $Q:(\mathbb{F}_q)^n\to\mathbb{F}_q$ similarly. Define $m'$ and rearrange $v_1,\dots,v_m$ similarly. By induction, we may assume that there exists a linear function $g:(\mathbb{F}_q)^{n-1}\to\mathbb{F}_q$, such that for $f_j=g\circ P+j\cdot Q$, $0\le j<q$, $$|\{1\le i\le m'\mid f_j(v_i)\ne 0\}|\ge\frac{(q-1)m'}{q}.\tag{6}$$ By definition of $m'$ and $f_j$, for every $m'<i\le m$, $f_0(v_i),\dots, f_{q-1}(v_i)$ are pairwise different, so one and only one of them is $0$. It follows that for some $0\le j<q$, $$|\{m'< i\le m\mid f_j(v_i)\ne 0\}|\ge\frac{(q-1)(m-m')}{q}.\tag{7}$$ $(5)$ follows from $(6)$ and $(7)$ for $f=f_j$. | {
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-
This is just lovely. – Kevin Carlson May 10 '13 at 9:00
@KevinCarlson: Thank you! The question itself is lovely. – 23rd May 10 '13 at 9:08
How does this generalize to other finite fields, I wonder? – Pete L. Clark May 10 '13 at 13:34
@PeteL.Clark: A simple observation based on this argument could show that for finite field $\mathbb{F}_q$, a lower bound is $\frac{(q-1)m}{q}$. Sketch of proof: we can define $f_j$, $0\le j\le q-1$ similarly as in $(2)$. By induction, we can assume that $(3)$ holds when $\frac{m'}{2}$ is replaced by $\frac{(q-1)m'}{q}$. When $i>m'$, $f_j(v_i)$ are pairwise different, so exactly only one of them is $0$. Then we can conclude that there exists $0\le j\le q-1$, such that $|\{m'<i\le m:f_j(v_i)\ne 0\}|\ge\frac{(q-1)(m-m')}{q}$, which completes the induction. – 23rd May 10 '13 at 14:01
@PeteL.Clark: Sorry, please ignore my first reply and see the current one. – 23rd May 10 '13 at 14:04
My coauthors and I needed this result recently (to prove something about pseudo-Anosov dilatations), and I was googling to see if it was known. I'm amazed that it was asked here so recently!
I'm answering now to record a short alternative proof that we found. I'll prove the same more general result that Landscape proved, namely:
Claim : Let $\vec{v}_1,\ldots,\vec{v}_m \in \mathbb{F}_q^n$ be nonzero vectors (not necessarily distinct). Then there exists a linear map $f : \mathbb{F}_q^n \rightarrow \mathbb{F}_q$ such that $f(\vec{v}_i) = 1$ for at least $\frac{q-1}{q}$ of the $\vec{v}_i$, i.e. such that $\{\text{$i|1 \leq i \leq m$,$f(\vec{v}_i) \neq 0$}\}$ has cardinality at least $\frac{q-1}{q}m$. | {
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Proof : Let $\Omega$ be the probability space consisting of all linear maps from $\mathbb{F}_q^n \rightarrow \mathbb{F}_q$, each given equal probability. Let $\mathcal{X} : \Omega \rightarrow \mathbb{R}$ be the random variable that takes $f \in \Omega$ to the cardinality of the set $\{\text{$i|1 \leq i \leq m$,$f(\vec{v}_i) \neq 0$}\}$. We will prove that the expected value $E(\mathcal{X})$ of $\mathcal{X}$ is $\frac{q-1}{q} m$, which clearly implies that there exists some element $f \in \Omega$ such that $\mathcal{X}(f) \geq \frac{q-1}{q} m$.
To prove the desired claim, for $1 \leq i \leq m$ let $\mathcal{X}_i : \Omega \rightarrow \mathbb{R}$ be the random variable that takes $f \in \Omega$ to $1$ if $f(\vec{v}_i) \neq 0$ and to $0$ if $f(\vec{v}_i) = 0$. Viewing $\vec{v}_i$ as an element of the double dual $(\mathbb{F}_q^n)^{\ast \ast}$, the kernel of $\vec{v}_i$ consists of exactly $\frac{1}{q}$ of the elements of $(\mathbb{F}_q^n)^{\ast}$. This implies $E(\mathcal{X}_i) = \frac{q-1}{q}$. Using linearity of expectation (which recall does not require that the random variables be independent), we get that $$E(\mathcal{X}) = E(\sum_{i=1}^m \mathcal{X}_i) = \sum_{i=1}^m E(\mathcal{X}_i) = \sum_{i=1}^m \frac{q-1}{q} = \frac{q-1}{q} m,$$ as desired.
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# How many $1$s are in the first $1023$ binary numbers?
How many $1$s are in the first $1023$ binary numbers?
I'm not to sure how to approach this question. An idea, formula, or solution is appreciated!
• Have you noticed that $1023 = 2^{10}-1$? Feb 26 '17 at 0:20
• What is the "first" binary number? 0 or 1? Feb 26 '17 at 0:22
• Worth clarifying that we are talking about positive, unsigned integers.
– J...
Feb 26 '17 at 2:09
• @J... actually, can an unsigned integer be positive? Feb 26 '17 at 5:56
• @Joffan Cheeky bastard :P Yes, unsigned integers are positive because all of them (other than 0) are greater than 0. Feb 26 '17 at 9:35
Assuming the "first" binary number is $1$ note that the first $1023$ binary numbers, plus $0$, are all the binary numbers you can write with exactly $10$ binary digits or bits (prepending $0$s to "short" numbers, as in $0000101010_2$). Between all of them, you then have $1024 \cdot 10=10240$ bits, and for symmetry reasons exactly half of those, $5120$, are $1$s. | {
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• why 1024*10 digits? Feb 26 '17 at 0:28
• Ohh, the amount of 0's don't matter! I see! Feb 26 '17 at 0:29
• It does matter, because if we start at 0, then 1111111111 should not be counted among the the first 1023 binary numbers. The question did not say the first 1023 positive binary numbers. Feb 26 '17 at 2:23
• In computer science, typically indexing start at 0. I was biased . I should have read the comments below the question... The issue was raised before. Feb 26 '17 at 2:52
• @MarDev We're looking at all of the integers we can represent with 10 bits - for every integer there is exactly one other which has all of its bits inverted (the result of an XOR with 1111111111) and therefore the opposite amount of zeros and ones. Furthermore, this inversion always lies in the opposite half of the possible numbers (the operation is essentially equivalent to 1111111111 - n) so each of the first 2^9 numbers (half of the possible 2^10) there exists exactly one number in the second half with the opposite number of zeros and ones. Feb 26 '17 at 8:46
Hint: For how many of those numbers will the one's bit be a $1$ (in other words: how many of those numbers are odd)? For how many of them will the two's bit be a $1$? For how many of them will the four's bit be a $1$? And so on. Also, it will probably be advantageous to include $0$ (and thus look at a collection of $1024$ binary numbers) to make the counting a bit easier. Or, if $0$ is already included, include $1023$ initially, then correct for it when you're done counting.
• I'm struggling with that idea right now. I'm not entirely sure how to solve that. Is it like half half? So if there are x digits in total, half of them are 1? Feb 26 '17 at 0:27
• On average, yes, half of the bits will be a $1$. Or something. Depends on what you mean by "of the bits". Feb 26 '17 at 0:31 | {
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Any such number can be represented by a string of 10 0s and 1s. The number of such strings with $n$ ones is $10$ choose $n$. Thus, the number of ones which appear is $$\sum_{n=0}^{10} n{10 \choose n}=5\cdot 2^{10}.$$
As an addendum to Arthur's answer - if you count to, say, $2^{4} - 1 = 15$, you can easily figure out a pattern in the columns:
0000
0001
0010
0011
0100
0101
0110
0111
1000
1001
1010
1011
1100
1101
1110
1111
So if you count to $2^{n} - 1$, you have $2^{n} * n$ digits ("rows * columns"), of which 50% are 1's.
For your example: Counting to $1023 =2^{10} - 1$, gets you $2^{10} * 10 * \frac{1}{2} = 5120$ binary 1's in total.
It's easy to calculate. There is a pattern with any number used as pow of tow:
You have 2^x, where x> 0 and you will have a binary number as:
2^0 = 1 (Dec) = 1 (Binary)
2^1 = 2 (Dec) = 10 (Binary)
2^2 = 4 (Dec) = 100 (Binary)
...
2^10 = 1024 (Dec) = 100 0000 0000(Binary)
If you pay attention you will notice this:
2^10 - 1 = 1023 (Dec) = 011 1111 1111(Binary)
...
2^2 - 1 = 3 (Dec) = 011 (Binary)
2^1 = 1(Dec) = 01 (Binary)
2^0 - 1 = 0 (Dec) = 0 (Binary)
In this case, your question, the Number of ones matches with the number that is used for pow. | {
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# invertible matrix theorem | {
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Let two inverses of A be B and C Services; Math; Blog ; About; Math Help; Invertible Matrix and It’s Properties. This section consists of a single important theorem containing many equivalent conditions for a matrix to be invertible. Invertible Matrix Theorem. Section 3.5 Matrix Inverses ¶ permalink Objectives. * The determinant of $A$ is nonzero. * $A$ has only nonzero eigenvalues. : An matrix is invertible if and only if has only the solution . structure theorem for completely bounded module maps. Invertible matrix 2 The transpose AT is an invertible matrix (hence rows of A are linearly independent, span Kn, and form a basis of Kn). Showing any of the following about an $n \times n$ matrix $A$ will also show that $A$ is invertible. (If one statement holds, all do; if one statement is false, all are false.) Proof: Let there be a matrix A of order n×n which is invertible. Any nonzero square matrix A is similar to a matrix all diagonal elements of which are nonzero. According to WolframAlpha, the invertible matrix theorem gives a series of equivalent conditions for an n×n square matrix if and only if any and all of the conditions hold. e. The columns of A form a linearly independent set. The Invertible Matrix Theorem Let A be a square n×n matrix. Let A be a square n by n matrix over a field K (for example the field R of real numbers). 1. Understand what it means for a square matrix to be invertible. Then a natural question is when we can solve Ax = y for x 2 Rm; given y 2 Rn (1:1) If A is a square matrix (m = n) and A has an inverse, then (1.1) holds if and only if x = A¡1y. We define invertible matrix and explain many of its properties. 1 The Invertible Matrix Theorem Let A be a square matrix of size N × N. The following statement are equivalent: • A is an invertible matrix. (d)Show that if Q is invertible, then rank(AQ) = rank(A) by applying problem 4(c) to rank(AQ)T. (e)Suppose that B is n … Then the following statements are equivalent. Yes. We will append two | {
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(e)Suppose that B is n … Then the following statements are equivalent. Yes. We will append two more criteria in Section 6.1. An invertible matrix is sometimes referred to as nonsingular or non-degenerate, and are commonly defined using real or complex numbers. abelian group augmented matrix basis basis for a vector space characteristic polynomial commutative ring determinant determinant of a matrix diagonalization diagonal matrix eigenvalue eigenvector elementary row operations exam finite group group group homomorphism group theory homomorphism ideal inverse matrix invertible matrix kernel linear algebra linear combination linearly … Theorem 1. 5.The columns of A are linearly independent. The Invertible Matrix Theorem|a small part For an n n matrix A, the following statements are equivalent. A is column-equivalent to the n-by-n identity matrix In. Let A be an n n matrix. A is an invertible matrix. Usually, when a set is written as the span of one vector, it’s one dimensional. Let A 2R n. Then the following statements are equivalent. • The columns of A form a linearly independent set. The matrix A can be expressed as a finite product of elementary matrices. Theorem (The QR Factorization) If A is an mxn matrix with linearly independent columns, then A can be factored as A=QR, where Q is an mxn matrix whose columns form an orthonormal basis for Col A and R is an nxn upper triangular invertible matrix with positive entries on the main diagonal. Theorem1: Unique inverse is possessed by every invertible matrix. The following statements are equivalent, i.e., for any given matrix they are either all true or all false: A is invertible, i.e. Some theorems, such as the Neumann Series representation, not only assure us that a certain matrix is invertible, but give formulas for computing the inverse. I. row reduce to! 4. The uniqueness of the polar decomposition of an invertible matrix. W. Sandburg [8] and Wu and Desoer [ … A has an inverse or is nonsingular. Dave4Math » Linear | {
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matrix. W. Sandburg [8] and Wu and Desoer [ … A has an inverse or is nonsingular. Dave4Math » Linear Algebra » Invertible Matrix and It’s Properties. Skip to content. 6.The linear transformation T defined by T(x) = Ax is one-to-one. Note that finding this matrix B is equivalent to solving a system of equations. The following statements are equivalent: A is invertible, i.e. A has an inverse, is nonsingular, or is nondegenerate. tem with an invertible matrix of coefficients is consistent with a unique solution.Now, we turn our attention to properties of the inverse, and the Fundamental Theorem of Invert- ible Matrices. lie in the commutants of d and 59’. The number 0 is not an eigenvalue of A. A matrix that has no inverse is singular. Here’s the first one. The Invertible Matrix Theorem (Section 2.3, Theorem 8) has many equivalent conditions for a matrix to be invertible. Introduction and Deflnition. Furthermore, the following properties hold for an invertible matrix A: • for nonzero scalar k • For any invertible n×n matrices A and B. Learn about invertible transformations, and understand the relationship between invertible matrices and invertible transformations. Menu. A is row-equivalent to the n-by-n identity matrix In. 2 det(A) is non-zero.See previous slide 3 At is invertible.on assignment 1 4 The reduced row echelon form of A is the identity matrix. When the determinant value of square matrix I exactly zero the matrix is singular. 5. (c)Showthatif P isaninvertiblem ×m matrix, thenrank(PA) = rank(A) byapplying problems4(a)and4(b)toeachofPA andP−1(PA). its nullity is zero. • A has N pivot positions. A is invertible.. A .. A is row equivalent to the n×n identity matrix. The Invertible Matrix Theorem has a lot of equivalent statements of it, but let’s just talk about two of them. Matrix I exactly zero the matrix must be a square n by n matrix over a field K for! A $3\times 3$ matrix is detailed along with characterizations the span of one vector, ’! Following statements | {
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3$ matrix is detailed along with characterizations the span of one vector, ’! Following statements are equivalent services ; math help ; invertible matrix finite product of elementary.. Numerical analysis it, but let ’ s Properties with full rank, is! A matrix invertible matrix theorem invertible is that the null space is zero-dimensional × n identity matrix in must a... E. the columns of a single important theorem containing many equivalent conditions for a given a, the following are. Single important theorem containing many equivalent conditions for a matrix to be invertible let there be square. You to prove has an inverse, is nonsingular, or is nondegenerate for a square and! A field K ( for example the field R of real numbers ) the inverse of a be a n. It, but let ’ s just talk about two of them of. True or all false. is sometimes referred to as nonsingular or,!, 2019 by Dave can only happen with full rank only happen with full rank, ’... The following statements are equivalent track of the invertible matrix theorem ( Section 2.3, theorem invertible matrix theorem ) many! S one dimensional invertible matrix theorem answer if a matrix being invertible is that the null is... Of a Ax = has only the solution vector, it is invertible, how you... Zero the matrix must be a square n×n matrix row equivalent to E... Matrix I exactly zero the matrix is sometimes referred to as nonsingular or non-degenerate, and understand the between. Is intended to help you keep track of the invertible matrix and it ’ s just talk two. Non-Degenerate, and understand the relationship between invertible matrices and invertible transformations I... Reducedref E.. F a is invertible each numbered arrow = I =! Unique. commutants of d and 59 ’ the papers of 1 or all false ). Elementary matrices determinant of [ math ] a [ /math ] has only the trivial solution just! 2.3, theorem 8 ) has many equivalent conditions for a matrix to be invertible invertible. Nonsingular, or is nondegenerate is | {
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dominant matrix a of order n×n which is invertible eigenvalue of a dominant matrix is. Referred to as nonsingular or non-degenerate, and numerical analysis [ /math ] is nonzero matrix! Such applications in network theory, consult the papers of 1 matrix be... Prove the rest of the invertible matrix is square and has full rank ( for example field... And of invertible matrices elements of which are nonzero learn about invertible transformations, and understand relationship. Matrix all diagonal elements of which are nonzero brief explanation for each arrow! Let a 2R n. Then the following statements are equivalent a lot of statements! Of real numbers ) a lot of equivalent statements of it, but let ’ one. Be invertible when A~x = ~b has a lot of equivalent statements of it, but let ’ s dimensional... Of this equivalence and leave the other direction for you to prove the n × n matrix. Has only the solution a has an inverse, is nonsingular, or is nondegenerate theorem let a a., or is nondegenerate rev August 6, 2008 1 [ math ] [... Every invertible matrix is a square matrix to be invertible consult the papers of 1 as the span of vector... ) diagonally dominant matrix a is row equivalent to solving a system of equations n identity matrix.... Intended to help you keep track of the conditions and the relationships between.. Theorem, one of the invertible matrix invertible matrix theorem function theorems are much used in diverse... If and only if has only the trivial solution space is zero-dimensional the! Of the invertible matrix theorem conditions for a square matrix and its inverse is the identity matrix column-equivalent to the n n. Conditions for a matrix to be invertible diverse areas as network theory,,. Invertible if the matrix a, the following statements are equivalent: a is row-equivalent to the identity! That the null space is zero-dimensional other direction for you to prove S. Sawyer | September 7, by. Of them being invertible is that the null space is | {
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# Probability that the sum of 6 dice rolls is even
Question:
6 unbiased dice are tossed together. What is the probability that the sum of all the dice is an even number?
I think the answer would be 50%, purely by intuition. However, not sure if this is correct. How should I go about solving such a problem?
Notice that whatever the sum of the first 5 rolls, whether the outcome is odd or even is totally determined by the last die. It is even or odd with equal probability, so the probability of an even sum is exactly the same as the probability of an odd sum.
The first 5 dice don't matter.
Statement:
Let $X_1,\ldots,X_n$ be independent, integer-valued random variables, and let some $X_k$ satisfy $P(X_k \mbox{ odd}) = 0.5$. Then $P(\sum X_i \mbox{ odd}) = 0.5$.
Proof:
Without loss of generality let $k=n$, and let $S = \sum_i^{n-1} X_i$. Since $S$ and $X_n$ are independent random variables, $$\begin{eqnarray*} P\left(\sum X_i \mbox{ odd}\right) = P(S + X_n \mbox{ odd}) &=& P(S \mbox{ odd and } X_n \mbox{ even}) + P(S \mbox { even and } X_n \mbox{ odd}) \\ &=& P(S \mbox{ odd})P(X_n \mbox{ even}) + P(S \mbox{ even})P(X_n \mbox{ odd}) \\ &=& P(S \mbox{ odd})P(X_n \mbox{ odd}) + P(S \mbox{ even})P(X_n \mbox{ odd}) \\ &=& \left(P(S \mbox{ odd}) + P(S \mbox{ even})\right)P(X_n \mbox{ odd}) \\ &=& P(X_n \mbox{ odd}) \\ &=& 0.5 \end{eqnarray*}$$ | {
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• o.O That's also a nice approach. Will the same apply for a dice with 7 sides? – Gummy bears Sep 26 '15 at 5:18
• @Gummybears no, because then you have inequal probabilities coming from the last die, so the parity of the sum you had before you roll it matters. However, this sort of thinking does apply in a pretty wide variety of cases, and it saves a lot of work when you notice it happening. – Eric Tressler Sep 26 '15 at 5:22
• @Gummybears: sorry, I missed that. However, what is not mentioned is that with an odd number of seven sided dice, there is a higher chance of an odd sum. Whereas with an even number of seven sided dice, there is a higher chance of an even sum. – robjohn Sep 26 '15 at 5:50
• @JoseAntonioDuraOlmos, the first 5 dice matter as much as the last one. They all matter the same here. It is not our neglect of the first 5 dice which leads us to the conclusion that the probability is $1/2$ but it is the symmetry of odd/even tosses. There is a great danger in basing the argument on the neglect, as one might be tempting to write it for the case where all dice are biased. – zhoraster Sep 26 '15 at 8:53
• There is no neglect. If I am presented a box with an unknown number of dices of unknown shapes all of them labeled with integers. I take only one out of the box. The dice has 6 sides, from 1 to 6, and is unbiased. Then I can claim that if I take all the dices, roll them and add their results the sum has a 50% probability of being even. I can claim it without examining the other dices in the box. And the proof of such claim is like the one in this answer. As for the case where all dice are biased, yeah this proof scheme does not work for that, don't fall into those temptations. – Jose Antonio Reinstate Monica Sep 26 '15 at 9:08 | {
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The only thing about the numbers rolled that matters is their parity - whether they are even or odd. In order to get an even sum, an even number of the six dice must be even. In order to get an odd sum, an odd number of the six dice must be even.
Using O for odd and E for even, we can list out the possibilities.
Even sum:
• OOOOOO $$\binom{6}{6}=1 \text{ arrangement}$$
• OOOOEE $$\binom{6}{4}=15 \text{ arrangements}$$
• OOEEEE $$\binom{6}{2}=15 \text{ arrangements}$$
• EEEEEE $$\binom{6}{0}=1 \text{ arrangement}$$
This gives a total of $32$ arrangements with even sum.
Since there are $2^6 = 64$ total possibilities, we see that your intuition of $50\%$ is correct.
• Lovely answer. Although slightly lengthy, it is simple to understand. Thank you. – Gummy bears Sep 26 '15 at 5:24
• +1 Good answer. I just have a wording suggestion: instead of saying "to get an even/odd sum, an even/odd number of the six dice must be even", I think wording it in terms of the number of odd dice gives more intuition and can generalize to any number of dice. The number of even dice doesn't really matter, the number of odd dice does. – jadhachem Sep 28 '15 at 5:39
The generating function approach:
$$P(x)=(1+x+x^2+x^3+x^4+x^5)^6=\sum a_i x^i$$
Then $a_i$ counts the number of ways of getting a total of $i+6$ from $6$ dice.
Now, to find the even terms, you can compute $$\frac{P(1)+P(-1)}{2}=\sum_i a_{2i}.$$
But $P(1)=6^6$ and $P(-1)=0$. So $$\frac{P(1)+P(-1)}{2}=\frac{6^6}{2},$$ or exactly half, as you conjectured.
For another example, let $N_{i}$ be the number of ways to roll $6$ dice and getting a value $\equiv i\pmod{5}$. Then it turns out that if $z$ is a primitive $5$th root of unity, then the value can be counted by defining:
$$Q_i(x)=x^{6-i}(1+x+x^2+x^3+x^4+x^5)^6$$ then computing $$N_i=\frac{Q_i(1)+Q_i(z)+Q_i(z^2)+Q_i(z^3)+Q_i(z^4)}{5}$$
This gives the result:
$$N_i =\begin{cases}\frac{6^6+4}{5}&i\equiv 1\pmod 5\\ \frac{6^6-1}{5}&\text{otherwise} \end{cases}$$ | {
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More generally, if $N_{n,i}$ is the number of ways to get $\equiv i\pmod 5$ when $n$ dice are rolled, you get:
$$N_{n,i} =\begin{cases}\frac{6^n+4}{5}&i\equiv n\pmod 5\\ \frac{6^n-1}{5}&\text{otherwise} \end{cases}$$
It's this simple because of the fact that $6=5+1$.
If each die has $d$ sides, and you ask what are the number of ways to get a total $\equiv i\pmod {d-1}$, then you get:
$$N_{d,n,i} =\begin{cases}\frac{d^n+{d-2}}{d-1}=\frac{d^n-1}{d-1}+1&i\equiv n\pmod {d-1}\\ \frac{d^n-1}{d-1}&\text{otherwise} \end{cases}$$
• I'm afraid I lost you at the last sentence. – Gummy bears Sep 26 '15 at 5:03
• That's okay, it is an advanced approach, but a useful one for problems like this. If you rolled $n$ dice, each with $7$ sides, you'd get the number of even results is: $$\frac{7^n + (-1)^n}{2}$$ would be even. – Thomas Andrews Sep 26 '15 at 5:09
• Aha..... That's a good approach. However, I'm not sure if I completely get it. Not sure how you used the probability of 1 and $-1$ to calculate the even terms. – Gummy bears Sep 26 '15 at 5:16
• If $i$ is odd, then $a_i\cdot 1^i + a_i(-1)^{i} = 0$. If $i$ is even, then $a_u\cdot 1^i + a_i(-1)^i = 2a_i$. So $P(1)+P(-1)=\sum_{i} 2a_{2i}$, which is twice the value you are looking for. @Gummybears – Thomas Andrews Sep 26 '15 at 5:19
• Aha.... I'm stupid, aren't I? This is perfect! It holds true for any sided dice and for any number of dice. – Gummy bears Sep 26 '15 at 5:23
Using the same logic given by @Eric tessler but writing the maths behind it.
Let $P_x$ = We get an even sum on the roll of x dice and
$1 - P_x$ = We get an odd sum.
So we require $P_6$
Now $P_6 = P_5 \times$ P[even number on the last dice] + $(1 - P_5) \times$ P[odd number on the last dice]
$P_6 = P_5 \times 0.5 + (1-P_5) \times 0.5$
$P_6 = 0.5$ | {
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$P_6 = P_5 \times 0.5 + (1-P_5) \times 0.5$
$P_6 = 0.5$
• Exactly what @erictessler is missing - a rigurous proof. One by induction could work for all sorts of dice. – Tibos Sep 26 '15 at 23:22
• More upvotes needed! – Gummy bears Sep 27 '15 at 4:14
• Nicely done! +1 – Zubin Mukerjee Sep 27 '15 at 17:27
A lot of very informative answers but none of them do not explain your intuition, so I'll post mine.
There is a symmetry between all tosses with odd sum and those with even sum: just take the number $k$ on the first die and replace it with $7-k$.
So the numbers of odd tosses and even tosses are equal, therefore, the probability is $1/2$.
• I'm not getting you much.... elaborate please? – curiousbrain Sep 26 '15 at 5:22
• Same here. Please elaborate. – Gummy bears Sep 26 '15 at 5:23
• See the answer by @EricTressler, posted the same time. – zhoraster Sep 26 '15 at 5:24
Your intuition is correct. There could be a proof based on this intuition (using the fact that each die has 50% chance of being even and we're tossing 6 of them) but I'm not so sure.
But there are ways to prove it. For example, the only ways we could get an even sum are if:
1. All 6 dice show even numbers.
2. 4 dice show even and 2 dice show odd.
3. 2 dice show even and 4 dice show odd.
4. All 6 dice show odd.
The probability of cases 1 and 4 is $(\frac{1}{2})^6$ and the probability of cases 2 and 3 is $\binom{6}{2}(\frac{1}{2})^6=15(\frac{1}{2})^6$ (because we need to order the dice which show odd numbers.) Add them up and we get the probability is $$32(\frac{1}{2})^6=\frac{1}{2}.$$
The problem is about fair dice.
Whether the number of dice is 6 (even) or 7 (odd) $Pr = \dfrac12$
The logic is based on parity (odd/even). Since the probability that any particular throw is even or odd is equal at $\dfrac12$
(a) For an even number of dice, if you interchange odd and even throws, the parity remains the same, thus there will always be an equal number of odd and even sums.
EEEEEE OOOOOO even | {
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EEEEEE OOOOOO even
EEEEEO OOOOOE odd
EEEEOO OOOOEE even
EEEOOO OOOEEE odd
(b) For an odd number of dice, every such interchange changes the parity, but by symmetry, again there will be an equal number of odd and even sums.
EEEEEEE even OOOOOOO odd
EEEEEEO odd OOOOOOE even
EEEEEOO even OOOOOEE odd
EEEEOOO odd OOOOEEE even
All outcomes are equally likely. Half the outcomes will have an odd number of odd dice faces and half of the outcomes will have an even number of odd dice faces. The outcomes with an odd number will have an odd total and the outcomes with an even number of odd face will be even. As there are equal numbers of outcomes for each case the probability for each case is 50%.
Okay so how do we know that half the outcomes will have on odd number of odd faces and half the number will have an even number of even faces? Well,we can make a one to one correspondence between outcomes with an odd number of odd faces to outcomes to an even number of odd faces. If outcome has an even number of odd faces, map it to the exact same outcome but the 6th die is one number higher (let's assume 1 is one number higher than 6). This outcome has an odd number of odd faces. If an outcome has an odd number of odd faces map it the exact same outcome but the 6th die is one number lower. This is a one to one correspondence. So there are an equal number of outcomes with an odd number of odd faces and there are outcomes with an even number of odd faces.
And that's it. Equal number of odd comes as even outcomes, each outcome equally likely, odd outcomes and even outcomes equally likely. | {
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• Oops, Zhoraster beat me to it and did a much clearer and shorter job explaining it. Oh, well. – fleablood Sep 26 '15 at 23:27
• It was asked above about 7 sided die. I don't have enough reputation to comment there, so I'm going to comment here: If a 7-sided die has 4 odd faces and 3 even faces the probability of an even result with n die is 1/2 + (- 1/7)^n. This can be shown by induction by noting that each extra die has a 4/7 chance of changing parity. So if Pn is probability of an even total with n die. Then P(n+1) = Pn*3/7 + (1 - Pn)*4/7. as P1 = 3/7 = 1/2 + (- 1/7), the result follows by induction – fleablood Sep 26 '15 at 23:51 | {
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# Show that if $A$ and $B$ are sets, then $A \subseteq B$ if and only if $A \cap B = A$.
I'm working through a real analysis textbook, and it starts out with set theory. The first exercise is
Show that if $A$ and $B$ are sets, then $A \subseteq B$ if and only if $A \cap B = A$.
I think I proved it correctly but I'm not sure. Here's what I did. I proved that if $A \subseteq B$, then $A \cap B = A$ the same way as this answer did (https://math.stackexchange.com/a/446114/93114), but I want to make sure I proved the converse correctly because it seems really easy (yes it's the first problem in the book, but still) and math usually isn't this easy for me, even the basic stuff!
Proof of "If $A \cap B = A$, then $A \subseteq B$."
If $x \in A \cap B$, then $x \in A$ and $x \in B$, but this applies to all $x \in A$ because $A \cap B = A$. So, for any $x \in A$, we know that $x \in B$, so $A \subseteq B$.
Am I on the right track?
• Yes, although you're doing it in a convoluted way. Let $x\in A$; since $A=A\cap B$ we have $x\in B$. – egreg Oct 12 '13 at 22:27
• @egreg So if $x \in A$, then because $A = A \cap B$, we know that $x \in A \cap B$. By definition that implies that $x \in A$ and $x \in B$, which is enough to show that $A \subseteq B$. – M T Oct 12 '13 at 22:28
• @egreg Sorry I replied to your other comment that disappeared, but we're doing it the same way now. – M T Oct 12 '13 at 22:30
• I realized to have misread your argument, so I changed my first comment. – egreg Oct 12 '13 at 22:32
Since you are just starting, I would suggest to be verbose instead of pulling everything in a single sentence.
To prove $A \subseteq B$ iff $A \cap B = A$, you have to
1. show $A \cap B = A$ given $A \subseteq B$. That is to
1. show $A \cap B \subseteq A$.
2. show $A \subseteq A \cap B$.
2. show $A \subseteq B$ given $A \cap B = A$.
Proof:
1.1) It is trivially true. You don't need to be given $A \subseteq B$ for it to be true. | {
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Proof:
1.1) It is trivially true. You don't need to be given $A \subseteq B$ for it to be true.
1.2) If $x \in A$, then $x \in B$ since we are given $A \subseteq B$. Then $x \in A$ and $x \in B$ are both true. Therefore, $x \in A \cap B$.
2) From $A \cap B = A$, we know $x \in A$ and $x \in B$ whenever $x \in A$. If $x \in A$, then it must be the case that $x \in B$. Therefore, $A \subseteq B$.
• Thanks for the answer. I think I'm slowly starting to catch the hang of these proofs. – M T Oct 13 '13 at 1:58
Suppose $A\cap B=A$. To prove $A\subseteq B$, suppose $x\in A$. Since $A\cap B=A$, $x\in A\cap B$ and thus $x\in B$. Since $x$ was an arbitrary element of $A$, this holds for all $x$ and thus $A\subseteq B$.
The above proves $A\cap B=A\implies A\subseteq B$. You will then need to prove the reverse implication to establish $A\cap B=A\iff A\subseteq B$.
Here is another way to approach this type of problem: start with the most complex part (here: $\;A \cap B = A\;$), expand the definitions so that you get from the set level to the element level, then simplify using predicate logic, and finally go back to the set level as suggested by the shape of the formula.
In this case, \begin{align} & A \cap B = A \\ \equiv & \;\;\;\;\;\text{"set extensionality, i.e., the definition of $\;=\;$ for sets"} \\ & \langle \forall x :: x \in A \cap B \equiv x \in A \rangle \\ \equiv & \;\;\;\;\;\text{"definition of $\;\cap\;$"} \\ & \langle \forall x :: x \in A \land x \in B \equiv x \in A \rangle \\ \equiv & \;\;\;\;\;\text{"logic: $\;p \equiv p \land q\;$ is one way to write $\;p \Rightarrow q\;$"} \\ & \langle \forall x :: x \in A \Rightarrow x \in B \rangle \\ \equiv & \;\;\;\;\;\text{"definition of $\;\subseteq\;$"} \\ & A \subseteq B \\ \end{align} | {
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# Show that a certain set of positive real numbers must be finite or countable
Let $B$ be a set of positive real numbers with the property that adding together any finite subset of elements from $B$ always gives a sum of $2$ or less. Show that $B$ must be finite or at most countable.
$B$ = {$x \in R:x>0\}$, $x_1,x_2...x_n \in B$ such that $x_1+x_2+...+x_n \le 2$.
Question: for any $a,b$ $(a,b)$~$R$, but $B$ is $(0,+\infty)$ so why $B$ is not uncountable (taking as $a = 0$, and letting $b$->$\infty$)?
And why for $B$ being countable doesn't contradict: for any $a,b$ $(a,b)$~$R$?
P.S. I read Showing a set is finite or countable and understood it. | {
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P.S. I read Showing a set is finite or countable and understood it.
• It sounds like you are confused about what the question is asking. It is not saying that B is the set of positive real numbers; it is a set of positive real numbers. This means that B is a subset of the set of all positive real numbers. Aug 8 '18 at 17:23
• No. B is some unspecified subset of the positive real numbers such that if you add up a finite number of the elements of B the sum is always less than 2. For example, B could be {0.1, 0.2, 0.5, 1.1} or B could be the infinite set {1, 1/2, 1/4, 1/8, 1/16, ...}. Note that B cannot be {1, 1.4, 1.8} and B cannot be {1, 1/2, 1/3, 1/4, ...} Aug 8 '18 at 17:47
• B does not have to be an interval (0,a). In fact, it cannot be such an interval. Your job is to prove that B must be finite or countable. Intervals are uncountable. Aug 8 '18 at 17:49
• Well, I'm just trying to help clarify the question. I haven't given a solution. I think the hints and discussion below do that for you. But the point is you cannot assume anything about B except: 1. its elements are all positive real numbers and 2. if you take a finite number of elements from B and add them up then you get a sum less than two. Given any such B you have to prove that B is either finite or countable. Aug 8 '18 at 17:59
• Here is a proof that B cannot be the open interval (0,1). If it were then these numbers would be in B: 0.5, 0.51, 0.52, and 0.53. These numbers add up to more than 2. Do you see? You could do something similar to show that B cannot be (0,a) for any positive a. But you have to show more than that. You have to show B is finite or countable. Aug 8 '18 at 18:00
Hint 1: How many elements of $B$ can be in the set $[2,\infty)$?
Hint 2: How many elements of $B$ can be in the set $[1,2)$?
Hint 3: How many elements of $B$ can be in the set $[0.5,1)$? | {
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Hint 3: How many elements of $B$ can be in the set $[0.5,1)$?
• At this rate you are going to need countably many hints :) Aug 8 '18 at 17:04
• @ArnaudMortier Can't he give a hint schema? Aug 8 '18 at 17:07
• 1) One $x = 2$, 2) set $x_1=1$ so one for sure, then I don't know how many $x_2,x_3,...x_n$ to choose so that they sum up to $1$. 3) The same, I don't know. Aug 8 '18 at 17:10
• Obviously $B\cap[1,\infty)$ has at most two elements, because if $x,y,z\ge1$ then $x+y+z>2$. Similarly $B\cap[2/n,\infty)$ has at most $n$ elements. (So $B$ is the union of countably many finite sets...) Aug 8 '18 at 17:17
• No, we certainly can't write $B$ as $(0,a)$. If $B=(0,a)$ then there are finite subsets of $B$ with sum larger than $2$. Aug 8 '18 at 18:56
Not only can we show that it's countable, it's not too difficult to construct an enumeration: given an element $b$, just count how many other elements of $B$ are larger. This has to be a finite integer, since if there were an infinite number of elements greater than $b$, then the sum of $n$ such elements would be greater than $nb$, so picking an $n>2/b$ would give a sum greater than 2.
For any $a\in B$, define $B_a\equiv\{b\in B:a<b\}$. The cardinality of $B_a$ must be less than $\lceil2/a\rceil$, otherwise the sum of any $\lceil2/a\rceil$ elements of $B_a$ would exceed 2. That is, $\sum_{i=1}^{\lceil2/a\rceil}x_i>\sum_{i=1}^{\lceil2/a\rceil}a=a\lceil2/a\rceil\ge2$ for any sequence $x_i$ in $B_a$, contradicting the requirements on $B$.
Define the function $f:B\rightarrow\mathbb{N}$ as $f(a)=\left|B_a\right|$.
Homework: If you can show that $f$ is one-to-one then you can conclude that $|B|\le|\mathbb{N}|$. That is, $B$ is either finite or countable.
Credit goes to Acccumulation for the outline of this proof.
• What if $a =$ sup$B$? Then $B_a$ is the empty set. If $B_a \equiv \{b \in B: a \leq b \}$, then the sum of any $\lceil r/a \rceil$ for $r > 2$ elements of $B_a$ would exceed 2. Jul 18 '19 at 23:08 | {
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Converse of Uniqueness of Universal Property of Tensor Products?
In Atiyah Macdonald, for a commutative and unital ring $A$ the tensor product of $A$-modules, $M,N$ is defined by the usual universal property:
A pair $(T,g)$ with $T$ an $A$-module and $g$ an $A$-bilinear map $g\colon M \times N \to T$ such that for any $A$-module $P$ with an $A$-bilinear map $f\colon M \times N \to P$ there exists a unique $A$-module homomorphism $h\colon T \to P$ such that $h \circ g = f$.
And then, as usually goes with universal properties:
Moreover, if $(T,g), (T',g')$ are any two pairs satisfying this condition, then there is a unique isomorphism $j \colon T \to T'$ such that $j \circ g = g'$
My question is, in the second highlighted selection, is the converse true? If I have already established the existence of the usual tensor product, $M \otimes_A N$, via quotient of free module, and I encounter some other pair $(T', g')$ and a unique isomorphism $j: T \to T'$ such that $j \circ g = g'$, can I conclude that $T'$ also satisfies the universal property defining the tensor product, thus I can regard both $T$ and $T'$ as the tensor product of this $M$ and $N$?
I believe I proved that I can indeed do this, but I need reassurance because I have overlooked small details with tensor products before, and don't want to make that mistake again. My proof was as follows:
Suppose $(T,g)$ satisfies the universal property and $(T',g')$ is a pair where $T'$ is an $A$-mod and $g'$ an $A$-bilinear map $g' \colon M \times N \to T'$. Suppose $j$ is the unique isomorphism $j\colon T \to T'$ such that $j \circ g = g'$, or $g = j^{-1} \circ g'$. Suppose $P$ is any $A$-mod with an $A$-bilinear map $f\colon M \times N \to P$. We wish to show there exists a unique $A$-module homomorphism $\phi\colon T' \to P$ such that $\phi \circ g' = f$. | {
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Since $(T,g)$ already satisfies the property by hypothesis, we know there exists a unique $A$-module homomorphism $\varphi \colon T \to P$ such that $\varphi \circ g = f$. So define the map $\phi \colon T' \to P$ by $\varphi \circ j^{-1}$. Then $\phi \circ g' = \varphi \circ j^{-1} \circ g' = \varphi \circ g = f$. So the diagram commutes as desired and the map $\phi$ is defined as the composition of two unique maps so is unique as well, thus $(T',g')$ satisfies the universal property.
Sorry that got a little (unnecessarily) complicated, I don't know how to draw diagrams on here, which really makes everything easier. As an aside, I feel like this probably is true and Atiyah & Macdonald didn't mention it in the text because they deemed it trivial.
• I think this is the first time I've ever seen someone use $\phi$ and $\varphi$ as distinct variable names at the same time. – Eric Wofsey Mar 12 '18 at 6:39
• I do it all the time! – Prince M Mar 12 '18 at 15:08 | {
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This is correct, although your argument for uniqueness of $\phi$ is not very clear: you chose to define it as a composition with certain maps that are unique, but how do you know you couldn't get a $\phi'$ that works and is not given by those compositions? What you want to say is that if $\phi'$ makes the diagram commute for $(T',g')$, then the map $\varphi'=\phi'\circ j$ would make the diagram commute for $(T,g)$ by just swapping the roles of $(T,g)$ and $(T',g')$ in your argument. Therefore by uniqueness of $\varphi$, $\varphi=\varphi'$, and so $\phi'=\varphi'\circ j^{-1}=\varphi\circ j^{-1}=\phi$.
Intuitively, what's going on here is that the isomorphism $j$ says that $(T,g)$ and $(T',g')$ have exactly the same structure (just the elements of $T$ have to be renamed via $j$ to get $T'$). So any property of $(T,g)$ (which is defined using only the module structure on $T$ together with the map $g$) will also be true of $(T',g')$. In particular, this applies to the universal property of the tensor product.
• Ok, thanks. One last question I had was, I just now wrote the whole argument out to ensure I comprehend the entirety of it and I cannot identify where the uniqueness (w/ respect to commuting diagrams) of the isomorphism j is invoked? I see plainly that this theorem fails if it is not unique, because there will be at least 2 distinct ways to construct the map $/Phi$, but I can’t seem to identify, where in the uniqueness argument, we actually invoke the assumption that j is unique? – Prince M Mar 12 '18 at 22:16
• You don't need uniqueness of $j$, just its existence. If $j$ exists, it turns out to automatically be unique. – Eric Wofsey Mar 12 '18 at 22:17 | {
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# the number of arrangements of N players around a round table, where each player can sit on one of 3 contiguous chairs
Consider the fact that each player can either sit on their desired chair or on the neighbouring chair. Two configurations are distinct if at least one person is sitting in another chair. My attempt For example, if there are 4 players sitting around a circular table there can be 9 distinct arrangements.
1. F1,F2,F3,F4
2. F2,F1,F3,F4
3. F2,F1,F4,F3
4. F4,F2,F3,F1
5. F1,F3,F2,F4
6. F1,F2,F4,F3
7. F4,F1,F2,F3
8. F2,F3,F4,F1
9. F4,F3,F2,F1
Now, I am stuck here. How can calculate this and what about larger values of N such as 50?
I tried using vectors and pushing back every element and do the reverse operators and swap elements but I am not sure whether this is correct or not.
• Do you want to enumerate the arrangements or simply get the number? What have you tried? Jan 4, 2021 at 20:34
• Can you credit the source where you originally encountered this task?
– D.W.
Jan 4, 2021 at 20:36
• @D.W. No I cant this is a competetive programming question that was asked in a contest Jan 4, 2021 at 21:59
• Is the contest currently ongoing? When was the deadline for the contest? What prevents you from providing a link and/or reference to the programming contest? Sometimes helping us see the original statement of the question helps us understand what you are asking better. Also, please don't leave clarifications in the comments (such as regarding enumerating vs counting): instead, edit the question to clarify it.
– D.W.
Jan 4, 2021 at 22:08
• Please edit the question to make it clear whether the desired chairs of different players can be the same or not. Jan 6, 2021 at 6:09
Consider the variant of the problem in which the people are arranged in a segment instead of circle and let $$S(n)$$ be the number feasible arrangements for $$n$$ people.
Now lets go back to your original problem, and let's name $$C(n)$$ the number of feasible arrangements of $$n$$ people. | {
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Clearly $$C(0)=C(1)=1$$ and $$C(2)=2$$, so we will henceforth suppose that $$n \ge 3$$.
Let's name the people $$p_1, \dots, p_n$$ where the assigned position of $$p_i$$ is $$i$$. If $$p_1$$ is sitting in position 1 then the number of possible arrangements of $$p_2, \dots, p_n$$ is exactly $$S(n-1)$$. Otherwise, $$p_1$$ is sitting either in position $$2$$ or in position $$n \neq 2$$. We will only consider the former case since the latter one is symmetric. Clearly $$p_2$$ cannot sit in position $$2$$, so he/she must either sit in position $$1$$ or in position $$3$$. If $$p_2$$ is sitting in position $$1$$, then the number of possible arrangements of $$p_3, \dots, p_n$$ is exactly $$S(n-2)$$. Otherwise, $$p_2$$ must be sitting in position $$3$$, which means that $$p_3$$ must be sitting in position $$4$$ (since position $$2$$ and $$3$$ are occupied) and, in general, $$p_i$$ must be sitting in position $$(i \bmod n) + 1$$. Since this completely determines everyone's position, this case only contributes $$1$$ to the total number of configurations.
To summarize, we have that: $$C(n) = S(n-1) + 2(S(n-2)+1) = S(n-1) + 2S(n-2) + 2.$$
We are left with figuring out what $$S(n)$$ is. Clearly $$S(0) = S(1) = 1$$, so we consider $$n \ge 2$$. In this case $$p_1$$ can only sit in position $$1$$ or $$2$$. If $$p_1$$ is sitting in position $$1$$ then there are $$S(n-1)$$ possible arrangements of $$p_2, \dots, p_n$$. If $$p_1$$ is sitting in position $$2$$, then $$p_2$$ must be sitting in position $$1$$ (as it is the only other person that can sit there) and there are $$S(n-2)$$ possible arrangements for $$p_3, \dots, p_n$$. We hence have: $$S(n) = \begin{cases} 1 & \mbox{if } n \in \{0,1\} \\ S(n-1) + S(n-2) & \mbox{otherwise} \end{cases} = \mathcal{F}_{n+1},$$ where $$\mathcal{F}_{i}$$ denotes the $$i$$-th Fibonacci number. Substituting back, we obtain: | {
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$$C(n) = \begin{cases} 1 & \mbox{if } n \in \{0,1\} \\ 2 & \mbox{if } n = 2 \\ \mathcal{F}_n + 2 \mathcal{F}_{n-1} + 2 & \mbox{otherwise} \end{cases}.$$
• we can also say that C(n) = F_(n+1) + F_(n-1) + 2 since C(n) = S(n) + S(n-2) + 2 Jan 6, 2021 at 13:51
• Sure, they are equivalent :) Jan 6, 2021 at 15:23 | {
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# Math Help - [HELP!] Surface Integrals
1. ## [SOLVED] Surface Integrals
I don't know how to use the MATH code here...and I'm running out of time for reviewing for finals, so forgive me if it isn't clear for you... >.<
OK...here's my question:
DoubleInt(s) (z+x^2y)dS, S is the part of the cylinder y^2 + z^2 = 1 that lies between the planes x = 0 and x = 3 in the first octant.
I know that maybe...I should use
y = rcos(theta) = cos(theta)
z = rsin(theta) = sin(theta)
x = x
and find r_(theta) x r_(z) = < cos(theta), sin(theta), o> (I'm not sure if this way is the right path...)
and find the magnitude of |r_(theta) x r_(z)| = 1
I don't know what to do next, which formula to use (there were two...) and what boundaries? How can I get the numbers for those boundaries?
Can anyone solve this question for me please? I know the final answer, which is 12.
2. Hi. By definition:
$S=\mathop\int\int\limits_{\hspace{-15pt}S} g(x,y,z)dS=\mathop\int\int\limits_{\hspace{-15pt}R} g(x,y,f(x,y))\sqrt{f_x^2+f_y^2+1} dA$
So you're integrating the function $g(x,y,z)=z+x^2 y$ over the surface $f(x,y)=z=\sqrt{1-y^2}$ between the planes x=0 and x=3 in the first quadrant. That's just the top one-quarter of the cylinder along the x-axis shown in red below. Now, the region in the x-y plane below that part, since the cylinder has a radius of one, would be 0 to 3 in the x-direction and 0 to 1 in the y-direction. Now, turn the crank:
$\int_0^3\int_0^1 (\sqrt{1-y^2}+x^2y)\sqrt{f_x^2+f_y^2+1}dydx$
3. Originally Posted by violet8804
I don't know how to use the MATH code here...and I'm running out of time for reviewing for finals, so forgive me if it isn't clear for you... >.<
OK...here's my question:
DoubleInt(s) (z+x^2y)dS, S is the part of the cylinder y^2 + z^2 = 1 that lies between the planes x = 0 and x = 3 in the first octant. | {
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I know that maybe...I should use
y = rcos(theta) = cos(theta)
z = rsin(theta) = sin(theta)
x = x
and find r_(theta) x r_(z) = < cos(theta), sin(theta), o> (I'm not sure if this way is the right path...)
and find the magnitude of |r_(theta) x r_(z)| = 1
I don't know what to do next, which formula to use (there were two...) and what boundaries? How can I get the numbers for those boundaries?
Can anyone solve this question for me please? I know the final answer, which is 12.
Here's a general concept: if you write the surface as $\vec{r}(u,v)= x(u,v)\vec{i}+ y(u,v)\vec{j}+ z(u,v)\vec{k}$, then the two derivatives, $\vec{r}_u= x_u \vec{i}+ y_u\vec{j}+ z_u\vec{k}$ and $\vec{r}_v= x_v\vec{i}+ y_v\vec{j}+ z_v\vec{k}$ are tangent to the surface and their cross product (called the "fundamental vector product" of the surface), $\vec{r}_u\times\vec{r_v}$, is perpendicular to the surface and its length gives the "differential of surface area": $||\vec{r}_u\times\vec{r}_v||dudv$.
Yes, you can use $y= cos(\theta)$, $z= sin(\theta)$, x= x with $\theta$ and x as parameters u and v. That is, $\vec{r}= x\vec{i}+ cos(\theta)\vec{j}+ sin(\theta)\vec{k}$ so $\vec{r}_x= \vec{i}$ and $\vec{r}_\theta= -sin(\theta)\vec{j}+ cos(\theta)\vec{k}$ so that the "fundamental vector product" is
$\vec{r}_x\times\vec{r}_\theta= -cos(\theta)\vec{i}- sin(\theta)\vec{k}$.
Its length is, of course, 1 so $dV= dxd\theta$. The function to be integrated is $z+ x^2y= sin(\theta)+ x^2cos(\theta)$. In the first octant, both y and z are positive so $\theta$ goes from 0 to $\pi/2$ and, of course, x goes from 0 to 1. The integral is
$\int_{x=0}^1\int_{\theta= 0}^{\pi/2} sin(\theta)+ x^2 cos(\theta) dxd\theta$.
By the way, if you click on a mathematical expression in any post, you can see the LaTex code used. | {
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By the way, if you click on a mathematical expression in any post, you can see the LaTex code used.
4. Originally Posted by HallsofIvy
Here's a general concept: if you write the surface as $\vec{r}(u,v)= x(u,v)\vec{i}+ y(u,v)\vec{j}+ z(u,v)\vec{k}$, then the two derivatives, $\vec{r}_u= x_u \vec{i}+ y_u\vec{j}+ z_u\vec{k}$ and $\vec{r}_v= x_v\vec{i}+ y_v\vec{j}+ z_v\vec{k}$ are tangent to the surface and their cross product (called the "fundamental vector product" of the surface), $\vec{r}_u\times\vec{r_v}$, is perpendicular to the surface and its length gives the "differential of surface area": $||\vec{r}_u\times\vec{r}_v||dudv$.
Yes, you can use $y= cos(\theta)$, $z= sin(\theta)$, x= x with $\theta$ and x as parameters u and v. That is, $\vec{r}= x\vec{i}+ cos(\theta)\vec{j}+ sin(\theta)\vec{k}$ so $\vec{r}_x= \vec{i}$ and $\vec{r}_\theta= -sin(\theta)\vec{j}+ cos(\theta)\vec{k}$ so that the "fundamental vector product" is
$\vec{r}_x\times\vec{r}_\theta= -cos(\theta)\vec{i}- sin(\theta)\vec{k}$.
Its length is, of course, 1 so $dV= dxd\theta$. The function to be integrated is $z+ x^2y= sin(\theta)+ x^2cos(\theta)$. In the first octant, both y and z are positive so $\theta$ goes from 0 to $\pi/2$ and, of course, x goes from 0 to 1. The integral is
$\int_{x=0}^1\int_{\theta= 0}^{\pi/2} sin(\theta)+ x^2 cos(\theta) dxd\theta$.
By the way, if you click on a mathematical expression in any post, you can see the LaTex code used. | {
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By the way, if you click on a mathematical expression in any post, you can see the LaTex code used.
Thank you!
And, by the way...
Is the (theta) always from 0 to 2*Pi if it's in the 1st octant (after change to polar coordinates)?
Also, can you please teach me when to use which of the two formula, is there a trick to see from the question when to use which formula?
I always get confused which one to use, they are all the homework questions, and i got partial solutions (the library only have the older version of the solution manual...); so, when I do the question myself, and checked from the solution manual, it used a different formula (or different method) to solve the surface integral...I'm so confused T_T | {
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# Prove $\ \forall n \ge 2, \ 2^{n} + 3^{n} < 4^{n}$
Question:
Prove $\ \forall n \ge 2, \ 2^{n} + 3^{n} < 4^{n}$
My attempt:
Base case is trivial.
Suppose $\ n \ge 2$ and $\ 2^{n} + 3^{n} < 4^{n}$
Then,
$2^{n+1} + 3^{n+1} = 2.2^{n} + 3.3^{n} = 2.2^{n} + 2.3^{n} + 3^{n} = 2(2^{n} + 3^{n}) + 3^{n} <2(4^{n}) + 3^{n}$, by I.H.
I am stuck here. how do I show that this expression is $< 4^{n+1}$?
• Hints: Is $3^n<4^n$? Is $3(4^n)<4(4^n)$? – Michael Burr Sep 16 '17 at 20:25
• Another proof could use that the equivalent inequality $(2/4)^n+(3/4)^n<1$ after dividing both sides by $4^n$, where $4^n>0$, holds because it holds for $n=2$ and $(2/4)^n$ and $(3/4)^n$ are both strictly decreasing. – user236182 Sep 16 '17 at 20:32
You can say $$2\cdot 2^n + 3\cdot 3^n < 4\cdot 2^n + 4\cdot 3^n = 4(2^n+3^n) < 4\cdot 4^n = 4^{n+1}.$$
multiplying $$2^n+3^n<4^n$$ by $$4^n$$ we get $$2^n\cdot 2^{2n}+3^n\cdot 2^{2n}<4^{n+1}$$ and we have $$2^n\cdot 2<2^{3n}$$ since $$1<2^{2n-1}$$ for $n\geq 2$ and $$3^n\cdot 3<3^n\cdot 2^{2n}$$ since $$3<2^{2n}$$ therefore we have $$2^{n+1}+3^{n+1}<2^n\cdot 2^{2n}+3^n\cdot 2^{2n}<4^{n+1}$$
Hint: for $\,0 \lt a \lt 1\,$ the sequence $\,a^n\,$ is decreasing with $\,n\,$ since $\,a^{n+1} = a \cdot a^n \lt 1 \cdot a^n\,$, so:
$$\left(\frac{2}{4}\right)^n+\left(\frac{3}{4}\right)^n \le \left(\frac{2}{4}\right)^2+\left(\frac{3}{4}\right)^2 = \frac{13}{16} \lt 1 \quad\quad\text{for}\;\; \forall n \ge 2$$ | {
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# Minimal specification of isometry in terms of norm preservation
Let $V,W$ be $n$-dimensional (real) inner product spaces, and let $T:V \to W$ be a linear map.
Let $v_1,...,v_n$ be a basis of $V$. It is easy to see that if $|T(v)|_W=|v|_V$ for every $v \in \{v_1,...,v_n,v_1+v_2,v_1+v_3,...,v_{n-1}+v_n\}$, then $T$ is an isometry (a proof is provided below).
In other words, after choosing wisely $k(n):=\frac{n(n+1)}{2}$ vectors, it is enough to verify $T$ preserves the norms of these special vectors, in order to conclude it's an isometry.
Question: Is there no way to choose less than $k(n)$ vectors, in such a way that every linear map which preserves their norms is an isometry?
I believe we cannot choose less vectors. I have some "convincing evidence" for the cases $n=1,2,3$ (see below), but I am not sure how to give a rigorous argument.
Note that a "wise choice" of vectors does not have to be of the form of some vectors, and linear combinations of them (I do feel this it the most efficient method, but I don't see how to prove this). Even if we prove that this is the case, than we need to show we cannot do better than to work with only orthonormal bases.
The partial "evidence":
$n=1: k=1$. Obvious
$n=2: k=3$. Take $V=W=\mathbb{R}^n$ with its standard inner product. Then, $T(e_1)=e_1, T(e_2)=\frac{e_1+e_2}{\sqrt 2}$ is a counter example.
$n=3: k=6$. Then any matrix of the form $$\begin{pmatrix} c & s & x \\ -s & c & y \\ 0 & 0 & z \\\end{pmatrix}$$ where $c^2+s^2=1,x^2+y^2+z^2=1, sx+cy=0$ preserves the norms $e_1,e_2,e_1+e_2,e_3,e_2+e_3$ but it's an isometry only if $|z|=1,x=y=0$.
Proof that $k(n)=\frac{n(n+1)}{2}$ vectors are enough:
Noting that $$\langle u,v \rangle = \frac{1}{2}(|u+v|^2 - |u|^2 - |v|^2) ,$$ we obtain
$$\langle Tv_i,Tv_j \rangle = \frac{1}{2}(|Tv_i+Tv_j|^2 - |Tv_i|^2 - |Tv_j|^2) = \frac{1}{2}(|T(v_i+v_j)|^2 - |v_i|^2 - |v_j|^2)$$ $$= \frac{1}{2}(|v_i+v_j|^2 - |v_i|^2 - |v_j|^2) = \langle v_i,v_j \rangle,$$
thus $T$ is an isometry. | {
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thus $T$ is an isometry.
• I think there is not a way to choose less. Since it must preserve the scalar product and then all products $<v_i,v_j>$ of the basis must be preserved. By symmetry they are exactly the number you proposed. – Harnak Dec 25 '16 at 11:02
• @Harnak, please make your comment into an answer, I'd like to +1 it. – Andreas Caranti Dec 25 '16 at 11:09
• @Harnak I am not sure I am convinced by your argument. Can you please elaborate? (I also think there is no way to choose less). – Asaf Shachar Dec 25 '16 at 11:13
Let $f:\mathbb R^n\to V$ and $g:W\to\mathbb R^n$ be any two linear isometries. Then $T$ is an isometry if and only if $g\circ T\circ f$ is an isometry. So, we may assume without loss of generality that $V=W=\mathbb R^n$ equipped with the Euclidean inner product.
Let $v_1,\ldots,v_k\in\mathbb R^n$ be $k<\frac12n(n+1)$ arbitrarily chosen vectors. It suffices to exhibit the existence of a non-isometric linear transformation $T$ on $\mathbb R^n$ that preserves the norm of each $v_j$. Consider the system of homogeneous linear equations $$v_j^\top Av_j=0,\quad j=1,\ldots,k,\tag{1}$$ where the $n^2$ entries of $A\in M_n(\mathbb R)$ are unknown. Since the subspace of all $n\times n$ skew-symmetric matrices has dimension $\frac12n(n-1)<n^2-k$, the system $(1)$ must admit a nontrivial solution $A$ that is not skew-symmetric. However, if $A$ is a solution, so is $A+A^\top$. Therefore, $(1)$ admits a nontrivial symmetric solution $A$.
Let $P=I+\varepsilon A$, where $\varepsilon>0$ is sufficiently small. Then $P$ is positive definite. Define $Tx=\sqrt{P}x$. Then $T$ is not an isometry because $\sqrt{P}$ is not real orthogonal. However, for each $j$ we have $$\|Tv_j\|^2=v_j^\top Pv_j=v_j^\top(I+\varepsilon A)v_j=v_j^\top v_j=\|v_j\|^2.$$ | {
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• Wow! your solution is great and provides exactly what I was interested in, thanks. By the way, I am curious how did you think about using symmetric matrices. (I guess you started from considering perturbations of the identity, but then noticed that if you do not take a square root, the constraint you get is a quadratic equation, not linear, hence the necessity to consider roots...). – Asaf Shachar Dec 28 '16 at 12:53
• @AsafShachar Once you recognise that the problem boils down to finding a positive semidefinite matrix $P\ne I$ such that $v_j^\top Pv_j=v_j^\top v_j$ for each $j$, it's natural to consider symmetric matrices, because $I-P$ is symmetric. – user1551 Dec 28 '16 at 13:27
• I agree, but it was not immediate for me to consider the "square root" trick. (i.e I thought of starting with $Tx=Px$ where $P=I+\epsilon A$, but then the equation on $P$ is quadratic...). Anyway, your explanation is very reasonable, thanks again. – Asaf Shachar Dec 28 '16 at 16:52
I'll try to elaborate on my idea in the comment. Preserving the scalar product means that the two symmetric bilinear forms must coincide: $$(x,y) \mapsto \langle x,y \rangle$$ $$(x,y) \mapsto \langle Tx,Ty \rangle$$ This is equivalent to saying that their matrices must coincide. But a symmetric matrix has $\frac{n(n+1)}{2}$ degrees of freedom.
This means, once we've chosen the basis in $V$ we must choose the products $\langle v_i,v_j \rangle$ for $i \geq j$.
• Is it enough? Yes, because we've chosen the coefficients of the matrix relative to its canonic basis (here I'm taking about the basis of matrices which have $1$ in one component and $0$ in the others, but only for $i \geq j$ since the matrix is symmetric).
• Can we take less? No, because the symmetric matrices space has dimension $\frac{n(n+1)}{2}$ as we've said. | {
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How does this relate to our question? Choosing a basis for $V$ and then such products is equivalent to choosing the list of vectors you proposed (along with their norms).
(I.e. Choosing a bilinear form is equivalent to choosing the quadratic forum associated to it, which in the case of a scalar product is the norm)
• Thanks for your elaborate answer. However, I must say I am not entirely convinced by your argument. Indeed, I am also thinking that we cannot choose less vectors, and your observation that a (symmetric) bilinear form is determined by $\frac{n(n+1)}{2}$ vectors is also appealing and looks connected to this issue somehow. In spite all of that I am still not convinced. Suppose we know $\langle v_i,v_j \rangle = \langle Tv_i,Tv_j \rangle$ for all $i \le j$ except for a single pair $(i^*,j^*)$... – Asaf Shachar Dec 26 '16 at 15:34
• I would like to see a more explicit demonstration for how to build such a map $T$ which is not an isometry, i.e does not satisfy $\langle v_i,v_j \rangle = \langle Tv_i,Tv_j \rangle$ for $(i,j)=(i^*,j^*)$. I am not sure this follows so clearly from your "heuristic" argument. – Asaf Shachar Dec 26 '16 at 15:34
• So, you want to prove that there exists a map $T$ that preserves all those couples except for one? – Harnak Dec 26 '16 at 16:49
• Yes, for a start. My intention in the question was to show that no matter how cleverly you choose $k < \frac{n(n+1)}{2}$ vectors, and require they will be mapped to vectors of the same norm, you will be able to do this with a map which is not an isometry. However, in some sense this will only show the scheme of choosing basis vectors (and certain linear combinations of them) cannot be successful with less vectors, we also need to show we cannot do better by any choice. – Asaf Shachar Dec 26 '16 at 19:50 | {
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D
#### What is the probability of this problem?
63 viewed last edited 2 years ago
Anshul Malik
1
A and B roll a dice each. A wins if his dice number is greater than B. B wins if his dice number is greater than or equal to A.
If A gets a 1, then B will win for sure. So B makes up a rule that A will play the same game again if A rolls a 1 in the first throw. The winner of the second game (if it reaches that stage) will be the final winner even if A rolls a 1 again.
What is the probability that A will win this game?
Mahesh Godavarti
3
I think the problem statement can be worded better for easier understanding. However, I think I got it.
Probability of A winning = 1 - Probability of B winning.
Probability of B winning =
Probability of B throwing an equal or greater than when A throws a number greater than 1 in the first throw +
Probability that A throws a 1 in the first throw X Probability of B getting an equal or higher number in the second throw than A =
\frac{1}{6 } \times (\frac{5}{6} + \frac{4}{6} + \frac{3}{6} + \frac{2}{6} + \frac{1}{6}) + \frac{1}{6} \times \frac{1}{6} \times (\frac{6}{6} + \frac{5}{6} + \frac{4}{6} + \frac{3}{6} + \frac{2}{6} + \frac{1}{6}) =
\frac{1}{6} \times \frac{1}{6} \times 5 \times 6 \times \frac{1}{2} + \frac{1}{6} \times \frac{1}{6} \times \frac{1}{6} \times 6 \times 7 \times 1/2 = 1/6 \times 1/6 \times 1/2 \times (30 + 7) = \frac{37}{72}
Therefore, probability of A winning = 35/72.
Earlier, if B had not changed the rule then the probability of B winning would be
1/6 \times (6/6 + 5/6 + 4/6 + 3/6 + 2/6 + 1/6) = 1/6 \times 1/6 \times 6 \times 7 \times 1/2 = 7/12 .
And the probability of A winning would be 5/12.
Essentially, B leveled the playing field a little bit by changing the rules of the game. | {
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Essentially, B leveled the playing field a little bit by changing the rules of the game.
Vivekanand Vellanki
0
It would be good to explain how the probability of B throwing a number greater than or equal to A is determined.
Anshul Malik
0
But the cases in which A rolls a 1 in the first roll should not be counted as sample cases right? Also, the probability that the game is decided in the first roll is 5/6? If I'm understanding conditional probability correctly, then this should also be factored in right?
Mahesh Godavarti
0
I just swt up the sample space and added up all the cases that end up in B's victory. You could set it up as a conditional as well, with A rolling a 1 or not rolling a 1. Or the game finishing in one throw or finishing in two throws, you will get the same answer. In fact, I will set it up and show you that it is the same. Look for my second answer.
Mahesh Godavarti
0
Let's do it another way. Let's condition it first on the event that the game will finish in one throw or two throws.
Then P(B wins) = P(one throw) P(B wins | one throw) + P(two throws) P(B wins | two throws).
P(one throw) = P(A throws a number other than 1) = 5/6
P(two throws) = P(A throws a 1) = 1/6.
P(B wins | one throw) = P(A throws a 2 | A threw a number from 2 to 5) X P(B throws 2 or higher) + P(A throws a 3 | A threw a number from 2 to 5) X P(B throws 3 or higher) + and so on till A throws a 6 = 1/5 * 5/6 + 1/5 * 4/6 + 1/5 * 3/6 + 1/5 *2/6 + 1/5 * 1/6.
Therefore, P(one throw) P (B wins | one throw) = 5/6 * [1/5 * 5/6 + 1/5 * 4/6 + 1/5 *3/6 + 1/5 *2/6 + 1/5 * 1/6 ] = 1/6 * 5/6 + 1/6 *4/6 + 1/6 * 3/6 + 1/6 * 2/6 + 1/6 *1/6 = 1/6 * [5/6 + 4/6 + 3/6 + 2/6 + 1/6]. | {
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P(B wins | two throws) = P(A throws a 1 in the second throw | A threw a number from 2 to 5 in the first throw) X P(B throws 1 or higher in the second throw) + P(A throws a 2 in the second throw | A threw a number from 2 to 5 in the first throw) X P(B throws 2 or higher in the second throw) + and so on till A throws a 6 in the second throw
Since, the throws are independent what A throws in the second is independent of what is thrown in the first. Therefore, the above expression is the same as:
P(B wins | two throws) = P(A throws a 1 in the second throw) X P(B throws 1 or higher in the second throw) + P(A throws a 2 in the second throw) X P(B throws 2 or higher in the second throw) + and so on till A throws a 6 in the second throw = 1/6 * 6/6 + 1/6 *5/6 + 1/6 * 4/6 + 1/6 * 3/6 + 1/6 * 2/6 + 1/6 * 1/6
Therefore, P (two throws) P(B wins | two throws) = 1/6 * [1/6 * 6/6 + 1/6 *5/6 + 1/6 * 4/6 + 1/6 * 3/6 + 1/6 * 2/6 + 1/6 * 1/6] = 1/6 * 1/6 * [6/6 + 5/6 + 4/6 + 3/6 + 2/6 + 1/6]
Therefore, P(B wins) = 1/6 * [5/6 + 4/6 + 3/6 + 2/6 + 1/6] + 1/6 * 1/6 * [6/6 + 5/6 + 4/6 + 3/6 + 2/6 + 1/6].
Which is the same as what we had in the other answer. | {
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# Using induction, prove that $\sum_{k=1}^n\frac 1 {(2k-1)(2k+1)}= \frac 12-\frac 1 {4n+2}$ when $n$ is in $\mathbb{N}$
I can prove the base case and get that $1/3=1/3$, but i can't get any further with the $n+1$ case. Can someone help me?
I was told to conjecture a formula for the sum $$\frac{1}{3} + \frac{1}{{3 \cdot 5}} + \cdots + \frac{1}{{\left( {2n - 1} \right)\left( {2n + 1} \right)}}$$ I thought I figured out that this sum was equal to $$\frac{1}{2} { - \frac{1}{{4n + 2}}}$$ but I'm starting to think I'm wrong about that. After we have the formula, we were told to prove our conjecture using induction.
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I really can't understand what is the identity the you want to prove. Could you try to typeset it (better if in LaTeX), with all the necessary parentheses, please? – Andrea Orta Sep 18 '12 at 18:25
Are you sure you have the right expression? For $n = 2$ you get: $\frac{1}{2 \cdot 6} = \frac{1}{2} - \frac{1}{10}$ which is clearly untrue. – Feanor Sep 18 '12 at 18:25
If $n=1$ you are dividing by $0$. Also please check your parentheses: one extra left on the left side, it is not clear whether $(2n+2)$ should be in the numerator, and you must mean $1/(4n+2)$, not $(1/4)n+2$ or $1/(4n)+2$ on the right. \frac is your friend in this regard. – Ross Millikan Sep 18 '12 at 18:25
Also, unless there was supposed to be some sum involved, there is no need for induction. Just multiply everything out so as to get rid of fractions ;) – Feanor Sep 18 '12 at 18:27
Based on your added comment, you want the left to be $\sum_{i=1}^n \frac 1{2i-1}\frac 1{2i+1}=\frac 12 - \frac 1{4n+2}$. Note the $1$'s, not $2$'s and the sum on the left. – Ross Millikan Sep 18 '12 at 18:37
Hint $\$ The sum telescopes since for $\rm\:f(k) = 1/(4k-2)\:$ we have
$$\rm f(k+1)-f(k)\ =\ \frac{1}{4k+2} - \frac{1}{4k-2}\ =\ -\frac{1}{(2k-1)(2k+1)}$$
so an inductive proof is a special case of the inductive proof of the closed form for a telescopic sum: | {
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$$\rm\sum_{k=1}^n f(k\!+\!1)-f(k) = - f(1) + \color{#C00}{f(2) - f(2) + \cdots + f(n)-f(n)} + f(n\!+\!1)\, =\, f(n\!+\!1)-f(1)$$
But it is easy to turn the above ellipses into a rigorous inductive proof. Then your problem is simply a corollary of this general telescopy lemma, for the special value of $\rm\:f(k)\:$ given above.
The advantage of proving it this way is that - with no extra effort - you now have a general lemma that works to prove (by induction!) all summation identities of this form. Note that even though the induction has been abstracted out into a proof of a more general telescopy lemma, a proof invoking the telescopy lemma still counts as a proof by induction. The induction simply has been encapsulated in the proof of the lemma, which need not be repeated inline every time it is invoked. DivideAbstract and conquer - once you've seen one telescopic induction proof you've seen them all!
You can find more examples and further discussion in my prior posts on telescopy.
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@Peter Telescopy can be viewed as a form of induction specifically tailored for computing certain sums and products. – Bill Dubuque Sep 18 '12 at 19:00
Yes, I was just being silly, Bill! – Pedro Tamaroff Sep 18 '12 at 19:03
What you want to prove, apparently, is that
$$\sum_{k=1}^n \frac{1}{(2k-1)(2k+1)}=\frac 1 2 - \frac 1 {2(2n+1)}$$
As you say, the base case, $n=1$ is true. Suppose true for $n$, and analyze $n+1$. We get
$$\sum_{k=1}^{n+1} \frac{1}{(2k-1)(2k+1)}=\frac 1 2 - \frac 1 {2(2(n+1)+1)}$$
$$\sum_{k=1}^{n} \frac{1}{(2k-1)(2k+1)}+\frac{1}{(2(n+1)-1)(2(n+1)+1)}=\frac 1 2 - \frac 1 {2(2(n+1)+1)}$$
By the inductive hypothesis, this is, $$\frac 1 2 - \frac 1 {2(2n+1)}+\frac{1}{(2(n+1)-1)(2(n+1)+1)}=\frac 1 2 - \frac 1 {2(2(n+1)+1)}$$
$$- \frac{1}{{2(2n + 1)}} + \frac{1}{{(2n + 1)(2n + 3)}} = - \frac{1}{{2(2n + 3)}}$$
Multiply through ${(2n + 1)(2n + 3)}$ to get
$$- \frac{{(2n + 3)}}{2} + 1 = - \frac{{(2n + 1)}}{2}$$
In the end, you get | {
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$$- \frac{{(2n + 3)}}{2} + 1 = - \frac{{(2n + 1)}}{2}$$
In the end, you get
$$- 2n - 3 + 2 = - 2n - 1$$ $$- 2n -1 = - 2n - 1$$
which is true, so the inductive step is complete, and the theorem is proven.
-
Hint: This is a telescoping series. Note that $\frac 1{2i-1}\frac 1{2i+1}=\frac 12 \left(\frac 1{2i-1}-\frac 1{2i+1}\right)$ so neighboring terms cancel.
-
While this is the easiest approach, is not solving the problem the way the problem asks ;) – N. S. Sep 18 '12 at 18:43
@N.S.: I think it is a step toward the induction. – Ross Millikan Sep 18 '12 at 18:45
@RossMilikan he already guessed the right result, so he has a very easy induction problem, where he doesn't really need telescoping. – N. S. Sep 18 '12 at 18:49
@N.S. Telescopy does use induction - it's simply encapsulated (as a lemma) in the general closed form for a telescopic sum (whose inductive proof is easier in the abstract than in any specific case). So the natural way to prove this is to prove by induction the closed form for a telescopic sum, then specialize it to the case at hand - see my answer. Then, later, one can reuse the telescopy lemma to quickly tackle all such problems. – Bill Dubuque Sep 18 '12 at 19:09
@BillDubuque True, and this is the easiest way for us... But the problem asks the student to conjecture a closed formula, and prove it by induction. If you telescope it, you already get the answer... Is like the problem of calculating $\int_0^1 x^2 dx$... We typically expect the students to solve it the easy way, but if the Question explicitly asks to evaluate it using a Riemann sum, the easy solution is not good anymore... – N. S. Sep 18 '12 at 20:53
show 1 more comment
Saw this type of problem before.
You need to find a closed for for
$$\frac{1}{1 \cdot 3} + \frac{1}{3 \cdot 5}+...+\frac{1}{(2n-1)(1n+2)}$$
and then prove it by induction.
Your guess looks right, now you can do the induction: | {
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and then prove it by induction.
Your guess looks right, now you can do the induction:
$$\frac{1}{1 \cdot 3} + \frac{1}{3 \cdot 5}+...+\frac{1}{(2n-1)(1n+2)} =\frac{1}{2}-\frac{1}{2n+1} \,.$$
Alternately, you can use partial fraction decomposition to find a simple formula for $$\frac{1}{(2k-1)(2k+1)} \,,$$ and then get a telescopic sum. There is no need to prove this partial fraction decomposition by induction, and this approach is not really a direct induction proof.
- | {
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# When does the two cars meet
At 10:30 am car $A$ starts from point $A$ towards point $B$ at the speed of $65$ km/hr, at the same time another car left from point $B$ towards point $A$ at the speed of $70$ km/hr, the total distance between two points is $810$ km, at what time does these two cars meet ?
This is what I have tried,
$t_1 = \frac{810}{65} \,$ km/hr $= 12.46 \,$hr
$t_2 = \frac{810}{70} \,$ km/hr $= 11.57\,$ hr
$t_1-t_2 = 0.89 \,$ hr
$t_1-t_2 = 0.89\cdot 60 = 53.4 \,$ minutes
but this couldn't be the answer because how could these two cars could meet after $53.4$ minutes ?
@Joe, when they start they are 810km apart; after 1 hour the person leaving from point A would have traveled 65km, so if the other person had not left from point B, then the distance between them would be 810-65=745km. But the other person also helps close the distance between them by traveling 70km in the 1st hour from point B towards A, so the distance between them is 810-(65+75)=810-135=675km. So they are effectively subtracting 135km worth of distance per hour from the original distance between them.
So the 1st question you needed to answer was how much distance do they cover together per hour (we just did that), now the next question to answer is how long it takes to cover 810km in total distance at their effective speed. | {
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• it means that if they are covering 135km each our then they would be meet together in 135xP = 810km right ? Where P is the number of hours they will meet after. like 810km/135km/hr = hr6 so, 10:30AM + 6Hours = 4:30PM right ? – Joe Jan 19 '15 at 18:52
• Yes, this is exactly right. – Acemanhattan Jan 19 '15 at 18:56
• But that's what the question was asking. When will they meet. If they were asking when will each cover 810km you would divide 810 by the slower train's speed. – turkeyhundt Jan 19 '15 at 19:00
• Suppose for a second that they each did travel 810km, where would they each end up? Wouldn't they have just traded places with the person who started at A being at B and the person who started at B being at A? In order for that to happen, at some point they would have had to pass each other along the way, and when they passed each other neither would have traveled 810km individually, but the distance between them would be zero. – Acemanhattan Jan 19 '15 at 19:02
• Thanks a lot for great explanation. =) – Joe Jan 19 '15 at 19:03
Hint: Think about how much closer the cars get each hour.
They are approaching each other at an effective speed of 135 km/hr...
• but at what time does these two cars meet ? would that be 135/2 ? – Joe Jan 19 '15 at 18:27
• like 135/2 = 67.5 and then 810/67.5 = 12 ? – Joe Jan 19 '15 at 18:29
• no. it would be how long it would take to travel 810km at a rate of 135km/hr. Because every hour, they get 135km closer to each other. – turkeyhundt Jan 19 '15 at 18:50
• it means that if they are covering 135km each our then they would be meet together in 135xP = 810km right ? Where P is the number of hours they will meet after. like 810km/135km/hr = hr6 so, 10:30AM + 6Hours = 4:30PM right ? – Joe Jan 19 '15 at 18:54
• Yes. very good. – turkeyhundt Jan 19 '15 at 18:59 | {
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Hint: The total distance between the two cars is initially $810~\text{km}$. Thus, in order to meet, the combined distance they travel is $810~\text{km}$. Since they are travelling toward each other, they get $65~\text{km} + 70~\text{km} = 135~\text{km}$ closer to each other each hour.
• so then I have two divide 135 by 2 and then divide the answer by 810, like this, 135/2 = 67.5 then 810/67.5 = 12hours right ? so they will meet at 10:30 pm right ? – Joe Jan 19 '15 at 18:48
• No. What you did is compute the time it would take for the average distance travelled by the cars to be $810~\text{km}$, by which time they would have long passed each other. The cars get $135~\text{km}$ closer to each other each hour. You want to reduce the distance between them, which is initially $810~\text{km}$, to $0~\text{km}$. – N. F. Taussig Jan 19 '15 at 18:51
• it means that if they are covering 135km each our then they would be meet together in 135xP = 810km right ? Where P is the number of hours they will meet after. like 810km/135km/hr = hr6 so, 10:30AM + 6Hours = 4:30PM right ? – Joe Jan 19 '15 at 18:53
• That is correct. – N. F. Taussig Jan 19 '15 at 18:54
• They do not have to cover $810~\text{km}$ individually. After six hours, the slower car has travelled $$65~\frac{\text{km}}{\text{h}} \cdot 6~\text{h} = 390~\text{km}$$ so it is now $810~\text{km} - 390~\text{km} = 420~\text{km}$ from where the faster car began. Since the faster car has travelled $$70~\frac{\text{km}}{\text{h}} \cdot 6~\text{h} = 420~\text{km}$$ toward the slower car, they meet at 4:30 pm. Draw a picture. – N. F. Taussig Jan 19 '15 at 19:03 | {
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Lucas numbers and fibonacci
This is a question straight from the Applied Combinatorics book.
Suppose that chairs are arranged in a circle. Let $L_n$ count the number of subsets of $n$ chairs which don't contain consecutive chairs. show that $$L_{n+1} = F_n + F_{n+2}.$$
What is meant here? Could some one explain this to me?
• Do you need a proof, or do you have difficulties to understand the statement ? Feb 26, 2015 at 10:35
• What confuses me is that the number of chairs is not given. Feb 26, 2015 at 10:37
• Proof and explanation please. I don't understand what is meant by the question. Feb 26, 2015 at 10:56
If $n$ chairs are arranged in a circle, let $a_n$ be the number of subsets of the chairs that don’t contain consecutive chairs. (I’ll call these good subsets.) Clearly $a_0=1$, $a_1=2$, $a_2=3$, and $a_3=4$, since the empty subset satisfies the condition.
Now suppose that $n\ge 4$, and number the chairs from $1$ to $n$ around the circle. Temporarily remove chair $n$; the circle of remaining chairs has $a_{n-1}$ good subsets, none of which contains both chair $1$ and chair $n-1$. Of course none of these contains chair $n$, either. Moreover, every good subset of the $n$ chairs that does not contain chair $n$ and contains at most one of chairs $1$ and $n-1$ is counted here, so there are $a_{n-1}$ such subsets. | {
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Now we’ll count the good subsets that contain either contain chair $n$ or contain both chair $1$ and chair $n-1$. Start by removing chairs $n-1$ and $n$; the remaining circle has $a_{n-2}$ good subsets. Let $S$ be one of these good subsets. If chair $1$ is in $S$, form $S'$ by adding chair $n-1$ to $S$; if chair $1$ is not in $S$, form $S'$ by adding chair $n$ to $S$. It’s not hard to check that in each case $S'$ is a good subset of the ring of $n$ chairs, and that every such subset that contains either chair $n$ or both of chairs $1$ and $n-1$ is obtained in this way. (It’s here that we need to have $n\ge 4$: in order to form $S'$ this way when chair $1$ is in $S$, we need to know that $n-1>2$.)
This argument shows that the sequence $\langle a_n:n\in\Bbb N\rangle$ satisfies the recurrence
$$a_n=a_{n-1}+a_{n-2}\tag{0}$$
for $n\ge 4$. By inspection $a_2=3=2+1=F_3+F_1$ and $a_3=4=3+1=F_4+F_2$, so a straightforward induction shows that $$a_n=F_{n+1}+F_{n-1}\tag{1}$$ for $n\ge 2$:
\begin{align*} a_{n+1}&=a_n+a_{n-1}\\ &=\left(F_{n+1}+F_{n-1}\right)+\left(F_n+F_{n-2}\right)&\text{induction hypothesis}\\ &=\left(F_{n+1}+F_n\right)+\left(F_{n-1}+F_{n-2}\right)\\ &=F_{n+2}+F_n\;. \end{align*}
Apart from the fact that I used $a_n$ instead of $L_n$ for the number of good subsets of a ring of $n$ chairs, $(1)$ is exactly relationship that you were to prove. | {
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Nothing in the wording of the question or the original tagging indicates that $L_n$ is supposed to be the $n$-th Lucas number: $L_n$ is explicitly defined to be the number of good subsets of a ring of $n$ chairs. (The tag was added by Martin Sleziak.) The choice of the notation $L_n$ for my $a_n$ in the problem statement was very likely motivated by the fact that $a_n$ is the $n$-th Lucas number, $L_n$, for $n>1$. This follows immediately from the fact that the sequence $\langle L_n:n\in\Bbb Z^+\rangle$ of Lucas numbers by definition satisfies the same recurrence $(0)$ as the sequence $\langle a_n:n\in\Bbb N\rangle$, with $L_1=1$ and $L_2=3=a_2$: $L_3=4=a_3$, and the recurrence then ensures that $a_n=L_n$ for all $n\ge 2$.
• Thanks for the clear explaination! Mar 1, 2015 at 11:41
• @Discreteballoons: My pleasure! Mar 1, 2015 at 11:55
The Lucas numbers are defined recursively by \begin{align*} L_1 & = 1\\ L_2 & = 3\\ L_n & = L_{n - 1} + L_{n - 2}, n \geq 3 \end{align*} The first few terms of the Lucas sequence are $\{1, 3, 4, 7, 11, 18, 29, 47, 76, 123, \ldots\}$.
The Fibonacci numbers are defined recursively by \begin{align*} F_1 & = 1\\ F_2 & = 1\\ F_n & = F_{n - 1} + F_{n - 2}, n \geq 3 \end{align*} The first few terms of the Fibonacci sequence are $\{1, 1, 2, 3, 5, 8, 13, 21, 34, 55, \ldots\}$.
We can prove the assertion that $L_{n + 1} = F_n + F_{n + 2}$ by induction on $n$. Let $P(n)$ be the statement that $L_{n + 1} = F_n + F_{n + 2}$.
If $n = 1$, then $L_2 = L_{1 + 1} = 3 = 1 + 2 = F_1 + F_3$. Thus, $P(1)$ holds.
IF $n = 2$, then $L_3 = L_{2 + 1} = 4 = 1 + 3 = F_2 + F_4$. Thus, $P(2)$ holds. | {
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IF $n = 2$, then $L_3 = L_{2 + 1} = 4 = 1 + 3 = F_2 + F_4$. Thus, $P(2)$ holds.
Assume $L_{m + 1} = F_m + F_{m + 2}$ for each $n \leq m$, where $m \geq 2$. Let $n = m + 1$. Then \begin{align*} L_{m + 2} & = L_{m + 1} + L_{m} && \text{by definition of the Lucas sequence}\\ & = F_{m} + F_{m + 2} + F_{m - 1} + F_{m + 1} && \text{by the induction hypothesis}\\ & = F_{m} + F_{m - 1} + F_{m + 2} + F_{m + 1}\\ & = F_{m + 1} + F_{m + 3} && \text{by definition of the Fibonacci sequence} \end{align*} so $P(m) \Rightarrow P(m + 1)$. Hence, $P(n)$ holds for each positive integer $n$.
Let $a_m$ denote the number of subsets of $m$ chairs in which no two chairs are consecutive. We will consider the first few cases, where the chairs are numbered from $1$ to $m$.
\begin{align*} & m = 1: \emptyset, \color{blue}{\{1\}}\\ & m = 2: \color{red}{\emptyset, \{1\}}, \color{blue}{\{2\}}\\ & m = 3: \color{red}{\emptyset, \{1\}, \{2\}}, \color{blue}{\{3\}}\\ & m = 4: \color{red}{\emptyset, \{1\}, \{2\}, \{3\}}, \color{blue}{\{4\}}, \color{green}{\{1, 3\}}, \color{blue}{\{2, 4\}}\\ & m = 5: \color{red}{\emptyset, \{1\}, \{2\}, \{3\}, \{4\}}, \color{blue}{\{5\}} \color{red}{\{1, 3\}}, \color{green}{\{1, 4\}}, \color{red}{\{2, 4\}}, \color{blue}{\{2, 5\}, \{3, 5\}}\\ & m = 6: \color{red}{\emptyset, \{1\}, \{2\}, \{3\}, \{4\}, \{5\}}, \color{blue}{\{6\}}, \color{red}{\{1, 3\}, \{1, 4\}}, \color{green}{\{1, 5\}}, \color{red}{\{2, 4\}, \{2, 5\}}, \color{blue}{\{2, 6\}}, \color{red}{\{3, 5\}}, \color{blue}{\{3, 6\}, \{4, 6\}}, \color{green}{\{1, 3, 5\}}, \color{blue}{\{2, 4, 6\}}\\ \end{align*}
Based on these examples, we see that \begin{align*} a_1 & = 2\\ a_2 & = 3\\ a_3 & = 4\\ a_4 & = 7\\ a_5 & = 11\\ a_6 & = 18 \end{align*} Thus, the assertion that $a_1 = L_1$ is false. However, it appears that $a_m = L_m$ if $m > 1$.
Let $m = n + 1$ for $n \geq 1$. | {
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Let $m = n + 1$ for $n \geq 1$.
I have used three colors to highlight the subsets of $m = n + 1$ chairs that do not contain consecutive chairs. Subsets marked in blue contain chair $n + 1$. Subsets marked in red in row $n + 1$ are those subsets that appeared in row $n$. Subsets marked in green contain both $1$ and $n$.
Observe that the elements in blue in row $n + 1$ are formed by taking the union of elements in row $n - 1$ that do not contain $1$ with the set $\{n + 1\}$, while the elements in green are formed by taking the union of elements in row $n - 1$ that do contain $1$ with the set $\{n\}$. Hence, the total number of elements in row $n + 1$ that are either blue or green is $L_{n - 1}$. The number of elements in row $n + 1$ that are red is the number of elements in row $n$, which is $L_n$. Hence, for each $n \geq 1$, \begin{align*} L_{n + 1} & = L_n + L_{n - 1}\\ & = F_{n - 1} + F_{n + 1} + F_{n - 2} + F_n\\ & = F_{n - 1} + F_{n - 2} + F_{n + 1} + F_n\\ & = F_{n} + F_{n + 2} \end{align*} We can see that the assertion is false when $n = 0$ since the empty set in row $n + 1 = m = 1$ is not a member of one of the three types of highlighted sets.
While I did not use this observation in my proof, it appears that for $n \geq 1$ that the number of sets containing $n + 1$ in row $n + 1$ is $F_n$, while the number of sets containing $1$ and $n$ in row $n + 1$ is $F_{n - 2}$ (where we define $F_0 = F_2 - F_1 = 0$ and $F_{-1} = F_1 - F_0 = 1 - 0 = 1$).
Are you sure of the formulation of your problem? I am wondering if the good one would not be: "Suppose that $n$ chairs are arranged in a circle. Let $L_n$ count the number of subsets of $k \geq 1$ chairs which don't contain consecutive chairs, show that..."
• Please see this tutorial on how to format mathematics on this site. Feb 26, 2015 at 12:46
• I have just checked it, thank you. Feb 26, 2015 at 13:50 | {
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# Calculating the surface area of revolution for parametric equation.
I solved a problem using a method that's completely different from the mark scheme and I got the right answer, but I'm unsure whether or not it's just some coincidence. Here's the question:
The curve $C$ has parametric equations $$x=t^2, y=\frac{1}{4}t^4-\ln{t}$$ for $1\leq t\leq 2$. Find the are of the surface generated when $C$ is rotated $2\pi$ radians about the y-axis.
Here's what I did ( not very rigorous):
I thought cutting the curve with horizontal cuts should do the trick. So I just evaluated $2\pi\int _{t=1}^2xdy$ since $dy=(t^3-\frac{1}{t})dt$ it's just $$2\pi\int_1^2(t^5-t)dt$$ which is fairly easy to find.
Here's what's in the mark scheme:
$x'=2t$ and $y'=t^3-\frac{1}{t}$
$(\frac{ds}{dt})^2=x'^2+y'^2=(t^3-\frac{1}{t})^2$
$S=\int 2\pi x ds=2\pi\int_1^2(t^5-t)dt$
My understanding is that both answers slice $C$ differently and so maybe this was just some coincidence. I can't prove that both methods are equivalent though. So if someone can give me an idea as to why they are that'd be great. Also, it seems to me that the method I used was way easier, is there a place where it's better to use that other method?
• If you examine the mark scheme's computation, you'll find a difference of sign with what's in your post: $x' = 2t$ and $y' = (t^{3} - \frac{1}{t})$, so $$\left(\frac{ds}{dt}\right)^{2} = x'(t)^{2} + y'(t)^{2} = \left(t^{3} + \tfrac{1}{t}\right)^{2},$$ n.b., not $\left(t^{3} - \tfrac{1}{t}\right)^{2}$, which would be equal to $y'(t)^{2}$ itself. If the mark scheme really has a minus sign on the right, that's an error. | {
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# Tag Info
## Hot answers tagged nonlinear-equations
19
There are two issues that you are likely to be encountering. Ill-conditioning First, the problem is ill-conditioned, but if you only provide a residual, Newton-Krylov is throwing away half your significant digits by finite differencing the residual to get the action of the Jacobian: $$J[x] y \approx \frac{F(x+\epsilon y) - F(x)}{\epsilon}$$ If you ...
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Short answer If you only want second order accuracy and no embedded error estimation, chances are that you'll be happy with Strang splitting: half-step of reaction, full step of diffusion, half step of reaction. Long answer Reaction-diffusion, even with linear reaction, is famous for demonstrating splitting error. Indeed, it can be much worse, including "...
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This is the stochastic root-finding problem, as in The stochastic root-finding problem: Overview, solutions, and open questions.
11
Is the shooting method the only general numerical method for solving BVP of nonlinear ODE(s)? No. Most other methods consist of three parts: Discretization. This may be done with finite differences, finite volumes, finite elements (Galerkin or collocation), spectral methods, and so forth. This reduces the problem from an infinite-dimensional one to a ...
9
If you can stand to use complex arithmetic, simultaneous iteration methods might be preferable for computing all the roots of your polynomial. The simplest simultaneous iteration method, the (Weierstrass-)Durand-Kerner method, is effectively equivalent to applying Newton-Raphson to the Vieta relations relating the coefficients and roots of a polynomial, ...
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I assume, that you have conducted a space discretization, so that you are about solving the (vector-valued) ODE $$\dot u_h(t) = F_h(t,u_h(t)), \text{ on [0,T] }, u_h(0) = \alpha.$$ via a numerical scheme $\Phi$ that advances the approximation $u_h^n$ at the current time instance $t=t^n$ to the next value $u_h^{n+1}$ at $t=t^{n+1}:=t^n+\tau$. Then your ...
8
You can solve this numerically in Python without symbolic computation. from __future__ import print_function, division import numpy as np from numpy import exp from scipy.integrate import quad from scipy.optimize import root def f1(a1, a2, x): return exp(a1 * x + a2 * x * x * x) / (1 + x * x) def f2(a1, a2, x): return exp(a1 * x + a2 * x * x * x) *...
8
PDEs are a form of dynamical system where there is another continuous variable. Usually this is space, so you're looking at how things over time and space instead of just over time. Here's an example of generalizing an ODE to a PDE. Take your ODE model of chemical reactions which models the concentration of certain chemicals over time. Now generalize the ...
8
Julia has a whole ecosystem for generating sparsity patterns and doing sparse automatic differentiation in a way that mixes with scientific computing and machine learning (or scientific machine learning). Tools like SparseDiffTools.jl, ModelingToolkit.jl, and SparsityDetection.jl will do things like: Automatically find sparsity patterns from code Generate ...
7
To answer your questions in order: Any implicit method for solving an ordinary differential equation involves solving a system of nonlinear equations. You can do this through variants of Newton's method, successive substitution, full approximation scheme, or any other approach that solves systems of nonlinear equations. (The caveat is, of course, depending ...
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7
What you describe as your time discretization is called the Crank-Nicolson scheme. For nonlinear differential equations it leads, as you have observed, to a nonlinear algebraic system that needs to be solved at each time step. The typical approach is to solve it with a Newton method -- in your case, that requires to solve a nonlinear system in 5 variables. ...
7
As pointed out by David Ketcheson, one method is to use the companion matrix and find its eigenvalues (this is what Matlab does for the roots function). However, if you want to code everything by yourself, you can try to use the Sturm sequences, which you use as a first step to find an interval with only one zero. Then, you can apply one standard methods ...
7
For your example equation, taking the average approach, the local consistency error $$\frac{1}{h^2}[u(x-h) - 2 u(x) + u(x+h)]-f(\frac{1}{2}[u(x-h)+u(x+h)]) = \frac{1}{2}f_uu_{xx}h^2 + hot.$$ will be of order $2$ (instead of order $3$). ($hot.$ means higher order terms) Therefore, if your overall approximation is of order $1$, e.g. if you use upwind ...
7
This is somehow unexpected, but my recent experience with solving a system of nonlinear equations is that treating them as the right hand side of a system of ordinary equations and then evolve the system with an ODE solver can be considerably faster than with the usual Newton-Raphson iteration. It sounds like you're doing some sort of pseudotransient ...
7
The issues you're running into now are not a failing of Newton-Raphson, but a question of coupling. You're doing iterated sequential coupling -- solving each equation sequentially and then iterating until (hopeful) convergence. No solver choice in place of NR is going to fix this lack of convergence, as long as you are doing iterated sequential coupling. ...
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It's a bit easier to see if you write your equation in the a semi-discretised system of the form $u^{\prime}(t) = F(u(t))$ and with the application of the $\theta$-method and approximating $u^{\prime}(t) \approx (w^{n+1} - w^{n})/\tau$ this gives, $$w^{n+1} - w^{n} - (1-\theta) \tau F(w^n) - \theta\tau F(w^{n+1}) = 0$$ with unknown vector $w^{n+1}$ and ...
7
$s_k$ is the "approximate Newton" search direction. So in essence, when they say Choose $s_k$ such that $\|F(x_k)+F'(x_k)s_k\| \le \eta_k \|F(x_k)\|$ they are saying: Solve the Newton system $F'(x_k)s_k = - F(x_k)$ inexactly for $s_k$ until the norm of the residual $F(x_k)+F'(x_k)s_k$ is smaller than the norm of the right hand side $-F(x_k)$ by a ...
6
If you need Jacobian matrix information for a numerical method, you should calculate the Jacobian matrix of the discretized form of the equations, since that will be consistent with the discretized equations you are solving.
6
You should rather think of what you will need in the following step, which is probably the numerical time integration of the semi-discrete equations. If you are going to use a (semi) explicit time stepping scheme, all you need is a function that for a given $\phi_0$ assembles the vector $\langle (\nabla \phi_0)^2, v \rangle$, where $v$ are your test ...
6
$K u$ equals the internal forces only in the linear case. The tangent stiffness matrix, $K$, in a nonlinear problem is normally used in a Newton-Raphson algorithm to calculate updates to the displacement vector as follows: $$K \Delta u = f - f_{internal}$$ $$u_{i+1} = u_i + \Delta u$$ The vector of internal forces, $f_{internal}$ must be calculated from ...
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If you change variables to optimize for the residual of the linear part, then the Hessian will be a low-rank update to the identity. Then L-BFGS would work very well. Specifically, your problem takes the form $$\min_x \frac{1}{2}\|Ax-b\|^2 + \frac{\mu}{2}\|g(x)\|^2$$ where $Ax=b$ is the linear PDE and $g$ is the nonlinear part, and $\mu$ is a tradeoff ...
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The reason is that GMRES can only be used for solving linear equations, i.e. equations of the form $Ax=b$, where $A$ is some matrix and $x,b$ are vectors. What GMRES does, essentially, is it approximates multiplication by the matrix $A^{-1}$ using a matrix polynomial of $A$. In this case (I assume) $f(y^{n+1},t)$ is not necessarily linear in the vector $y^{... 6 Two systematic ways of smoothing a function$h$would be: 1. Join the piecewise smooth parts of your function using Hermite interpolation so that the derivatives are matched to your satisfaction. 2. Convolve your function$h(x)$with a heat kernel of the form$f(x) = \frac{\exp\left\{-\frac{x^2}{2 \sigma^2}\right\}}{\sqrt{2 \pi \sigma^2}}$so that instead ... 5 That's actually very easy to do in Dolfin: from dolfin import * # define mesh, function space (piecewise linear) mesh = UnitSquareMesh(64,64) V = FunctionSpace(mesh,'CG',1) # inhomogeneous boundary conditions (otherwise the solution is trivial) bc = DirichletBC(V, Constant(1.0), lambda x,on_boundary: on_boundary) # define bi(non)linear form # note that ... 5 Your observed quadratic convergence indicates that the Jacobian is likely correct. Have you looked at the solutions for your under-resolved configurations? Galerkin optimality uses the operator norm, which contains only the symmetric part, thus the solution of the discrete system could be quite different from the projection of the exact solution. This ... 5 As it looks to me your formulation of the discrete system is incomplete. You need to state your equation at each (grid?) point associated with$n$put back the summation over$l$(grid) points in your last equation assign a time index$i+1$or$i$to your$c_l$s In this way, you get a nonlinear (if you take$c_l^{n+1}$) or linear (if you take$c_l^{n}$) ... 5 A common strategy is to employ a damping strategy, i.e., to compute$\vec{w}^{\ast} = F \left( \vec{w}^{(n)} \right)$and then set$\vec{w}^{(n+1)} = \alpha \vec{w}^{(\ast)} + | {
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= F \left( \vec{w}^{(n)} \right)$and then set$\vec{w}^{(n+1)} = \alpha \vec{w}^{(\ast)} + (1-\alpha)\vec{w}^{(n)}$where$\alpha\in[0,1]$. You typically choose$\alpha$in such a way that it minimizes$\|[\alpha \vec{w}^{(\ast)} + (1-\alpha)\vec{w}^{(n)}] - F(\alpha \vec{w}^... | {
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5
Certified homotopy continuation methods are used both for finding roots and for proving that they indeed exist (inside a certain interval). A quick web search turned out this paper: Reliable homotopy continuation by Joris van der Hoeven.
5
No it is not. There is also multiple shooting collocation finite differences fixed point iterations and probably some more.
5
Since $\phi$ is a scalar between $0$ and $1$, the easiest method for finding a root is bisection. If you cannot calculate the expectation of the nonlinear function $$f_\phi(\theta) = \left(\phi r_z +(1-\phi)(r_k+\theta)\right)^{1-\gamma}(r_k+\theta-r_z)$$ in terms of $\phi$ analytically, you can use quadrature to approximate. For the first variant, this ...
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• ### $450 Tuition Credit & Official CAT Packs FREE December 15, 2018 December 15, 2018 10:00 PM PST 11:00 PM PST Get the complete Official GMAT Exam Pack collection worth$100 with the 3 Month Pack ($299) • ### FREE Quant Workshop by e-GMAT! December 16, 2018 December 16, 2018 07:00 AM PST 09:00 AM PST Get personalized insights on how to achieve your Target Quant Score. # Is 3^(a^2/b) < 1? new topic post reply Question banks Downloads My Bookmarks Reviews Important topics Author Message TAGS: ### Hide Tags Manager Joined: 27 Apr 2012 Posts: 57 Location: United States GMAT Date: 06-11-2013 GPA: 3.5 WE: Marketing (Consumer Products) Is 3^(a^2/b) < 1? [#permalink] ### Show Tags 21 Jun 2012, 00:59 1 15 00:00 Difficulty: 95% (hard) Question Stats: 32% (00:58) correct 68% (01:01) wrong based on 378 sessions ### HideShow timer Statistics Is 3^(a^2/b) < 1? (1) a<0 (2) b<0 Hi, request your help to please understand the fundamental concept behind this question. ##### Most Helpful Expert Reply Math Expert Joined: 02 Sep 2009 Posts: 51218 Re: Is 3^(a^2/b) < 1? [#permalink] ### Show Tags 21 Jun 2012, 01:17 7 6 Is 3^(a^2/b) < 1? Notice that $$3^{\frac{a^2}{b}} < 1$$ to hold true, the power of 3 must be less than 0. So, the question basically asks whether $$\frac{a^2}{b}<0$$. This will happen if $$a\neq{0}$$ AND $$b<0$$ (if $$a=0$$ then $$\frac{a^2}{b}=0$$). (1) a<0. The first condition is satisfied ($$a\neq{0}$$) but we don't know about the second one. Not sufficient. (2) b<0. The second condition is satisfied ($$b<0$$) but we don't know about the first one (again if $$a=0$$ then $$\frac{a^2}{b}=0$$). Not sufficient. (1)+(2) Both condition are satisfied. Sufficient. Answer: C. For more on number theory and exponents check: math-number-theory-88376.html DS questions on exponents: search.php?search_id=tag&tag_id=39 PS questions on exponents: search.php?search_id=tag&tag_id=60 Tough and tricky DS exponents and roots questions with detailed solutions: | {
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Tough and tricky DS exponents and roots questions with detailed solutions: tough-and-tricky-exponents-and-roots-questions-125967.html Tough and tricky PS exponents and roots questions with detailed solutions: tough-and-tricky-exponents-and-roots-questions-125956.html Hope it helps. _________________ ##### General Discussion Manager Joined: 27 Apr 2012 Posts: 57 Location: United States GMAT Date: 06-11-2013 GPA: 3.5 WE: Marketing (Consumer Products) Re: Is 3^(a^2/b) < 1? [#permalink] ### Show Tags 21 Jun 2012, 02:10 This is great.Thanks a ton! Math Expert Joined: 02 Sep 2009 Posts: 51218 Re: Is 3^(a^2/b) < 1? [#permalink] ### Show Tags 25 Jun 2013, 03:47 Bumping for review and further discussion*. Get a kudos point for an alternative solution! *New project from GMAT Club!!! Check HERE Theory on Exponents: math-number-theory-88376.html All DS Exponents questions to practice: search.php?search_id=tag&tag_id=39 All PS Exponents questions to practice: search.php?search_id=tag&tag_id=60 Tough and tricky DS exponents and roots questions with detailed solutions: tough-and-tricky-exponents-and-roots-questions-125967.html Tough and tricky PS exponents and roots questions with detailed solutions: tough-and-tricky-exponents-and-roots-questions-125956.html _________________ Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 6639 GMAT 1: 760 Q51 V42 GPA: 3.82 Re: Is 3^(a^2/b) < 1? [#permalink] ### Show Tags 27 Jan 2016, 17:46 Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution. Is 3^(a^2/b) < 1? (1) a<0 (2) b<0 When you modify the original condition and the question, they become 3^(a^2/b) < 1? --> 3^(a^2/b) < 3^0? --> a^2/b>0?. Multiply b^2 on the both equations(since b^2 is positive, even if it’s multiplied, the sign of inequality doesn’t change.) and it becomes a^2(b)>0?. There | {
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even if it’s multiplied, the sign of inequality doesn’t change.) and it becomes a^2(b)>0?. There are 2 variables(a,b), which should match with the number of equations. So you need 2 equations. For 1) 1 equation, for 2) 1 equation, which is likely to make C the answer. When 1) & 2), since a<0, it can’t be 0. Divide the both equations with a^2, they become a^2(b)<0?-->b<0?. Since 2) is b<0, it is yes and sufficient. Therefore, the answer is C. For cases where we need 2 more equations, such as original conditions with “2 variables”, or “3 variables and 1 equation”, or “4 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 70% chance that C is the answer, while E has 25% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, D or E. _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$99 for 3 month Online Course" | {
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Posts: 12
### Show Tags
25 Oct 2018, 05:33
Bunuel wrote:
Is 3^(a^2/b) < 1?
Notice that $$3^{\frac{a^2}{b}} < 1$$ to hold true, the power of 3 must be less than 0. So, the question basically asks whether $$\frac{a^2}{b}<0$$. This will happen if $$a\neq{0}$$ AND $$b<0$$ (if $$a=0$$ then $$\frac{a^2}{b}=0$$).
(1) a<0. The first condition is satisfied ($$a\neq{0}$$) but we don't know about the second one. Not sufficient.
(2) b<0. The second condition is satisfied ($$b<0$$) but we don't know about the first one (again if $$a=0$$ then $$\frac{a^2}{b}=0$$). Not sufficient.
(1)+(2) Both condition are satisfied. Sufficient.
For more on number theory and exponents check: http://gmatclub.com/forum/math-number-theory-88376.html
DS questions on exponents: http://gmatclub.com/forum/search.php?se ... &tag_id=39
PS questions on exponents: http://gmatclub.com/forum/search.php?se ... &tag_id=60
Tough and tricky DS exponents and roots questions with detailed solutions: http://gmatclub.com/forum/tough-and-tri ... 25967.html
Tough and tricky PS exponents and roots questions with detailed solutions: http://gmatclub.com/forum/tough-and-tri ... 25956.html
Hope it helps.
Bunuel ...if a=-1 and b=-9 then we get 3 as our answer and if a=-1 and b=-2 in that case we get .66
Should the answer not be E.
Math Expert
Joined: 02 Sep 2009
Posts: 51218
Re: Is 3^(a^2/b) < 1? [#permalink]
### Show Tags
25 Oct 2018, 06:17
angarg wrote:
Bunuel wrote:
Is 3^(a^2/b) < 1?
Notice that $$3^{\frac{a^2}{b}} < 1$$ to hold true, the power of 3 must be less than 0. So, the question basically asks whether $$\frac{a^2}{b}<0$$. This will happen if $$a\neq{0}$$ AND $$b<0$$ (if $$a=0$$ then $$\frac{a^2}{b}=0$$).
(1) a<0. The first condition is satisfied ($$a\neq{0}$$) but we don't know about the second one. Not sufficient.
(2) b<0. The second condition is satisfied ($$b<0$$) but we don't know about the first one (again if $$a=0$$ then $$\frac{a^2}{b}=0$$). Not sufficient. | {
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(1)+(2) Both condition are satisfied. Sufficient.
For more on number theory and exponents check: http://gmatclub.com/forum/math-number-theory-88376.html
DS questions on exponents: http://gmatclub.com/forum/search.php?se ... &tag_id=39
PS questions on exponents: http://gmatclub.com/forum/search.php?se ... &tag_id=60
Tough and tricky DS exponents and roots questions with detailed solutions: http://gmatclub.com/forum/tough-and-tri ... 25967.html
Tough and tricky PS exponents and roots questions with detailed solutions: http://gmatclub.com/forum/tough-and-tri ... 25956.html
Hope it helps.
Bunuel ...if a=-1 and b=-9 then we get 3 as our answer and if a=-1 and b=-2 in that case we get .66
Should the answer not be E.
If a=-1 and b=-9, then 3^(1/(-9)) = ~0.9.
_________________
Re: Is 3^(a^2/b) < 1? &nbs [#permalink] 25 Oct 2018, 06:17
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}}{{k!\left( {n - k} \right)!}}. As you see, the command \binom{}{}will print the binomial coefficient using the parameters passed inside the braces. If the sampling is carried out without replacement, the draws are not independent and so the resulting distribution is a hypergeometric distribution, not a binomial one. In mathematics, the Gaussian binomial coefficients (also called Gaussian coefficients, Gaussian polynomials, or q-binomial coefficients) are q-analogs of the binomial coefficients.The Gaussian binomial coefficient, written as () or [], is a polynomial in q with integer coefficients, whose value when q is set to a prime power counts the number of subspaces of dimension k in a vector … Pascal's triangle can be extended to find the coefficients for raising a binomial to any whole number exponent. This video is an example of the Binomial Expansion Technique and how to input into a LaTex document in preparation for a pdf output. \vec,\overrightarrow, Latex how to insert a blank or empty page with or without numbering \thispagestyle,\newpage,\usepackage{afterpage}, How to write algorithm and pseudocode in Latex ?\usepackage{algorithm},\usepackage{algorithmic}, How to display formulas inside a box or frame in Latex ? One can drop one of the numbers in the bottom list and infer it from the fact that sum … binomial The combination (n r) (n r) is called a binomial coefficient. }}{{k!\left( {n - k} \right)!}} The binomial coefficient is the number of ways of picking unordered outcomes from possibilities, also known as a combination or combinatorial number. Binomial coefficients are a family of positive integers that occur as coefficients in the binomial theorem. C — All combinations of v matrix. = \binom {n} {k} This is the binomial coefficient. The binomial coefficient $\binom{n}{k}$ can be interpreted as the number of ways to choose k elements from an n-element set. where A is the permutation, A_n^k = \frac{n!}{(n-k)! In latex mode we must use \binom fonction | {
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set. where A is the permutation, A_n^k = \frac{n!}{(n-k)! In latex mode we must use \binom fonction as follows: \frac {n!} Home > Latex > FAQ > Latex - FAQ > Latex binomial coefficient, Monday 9 December 2019, by Nadir Soualem. In Counting Principles, we studied combinations.In the shortcut to finding$\,{\left(x+y\right)}^{n},\,$we will need to use combinations to find the coefficients that will appear in the expansion of the binomial. LaTeX provides a feature of special editing tool for scientific tool for math equations in LaTeX. Then it's a good reason to buy me a coffee. Binomial coefficients have been known for centuries, but they're best known from Blaise Pascal's work circa 1640. k-combinations of n-element set. Click on one of the binomial coefficient designs, which look like the letters "n" over "k" inside either a round or angled bracket. Gerhard "Ask Me About System Design" Paseman, 2010.03.27 \endgroup – Gerhard Paseman Mar 27 '10 at 17:00 Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share … This article explains how to typeset them in LaTeX. The binomial coefficient is defined by the next expression: \binom {n}{k} = \frac {n ! }}{{k!\left( {n - k} \right)!}} Binomial coefficient denoted as c (n,k) or n c r is defined as coefficient of x k in the binomial expansion of (1+X) n. The Binomial coefficient also gives the value of the number of ways in which k items are chosen from among n objects i.e. In this article, you will learn how to write basic equations and constructs in LaTeX, about aligning equations, stretchable horizontal lines, operators and delimiters, fractions and binomials. Latex binomial coefficient Definition. If your equation requires specific numbers in place of the "n" or "k," click on a letter to select it, press "Delete" and enter a number in its place. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, | {
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in its place. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share … The binomial coefficient is defined by the next expression: \ [ \binom{n} {k} = \frac{n!} (adsbygoogle = window.adsbygoogle || []).push({}); All the versions of this article: Blog template built with Bootstrap and Spip by Nadir Soualem @mathlinux. Knowledge base dedicated to Linux and applied mathematics. The binomial coefficient is the number of ways of picking unordered outcomes from possibilities, also known as a combination or combinatorial number. The binomial coefficient can be interpreted as the number of ways to choose k elements from an n-element set. In latex mode we must use \binom fonction as follows: \frac{n!}{k! The order of selection of items not considered. (adsbygoogle = window.adsbygoogle || []).push({}); Thank you ! The possibility to insert operators and functions as you know them from mathematics is not possible for all things. An example of a binomial coefficient is (5 2)= C(5,2)= 10 (5 2) = C (5, 2) = 10. \boxed, How to write table in Latex ? Using fractions and binomial coefficients in an expression is straightforward. The Texworks shows … Binomial coefficients are common elements in mathematical expressions, the command to display them in LaTeXis very similar to the one used for fractions. ( n - k )! Here's an equation: math \frac {n!} (n−k)! Mathematical Equations in LaTeX. = \binom{n}{k} This will give more accuracy at the cost of computing small sums of binomial coefficients. The symbols and are used to denote a binomial coefficient, and are sometimes read as "choose.". n! This same array could be expressed using the factorial symbol, as shown in the following. And of course this command can be included in the normal text flow \ (\binom{n} {k}\). All combinations of v, returned as a matrix of the same type as v. Specially useful for continued fractions. The second | {
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v, returned as a matrix of the same type as v. Specially useful for continued fractions. The second fraction displayed in the previous example uses the command \cfrac{}{} provided by the package amsmath (see the introduction), this command displays nested fractions without changing the size of the font. Binomial Coefficient. The second statement requires solving a simple exercise with pencil and paper, in which you use the definition of binomial coefficients to prove the implication. In UnicodeMath Version 3, this uses the \choose operator ⒞ instead of the \atop operator ¦. The symbols and are used to denote a binomial coefficient, and are sometimes read as " choose." Accordingly the binomial coefficient in the binomial theorem above can be written as “n\choose k”, assuming that you type a space after the k. This Then it's a good reason to buy me a coffee. Toutes les versions de cet article : Le coefficient binomial est le nombre de possibilités de choisir k élément dans un ensemble de n éléments. b is the same type as n and k. If n and k are of different types, then b is returned as the nondouble type. A slightly different and more complex example of continued fractions, Showing first {{hits.length}} results of {{hits_total}} for {{searchQueryText}}, {{hits.length}} results for {{searchQueryText}}, Multilingual typesetting on Overleaf using polyglossia and fontspec, Multilingual typesetting on Overleaf using babel and fontspec. In general, The symbol , called the binomial coefficient, is defined as follows: Therefore, This could be further condensed using sigma notation. binomial coefficient Latex. LaTeX forum ⇒ Math & Science ⇒ Expression like binomial Coefficient with Angle Delimiters Topic is solved Information and discussion about LaTeX's math and science related features (e.g. Asking for help, clarification, or responding to other answers. }{k ! therefore gives the number of k -subsets possible out of a set of distinct items. It is especially useful for | {
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gives the number of k -subsets possible out of a set of distinct items. It is especially useful for reasoning about recursive methods in programming. All combinations of v, returned as a matrix of the same type as v. Binomial coefficient denoted as c(n,k) or n c r is defined as coefficient of x k in the binomial expansion of (1+X) n.. The following are the common definitions of Binomial Coefficients.. A binomial coefficient C(n, k) can be defined as the coefficient of x^k in the expansion of (1 + x)^n.. A binomial coefficient C(n, k) also gives the number of ways, disregarding order, that k objects can be chosen from among n objects more formally, the number of k-element subsets (or k-combinations) of a n-element set. As you may have guessed, the command \frac{1}{2} is the one that displays the fraction. Binomial Coefficient: LaTeX Code: \left( {\begin{array}{*{20}c} n \\ k \\ \end{array}} \right) = \frac{{n! Open an example in Overleaf (n - k)!} formulas, graphs). So The combination (nr)\displaystyle \left(\begin{array}{c}n\\ r\end{array}\right)(nr) is calle… Binomial coefficient, returned as a nonnegative scalar value. Any coefficient $a$ in a term $ax^by^c$ of the expanded version is known as a binomial coefficient. Fractions and binomial coefficients are common mathematical elements with similar characteristics - one number goes on top of another. } ... Pascal’s triangle. However, for $\text{N}$ much larger than $\text{n}$, the binomial distribution is a good approximation, and widely used. Identifying Binomial Coefficients. It will give me the energy and motivation to continue this development. Ak n = n! The usual binomial coefficient can be written as $\left({n \atop {k, {n-k}}}\right)$. (n-k)!} I agree. The binomial coefficient is defined by the next expression: \[\binom {n}{k} = \ frac {n!}{k!(n-k)! Binomial coefficients are common elements in mathematical expressions, the command to display them in LaTeX is very similar to the one used for fractions. In | {
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