MATH1251 Questions HELP (4 Viewers)

Paradoxica · Oct 23, 2016

leehuan said:
This obviously converges but the ratio test is inconclusive, so how would you justify it

$\sum_{k=1}^\infty \frac{k!}{k^k}$

If you accept Stirling's Inequality:

$\frac{n!}{n^n} < e\sqrt{n} e^{-n}$

The root test gives convergence of the RHS, thus giving convergence of the LHS.

seanieg89 · Oct 23, 2016

leehuan said:
This obviously converges but the ratio test is inconclusive, so how would you justify it

$\sum_{k=1}^\infty \frac{k!}{k^k}$

Check your calculations, the ratio test is conclusive. (The ratio tends to exp(-1).)

leehuan · Oct 23, 2016

Oh bad example. My fault.

Nevermind. For starters I memorised his question wrong. My friend asked me about this one which was easy: $\sum_{k=1}^\infty \frac{k^2}{2^k}$

It goes to 1/2. But then I started wondering if similar forms tended to 1. I forgot that 1^k=1 here and thus the ratio test wouldn't be necessary for say, k/1^k

seanieg89 · Oct 23, 2016

On this topic, a good exercise for your intuition is trying to prove that the ratio test is strictly weaker than the root test.

I.e. let x_n be a given positive sequence.

a) Show that if the ratio test asserts that the sum of x_n is convergent, then the root test does too.

b) Show (by way of example) that there exist sequences x_n such that the root test tells you that the sum of x_n is convergent, but the ratio test is inconclusive.

1008 · Oct 25, 2016

leehuan said:
This obviously converges but the ratio test is inconclusive, so how would you justify it

$\sum_{k=1}^\infty \frac{k!}{k^k}$

Could you use something like this?

seanieg89 · Oct 25, 2016

1008 said:
Could you use something like this?

Huh?

Can you elaborate on what you are trying to do by splitting the sum in this way?

This is not a straightforward application of the p-test, because of the presence of the factorial and also because the exponent is not fixed. It is a basic application of the ratio test though. (As questions involving factorials often will be).

leehuan · Oct 27, 2016

$\text{So the question was }\int \int_T e^{\frac{y-x}{y+x}}dx\,dy\\ \text{ where }T\text{ is the triangle bound by }x+y=2$

$\text{And I don't need it done. Tutor did it already. But just confused about the Jacobian.}\\ \text{He made the change of variables }x=u+v\text{ and }y=u-v$

$\text{So the determinant of the Jacobian is given by }\begin{vmatrix}\frac{\partial x}{\partial u}&\frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u}&\frac{\partial y}{\partial v}\end{vmatrix}=-2\\ \text{But why is it that, and not say the rows or columns swapped?}$

Because say we swap the rows, then the determinant becomes +2. But it still "looks like" a Jacobian.

$\text{If it's of any use, his integral became }\int_0^1 \int_{-u}^u -2e^{-\frac{v}{u}}dv\,du$

Paradoxica · Oct 27, 2016

leehuan said:
$\text{So the question was }\int \int_T e^{\frac{y-x}{y+x}}dx\,dy\\ \text{ where }T\text{ is the triangle bound by }x+y=2$

$\text{And I don't need it done. Tutor did it already. But just confused about the Jacobian.}\\ \text{He made the change of variables }x=u+v\text{ and }y=u-v$

$\text{So the determinant of the Jacobian is given by }\begin{vmatrix}\frac{\partial x}{\partial u}&\frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u}&\frac{\partial y}{\partial v}\end{vmatrix}=-2\\ \text{But why is it that, and not say the rows or columns swapped?}$

Because say we swap the rows, then the determinant becomes +2. But it still "looks like" a Jacobian.

$\text{If it's of any use, his integral became }\int_0^1 \int_{-u}^u -2e^{-\frac{v}{u}}dv\,du$

According to a quick evaluation in Mathematica, the two integrals provided above have opposite signs.

Not to be of any disrespect, but I question the process your tutor used.

leehuan · Oct 27, 2016

Only adds to confusion really because I wouldn't know where the mistake is.

kawaiipotato · Oct 27, 2016

leehuan said:
Only adds to confusion really because I wouldn't know where the mistake is.

I believe we multiply by the absolute value of the Jacobian in a change of variable substitution.
Btw, when did we learn this change of variable substitition in 1251? Didn't see any questions on it.

leehuan · Oct 27, 2016

kawaiipotato said:
I believe we multiply by the absolute value of the Jacobian in a change of variable substitution.
Btw, when did we learn this change of variable substitition in 1251? Didn't see any questions on it.

I legit did not know that about the abs. value

Also, this was Q124 of the calculus pack. And yes we had a lecture on Jacobians; I think it was the second last one.

seanieg89 · Oct 28, 2016

There is an absolute value in the change of variables for the Jacobian because of the subtle issue of orientation.

When we write something like

$\int_S f \, dx$

for S a subset of R^n and f: S -> R, this is an unoriented notion of integration. (Because S is a subset, without choice of orientation specified. (*))

Compare this to the integral

$\int_a^b f \, dx$

which is the integral of f along the oriented line segment from a to b in R.

In particular, if f is non-negative, the first notion of integration will always be non-negative, whilst the second need not be (if a > b ).

(*) This is the same as the oriented integration over S such that the canonical basis in R^n is positively oriented. The situation gets a bit more interesting when we are learning to integrate things on manifolds (surfaces, etc), because these need not be orientable in general! (Think mobius strip.)

RenegadeMx · Oct 29, 2016

leehuan said:
I legit did not know that about the abs. value

Also, this was Q124 of the calculus pack. And yes we had a lecture on Jacobians; I think it was the second last one.

lol it depends on the lecturer, mine always called them Yacobians

leehuan · Nov 4, 2016

$\text{In part a), I found the eigenvalues and eigenvectors of }A=\begin{pmatrix}3&-15\\1&-5\end{pmatrix}\\ \text{to be }0, \begin{pmatrix}5\\1\end{pmatrix}, \quad -2, \begin{pmatrix}3\\1\end{pmatrix}$

$\text{In part b) I found the solution to the set of ODEs}\\ \begin{align*}\frac{dy_1}{dt}&=3y_1-15y_2\\ \frac{dy_2}{dt}&=y_1-5y_2\end{align*}\\ \text{to be:}\\ y_1=-50+60e^{-2t}, \quad y_2=-10+20e^{-2t}$
(Initial values: y1(0)=y2(0)=10)

$\text{c) Suppose that }\textbf{x}(k)\in \mathbb{R}^2, k=0,1,2,\dots \text{ satisfies the recurrence relation}\\ \textbf{x}(k+1)=A\textbf{x}(k)\\ \text{Find all vectors }\textbf{x}(0)\text{ for which the limit }\lim_{k\to \infty}\textbf{x}(k)=\lim_{k\to \infty}A^k\textbf{x}_(0)\text{ exists}$

InteGrand · Nov 4, 2016

leehuan said:
$\text{In part a), I found the eigenvalues and eigenvectors of }A=\begin{pmatrix}3&-15\\1&-5\end{pmatrix}\\ \text{to be }0, \begin{pmatrix}5\\1\end{pmatrix}, \quad -2, \begin{pmatrix}3\\1\end{pmatrix}$

$\text{In part b) I found the solution to the set of ODEs}\\ \begin{align*}\frac{dy_1}{dt}&=3y_1-15y_2\\ \frac{dy_2}{dt}&=y_1-5y_2\end{align*}\\ \text{to be:}\\ y_1=-50+60e^{-2t}, \quad y_2=-10+20e^{-2t}$
(Initial values: y1(0)=y2(0)=10)

$\text{c) Suppose that }\textbf{x}(k)\in \mathbb{R}^2, k=0,1,2,\dots \text{ satisfies the recurrence relation}\\ \textbf{x}(k+1)=A\textbf{x}(k)\\ \text{Find all vectors }\textbf{x}(0)\text{ for which the limit }\lim_{k\to \infty}\textbf{x}(k)=\lim_{k\to \infty}A^k\textbf{x}_(0)\text{ exists}$

$\noindent Note that since $\mathbf{x}(k+1) = A\mathbf{x}(k)$ and $A$ is diagonalisable with eigenvalues $\lambda_{1} = 0, \lambda_{2} = -2$, with corresponding eigenvectors $\mathbf{v}_{1} = \begin{pmatrix}5\\1\end{pmatrix}$ and $\mathbf{v}_{2} = \begin{pmatrix}3\\1\end{pmatrix}$ respectively, we have$

$\begin{align*}\mathbf{x}(k) = A^{k}\mathbf{x}\left(0\right) &= \alpha_{1}\lambda_{1}^{k}\mathbf{v}_{1} + \alpha_{2}\lambda_{2}^{k}\mathbf{v}_{2}\\ &= \alpha_{2} \left(-2\right)^{k}\begin{pmatrix}3 \\ 1 \end{pmatrix} ,\end{align*}$

$\noindent where $\begin{pmatrix}\alpha_{1}\\ \alpha_{2}\end{pmatrix}$ is the \emph{coordinate vector of $\mathbf{x}(0)$ with respect to the eigenbasis $\left\{\mathbf{v}_{1},\mathbf{v}_{2}\right\}$}. Clearly, there'll be a limiting vector if and only if $\alpha_{2} = 0$ (because if this is the case, the limit is the zero vector as we can see from above, and if $\alpha_{2} \neq 0$, we have no limit since the $\left(-2\right)^{k}$ does not settle down as $k \to \infty$). Since $\mathbf{x}(0) = \alpha_{1} \mathbf{v}_{1} + \alpha_{2} \mathbf{v}_{2}$ and we have a limiting vector iff $\alpha_{2} = 0$, we have a limiting vector iff $\mathbf{x}(0) = \alpha \mathbf{v}_{1}$ for some scalar $\alpha$ (in other words, iff the initial state vector happens to be in the eigenspace for eigenvalue $0$ of $A$, which is also the kernel of $A$).$

leehuan · Nov 11, 2016

This is a set of equations I got from using the Lagrange multiplier theorem. What would be a quick way of solving them?

$\begin{align*}2x&=\lambda(4x-4y)\\ 2y&=\lambda(10y-4x)\\ 0&=2x^2-4xy+5y^2-54\end{align*}$

leehuan · Nov 11, 2016

$\sum_{n=1}^\infty \frac{(-1)^n}{n^{n+\frac1n}}$

I want to use the alternating series test to show that this is conditionally convergent (it already fails absolute convergence). It's clear that the relevant terms will be positive and I can prove that they limit off to 0, but how do I prove that

$\left\{\frac{1}{n^{n+\frac1n}} \right\}_{n=1}^\infty \text{ is decreasing?}$

Paradoxica · Nov 11, 2016

leehuan said:
$\sum_{n=1}^\infty \frac{(-1)^n}{n^{n+\frac1n}}$

I want to use the alternating series test to show that this is conditionally convergent (it already fails absolute convergence). It's clear that the relevant terms will be positive and I can prove that they limit off to 0, but how do I prove that

$\left\{\frac{1}{n^{n+\frac1n}} \right\}_{n=1}^\infty \text{ is decreasing?}$

The sequence is strictly less than n^-n

I guess you could work on that instead.

leehuan · Nov 11, 2016

Paradoxica said:
The sequence is strictly less than n^-n

I guess you could work on that instead.

Yeah I'll buy that if nobody comes with a better idea

InteGrand · Nov 11, 2016

leehuan said:
$\sum_{n=1}^\infty \frac{(-1)^n}{n^{n+\frac1n}}$

I want to use the alternating series test to show that this is conditionally convergent (it already fails absolute convergence). It's clear that the relevant terms will be positive and I can prove that they limit off to 0, but how do I prove that

$\left\{\frac{1}{n^{n+\frac1n}} \right\}_{n=1}^\infty \text{ is decreasing?}$

It's decreasing because the denominator is increasing. The denominator is increasing because

(n+1)^{(n+1) + 1/(n+1)} > n^{(n+1) + 1/(n+1)} (*)

> n^{n + 1/n} (**).

Note (*) follows since the function x^{(c+1) + 1/(c+1)} is increasing on the positive reals for any given positive integer c, and (**) follows because the function c^{x} is increasing on the positive reals for any given c large enough, and since (n+1) + 1/(n+1) > n + 1/n for all n.

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