First Year Mathematics B (Integration, Series, Discrete Maths & Modelling) (3 Viewers)

He-Mann · Oct 27, 2016

Re: MATH1231/1241/1251 SOS Thread

iforgotmyname said:
=Int(Coshx+coshx*sinh^2x )

Since coshx*sinh^2x is of the form int(y'*y)
We can treat it as x^2 so int(x^2) is (x^3)/3

Thus Int(Coshx+coshx*sinh^2x )=sinhx+sinh^3x/3

Just saying, that's one way to confuse people. Explanation is not clear.

leehuan · Oct 29, 2016

Re: MATH1231/1241/1251 SOS Thread

$\text{For what values of }a\text{ does this series converge?}\\ \sum_{k=2}^\infty \frac{1}{k^a \ln k}$

I can tell by inspection that a > 1 but how do you prove it? Is there a way to get out of using the comparison test twice?

leehuan · Oct 29, 2016

Re: MATH1231/1241/1251 SOS Thread

Also why is it that if I have a 3x3 matrix with three distinct eigenvectors, these eigenvectors form a basis over R³?

Paradoxica · Oct 29, 2016

Re: MATH1231/1241/1251 SOS Thread

leehuan said:
$\text{For what values of }a\text{ does this series converge?}\\ \sum_{k=2}^\infty \frac{1}{k^a \ln k}$

I can tell by inspection that a > 1 but how do you prove it? Is there a way to get out of using the comparison test twice?

Reverse AM-GM should work.

leehuan · Oct 29, 2016

Re: MATH1231/1241/1251 SOS Thread

Paradoxica said:
Reverse AM-GM should work.

What is this reverse AM-GM?

Paradoxica · Oct 29, 2016

Re: MATH1231/1241/1251 SOS Thread

$\frac{2}{x^a \log{x}} < \frac{1}{x^{2a}} + \frac{1}{\log^2{x}}$

leehuan · Oct 29, 2016

Re: MATH1231/1241/1251 SOS Thread

Paradoxica said:
$\frac{2}{x^a \log{x}} < \frac{1}{x^{2a}} + \frac{1}{\log^2{x}}$

Never seen that before although it does make intuitive sense. Except, doesn't log^2 still diverge?

Paradoxica · Oct 29, 2016

Re: MATH1231/1241/1251 SOS Thread

leehuan said:
Never seen that before although it does make intuitive sense. Except, doesn't log^2 still diverge?

In that case just rearrange the powers so there's an x² with the log.

leehuan · Oct 29, 2016

Re: MATH1231/1241/1251 SOS Thread

Paradoxica said:
In that case just rearrange the powers so there's an x² with the log.

Rearrange the powers? I can't change the expression of my series.

InteGrand · Oct 29, 2016

Re: MATH1231/1241/1251 SOS Thread

leehuan said:
Also why is it that if I have a 3x3 matrix with three distinct eigenvectors, these eigenvectors form a basis over R³?

$\noindent In general, any $n\times n$ real matrix with $n$ distinct eigenvalues is diagonalisable (which is equivalent to saying there is a basis for $\mathbb{R}^n$ consisting entirely of eigenvectors of $A$ (such a basis is called an \textsl{eigenbasis})). This is because recall eigenvectors from distinct eigenvalues are linearly independent, so if there's $n$ distinct eigenvalues, we clearly have $n$ independent eigenvectors, which means we have a basis for $\mathbb{R}^{n}$ consisting of eigenvectors. (This is also all true for matrices over general fields $\mathbb{F}$, replacing $\mathbb{R}^{n}$'s with $\mathbb{F}^{n}$'s. So for example also true for complex matrices: any $n\times n$ matrix over $\mathbb{C}$ that has $n$ distinct eigenvalues has its eigenvectors form a basis for $\mathbb{C}^{n}$.)$

InteGrand · Oct 29, 2016

Re: MATH1231/1241/1251 SOS Thread

leehuan said:
$\text{For what values of }a\text{ does this series converge?}\\ \sum_{k=2}^\infty \frac{1}{k^a \ln k}$

I can tell by inspection that a > 1 but how do you prove it? Is there a way to get out of using the comparison test twice?

$\noindent If $a>1$, the series converges using the comparison test and `$p$-test', since $0\leq \frac{1}{k^a \ln k} \leq \frac{1}{k^a}$ for all $k$ sufficiently large (say $k\geq 3$) and $\sum_{k=2}^{\infty} \frac{1}{k^a}$ converges if $a>1$.$

$\noindent If $a\leq 1$, the given sum diverges. If $a < 0$, this is obvious since then $\frac{1}{k^a \ln k}\to \infty$ as $k\to \infty$ (rather than going to $0$ as $k\to \infty$). If $0\leq a \leq 1$, the sum still diverges. E.g. use the integral test or whatever other method you like to show that $\sum \frac{1}{k \ln k}$ (so when $a=1$) diverges. Then if $0<a < 1$, we have $0\leq \frac{1}{k \ln k} \leq \frac{1}{k^a \ln k}$ for all $k\geq 2$ (as $k^a < k$ for all $k \geq 2$), so by comparison test, we have divergence.$

leehuan · Oct 29, 2016

Re: MATH1231/1241/1251 SOS Thread

InteGrand said:
$\noindent In general, any $n\times n$ real matrix with $n$ distinct eigenvalues is diagonalisable (which is equivalent to saying there is a basis for $\mathbb{R}^n$ consisting entirely of eigenvectors of $A$ (such a basis is called an \textsl{eigenbasis})). This is because recall eigenvectors from distinct eigenvectors are linearly independent, so if there's $n$ distinct eigenvalues, we clearly have $n$ independent eigenvectors, which means we have a basis for $\mathbb{R}^{n}$ consisting of eigenvectors. (This is also all true for matrices over general fields $\mathbb{F}$, replacing $\mathbb{R}^{n}$'s with $\mathbb{F}^{n}$'s. So for example also true for complex matrices: any $n\times n$ matrix over $\mathbb{C}$ that has $n$ distinct eigenvalues has its eigenvectors form a basis for $\mathbb{C}^{n}$.)$

Oh right tbh I forgot about the theorem at the start. But now I got another problem

$\text{Say }\textbf{v}\text{ is an eigenvector of }A\text{ with eigenvalue }\lambda,\text{ so }A\textbf{v}=\lambda \textbf{v}\\ \text{Then why is this true?}$

$A^k\textbf{v}=\lambda^k\textbf{v}$

seanieg89 · Oct 29, 2016

Re: MATH1231/1241/1251 SOS Thread

If applying the linear operator A to the vector v scales it by lambda each time, and we apply A to v k times...

InteGrand · Oct 29, 2016

Re: MATH1231/1241/1251 SOS Thread

leehuan said:
Oh right tbh I forgot about the theorem at the start. But now I got another problem

$\text{Say }\textbf{v}\text{ is an eigenvector of }A\text{ with eigenvalue }\lambda,\text{ so }A\textbf{v}=\lambda \textbf{v}\\ \text{Then why is this true?}$

$A^k\textbf{v}=\lambda^k\textbf{v}$

$\noindent An easy induction! If $k=1$, it's true by definition. Then assuming it's true for some particular $k\in \mathbb{Z}^{+}$, i.e. that $A^k \mathbf{v} = \lambda^{k}\mathbf{v}$, we have$

$\begin{align*}A^{k+1}\mathbf{v} &= A\left(A^k \mathbf{v}\right)\\ &= A\left(\lambda^k \mathbf{v}\right) \quad (\text{inductive hypothesis}) \\ &= \lambda^k A\mathbf{v} \\ &= \lambda^k \lambda \mathbf{v} \\ &= \lambda^{k+1}\mathbf{v},\end{align*}$

$\noindent and so the result is true by induction.$

leehuan · Oct 29, 2016

Re: MATH1231/1241/1251 SOS Thread

InteGrand said:
$\noindent An easy induction! If $k=1$, it's true by definition. Then assuming it's true for some particular $k\in \mathbb{Z}^{+}$, i.e. that $A^k \mathbf{v} = \lambda^{k}\mathbf{v}$, we have$

$\begin{align*}A^{k+1}\mathbf{v} &= A\left(A^k \mathbf{v}\right)\\ &= A\left(\lambda^k \mathbf{v}\right) \quad (\text{inductive hypothesis}) \\ &= \lambda^k A\mathbf{v} \\ &= \lambda^k \lambda \mathbf{v} \\ &= \lambda^{k+1}\mathbf{v},\end{align*}$

$\noindent and so the result is true by induction.$

Woah. And right when I was about to say "intuitively makes sense but couldn't convince myself of it mathematically". That's neat!

One last one for the time being please

$\noindent\text{As far as I understand it, the Cauchy-Schwarz inequality is the statement }\\|\textbf{u}\cdot\textbf{v}| \le |\textbf{u}||\textbf{v}|\\ \text{So I'm not sure how to apply it to this series question.}$

$\text{Suppose }\left\{a_k\right\}_{k=1}^\infty\text{ is a sequence of +'ve numbers for which }\sum_{k=1}^\infty a_k^2\text{ converges.}$

$\text{Let }s_n=\sum_{k=1}^n\frac{a_k}{k}\text{ for }n=1,2,\dots$

$\text{Use C-S to prove that the sequence }\{s_n\}_{n=1}^\infty\text{ is bounded.}$

InteGrand · Oct 29, 2016

Re: MATH1231/1241/1251 SOS Thread

leehuan said:
Woah. And right when I was about to say "intuitively makes sense but couldn't convince myself of it mathematically". That's neat!

One last one for the time being please

$\noindent\text{As far as I understand it, the Cauchy-Schwarz inequality is the statement }\\|\textbf{u}\cdot\textbf{v}| \le |\textbf{u}||\textbf{v}|\\ \text{So I'm not sure how to apply it to this series question.}$

$\text{Suppose }\left\{a_k\right\}_{k=1}^\infty\text{ is a sequence of +'ve numbers for which }\sum_{k=1}^\infty a_k^2\text{ converges.}$

$\text{Let }s_n=\sum_{k=1}^n\frac{a_k}{k}\text{ for }n=1,2,\dots$

$\text{Use C-S to prove that the sequence }\{s_n\}_{n=1}^\infty\text{ is bounded.}$

$\noindent By Cauchy-Schwarz, we have $\frac{a_1}{1} + \frac{a_2}{2} + \cdots + \frac{a_n}{n} \leq \sqrt{a_1 ^2 + a_2 ^2 + \cdots + a_n ^2}\sqrt{\frac{1}{1^2} + \frac{1}{2^2} + \cdots + \frac{1}{n^2}}$. By the Basel problem series and assumption of convergence of $\sum a_n ^2$, the RHS is bounded above by a finite number, and the LHS is thus too, as required. (Also the LHS is an increasing sequence. So the series in question has its partial sums bounded above and increasing, so it in fact converges.)$

seanieg89 · Oct 29, 2016

Re: MATH1231/1241/1251 SOS Thread

leehuan said:
Woah. And right when I was about to say "intuitively makes sense but couldn't convince myself of it mathematically". That's neat!

One last one for the time being please

$\noindent\text{As far as I understand it, the Cauchy-Schwarz inequality is the statement }\\|\textbf{u}\cdot\textbf{v}| \le |\textbf{u}||\textbf{v}|\\ \text{So I'm not sure how to apply it to this series question.}$

$\text{Suppose }\left\{a_k\right\}_{k=1}^\infty\text{ is a sequence of +'ve numbers for which }\sum_{k=1}^\infty a_k^2\text{ converges.}$

$\text{Let }s_n=\sum_{k=1}^n\frac{a_k}{k}\text{ for }n=1,2,\dots$

$\text{Use C-S to prove that the sequence }\{s_n\}_{n=1}^\infty\text{ is bounded.}$

$s_n^2 \leq \left(\sum_{k=1}^n a_k^2\right)\left(\sum_{k=1}^n \frac{1}{k^2}\right).$

Each of these factors is a partial sum of a convergent positive series and so is bounded above by its limit. Hence s_n is bounded.

leehuan · Oct 29, 2016

Re: MATH1231/1241/1251 SOS Thread

$\text{So with those eigenvalue questions, they arose out of this markov system:}\\ \textbf{x}(k+1)=A\textbf{x}(k)\\ A=\begin{pmatrix}0.50&0.15&0.30\\0.10&0.60&0.30\\0.40&0.25&0.40\end{pmatrix}\\ \text{And let }B=\lim_{k\to \infty}A^k$

In a MATLAB output they computed the eigenvectors and eigenvalues of A for me. In part a) I had to write down a basis for ker(B) and im(B)

Part c) gave an initial condition
$\textbf{x}(0)=\begin{pmatrix}-1\\2\\-1\end{pmatrix}\\ \text{and asked me to find }\lim_{k\to \infty}\textbf{x}(k)$

$\text{The answers secondly say that }\lim_{k\to \infty}\textbf{x}(k)\text{ is proportional to }\begin{pmatrix}0.5\\0.1\\0.4\end{pmatrix}\\ \text{Which I get from part a) calculations.}\\ \text{But then part c) says the sum of the components of }\textbf{x}(k)\text{ is preserved.}\\ \text{Is this something about state vectors that i should know about?}$

InteGrand · Oct 29, 2016

Re: MATH1231/1241/1251 SOS Thread

leehuan said:
$\text{So with those eigenvalue questions, they arose out of this markov system:}\\ \textbf{x}(k+1)=A\textbf{x}(k)\\ A=\begin{pmatrix}0.50&0.15&0.30\\0.10&0.60&0.30\\0.40&0.25&0.40\end{pmatrix}\\ \text{And let }B=\lim_{k\to \infty}A^k$

In a MATLAB output they computed the eigenvectors and eigenvalues of A for me. In part a) I had to write down a basis for ker(B) and im(B)

Part c) gave an initial condition
$\textbf{x}(0)=\begin{pmatrix}-1\\2\\-1\end{pmatrix}\\ \text{and asked me to find }\lim_{k\to \infty}\textbf{x}(k)$

$\text{The answers secondly say that }\lim_{k\to \infty}\textbf{x}(k)\text{ is proportional to }\begin{pmatrix}0.5\\0.1\\0.4\end{pmatrix}\\ \text{Which I get from part a) calculations.}\\ \text{But then part c) says the sum of the components of }\textbf{x}(k)\text{ is preserved.}\\ \text{Is this something about state vectors that i should know about?}$

$\noindent Note that $A$ has columns that sum to 1. You can show as an exercise that this implies that the sequence $\mathbf{x}(0), \mathbf{x}(1), \mathbf{x}(2),\ldots$ has \emph{constant entry sum}$.$

$\noindent So basically the key result is \emph{if a square matrix $A$ has all columns summing to $1$, then that sequence $\left(\mathbf{x}(k)\right)$ has constant entry sum}. To show this, all you need to do is show that for such a matrix $A$ (one whose columns sum to $1$), for any vector $\mathbf{v}$, we have that the entry sum of $\mathbf{v}$ equals that of $A\mathbf{v}$.$

Flop21 · Oct 30, 2016

Re: MATH1231/1241/1251 SOS Thread

Could someone help me do this thanks, I'm keen to see the working out so I can get the general idea of things (I'm getting confused with what happens when you get down to the n limit).

First Year Mathematics B (Integration, Series, Discrete Maths & Modelling) (3 Viewers)

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