MATH1251 Questions HELP (5 Viewers)

seanieg89 · Nov 11, 2016

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

What don't you find appealing about that idea? It is a fundamental skill in analysis to isolate the part of of an ugly expression that dominates behaviour (in this case n^{-n}), and to bound uglier expressions by nicer expressions.

leehuan · Nov 11, 2016

seanieg89 said:
What don't you find appealing about that idea? It is a fundamental skill in analysis to isolate the part of of an ugly expression that dominates behaviour (in this case n^{-n}), and to bound uglier expressions by nicer expressions.

It's easier to deal with, but I was hoping for a more obvious method

Paradoxica · Nov 11, 2016

leehuan said:
It's easier to deal with, but I was hoping for a more obvious method

That is the obvious method.

leehuan · Nov 11, 2016

Well ok, I didn't want to use the word "trivial" because I figured there wouldn't be one. But otherwise I don't know what word would be appropriate to use

seanieg89 · Nov 11, 2016

Its way more natural IMO. The sequence decreases so rapidly that the sign of terms is irrelevant (i.e. it is absolutely convergent contrary to your original claim).

Bounding it above by the sequence n^(-n) makes this abundantly clear.

The alternating series test is more useful for things that do not converge absolutely (like the alternating harmonic series), because the conclusion is weaker.

leehuan · Nov 11, 2016

seanieg89 said:
Its way more natural IMO. The sequence decreases so rapidly that the sign of terms is irrelevant (i.e. it is absolutely convergent contrary to your original claim).

Bounding it above by the sequence n^(-n) makes this abundantly clear.

The alternating series test is more useful for things that do not converge absolutely (like the alternating harmonic series), because the conclusion is weaker.

Are you sure that it converges absolutely? Or were you not referring to my original question and instead the one Paradoxica used

Further reference: I applied the limit form of the comparison test with 1/n

seanieg89 · Nov 11, 2016

You wrote n^(-n-1/n) in your original post and all subsequent posts until this wolfram comment, in which you have used n^(-1-1/n).

The absolute convergence of the series paradoxica mentioned implies the absolute convergence of the former series you did, and yes I am 100% certain of this fact.

The latter does not converge absolutely and is a more appropriate candidate for the alternating series test.

leehuan · Nov 11, 2016

Ah never mind. I don't have that paper with me right now, so I'm gonna apologise - I think I typed the question wrong this whole time..

Sent from my iPhone using Tapatalk

leehuan · Nov 12, 2016

$\text{From my understanding of the integral test:}\\ \text{The convergence/divergence of }\int_{c}^\infty f(x)dx\text{ implies the convergence/divergence of }\sum_{k=c}^\infty a_k\\ \text{where }a_k=f(k) \forall k=c, c+1, \dots$

$\text{Is the converse necessarily true? i.e. does the convergence of the sum}\\ \text{imply the convergence of the integral?}\\ \text{Because I'm having second doubts for the solutions to this question.}$

Tsonga · Nov 12, 2016

Can someone explain the working in finding the matrix when they give a basis

Thanks

kawaiipotato · Nov 12, 2016

Tsonga said:
View attachment 33588
Can someone explain the working in finding the matrix when they give a basis

Thanks

The general method for this is
For a linear mapping T: V -> W
with basis B = {v1,v2,v3,...,vn} for the domain and C = {u1,u2,u3,...,un} for the codomain then the matrix of T is A = [a1 a2 a3 ... an] (where ak is the k'th column vector of the matrix A) and
ak = [T(vk)]_C (the coordinate vector of vk with respect to the ordered basis C in the codomain)
Using this, we just find
a1 = [T(1)]_C
a2 = [T(e^(itheta)]_C
a3 = [T(e^(-itheta)]_C and put this into a matrix as columns

Drsoccerball · Nov 12, 2016

Tsonga said:
View attachment 33588
Can someone explain the working in finding the matrix when they give a basis

Thanks

InteGrand said:
$\noindent It's because the map $T$ is the transpose map. Remember, the matrix of a linear map $T$ with respect to a basis $\mathcal{B} = \left\{\vec{b}_1, \vec{b}_{2},\ldots , \vec{b}_n\right\}$ in the domain and $\mathcal{C}$ in the codomain (for our case $\mathcal{B}$ and $\mathcal{C}$ were the same) is found as follows: column $j$ ($j=1,2,\ldots n$) of the desired matrix is \emph{the coordinate vector of $T\left(\vec{b}_j\right)$ with respect to the basis $\mathcal{C}$}. In other words, evaluate $T\left(\vec{b}_1\right)$, find the coordinates of this wrt the specified codomain basis, and this is our column $1$ of the desired matrix of $T$. Then repeat for the other columns using the other domain basis vectors.$

$\noindent So for our example, for first column, we compute that $T(P) = P$, which has coordinates $\begin{bmatrix}1\\0\\0\\0\end{bmatrix}$ wrt the codomain basis (which is also the domain basis here) $ \mathcal{B} = \left\{P,Q,R,S\right\}$. So this is column $1$ of our desired matrix.$

$\noindent Now apply $T$ to the second domain basis vector: $T(Q) = R$, and $R$ has coordinates $\begin{bmatrix}0\\0\\1\\0\end{bmatrix}$ wrt codomain basis $\mathcal{B}$, so this is our second column.$

$\noindent Now apply $T$ to the third domain basis vector: $T(R) = Q$, and $Q$ has coordinates $\begin{bmatrix}0\\1\\0\\0\end{bmatrix}$ wrt codomain basis $\mathcal{B}$, so this is our third column.$

$\noindent Finally, apply $T$ to the fourth domain basis vector: $T(S) = S$, and $S$ has coordinates $\begin{bmatrix}0\\0\\0\\1\end{bmatrix}$ wrt codomain basis $\mathcal{B}$, so this is our fourth column.$

Now that we've found all the columns of the desired matrix, just put them together to get said matrix.

.

InteGrand · Nov 13, 2016

leehuan said:
$\text{From my understanding of the integral test:}\\ \text{The convergence/divergence of }\int_{c}^\infty f(x)dx\text{ implies the convergence/divergence of }\sum_{k=c}^\infty a_k\\ \text{where }a_k=f(k) \forall k=c, c+1, \dots$

$\text{Is the converse necessarily true? i.e. does the convergence of the sum}\\ \text{imply the convergence of the integral?}\\ \text{Because I'm having second doubts for the solutions to this question.}$

Yeah in the integral test, the sum and integral have the same convergence/divergence status. This is because when proving the integral test, we bound the sum between two integral estimates using the integral in question, and then take the limit and use the squeeze law.

Besides, if you believe that convergence/divergence of integral implies convergence/divergence of the sum, then the sum implying the integral follows automatically from this (basically by using contrapositive).

leehuan · Nov 13, 2016

I always get confused when they map to a matrix

$\text{Let }T:\mathbb{R}^3\to M_{22}(\mathbb{R})\text{, a linear mapping, be defined}\\ T(\textbf{x})=\begin{pmatrix}x_1-x_2&x_2-x_3\\x_3-x_1&0\end{pmatrix}$

$\text{c) Find a basis for im}(T)$

InteGrand · Nov 13, 2016

leehuan said:
I always get confused when they map to a matrix

$\text{Let }T:\mathbb{R}^3\to M_{22}(\mathbb{R})\text{, a linear mapping, be defined}\\ T(\textbf{x})=\begin{pmatrix}x_1-x_2&x_2-x_3\\x_3-x_1&0\end{pmatrix}$

$\text{c) Find a basis for im}(T)$

$\noindent Note that for $\mathbf{x} = \begin{bmatrix}x_{1}\\ x_{2}\\ x_{3}\end{bmatrix} \in \mathbb{R}^{3}$, we have $T\left(\mathbf{x}\right) = x_{1} \begin{bmatrix}1 & 0 \\ -1 & 0\end{bmatrix} + x_{2} \begin{bmatrix}-1 & 1 \\ 0 & 0\end{bmatrix} + x_{3} \begin{bmatrix}0 & -1 \\ 1 & 0\end{bmatrix}$. So the image is at most three-dimensional (being the span of the three matrices on the RHS). The three matrices on the RHS are also clearly linearly dependent (e.g. note that those matrices sum to $O$, the zero matrix). So the image is actually at most two-dimensional. If we take (for example) the first two of those three matrices, these are linearly independent (they're not multiples of each other), and then the third is a linear combination of these. So a basis for the image is $\left\{\begin{bmatrix}1 & 0 \\ -1 & 0\end{bmatrix}, \begin{bmatrix}-1 & 1 \\ 0 & 0\end{bmatrix} \right\}$.$

leehuan · Nov 13, 2016

I've forgotten how to do this, because it's for a sequence and an integer N, so I forgot if we need floor or ceiling function (or rather tired and can't work it out from my intuition)

$\text{Prove, from the definition, for all }n\in \mathbb{N}\\ \lim_{n\to \infty}{\frac{n\sin(n^2) }{\sqrt{1+n^3 } } }=0$

Some progress:

$\begin{align*}|f(x)-L| &=\left| \frac{n\sin(n^2) }{\sqrt{1+n^3} } \right| \\ &\le \left| \frac{n}{\sqrt{1+n^3} } \right| \\ & < \frac{1}{\sqrt{n} }\end{align*}$

InteGrand · Nov 13, 2016

leehuan said:
I've forgotten how to do this, because it's for a sequence and an integer N, so I forgot if we need floor or ceiling function (or rather tired and can't work it out from my intuition)

$\text{Prove, from the definition, for all }n\in \mathbb{N}\\ \lim_{n\to \infty}{\frac{n\sin(n^2) }{\sqrt{1+n^3 } } }=0$

Some progress:

$\begin{align*}|f(x)-L| &=\left| \frac{n\sin(n^2) }{\sqrt{1+n^3} } \right| \\ &\le \left| \frac{n}{\sqrt{1+n^3} } \right| \\ & < \frac{1}{\sqrt{n} }\end{align*}$

$\noindent Given $\varepsilon > 0$, take $N= \frac{1}{\varepsilon^{2}}$, then $|f(n) - 0| < \varepsilon$ for all $n >N$, by your working.$

$\noindent Don't need floor or ceiling functions because there's no need for the $N$ above to be an integer, it suffices for it to just be a positive real number, given $\varepsilon > 0$. If you wanted it to be an integer though, we'd take the ceiling, i.e. $\left\lceil \frac{1}{\varepsilon^{2}}\right\rceil$.$

leehuan · Nov 13, 2016

$Tf = f^{\prime \prime} - f^\prime\text{ is a linear map and consider}\\ \text{the ordered basis }B=\{ \textbf{1}, \sinh, \cosh\}$

$\text{Note that }\textbf{1}(x)\equiv 1 \forall x$

$\text{In c) A matrix representation of }T\text{ w.r.t. }B\text{ is}\\ A=\begin{pmatrix}0&0&0\\ 0&1&-1\\ 0&-1&1\end{pmatrix}$

$\text{d) So it was blatantly obvious that nullity}(T)=2\\ \text{and }\textbf{1}\text{ is in the basis for a kernel for }T$

$\text{It only makes sense if I work backwards.}\\ \text{How would you, from some first principles method, show that:}\\ \ker(T)=\text{span} \{1, \sinh+\cosh\}\\ \text{im}(T)=\text{span} \{\sinh-\cosh\}$

Cause I get the intuition method but I might miss it in the exam

InteGrand · Nov 13, 2016

leehuan said:
$Tf = f^{\prime \prime} - f^\prime\text{ is a linear map and consider}\\ \text{the ordered basis }B=\{ \textbf{1}, \sinh, \cosh\}$

$\text{Note that }\textbf{1}(x)\equiv 1 \forall x$

$\text{In c) A matrix representation of }T\text{ w.r.t. }B\text{ is}\\ A=\begin{pmatrix}0&0&0\\ 0&1&-1\\ 0&-1&1\end{pmatrix}$

$\text{d) So it was blatantly obvious that nullity}(T)=2\\ \text{and }\textbf{1}\text{ is in the basis for a kernel for }T$

$\text{It only makes sense if I work backwards.}\\ \text{How would you, from some first principles method, show that:}\\ \ker(T)=\text{span} \{1, \sinh+\cosh\}\\ \text{im}(T)=\text{span} \{\sinh-\cosh\}$

Cause I get the intuition method but I might miss it in the exam

$\noindent The kernel is just the set of solutions to $y'' - y' = 0$. The characteristic equation is $r^{2} - r = 0$, which has roots $0,1$. So the solutions to this ODE (hence the kernel) are the span of $1$ and $e^{x}$ as required (note that $\sinh x + \cosh x = e^{x}$).$

$\noindent For the image, looking at the matrix $A$, that matrix has clearly image equal to the span of its second column. The second column represents $\sinh - \cosh$, so this is a basis for the image of $T$. (We could have similarly found the kernel by just finding the kernel of $A$ and interpreting the results as functions from the given basis.)$

leehuan · Nov 13, 2016

MATH1251 Questions HELP (5 Viewers)

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