MATH1251 Questions HELP (4 Viewers)

InteGrand · Nov 13, 2016

leehuan said:
$\text{Let }f(x)=e^{2x}(1-x)^{-1}\\ \text{a) By multiplying the Maclaurin series for }e^{2x}\text{ and }(1-x)^{-1}\\ \text{obtain an expression for the coefficient }a_n\text{ in the Maclaurin series}\\ f(x)=\sum_{k=1}^\infty e^{2x}(1-x)^{-1}$

$\text{I don't remember the formula so I just did it from first principles. Where'd I screw up?}\\ \begin{align*}f(x)&=e^{2x}\left(\frac1{1-x}\right)\\ &= \left(\sum_{k=0}^\infty \frac{(2x)^k}{k!}\right)\left(\sum_{k=0}^\infty x^k\right)\\ &= \left(\frac{2^0}{0!}+\frac{2^1}{1!}x+\frac{2^2}{2!}x^2+\frac{2^3}{3!}x^3+\dots\right)\left(1+x+x^2+x^3+\dots\right)\\ &= \frac { 2^{ 0 } }{ 0! } +\left( \frac { 2^{ 1 } }{ 1! } +\frac { 2^{ 0 } }{ 0! } \right) x+\left( \frac { 2^{ 2 } }{ 2! } +\frac { 2^{ 1 } }{ 1! } +\frac { 2^{ 0 } }{ 0! } \right) x^{ 2 }+\left( \frac { 2^{ 3 } }{ 3! } +\frac { 2^{ 2 } }{ 2! } +\frac { 2^{ 1 } }{ 1! } +\frac { 2^{ 0 } }{ 0! } \right) x^{ 3 }+\dots\\ \therefore a_n&=\sum_{k=0}^n\frac{2^k}{k!}\end{align*}$

$\text{Given answer: }a_n=\sum_{n=0}^\infty \frac{2^n}{n!}$

$\noindent I assume you want the coefficients $a_n$ in the Maclaurin series for $e^{2x}\cdot (1-x)^{-1}$. Since $e^{2x} = \sum _{k = 0}^{\infty}b_{k}x^{k}$, where $b_{k} = \frac{2^{k}}{k!}$, and for $(1-x)^{-1}$, it's $\sum_{k=0}^\infty c_k x^k$, where $c_{k} \equiv 1$, the desired $a_{n}$ is $a_{n} = \sum_{j=0}^{n}b_{j}c_{n-j} = \sum_{j=0}^{n} \frac{2^j}{j!}$, for $n\geq 0$.$

$\noindent Your answer is right, assuming this is what you were after. The given answer is just $e^2$.$

Paradoxica · Nov 13, 2016

leehuan said:
$\text{Let }f(x)=e^{2x}(1-x)^{-1}\\ \text{a) By multiplying the Maclaurin series for }e^{2x}\text{ and }(1-x)^{-1}\\ \text{obtain an expression for the coefficient }a_n\text{ in the Maclaurin series}\\ f(x)=\sum_{k=1}^\infty e^{2x}(1-x)^{-1}$

$\text{I don't remember the formula so I just did it from first principles. Where'd I screw up?}\\ \begin{align*}f(x)&=e^{2x}\left(\frac1{1-x}\right)\\ &= \left(\sum_{k=0}^\infty \frac{(2x)^k}{k!}\right)\left(\sum_{k=0}^\infty x^k\right)\\ &= \left(\frac{2^0}{0!}+\frac{2^1}{1!}x+\frac{2^2}{2!}x^2+\frac{2^3}{3!}x^3+\dots\right)\left(1+x+x^2+x^3+\dots\right)\\ &= \frac { 2^{ 0 } }{ 0! } +\left( \frac { 2^{ 1 } }{ 1! } +\frac { 2^{ 0 } }{ 0! } \right) x+\left( \frac { 2^{ 2 } }{ 2! } +\frac { 2^{ 1 } }{ 1! } +\frac { 2^{ 0 } }{ 0! } \right) x^{ 2 }+\left( \frac { 2^{ 3 } }{ 3! } +\frac { 2^{ 2 } }{ 2! } +\frac { 2^{ 1 } }{ 1! } +\frac { 2^{ 0 } }{ 0! } \right) x^{ 3 }+\dots\\ \therefore a_n&=\sum_{k=0}^n\frac{2^k}{k!}\end{align*}$

$\text{Given answer: }a_n=\sum_{n=0}^\infty \frac{2^n}{n!}$

Not only is that answer wrong, that is a horrible abuse of notation (making the index variable the same as the defining subscript.)

Drsoccerball · Nov 13, 2016

Paradoxica said:
Not only is that answer wrong, that is a horrible abuse of notation (making the index variable the same as the defining subscript.)

Paradoxica is more salty than the Pacific Ocean...

leehuan · Nov 13, 2016

Drsoccerball said:
Paradoxica is more salty than the Pacific Ocean...

Actually whilst maybe not as strongly felt, I did think the same way he did

leehuan · Nov 13, 2016

Parts a), b) and c) are purely for reference

Here's a proof to part d) (I) also for the sake of reference

$\text{Suppose w.l.o.g. }\gamma\text{ represents the unit circle on the Argand plane}\\ \text{with its centre }O\text{ being the origin.}$

$\text{Take }A\text{ as the point }re^{i\theta}\text{ and it follows from collinearity and }B\in \gamma\\ B\text{ represents }e^{i\theta}\\ \text{Then clearly }OA=r \implies AB=|1-r|=\min_{\alpha \in \mathbb{R}}f(\alpha)$

$\text{Since }X\in \gamma\text{, take }X=e^{i\alpha}\\ \text{Then }AX=|re^{i\theta}-e^{i\alpha}|=f(\alpha)\\ \implies AB \le AX \, \forall X \in \gamma$

Q: How would you go about (II)? I want to try proof by contradiction but I'm not sure how to go about anything

InteGrand · Nov 13, 2016

leehuan said:
Parts a), b) and c) are purely for reference

Here's a proof to part d) (I) also for the sake of reference

$\text{Suppose w.l.o.g. }\gamma\text{ represents the unit circle on the Argand plane}\\ \text{with its centre }O\text{ being the origin.}$

$\text{Take }A\text{ as the point }re^{i\theta}\text{ and it follows from collinearity and }B\in \gamma\\ B\text{ represents }e^{i\theta}\\ \text{Then clearly }OA=r \implies AB=|1-r|=\min_{\alpha \in \mathbb{R}}f(\alpha)$

$\text{Since }X\in \gamma\text{, take }X=e^{i\alpha}\\ \text{Then }AX=|re^{i\theta}-e^{i\alpha}|=f(\alpha)\\ \implies AB \le AX \, \forall X \in \gamma$

Q: How would you go about (II)? I want to try proof by contradiction but I'm not sure how to go about anything

$\noindent Hint: All you need to do is show that the unique closest point on the circle to $A$ is the point $B$. This can be done by showing that $g\left(\alpha\right) := 1+r^{2}-2r\cos\left(\theta - \alpha\right)$ has a \emph{unique} point $\alpha$ in $\left[0,2\pi)$ where it is minimised (actually by symmetry it suffices to set $\theta = 0$, and then this becomes trivial to show).$

leehuan · Nov 13, 2016

Interpreting the MATLAB output: It's just the RREF of the matrix representation augmented with the identity.

i.e. the RREF of $(\textbf{v}_1 \,\textbf{v}_2\, \textbf{v}_3\, \textbf{v}_4\mid \textbf{e}_1\, \textbf{e}_2\, \textbf{e}_3\, \textbf{e}_4)$

How is the output supposed to help me do c)?

Because I thought the image would just be $\text{span}\left\{ \textbf{v}_1,\textbf{v}_2 \right\}$

InteGrand · Nov 13, 2016

leehuan said:
Interpreting the MATLAB output: It's just the RREF of the matrix representation augmented with the identity.

i.e. the RREF of $(\textbf{v}_1 \,\textbf{v}_2\, \textbf{v}_3\, \textbf{v}_4\mid \textbf{e}_1\, \textbf{e}_2\, \textbf{e}_3\, \textbf{e}_4)$

How is the output supposed to help me do c)?

Because I thought the image would just be $\text{span}\left\{ \textbf{v}_1,\textbf{v}_2 \right\}$

The right-hand matrix's columns (which started off being the standard basis vectors) are all leading columns if placed next to the reduced row-echelon form of A. Hence none of the standard basis vectors are in the image of T.

(It's like augmenting A with a single vector to solve a system of equations, except we augmented with four vectors (the four standard basis vectors), which is like looking at solving four separate systems of equations.)

leehuan · Dec 11, 2016

Are there some subtleties that we need to consider here or is this actually just as easy to do in R?

$\text{Prove that }\frac{d}{d\theta}e^{i\theta}=ie^{i\theta}$

Paradoxica · Dec 12, 2016

leehuan said:
Are there some subtleties that we need to consider here or is this actually just as easy to do in R?

$\text{Prove that }\frac{d}{d\theta}e^{i\theta}=ie^{i\theta}$

It is one variable... Maybe differentiate it as the unit vector?

seanieg89 · Dec 12, 2016

leehuan said:
Are there some subtleties that we need to consider here or is this actually just as easy to do in R?

$\text{Prove that }\frac{d}{d\theta}e^{i\theta}=ie^{i\theta}$

The answer here mildly depends on:

a) your definition of the exponential function.

b) whether theta ranges over R or C. (and hence whether the differentiation refers to differentiation with respect to a complex variable or a real variable)

c) which facts along the lines of the chain rule you know and can assume.

leehuan · Dec 13, 2016

seanieg89 said:
The answer here mildly depends on:

a) your definition of the exponential function.

b) whether theta ranges over R or C. (and hence whether the differentiation refers to differentiation with respect to a complex variable or a real variable)

c) which facts along the lines of the chain rule you know and can assume.

Hmm. I was picturing theta to be a complex variable, because I didn't want to consider a real variable with complex coefficients or something.

Which definitions of the exponential function should I choose from?
(I actually forgot what definition we used in high school. Possibly the limit of (1+x/n)^n?)

Also chain rule: All I know is dy/dx = dy/du . du/dx, over R (not sure if result holds over C)

seanieg89 · Dec 13, 2016

leehuan said:
Hmm. I was picturing theta to be a complex variable, because I didn't want to consider a real variable with complex coefficients or something.

Which definitions of the exponential function should I choose from?
(I actually forgot what definition we used in high school. Possibly the limit of (1+x/n)^n?)

Also chain rule: All I know is dy/dx = dy/du . du/dx, over R (not sure if result holds over C)

In high school you don't define the exponential function per se, you define the real number e as the unique real number a such that a^x has slope 1 at the origin, and then you work with e^x from there. This is a terribly nonrigorous, because high school mathematics does not rigorously define the reals, let alone what it means to raise the real number a to the real power x, etc. It is best to largely forget high school definitions of things, and just recall your high school experience in computing things like integrals etc.

The most common and cleanest definition of the exponential for use in higher math is the power series $\exp(z):=\sum_{n\in \mathbb{N}} \frac{z^n}{n!}$ which is absolutely convergent on all of C. (Radius of convergence is infinity, you should be able to prove this.)

Now since complex power series are complex differentiable in the interior of their disk of convergence (*) (the derivative is obtained by differentiating the power series "term-by-term"), and the function z -> iz is complex differentiable everywhere, we can apply the (complex) chain rule to draw the desired conclusion.

The complex chain rule is that if f:X->Y and g:Y->Z are complex differentiable, then (gof) is complex differentiable with derivative g'(f(z))f'(z). Since you were not sure that this result held true, you should try to prove it if you go about this method of solution to the original question.

If you wanted to avoid using the chain rule, you could alternatively use the fact that the derivative of a power series is given by differentiating term by term. Writing exp(iz) as a power series, differentiate term by term to explicitly show that we arrive at i*exp(iz).

Note:
If you are unfamiliar with (*), the result that power series can be differentiated term by term in the interior of their disk of convergence, you should prove this too.

MATH1251 Questions HELP (4 Viewers)

InteGrand

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Paradoxica

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Drsoccerball

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leehuan

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leehuan

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InteGrand

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leehuan

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InteGrand

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leehuan

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Paradoxica

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seanieg89

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leehuan

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