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Messages - Jingxuan Zhang

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APM346--Lectures / potential
« on: March 11, 2018, 05:46:54 PM »

My question is concerning equations $(9),(10)$. First what is $u(y)$ in fact? is that some sort of charge? is "charge" and "potential" the same? are they sort of energy? (You must know the asker is barely exposed to Physics.) Secondly, if $u=u(y)$ then what is meant by $\Delta u$?

Also I think it will be good if the online text is open to edit. Very often I can find such trivial typos that do not worth of posting here, but are nonetheless quite obstentious.

Term Test 1 / Re: P1 Night
« on: March 11, 2018, 02:10:44 PM »
A) Clearly
$$du=e^{t^2/2}dx=C/xdx\implies u=x\ln x e^{-t^2/2}+\varphi(xe^{-t^2/2})$$
$$x=x\ln x + \varphi(x)\implies\varphi(x)=x - x\ln x\implies u = xe^{-t^2/2}+\frac{xt^2}{2}e^{-t^2/2}$$

I am fortunate enough that I didn't write night exam...this I spent more than half an hour.

Quiz-5 / Re: Quiz5 tut 0201
« on: March 08, 2018, 04:40:53 PM »
I just want to remind everyone who sees this so that he won't make the same mistake as I did: 
$$F(\Re f)\neq\Re F(f)\text{ and } F(\Im f )\neq\Im F(f)$$

APM346––Home Assignments / Re: What is this?
« on: March 08, 2018, 09:37:24 AM »
But I assume its just another way to write $F^{-1}$?

APM346––Home Assignments / What is this?
« on: March 08, 2018, 08:45:09 AM »
$$F^*$$The one that suffices $FF^*=F^*F=I$?

Quiz-5 / Re: Quiz 5, T5102
« on: March 08, 2018, 07:23:50 AM »
The reason that the absolute value is missing is that the contour you choose must depend on the sign of $k$ If $k<0$ then your contour should instead be upper semicircle and if $k>0$ then you are right. This is done to satisfy the hypothesis of Jordan's lemma, which you implicitly used to control the integral over the arc as it increasese.

Also what property do you refer to? I think $\hat{xf}=i\hat{f}'$?

Beside: don't we always use $\frac{1}{2\pi}$ for the scaling? and this for one thing at least better suit Residual Theorem.
$$\hat{f}(\omega)=\frac{1}{2\pi}\int_{-\infty}^\infty \frac{e^{-i\omega x}}{(x-ai)(x+ai)}=\left\{\begin{align*}\left.i\frac{e^{-i\omega x}}{x+ai}\right|_{x=ai}&&\omega<0\\ \left.-i\frac{e^{-i\omega x}}{x-ai}\right|_{x=-ai}&&\omega>0\end{align*}\right.=\frac{e^{-|\omega|a}}{2a} $$
And so immediately
$$\hat{xf}(\omega)=i\hat{f}'(\omega)=-i \text{sgn}(\omega)\frac{e^{-|\omega|a}}{2}$$

APM346--Lectures / Re: Fourier coefficients
« on: March 07, 2018, 06:58:39 AM »
As if when we do the projection of $b$ on $a$ in Euclidean space, the coefficient is given by $a\cdot b / ||a||^2$?

APM346––Home Assignments / Re: 5.2P, Question 3 approach
« on: March 07, 2018, 06:54:30 AM »
If you used Fisher's text in MAT344, you can find it on p. 344. Factor the denominator and use CIF/Residual theorem provided a proper orientation is taken to make sure the integral vanish on the arc of the "semicircle contour".

The following questions are not meant to be dealt starting from scratch. Referring to the properties list in section 5.2 I found all them follow in one way or another the first question, the "base" case.

APM346––Home Assignments / bounded assumption
« on: March 06, 2018, 08:16:19 PM »
Do we not need to know that the function remains bounded in order to solve the first parts in chap 6, question 3,4?

APM346--Lectures / Fourier coefficients
« on: March 06, 2018, 08:06:25 PM »
Could anyone kindly inform me one more time the origin of the scalar multiple in finding Fourier coefficients? I know these are supposed be inner product, but I don't remember a scaling there.
Also, chap 4.3 remark 2 has some obvious typo. (Somewhere 2's in $L^2$'s are missing).

Quiz-4 / Quiz 4 -- both sections
« on: March 01, 2018, 10:07:07 PM »
Note: Since problems for both sections are very similar I suggest to discuss them together, so certain parts of the solutions could be used without repetition

The only difference that in Wed section condition on the right end are $u_{xx}|_{x=l}=u_{xxx}|_{x=l}=0$,

This is problem 3 part 1,2 from

The associated eigenvalue problem is
\begin{align}X^{iv}-\omega^4 X=0\label{1}\\X(0)=X'(0)=0\label{2}\\X(l)=X'(l)=0
From (\ref{1}) we write
\begin{equation}X=A\cosh (\omega x) + B\sinh(\omega x)+C\cos(\omega x)+D\sin(\omega x)\label{4}\end{equation}
whence (\ref{2}) implies, if $\omega\neq 0 $,
and so (\ref{4}) becomes
\begin{equation}X=A(\cosh (\omega x) -\cos(\omega x)) +  B(\sinh(\omega x)-\sin(\omega x))\label{6}\end{equation}

Now the algebraic system in variable of $A,B$ obtained from (\ref{3}) has nontrivial solution if and only if the coefficient matrix is singular, that is:
\begin{equation}\left|\begin{array}{cc}\cosh (\omega l) -\cos(\omega l)&\sinh(\omega l)-\sin(\omega l)\\ \sinh (\omega l) +\sin(\omega l)&\cosh(\omega l)-\cos(\omega l)\end{array}\right|=2-2\cosh (\omega l)\cos(\omega l)=0\iff\cosh (\omega l)\cos(\omega l)=1\label{equation}.\end{equation}
The null space of this system is
\begin{equation}(A,B)'=t(-\sinh(\omega l)+\sin(\omega l),\cosh (\omega l) -\cos(\omega l))',t\in\mathbb{R}\label{8}\end{equation}
and so (\ref{6}) becomes
\begin{equation}X=(-\sinh(\omega l)+\sin(\omega l))(\cosh (\omega x) -\cos(\omega x))+(\cosh (\omega l) -\cos(\omega l))(\sinh(\omega x)-\sin(\omega x))\label{9}\end{equation}
up to a scalar multiple. This is the eigenspace.

The graphs are those of $1/\cosh$ and $\cos$, imagined to be in variable of $\omega l$. Their (infinitely many) intersections suffice (\ref{equation}).

APM346––Home Assignments / To determine eigenfunction
« on: March 01, 2018, 05:10:59 PM »
I am referring to What's the reason of change of coordinate? shouldn't everything come to an end when as soon as we found the eigenvalue equation, and obtain a formula for $X$ up to a scalar multiple? what causes the concern that different eigenvalue might associate to different form of $X$?

In short: why do we bother changing the coordinate and consider parity?

APM346--Lectures / Re: Take a correct branch
« on: February 27, 2018, 08:14:49 PM »
Indeed, changing $t\mapsto -t$, $b\mapsto -b$ preserves equation (2).

So (3) holds for $bt>0$. But taking complex conjugation and $b\mapsto -b$ also preserves (2). This operation with (3) brings
\frac{1}{\sqrt {4\pi |bt|}} e^{i\pi/4}e^{ix^2/4b t}.

Is this because (heuristically) since $u$ is real-valued
$$u_t=ibu_{xx}=\overline{i(-b)u_{xx}} \implies u = \frac{1}{\sqrt {4\pi bt}} e^{i (-\pi/4+x^2/4b t)} = \overline{\frac{1}{\sqrt {4\pi (-b)t}} e^{i (-\pi/4+x^2/4(-b) t)}}=\frac{1}{\sqrt {4\pi (-b)t}} e^{i (\pi/4+x^2/4bt)}?$$
I hesitated a lot before posting. This $-b$ under square root is really poignant. How would you then justify your "taking complex conjugation and $b\mapsto -b$ also preserves (2)"?

APM346--Lectures / Take a correct branch
« on: February 27, 2018, 10:12:26 AM »
In chap 5.2 repeatedly I find "we need to take a correct branch". For instance, for the Schrodinger equation with $t\gtrless 0$, what in fact is the reason of taking $\sqrt{i}=\pm e^{i\pi/4}$?

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