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### Messages - Jingxuan Zhang

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1
##### MAT334--Lectures & Home Assignments / Bounding gamma function
« on: January 27, 2019, 08:50:05 PM »
I have almost forgotten how to due with

Any hint?

2
##### MAT334--Lectures & Home Assignments / Re: What can I say about f'(0)?
« on: November 21, 2018, 08:17:44 PM »
O.k., but is there anything I can say about $f'(0)$?

If there is really nothing to say, then please consider the following situation: $f:\mathbb{R}\to\mathbb{C}$ is continuous, and $\lim_{t\to 0} t^{-1}(f(t)+f(t)^{-1}-2)$ exists. What can say about this limit? (In particular I would love it to be 0).

3
##### MAT334--Lectures & Home Assignments / Re: 2.3 Q7
« on: November 21, 2018, 07:51:46 PM »
Ye: $\pm$ is \pm.

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##### MAT334--Lectures & Home Assignments / What can I say about f'(0)?
« on: November 21, 2018, 04:45:22 PM »
Suppose $f:D\to\mathbb{C}$ is analytic near 0, such that $\|x\|=1\implies |f(x)|=1$. Does it follow that $f'(0)$ is purely imaginary?

5
##### Final Exam / Re: FE-P6
« on: April 14, 2018, 07:14:10 AM »
Alternatively, professor, you can consider it a bonus, which is rewarding for those who has worked out one of the previous year's final where this situation happened in almost exactly the same manner.

Observe indeed $u$ must be spherical symmetric as is the boundary. Let $v=ru$, then \eqref{6-1}-\eqref{6-2}become, once identity \eqref{6-3} is known,
\begin{align} &v_{tt}-v_{rr}  =0, \label{6-1'}\\ &v|_{t=0}=0, &&v_t|_{t=0}= \left\{\begin{aligned} &\sin(r) &&r<\pi,\\ &0 &&r\ge \pi,\end{aligned}\right.\qquad \label{6-2'} \end{align}
which is easily solved with a combined use of even continuation and D'Alembert's:
v=\left\{\begin{aligned}&0&&r>t+\pi, \\ &\sin r\sin t&&0<r<-t+\pi, \\ &\frac{\cos(r-t)+1}{2}&&|t-\pi|<r<t+\pi,\\&0&&0<r<t-\pi.\end{aligned}\right.\qquad\label{6-4}
So then
u=\left\{\begin{aligned}&0&&r>t+\pi, \\ &\frac{\sin r\sin t}{r}&&0<r<-t+\pi, \\ &\frac{\cos(r-t)+1}{2r}&&|t-\pi|<r<t+\pi,\\&0&&0<r<t-\pi.\end{aligned}\right.\qquad\label{6-5}

6
##### Final Exam / Re: FE-P3
« on: April 11, 2018, 07:23:56 PM »
I agree with George. Tristan: but \eqref{3-4}.

7
##### Final Exam / Re: FE-P7
« on: April 11, 2018, 07:21:38 PM »
To All,

I think you should have the minus sign.

8
##### Final Exam / Re: FE-P1
« on: April 11, 2018, 07:20:42 PM »
Tristan,

I think
$$(e^{-x} \psi(x))' = e^{x} + 2e^{x} + e^{x},$$
not minus.

9
##### Final Exam / Re: FE-P6
« on: April 11, 2018, 07:09:45 PM »
Andrew,

Heed your trig! Beside there should be cases.

10
##### Web Bonus Problems / Re: Week 13 -- BP3, 4, 5, 6
« on: April 08, 2018, 01:00:33 PM »
Zhongnan,
I think c) is alright as the way it stands. After all \eqref{eq-11.1.12} is deduced from \eqref{eq-11.1.13} so that is also fine...

11
##### Term Test 2 / Re: TT2--P3N
« on: April 06, 2018, 04:54:41 PM »
Andrew,

Your square is misplaced at the place where it is pointed out.

12
##### Web Bonus Problems / Re: Week 13 -- BP2
« on: April 05, 2018, 04:51:30 PM »
Adam: no you cannot and don't need to assume $\varphi$ is even, but rather you would need to use $\varphi$ is smooth. The gap in your argument is smoothed by
$$(\varphi(-\varepsilon)-\varphi(\varepsilon))\ln\varepsilon\to 0 \text{ as } \varepsilon\downarrow0.$$

13
##### Term Test 2 / Re: tt2-Q2 problem
« on: April 05, 2018, 04:36:45 PM »
Same question. The TA says $\alpha<0$ or $\alpha =n\pi$ which I don't really quite understand.

14
##### Quiz-B / Re: Quiz-B P2
« on: April 02, 2018, 09:46:14 PM »
Observe for nice $\varphi$
$$(\sin(x) \varphi(x))'''=-\cos(x) \varphi(x)-3\sin(x)\varphi'(x)+3\cos(x)\varphi''(x)+\sin(x)\varphi'''(x).$$
Therefore
$$\langle\sin(x)\delta'''(x),\varphi(x)\rangle=\langle\delta'''(x),\sin(x) \varphi(x)\rangle=-\langle\delta(x),(\sin(x) \varphi(x))'''\rangle=\varphi(0)-3\varphi''(0).$$
So $$\sin(x)\delta'''(x)=\delta(x)-3\delta''(x).$$

15
##### Quiz-B / Re: Quiz-B P1
« on: April 02, 2018, 09:33:04 PM »
Define
$$L(r,u_r)=r\sqrt{1+u_r^2}.$$
Then E.-L. $$(L_{u_r})_r=L_u$$ gives
$$\label{1-1}\Bigl(\frac{ru_r}{\sqrt{1+u_r^2}}\Bigr)_r=0,$$
from which we derive
$$\frac{ru_r}{\sqrt{1+u_r^2}}=A\implies u_r^2(r^2-A^2)=A^2\implies u_r=\frac{A}{\sqrt{r^2-A^2}}\implies u=A\cosh^{-1}(r/A)+B.$$

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