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Messages - Jingxuan Zhang

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To deal with incomplete series (sin or cosine) we use complete series on $[-l,l]$.

And imagine them to be even on $[-l,l]$, to deal with either case?

But then how would you justify the "double the integral" in computing the incomplete series coefficients?

APM346––Home Assignments / Fourier expansion not centering at origin
« on: February 24, 2018, 05:44:53 PM »
In Promblems for 4.3 questions 4,5 ask for expansion on $[0,\pi]$. Are the same formulae valid there? or do we have to first make the change of variable or other manipulation? In particular for function such as $x(\pi-x)$, is there any trick that can shorten the direct computation?


Web Bonus Problems / Re: Web Bonus problem -- Reading Week
« on: February 17, 2018, 03:26:11 PM »
Part (b): normalize the integrand we have
$$I=\hat{u}(0)=\int e^{-\frac{1}{2}a x^2}\,dx=\sqrt{\frac{2\pi}{a}}\cdot\underbrace{\frac{1}{\sqrt{2\pi}}\int e^{-z^2/2}\,dz}_{1}=\sqrt{\frac{2\pi}{a}}$$
But here really $\sqrt{a}$ is ill-defined: $a\in\mathbb{C}$ might result in this root having multiple value. The sign therefore depends on the branch chosen for $a$. In particular the principal value of $I$ will have a positive sign if $\Im(a)\geq0$ and a negative sign otherwise.

EDIT: I consulted Ahlfors' text on complex variable, which I have completely forgotten now. Write $a=\alpha+i\beta$ then

If $\alpha>0,\beta=0$ then of course this is just the simple formula on LS.

Web Bonus Problems / Re: Web Bonus problem -- Reading Week
« on: February 17, 2018, 03:10:50 PM »
I find the second part perplexing: what are we to compare $I$ with? and shouldn't $I=\hat{u}(0)$ be immediate once the formula for $\hat{u}$ is determined by Adam?

Term Test 1 / Re: P5
« on: February 16, 2018, 02:50:23 PM »
A direct computation yields:

u&=\frac{1}{\sqrt{4t\pi}}\int_{-\infty}^{\infty} \exp{-\frac{(y-x)^2}{4t}-|y|}\,dx\\
&=\frac{1}{\sqrt{4t\pi}}\int_0^{\infty} \exp(-\frac{(y+x)^2}{4t}-y)\,dx +\frac{1}{\sqrt{4t\pi}}\int_0^{\infty} \exp(-\frac{(y-x)^2}{4t}-y)\,dx\\
&=\frac{\exp(x+t)}{\sqrt{4t\pi}}\int_0^{\infty} \exp(-\frac{(y+x+2t)^2}{4t})\,dx + \frac{\exp(-x+t)}{\sqrt{4t\pi}}\int_0^{\infty} \exp(-\frac{(y-x+2t)^2}{4t})\,dx\\
&=\frac{\exp(x+t)}{\sqrt{\pi}}\int_{\frac{x+2t}{\sqrt{4t}}}^{\infty} e^{-z^2} \,dz+ \frac{\exp(-x+t)}{\sqrt{\pi}}\int_{\frac{-x+2t}{\sqrt{4t}}}^{\infty} e^{-z^2} \,dz\\
&=\frac{\exp(x+t)}{2}(1-\text{erf}(\frac{x+2t}{\sqrt{4t}})) + \frac{\exp(-x+t)}{2}(1-\text{erf}(\frac{-x+2t}{\sqrt{4t}}))

Term Test 1 / Re: P2
« on: February 16, 2018, 02:36:46 PM »
A direct computation yields:
u &=\frac{1}{4}\int_0^t\int_{x-2t+2t'}^{x+2t-2t'} \frac{8}{x'^2+1} \,dx'\,dt'\\
&=\frac{1}{4}\int_{x-2t}^x\int_{0}^{x'/2+x/2+t} \frac{8}{x'^2+1} \,dt'\,dx' +\frac{1}{4}\int_x^{x+2t}\int_{0}^{-x'/2+x/2+t} \frac{8}{x'^2+1} \,dt'\,dx'\\
&=\int_{x-2t}^x\frac{x'-x+2t}{x'^2+1} \,dx' + \int_x^{x+2t}\frac{x'+x+2t}{x'^2+1} \,dx' \\
&= \int_{x-2t}^x\frac{x'}{x'^2+1} \,dx + (x-2t) \int_x^{x-2t}\frac{1}{x'^2+1}\,dx'+
\int_x^{x+2t}\frac{x'}{x'^2+1} \,dx + (x+2t) \int_x^{x+2t}\frac{1}{x'^2+1}\,dx'\\

To my great shame, I failed to directly compute this during the actual sitting.

APM346––Home Assignments / Complete the square
« on: February 14, 2018, 05:59:39 AM »
in a heat equation question when the initial distribution is given of the form $e^{-ax^2}$, how am I supposed to complete the square? or do I use any alternative?
I don't think if I end up having $\sqrt{1+4kt}y$ terms it will bring any help, while there are still others that cannot be factored out at once.

APM346--Lectures / $4.5 is not loaded properly
« on: February 12, 2018, 04:37:16 PM »
I notice our online book $4.5 can only load the codes but not the actual symbols. Please see if this can be fixed, thanks!

Quiz-3 / Re: Q3-T0101
« on: February 11, 2018, 02:56:12 PM »
Write $u=\varphi(x+ct)+\psi(x-ct)$ then Instantly by D'Alembert' s on $x>ct$
On $0<x<ct$ solution $\varphi=\phi$ still works, whereas from boundary condition
$$\phi(-s)'+\psi'(s)+\alpha\phi(-s)+\alpha\psi(s)=0 \implies (e^{\alpha s}\psi(s))'=(e^{\alpha s}\phi(-s))'-2\alpha\phi(-s)
\implies \psi(s)=\phi(-s)+2\alpha e^{-\alpha s}\int^s e^{-\alpha s'}\phi(s)\, ds'.$$
So that
\begin{equation}u=\phi(ct+x)+\phi(ct-x)+2\alpha e^{-\alpha (ct-x)}\int^{ct-x} e^{-\alpha s}\phi(s)\, ds,0<x<ct.\end{equation}
In particular, if $\phi(x)=e^{ikx}$, then
&e^{ik(ct+x)}+e^{ik(ct-x)}-2\frac{ik\alpha+\alpha^2}{k^2+\alpha^2}e^{(ik-2\alpha) (ct-x)}&0<x<ct
Provided, indeed, $k^2+\alpha^2\neq 0$.

Technical Questions / Tag each equation in an IVP
« on: February 10, 2018, 04:14:10 PM »
I am trying to typeset in latex a system such as
Is there anyway that allows me to label each individual equation such as with \tag? True, with align I can label but i need also the brace. True, with array I can brace but then cannot label. Thanks.

APM346––Home Assignments / Problem set 2.6 Q4
« on: February 08, 2018, 06:42:42 AM »
So now the boundary condition amounts to an ODE with initial condition $\psi(0)=\phi(0)/2$? And then do we just leave the integral as it is after variation of parameter?
I am little diffident about the change of variable in the integral I made, so should I get
$$\psi(x-ct)=e^{\alpha(ct-x)}\int_0^{ct-x} e^{\alpha y}(\phi'(y)+\alpha \phi(y)) \,dy +e^{\alpha(ct-x)}\phi(0)/2 ?$$
It seems to me that I can be simplified further but I don't know how.

EDIT: This is solved in today's tutorial. So in fact it simplifies a little bit. But how about the integration constant, the initial condition for this ODE? On textbook it says (in our case) $\psi(0)=\phi(0)/2$, so then should I still have the term $e^{\alpha(ct-x)}\phi(0)/2$?

APM346--Lectures / Another erf
« on: February 07, 2018, 06:19:32 AM »
In a post last year erf was given by
$$\text{erf}_1(z)=\sqrt{\frac{2}{\pi}}\int_0^z e^{-s^2/2} \,ds$$
I assume this is not equivalent to our erf, and in fact $\text{erf}(z/\sqrt{2})=\text{erf}_1(z)$. Am I right? How does this difference affect the solution, if any at all?

APM346--Lectures / Re: Section 2.6 Example 1 correction
« on: February 06, 2018, 07:28:59 AM »
I noticed this too. I seems to me a problem as well.

Quiz-2 / Re: Q2-T5102
« on: February 02, 2018, 09:23:39 PM »
Plug in D'Alembert's instantly we obtain:

            &0 && x+ct\leq -\pi/2 \text{ or } x-ct \geq \pi/2\\
            &\cos(x-ct)/2 && x-ct <\pi/2 \leq x+ct\\
            &\cos(x)\cos(ct) && -\pi/2 < x-ct < x+ct < \pi/2\\
            &\cos(x+ct)/2 && x-ct \leq -\pi/2 < x+ct\\
\tag*{Part (a)}\\[5pt]
            &0 && x+ct< 0\\
            &x/2c+t/2 && x-ct <0 \leq x+ct\\
            &t && 0 \leq x-ct\\
\tag*{Part (b)}

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