### Recent Posts

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##### Quiz-5 / Qz5-THREE-E
« Last post by duoyizhang on March 16, 2021, 01:27:02 AM »
The question is:Decompose f(x) = xcos(x)  into full Fourier series on interval [0, pi].
My confusion is how to decompose it on an interval like  [0,𝑙] rather than[-l,l]
Firstly,I compute f(x) into  full Fourier series on interval [-pi, pi] by the formula,Which is $$f1=f(x)=-\frac{sinx}{2}+\sum_{n=2}^\infty\frac{2n(-1)^{n}}{n^{2}-1}$$
Then what should we do to compute f(x) on[0,pi],I tried to use the property that f(x)is an odd function but it seems to be wrong.
Any help will be appreciated!
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##### Test-2 / Re: Test-2 problem-1 confusion regarding boundary conditions
« Last post by Victor Ivrii on March 07, 2021, 04:12:41 AM »
Solution is allowed to be discontinuous.
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##### Test-2 / Test-2 problem-1 confusion regarding boundary conditions
« Last post by aryakim on March 06, 2021, 10:07:36 AM »
I had a question about problem $1$ of my Test$2$. (My version was called alternative-F, night section).There was something confusing about this problem that I realized during the test.
Here is the problem:

\nonumber
\left\{ \begin{aligned}
& u_{tt}-u_{xx}=0, &&0< t < \pi, 0 < x < \pi, &(1.1) \\\
&u|_{t=0}= 2\cos (x),   && 0< x < \pi, &(1.2)\\
&u_t|_{t=0}= 0,  && 0< x < \pi,  &(1.3)\\
&u|_{x=0}= u|_{x= \pi}=0, && 0< t < \pi.  &(1.4)
\end{aligned}
\right.

So, on the $t-x$ diagram, the lines $t = 0$ and $x = \pi$ intersect at ($t=0,x=\pi$), which will be a boundary point of the region where $0<t<x<\pi$.
If this point (i.e. ($t=0,x=\pi$)) on the diagram is approached by the line $t = 0$, equation $(1.2)$ is used to conclude that the value of $u(x,t)$ approaches $-2$.
On the other hand, if the point is approached from the line $x = \pi$,  $u(x,t)$ should become zero, which is in contradiction with the other boundary condition. It seems this would make it impossible to incorporate conditions $(1.2)$ and $(1.4$) at the same time. I was hoping someone could clear my confusion regarding this problem.

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##### Quiz-3 / QUIZ3 5301 TWO-C
« Last post by Jin Qin on February 19, 2021, 06:30:49 PM »
Hi, this is my answer for QUIZ3 TWO-C in section 5301. Hope this can help you out!
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##### Quiz 1 / Re: 0101 quiz1
« Last post by Jiaqi Bi on December 20, 2020, 01:15:56 PM »
$\text{Im}(2iz)=2x=7\Rightarrow x=\frac{7}{2}$ (Typos at this line.)
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##### Quiz 1 / Re: Quiz1-6101 A
« Last post by Jiaqi Bi on December 20, 2020, 01:08:17 PM »
The second line of the answer, left part of your equation should be $|x+iy-i|$ instead of $|x+iy-1|$
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##### Quiz 7 / Re: Lec 0101- Quiz 7 C
« Last post by Jiaqi Bi on December 17, 2020, 07:43:36 AM »
Question: Using argument principle along line on the picture, calculate the number of zeroes of the following function in the left half-plane:
$z+a=e^z, (a>0)$

(Plus: I think Professor Victor originally wants to give us $a>1$ rather than $a>0$?)
(The graph will be attached.)
Solution:
\begin{align} h(iy)&=iy+a-e^{iy}\\ &=iy+a-cos(y)-isin(y)\\ &=(a-cos(y))+i(y-sin(y)) \end{align}
If $a>0$, we cannot conclude anything for $\text{Re} h(iy)$, but if $a>1$, then $\text{Re} h(iy)$ is always positive because the range of $cos(y)$ is consistently from $-1$ to $1$.
$\text{Im}h(iy)$ will increase when $y$ goes from $-R$ to $R$.
\begin{align} h(\text{Re}^{it})=\text{Re}^{it}+a=e^{\text{Re}^{it}} \end{align}
Where t is from $\frac{\pi}{2}$ to $\frac{3\pi}{2}$, and $z$ goes from $iR$ to $-iR$. $h(z)$ in this circumstance has been travelled a counterclockwise circuit.
Therefore, the argument for $h(z)$ should be $2\pi$.
By The Argument Principle, $\frac{1}{2\pi}\cdot 2\pi=1$.
Hence, $h(z)$ has a total of one zero in this plane.
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##### Quiz 7 / Lec 0101- Quiz 7 C
« Last post by Jiaqi Bi on December 17, 2020, 06:51:38 AM »
Question: Using argument principle along line on the picture, calculate the number of zeroes of the following function in the left half-plane:
$z+a=e^z, (a>0)$

(The graph will be attached.)
(Plus: I think Professor Victor originally wants to give us $a>1$ rather than $a>0$?)
Solution:
\begin{align} h(iy)&=iy+a-e^{iy}\\ \nonumber &=iy+a-cos(y)-isin(y)\\ \nonumber &=(a-cos(y))+i(y-sin(y)) \end{align}
If $a>0$, we cannot conclude anything for $\text{Re} h(iy)$, but if $a>1$, then $\text{Re} h(iy)$ is always positive because the range of $cos(y)$ is consistently from $-1$ to $1$.
$\text{Im}h(iy)$ will increase when $y$ goes from $-R$ to $R$.
\begin{align} h(\text{Re}^{it})=\text{Re}^{it}+a=e^{\text{Re}^{it}} \end{align}
Where t is from $\frac{\pi}{2}$ to $\frac{3\pi}{2}$, and $z$ goes from $iR$ to $-iR$. $h(z)$ in this circumstance has been travelled a counterclockwise circuit.
Therefore, the argument for $h(z)$ should be $2\pi$.
By The Argument Principle, $\frac{1}{2\pi}\cdot 2\pi=1$.
Hence, $h(z)$ has a total of one zero in this plane.
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##### Final Assessment--Part II / MAIN-B-PARTII Q6
« Last post by Mingxin Wang on December 17, 2020, 03:43:07 AM »
This is my q6 question
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##### Final Assessment--Part II / Re: mat 244 final part2 B Q5
« Last post by Mingxin Wang on December 17, 2020, 03:36:17 AM »
My full picture is the same as yours. And the only small difference with me is the Jacobian.
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