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91
Chapter 7 / Re: Drawing phase portrait with complex eigenvalues question
« Last post by Victor Ivrii on December 01, 2020, 08:46:13 PM »
"How ellipses wouls look like" means the directions and relative size of their semi-axis. See frame 4 of MAT244_W8L3 handout
92
Test 4 / Spring 2020 Test 2 Monday Sitting Problem 3
« Last post by Xinxuan Lin on December 01, 2020, 01:42:57 PM »
f(z) = (z$^4$ -$\pi^4$)tan$^2$($\frac{z}{2}$)

Part b of this question is asking to determine the types of the singular point.

In solution, it says z=2n$\pi$ with n$\neq$ $\pm$ 1 are double zeros; z=(2n+1)$\pi$ with n$\neq$ $\pm$ 1 are double poles.

Could anyone explain why n$\neq$ $\pm$ 1 here? Why is not n$\neq$ -1, 0?

Thanks in advanced!

93
Quiz 6 / quiz 6 0201 Two -E
« Last post by yantian4 on November 28, 2020, 10:55:17 AM »
for this quiz the answer I got is -te^-t + 1/2e^-t
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Test 3 / LEC5101-TT3-ALT-F-Q2
« Last post by RunboZhang on November 27, 2020, 08:03:20 PM »
$\textbf{Problem2:} \\ \\
\text{Find the power series expansion at z=0 of}$
$$f(z) = \frac{1}{z} \int_{0}^{z}{\frac{1}{z}\int_{0}^{z}{cos(w^2)\,dw}\,dz}$$
$\text{What is the radius of convergence?}$

$\textbf{Solution:} \\ \\
\text{Firstly, we have power series,}$
$$cos(w) = \sum_{n=0}^{\infty}{(-1)^{n}\frac{w^{2n}}{(2n)!}} \\ $$
$\text{with a radius of convergence of } \infty \\$

$\text{Then substitute } w \text{ by } w^2$
$$cos(w^2) = \sum_{n=0}^{\infty}{(-1)^{n}\frac{w^{4n}}{(2n)!}}$$

$\text{Now substitute the series in the integral and calculate the integral}$

$
\begin{gather}
\begin{aligned}

f(z) &= \frac{1}{z} \int_{0}^{z}{\frac{1}{z}} \int_{0}^{z}{cos(w^2)\,dw}\,dz \\\\
&= \frac{1}{z} \int_{0}^{z}{\frac{1}{z}}(\int_{0}^{z} {\sum_{n=0}^{\infty}{\frac{(-1)^{n}w^{4n}}{(2n)!}\,}}\,dw) \,dz \\\\
&= \frac{1}{z} \int_{0}^{z}{\frac{1}{z}}\sum_{n=0}^{\infty}{
\int_{0}^{z}{(-1)^{n} \cdot \frac{w^{4n}}{(2n)!}}\,dw}\,dz \\\\
&=\frac{1}{z} \int_{0}^{z}{\frac{1}{z}} \sum_{n=0}^{\infty}{(-1)^{n}\cdot \frac{z^{4n+1}}{(2n)! \cdot (4n+1)}}\,dz \\\\
&= \frac{1}{z} \int_{0}^{z}{\sum_{n=0}^{\infty}{(-1)^{n} \frac{z^{4n}}{(2n)! \cdot (4n+1)}}} \,dz\\\\
&= \frac{1}{z} \sum_{n=0}^{\infty}{\int_{0}^{z}{(-1)^{n} \frac{z^{4n}}{(2n)! \cdot (4n+1)}}}\,dz \\\\
&= \frac{1}{z} \sum_{n=0}^{\infty}{(-1)^{n} \frac{z^{4n+1}}{(2n)! \cdot (4n+1)^{2}}} \\\\
&= \sum_{n=0}^{\infty}{(-1)^{n} \frac{z^{4n}}{(2n)! \cdot (4n+1)^{2}}}

\end{aligned}
\end{gather}
$

$\text{Hence the power series expansion is }$
$$f(z) = \sum_{n=0}^{\infty}{(-1)^{n} \frac{z^{4n}}{(2n)! \cdot (4n+1)^{2}}}$$

$\text{Since the radius of convergence of } cos(z) \text{ is } \infty \text{. Thus } f(z) \text{ has the same radius of convergence of  } R=\infty.$
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Test 3 / LEC5101-TT3-ALT-F-Q1
« Last post by RunboZhang on November 27, 2020, 07:40:12 PM »
$\textbf{Problem 1:} \\\\
\text{Using Cauchy's Integral Formula or Residue Theorem to calculate}$
$$I:= \int_{\gamma}{\frac{z-1}{z^2-4z+5}} \,dz $$
$\text{where } \gamma \text{ is a contour on the picture with the shown direction; poles are shown like colored dots.}$


$
\textbf{Solution:} \\\\

\text{Since contour } \gamma \text{ is not a simple closed curve, then denote the contour } c_1 ,\ c_2 \text{ as following (pic is attached below)}$

$\text{Let }$ $$f(z) = z^2-4z+5 = 0$$

$\text{Then }$$$z=\frac{4 \pm \sqrt{4^2-4\cdot5}}{2} = 2 \pm i$$

$\text{For clockwise, simple closed contour } c_2 \text{, there is no pole. By Cauchy Theorem: }$
$$\int_{c_2}{\frac{z-1}{z^2-4z+5}} \,dz = 0$$

$\text{For clockwise, simple closed contour } c_1 \text{, there is two poles} z_1 = 2 + i ,\ z_2 = 2-i \text{. Then by Residual Theorem, we have:}$

$
\begin{gather}
\begin{aligned}

\int_{c_1}{\frac{z-1}{z^2-4z+5}} \,dz &= -2 \pi i \sum{Res} \\\\
&= -2 \pi i \cdot (\frac{z-1}{z-2+i}|_{z=2+i} + \frac{z-1}{z-2-i}|_{z=2-i}) \\\\
&= -2 \pi i \cdot (\frac{1+i}{2i} + \frac{1-i}{-2i}) \\\\
&= -2 \pi i

\end{aligned}
\end{gather}
$

$\text{Therefore we have}$
$\begin{gather}
\begin{aligned}
\int_{\gamma}{\frac{z-1}{z^2-4z+5}} \,dz &=\int_{c_1}{\frac{z-1}{z^2-4z+5}} \,dz + \int_{c_2}{\frac{z-1}{z^2-4z+5}} \,dz \\\\ &= -2\pi i
\end{aligned}
\end{gather}
$
96
Quiz 6 / Quiz6 lec0201 TWO C
« Last post by Zhiyue Yu on November 27, 2020, 12:07:28 PM »
Here is the question and the answer attached below:
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Quiz 6 / quiz6A
« Last post by TONG ZHU on November 27, 2020, 08:35:26 AM »
This is my answer for quiz 6A :)
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Chapter 7 / Drawing phase portrait with complex eigenvalues question
« Last post by Jessica Wang on November 26, 2020, 06:46:11 PM »
I'm struggling with how to draw phase portrait with purely imaginary complex eigenvalues. I know that the phase portrait should be ellipses but I'm not sure how to determine how exactly the ellipses look like. In the lecture slide, we used the technique of finding the eigenvectors of B (I have attached a screenshot). Is this the only method to determine how exactly the ellipses look like? There is an example provided in one of the tutorials, where the sidenotes say we should "project the eigenvectors to real parts" (I have also attached a screenshot) but I don't quite understand what it means. Any help is appreciated! Thanks!
99
Test 2 / TT2- Main - Q2
« Last post by Jiayue Yin on November 26, 2020, 07:48:01 AM »
Hi, this is my answer for Main B question 2.
100
Test 2 / TT2- Main B- Q1
« Last post by Jiayue Yin on November 26, 2020, 07:44:47 AM »
Hi, this is my answer for Main B question 1.
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