### Author Topic: TT2B Problem 2  (Read 3600 times)

#### Victor Ivrii

• Elder Member
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##### TT2B Problem 2
« on: November 24, 2018, 05:20:09 AM »
(a) Find the decomposition into power series at ${z=0}$ of $$f(z)=(1-z)^{-1}.$$ What is the radius of convergence?

(b) Plugging in $-z^2$ instead of $z$ and integrating, obtain a decomposition at $z=0$ of  $\arctan (z)$.
« Last Edit: November 29, 2018, 07:27:48 AM by Victor Ivrii »

#### Yifei Wang

• Jr. Member
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##### Re: TT2 Problem 2
« Reply #1 on: November 24, 2018, 05:38:47 AM »
I think the power is -1/2 inside of the -1

Correct V.I.  It was actually Test2B
« Last Edit: November 29, 2018, 07:28:07 AM by Victor Ivrii »

#### Wanying Zhang

• Full Member
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##### Re: TT2 Problem 2
« Reply #2 on: November 24, 2018, 12:15:18 PM »
Here is the solution to problem 2.

You need to know decomposition of $(1-z)^{-1}$. The rest is simply wrong. V.I.
« Last Edit: November 29, 2018, 07:19:35 AM by Victor Ivrii »

#### Huanglei Ln

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##### Re: TT2 Problem 2
« Reply #3 on: November 25, 2018, 01:34:47 AM »
\begin{aligned} a)f(z)&=\frac{1}{1-z}=\sum^{\infty}_{n=0}z^n \\ \frac{1}{R}&=\lim_{n\rightarrow {\infty} }|\frac{1}{1}|=1\Rightarrow R=1 \end{aligned}
\begin{aligned} b)f(-z^2)&=\frac{1}{1+z^2}=\sum^{\infty}_{n=o}{-z^2}^n=\Sigma^{\infty}(-1)^nz^{2n} \\ \Rightarrow \int f(-z^2)dz&=\sum^{\infty}_{n=o}(-1)^n\int^{2n}dz\\ \Rightarrow \int \frac{1}{1+z^2}dz&=\sum^{\infty}_(n=0)(-1)^n\int z^{2n}dz\\ \Rightarrow \arctan(z)+c&=\sum^{\infty}_{n=0}(-1)^n \frac{z^{2n+1}}{2n+1}\\ \Rightarrow \arctan(z)&=\sum^{\infty}_{n=0}\frac{(-1)^n}{2n+1}z^{2n+1} +c \end{aligned}
« Last Edit: November 29, 2018, 07:24:02 AM by Victor Ivrii »

#### Huanglei Ln

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• Posts: 8
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##### Re: TT2 Problem 2
« Reply #4 on: November 25, 2018, 01:39:58 AM »
\begin{aligned}
a)f(z)&=\frac{1}{1-z}=\sum^{\infty}_{n=o}z^n \\
\frac{1}{R}&=lim_{n\rightarrow {\infty} }|\frac{1}{1}|=1\Rightarrow R=1
\end{aligned}
\end{displaymath}
\begin{displaymath}
\begin{aligned}
b)f(-z^2)&=\frac{1}{1+z^2}=\sum^{\infty}_{n=o}{-z^2}^n=\Sigma^{\infty}(-1)^nz^{2n} \\
\Rightarrow \int f(-z^2)dz&=\sum^{\infty}_{n=o}(-1)^n\int^{2n}dz\\
\Rightarrow \int \frac{1}{1+z^2}dz&=\sum^{\infty}_(n=0)(-1)^n\int z^{2n}dz\\
\Rightarrow artan(z)+c&=\sum^{\infty}_{n=0}(-1)^n \frac{z^{2n+1}}{2n+1}\\
\Rightarrow artan(z)&=\sum^{\infty}_{n=0}\frac{(-1)^n}{2n+1}z^{2n+1} +c
\end{aligned}

#### Victor Ivrii

Also as $z=0$ you'll see that $c=0$. In actual test missing this will lead to the mark reduction