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Author Topic: TT2B Problem 2  (Read 9463 times)

Victor Ivrii

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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)=(1z)1. What is the radius of convergence?

(b) Plugging in z2 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

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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

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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 (1z)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 »
a)f(z)=11z=n=0zn1R=limn|11|=1R=1
b)f(z2)=11+z2=n=oz2n=Σ(1)nz2nf(z2)dz=n=o(1)n2ndz11+z2dz=(n=0)(1)nz2ndzarctan(z)+c=n=0(1)nz2n+12n+1arctan(z)=n=0(1)n2n+1z2n+1+c
« Last Edit: November 29, 2018, 07:24:02 AM by Victor Ivrii »

Huanglei Ln

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Re: TT2 Problem 2
« Reply #4 on: November 25, 2018, 01:39:58 AM »
 a)f(z)=11z=n=ozn1R=limn|11|=1R=1
 \end{displaymath}
\begin{displaymath}
 b)f(z2)=11+z2=n=oz2n=Σ(1)nz2nf(z2)dz=n=o(1)n2ndz11+z2dz=(n=0)(1)nz2ndzartan(z)+c=n=0(1)nz2n+12n+1artan(z)=n=0(1)n2n+1z2n+1+c

Victor Ivrii

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Re: TT2 Problem 2
« Reply #5 on: November 29, 2018, 07:23:37 AM »
Huanglei

I fixed your LaTeX. Don't learn it from crappy sources!

Also as z=0 you'll see that c=0. In actual test missing this will lead to the mark reduction
« Last Edit: November 29, 2018, 07:25:10 AM by Victor Ivrii »