/sci/ - Science & Math » Thread #13195380

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Anonymous Thu 27 May 04:59:45 2021 No.13195380 View Reply Original Report

Quoted By: >>13195545

Any help on how I find the Laurent series to this?

Anonymous

Anonymous Thu 27 May 2021 06:03:04 No.13195545 Report

Quoted By: >>13195608

>>13195380
Start with the series for sin() then modify as needed.

Anonymous

Anonymous Thu 27 May 2021 06:26:22 No.13195608 Report

Quoted By:

>>13195545
That's what I did, but I can't reach the answer :(

Anonymous

Anonymous Thu 27 May 2021 06:53:42 No.13195681 Report

Quoted By: >>13196671

$sin(z) = \sum_{n=0}{\infty} \frac{(-1)^n}{(2n+1)!}z^{2n+1}$

so

$\frac{z-sin(z)}{z^3} = \frac{1}{z^2} - \sum_{n=0}{\infty} \frac{(-1)^n}{(2n+1)!}z^{2n+1-3}$

First term in the sum is (-1)^0/1! * z^-2 = 1/z^2, so the first term cancels and you're left with

$\sum_{n=1}{\infty} \frac{(-1)^{n+1}}{(2n+1)!}z^{2n-2}$

reindex the sum so it starts at n=0 by changing n to n+1 everywhere and it becomes

$\sum_{n=0}{\infty} \frac{(-1)^{n+2}}{(2n+3)!}z^{2n} = \sum_{n=0}{\infty} \frac{(-1)^{n}}{(2n+3)!}z^{2n}$

Second one should be even simpler

Anonymous

Anonymous Thu 27 May 2021 06:54:49 No.13195685 Report

Quoted By:

just pretend I did the sum symbols right...

$\sum_{n=0}^{\infty}$

Anonymous

Anonymous Thu 27 May 2021 14:08:01 No.13196671 Report

Quoted By:

>>13195681
Thank you

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