anonymous
  • anonymous
hey guys I need to turn 1/x² into a series representation. since i know that 1/x= sum n=0 to infinite of (1-x)^n , I thought of writing it down as a cauchy-product. Here's a link of a picture of how far I got: http://s7.directupload.net/images/130605/jmaqbvrg.jpg Unfortunately I have no Idea what the next step is
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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experimentX
  • experimentX
with what center you want to expand?
anonymous
  • anonymous
I think it was x0=1
anonymous
  • anonymous
|dw:1370458428449:dw|

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experimentX
  • experimentX
with center x=1, you can do this \[ \frac{1}{x^2} = \frac{1}{(1 - (1-x))^2 } = 1 - 2 (1-x) + (-2)(-3)/2! (1-x)^2 + .. \] expand binomially
experimentX
  • experimentX
to generalize it \[ \frac{1}{x^2} = \sum_{n=0}^\infty (-1)^n n(1-x)^n\] should have a radius of convergence of 1
experimentX
  • experimentX
woops!! looks like i made mistake |dw:1370459251343:dw|
experimentX
  • experimentX
to generalize it \[ \frac{1}{x^2} = \sum_{n=0}^\infty (-1)^n n(x-1 )^n \]
anonymous
  • anonymous
thx a lot !
experimentX
  • experimentX
yw
anonymous
  • anonymous
do you have any idea how to solve my attempt (in the picture at the first post) now that I have a solution it's not important, but I'm curious if it is solvable via cauchy product :)
experimentX
  • experimentX
yes you will get a Cauchy product of coefficients which is always one and sums to n
experimentX
  • experimentX
\[ (\sum_{n=0}^\infty (1-x)^n)^2 = \sum_{n=0}^\infty \sum_{k=0}^n a_{k}b_{n-k} (1-x0^n\]
experimentX
  • experimentX
put a_k = 1 and b_k = 1
experimentX
  • experimentX
|dw:1370460071471:dw| sorry on my earlier part \[ \frac{1}{x^2} = \sum_{n=0}^\infty (-1)^n (n+1)(x-1 )^n \] |dw:1370460197821:dw|
experimentX
  • experimentX
\[ (\sum_{n=0}^\infty (1-x)^n)^2 = \sum_{n=0}^\infty \sum_{k=0}^n a_{k}b_{n-k} (1-x)^n \] putting \(a_k = b_k = 1\) and summing \( k = 0 \to n \) you get \( n + 1 \) as index begins from 0
anonymous
  • anonymous
thx again mate :) \[\sum_{n=0}^{\infty} \sum_{k=0}^{n} (1-x)^n = \sum_{n=0}^{\infty} (n+1)*(1-x)^n=\sum_{n=0}^{\infty} (n+1)*(-1)^n*(-1+x)^n\]
experimentX
  • experimentX
yes .. you are welcome :) sorry for being ignorant and making mistakes.
anonymous
  • anonymous
no need to apologize for it mate have a nice day/night!
experimentX
  • experimentX
night!! have a good day/night!!

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