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What are the coefficients in the expansion of the following expression at x=0: (1 + x)^(-2) and how are they calculated?

Thank you for help.

 Oct 16, 2018
edited by Guest  Oct 16, 2018
 #1
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The most straightforward way to do this is to just do the division.

It's hard to show the process of long division on here.

 

\(\dfrac{1}{(1+x)^2}=\dfrac{1}{1+2x+x^2}\)

 

and just set this up as a usual long division, if you go through it a few steps you'll see it's

 

\(\dfrac{1}{1+2x+x^2} = 1 - 2x + 3x^2 - 4x^3 + 5x^4 \dots + (-1)^k (k+1)x^k,~\forall k \geq 0\)

 

hrm.. there might be another way to approach it that's pretty straightforward

 

\(\sum \limits_{k=0}^\infty ~x^k = \dfrac{1}{1-x}\)

 

\(\dfrac{1}{1+x}=\dfrac{1}{1-(-x)}\)

 

\(\dfrac{1}{1-(-x)} = \sum \limits_{k=0}^\infty ~(-1)^k x^k\)

 

\(\dfrac{d}{dx}~\dfrac{1}{1+x} = - \dfrac{1}{(1+x)^2}\)

 

\(\dfrac{1}{(1+x)^2} = -\dfrac{d}{dx} \left(\sum \limits_{k=0}^\infty~(-1)^k x^k\right) = \\ -\left(\sum \limits_{k=1}^\infty~(-1)^k k x^{k-1}\right) = \sum \limits_{k=0}^\infty~(-1)^k(k+1)x^k\)

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 Oct 16, 2018
 #2
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Thank you very much Rom. The 2nd method is very easy to follow. Thanks again.

 Oct 16, 2018

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