Finding the Power Series Representation for x/(1-x)^2

[chyeaman]

hey, this is my first time posting, my question is find the power series representation for x/(1-x)^2
I know the representation for 1/1-x is x^n so does that mean x/(1-x)^2 is x^n^2? could use some clarification please

 

[jeffreydk]

Just use the Taylor series to expand it about whatever point [itex]a[/itex];

[tex]\sum_{n=0}^{n=\infty}\frac{f^{(n)}(a)}{n!}(x-a)^n[/tex]

Looks like for [itex]\frac{x}{(1-x)^2}[/itex] this it will be something like

[tex]x+2x^2+3x^3+4x^4+\ldots+[/tex]

if you center it on 0.

 

[adriank]

How about this: You know that
[tex]\frac{1}{1 - x} = \sum_{n = 0}^{\infty} x^n = 1 + x + x^2 + x^3 + x^4 + \dotsb.[/tex]
Take the derivative of both sides to obtain
[tex]\frac{1}{(1 - x)^2} = \sum_{n = 0}^{\infty} (n - 1) x^n = 1 + 2x + 3x^2 + 4x^3 + \dotsb,[/tex]
and multiply by x to get
[tex]\frac{x}{(1 - x)^2} = \sum_{n = 1}^{\infty} n x^n = x + 2x^2 + 3x^3 + 4x^4 + \dotsb.[/tex]

 

[chyeaman]

Thank you so much for clarifying!