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

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Discussion Overview

The discussion revolves around finding the power series representation for the function x/(1-x)^2. Participants explore different methods for deriving this representation, including Taylor series expansion and manipulation of known series.

Discussion Character

  • Exploratory, Technical explanation, Mathematical reasoning

Main Points Raised

  • One participant inquires about the power series representation for x/(1-x)^2, suggesting a possible confusion with the representation for 1/(1-x).
  • Another participant suggests using the Taylor series expansion centered at a point a, providing a general formula for the series.
  • A different participant explains that by taking the derivative of the series for 1/(1-x), one can derive the series for 1/(1-x)^2, and then multiply by x to obtain the desired series for x/(1-x)^2.
  • A later reply expresses gratitude for the clarification provided regarding the series representation.

Areas of Agreement / Disagreement

Participants do not explicitly agree or disagree on a single method but present different approaches to derive the power series representation. The discussion remains open with multiple perspectives on how to achieve the result.

Contextual Notes

Some assumptions about the convergence of the series and the choice of the center for the Taylor series expansion are not explicitly stated, which may affect the applicability of the methods discussed.

chyeaman
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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
 
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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.
 
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]
 
Thank you so much for clarifying!
 

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