How Does the Sum of Alternating Series Lead to a Power Series for 2/(1-x^2)?

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Homework Help Overview

The discussion revolves around finding a power series representation for the function f(x) = 2/(1-x^2) centered at c = 0, and determining its interval of convergence. Participants are exploring the relationship between alternating series and power series.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to manipulate the series into a usable form but expresses confusion about the transition to the solution shown in the manual. Some participants suggest writing out the first few terms to clarify the pattern, while others note the importance of even indices in the series.

Discussion Status

Participants are actively engaging with the problem, offering insights about the nature of the series and the conditions under which terms contribute to the final result. There is a recognition of the need for clearer explanations in the source material, indicating a productive exploration of the topic.

Contextual Notes

Some participants express frustration with the clarity of the solution manual, suggesting that the algebraic steps may not be adequately detailed. There is an emphasis on understanding the behavior of the series terms based on their indices.

kdinser
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Find a power series for the function centered at c and determine the interval of convergence.
c = 0

[tex]f(x)=\frac{2}{1-x^2}[/tex]

After some partial fractions work and getting the partials in the form of

[tex]\frac{a}{1-r}[/tex]

I have

[tex]\sum x^n + \sum(-x)^n[/tex]

if I factor out the x^n's I get

[tex]\sum(1+(-1)^n)x^n[/tex]

This is where I'm stuck, the solution manual shows it then going to

[tex]\sum2x^{2n}[/tex]

I've been staring at this thing for 15 mins and can't see how it's possible. Could someone give me a little push in the right direction with this? Thanks.
 
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write out th efirst few terms and you will see the answer, n = 0, 1+1 = 2; n = 1, (1 + (-1))x = 0, etc...,
 
easy...

n has to be even so that the coefficient wouldn't be zero. Then you have 1+1 = 2 and the exponent is written as 2n to make sure it is even...



marlon
 
Thanks guys, I wish the book or the manual would make it a little clearer when they do stuff like this. The way it's written seems to imply that there's some algebra going on to get to the final result.
 

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