Power Series Representation of a Function when a is a polynomial

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Power Series Representation of a Function when "r" is a polynomial

Homework Statement


Find a power series representation for the function and determine the radius of convergence.
f(x)=[tex]\stackrel{(1+x)}{(1-x)^{2}}[/tex]


Homework Equations


a series converges when |x|<1;
[tex]\stackrel{a}{1-r}[/tex]=[tex]\sum[/tex]a(r)[tex]^{n}[/tex]

The Attempt at a Solution


I got the power series of [tex]\sum[/tex](1+x)(2x-x^2)[tex]^{n}[/tex] by expanding the denominator of the function and getting it into geometric series form; I then converted it to a series using the above equation. The answer given in my book is [tex]\sum(2n+1)x^{n}[/tex], which seems more correct; also, the radius of convergence is much easier to determine with their answer. How do they arrive at this answer?
 
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Your answer isn't a power series. You'd need to multiply everything out and collect terms so that you end up with each term in the series being a number multiplied by a single power of x.

To find the series of the function in the problem, start with the series for 1/(1-x) and note that 1/(1-x)2 is the derivative of 1/(1-x).
 


I see. So I reworked the problem by differentiating f(x) with the numerator factored out, and the answer I got was (1+x)*[tex]\Sigma(-nx^{n-1}[/tex]which is equal to [tex]\sum(-nx^{n-1}-nx^{n}[/tex], which is still drastically different to the answer given in the book. Is my answer right but reducible, or have I made a mathematical error?
 


AceK said:
I see. So I reworked the problem by differentiating f(x) with the numerator factored out
What?
and the answer I got was (1+x)*[tex]\Sigma(-nx^{n-1}[/tex]which is equal to [tex]\sum(-nx^{n-1}-nx^{n}[/tex], which is still drastically different to the answer given in the book. Is my answer right but reducible, or have I made a mathematical error?
You have to collect terms. Try working out the first four or five terms explicitly and see if you can spot a pattern.
 


Wow, I've got it now. I was making that way more difficult than it needed to be. Thanks so much for your help, you really elucidated the problem for me.