Power Series Representation of a Function when a is a polynomial

[AceK]

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)=\stackrel{(1+x)}{(1-x)^{2}}

Homework Equations

a series converges when |x|<1;
\stackrel{a}{1-r}=\suma(r)^{n}

The Attempt at a Solution

I got the power series of \sum(1+x)(2x-x^2)^{n} 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 \sum(2n+1)x^{n}, which seems more correct; also, the radius of convergence is much easier to determine with their answer. How do they arrive at this answer?

 

[vela]

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).

 

[AceK]

I see. So I reworked the problem by differentiating f(x) with the numerator factored out, and the answer I got was (1+x)*\Sigma(-nx^{n-1}which is equal to \sum(-nx^{n-1}-nx^{n}, 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?

 

[vela]

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)*\Sigma(-nx^{n-1}which is equal to \sum(-nx^{n-1}-nx^{n}, 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.

 

[AceK]

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.