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Hi could anyone help me answer this question

Find the power series representation for the function and determine the radius of convergence

$f(x)=\frac{x^2}{\left(1-2x\right)^2}$

After getting 1/(1-2x)^2 I do not know how to convert it to a power series.

yswong
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  • Can you do it if the $x^2$ wasn't there? If so just drop it out, then once you have the power series for $\frac{1}{(1 - 2x)^2}$, just multiply the whole thing by $x^2$ (i.e. just add $2$ to each $x$-term's exponent). – Jared Mar 28 '14 at 05:04
  • How do I then represent 1/(1-2x)^2 as a power series? – yswong Mar 28 '14 at 05:07
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    Can you integrate $\frac{1}{(1 - 2x)^2}$? See what that gives you. – Jared Mar 28 '14 at 05:10
  • Here is a very similar question: http://math.stackexchange.com/questions/728299/power-series-representation-at-a-given-value-a/728307#728307 – Jared Mar 28 '14 at 05:11
  • That gave me (1/2)(1/(1-2X)) – yswong Mar 28 '14 at 05:11
  • $\frac{1}{1 - 2x}$ should look very familiar. – Jared Mar 28 '14 at 05:12
  • Oh thanks so that would be summation of (2x)^r – yswong Mar 28 '14 at 05:15
  • yes, that is correct. First, think about how you can get back to $\frac{1}{(1 - 2x)^2}$ and second, don't forget about that $\frac{1}{2}$! – Jared Mar 28 '14 at 05:17
  • oh and don't forget about the $x^2$ at the end either! – Jared Mar 28 '14 at 05:20
  • And btw, you can use inline latex by surrounding it with $'s. You already are using basic latex syntax, for instance x^2, just place in inside $'s: $x^2$ = $x^2$. For fractions you use \frac{}{} where the first braces is the numerator and the second the denominator, for instance $\frac{x^2}{(1 - 2x)^2}$ = $\frac{x^2}{(1 - 2x)^2}$. – Jared Mar 28 '14 at 05:23

2 Answers2

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Some starting material: Using that $\sum_{n=0} ^\infty x^n = \frac{1}{1-x}$, we can write $$ f(x) = \frac{x^2}{1 - 4x + 4x^2} = x^2 \cdot \frac{1}{1 - (4x - 4x^2)} = x^2 \sum_{n=0} ^\infty 4^n x^n(1-x)^n = \sum_{n=0} ^\infty 4^n x^{n+2}(1-x)^n. $$ I imagine you can take it from here.

izœc
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    it looks a bit delicate to me to get to the result from your start... – mookid Mar 28 '14 at 05:18
  • @mookid At first I was not sure what you meant - then I realized that this was supposed to be a power series! No idea what I was thinking. Good point, I guess my answer turned out to be not particularly helpful. – izœc Mar 28 '14 at 05:25
  • you can always expand the sum and swap both sums, but you then have a tricky sum to compute exactly. It may not be the end of the world... – mookid Mar 28 '14 at 05:29
  • Hmm.... I started along that route. I am sure it is possible, however, I am not seeing it.. I reduced it all the way, but I don't see how to move it beyond this: $$ f(x) = \sum_{n=0} ^\infty \sum_{k=0} ^n \binom{n}{k} 4^n (-1)^k x^{k + n + 2} .$$ I think that for now I will accede that my answer was a little rash.. – izœc Mar 28 '14 at 05:44
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You have $$ |z|<1\implies \frac {1}{1-z} = \sum_{n=0}^\infty z^n;\\ \frac d{dx}\frac {1}{1-x} = \frac {1}{(1-x)^2}; |x|<1\implies \frac {1}{(1-x)^2} = \sum_{n=0}^\infty (n+1)x^n.\\ $$ Now make the substitution $x\to 2x$: $$|x|<\frac 12\implies \frac {1}{(1-2x)^2} = \sum_{n=0}^\infty (n+1)2^nx^n\\ \implies \frac {x^2}{(1-2x)^2} = \sum_{n=0}^\infty (n+1)2^nx^{n+2}. $$

Then you see that for $x=1/2$, $(n+1)2^nx^{n+2}\nrightarrow 0$: the radius is $1/2$.

mookid
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