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I am trying to represent the function, $f(x) = \frac{x^3}{(2-x)^3}$ as a power series and saw this on a different post.

$ {1\over (1-u)} =\sum_{n=0}^{\infty}u^{n}, \qquad|u|<1, \tag1 $

$-{1 \over (1-u)^2} =\sum_{n=1}^{\infty}nu^{n-1}, \qquad|u|<1,\tag2$

${2 \over (1-u)^3} =\sum_{n=2}^{\infty}n(n-1)u^{n-2}, \qquad|u|<1. \tag3$

I understand up to this part. I began with the general form

$f(x) = \frac{1}{(1-x)}$

and derived it twice to get:

$f''(x) = \frac{2}{(1-x)^3}$

which somewhat resembles the function I need to represent as a power series. However, this is as far as I got and keep hitting a dead-end.

1) $f(x)=\frac{x^3}{(2-x)^3} = [x(\frac{1}{2-x})]^3 = [ \frac{x}{2}(\frac{1}{1-x/2})] ^3, $

2) $\frac{x^3}{2^3} (\frac{1}{1-x/2})^3 = \frac{x^3}{2^3}\sum_{n=2}^{\infty}n(n-1)(x/2)^{n-2}, $

3)Therefore, $f(x) = \frac{x^3}{(2-x)^3}$ represented as a power series is:

$\sum_{n=2}^{\infty}n(n-1)(x/2)^{n+1}, \qquad|x|<2 $

I have no solution sheet for this but I don't know if I am right or wrong. I also have a different answer from what I saw other people post online so if I am wrong, please tell me why.

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    $\begingroup$ I think what you have done is fine. You can also write the final answer as $ \sum\limits_{n=2}^{\infty} 2^{-n} \binom {n} {2} x^{n+1}$. $\endgroup$ – Kavi Rama Murthy Jan 24 '20 at 9:36
  • $\begingroup$ i forgot to carry the 2 over from the double derivative to my original function.. was i supposed to multiply by an extra two into my series? $\endgroup$ – swordlordswamplord Jan 24 '20 at 9:41
  • $\begingroup$ Strart with $x=2t$ to make life easier. $\endgroup$ – Claude Leibovici Jan 24 '20 at 10:08
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$$f(x)=\frac{x^3}{(2-x)^3}=\frac{x^3}{8}(1-\frac{x}{2})^{-3}=\frac{x^3}{8}[1-{-3 \choose 1}\frac{x}{2}+{-3 \choose 2} (\frac{x}{2})^2 -{-3 \choose 3}(\frac{x}{2})^3$$ $$+{-3 \choose k} (\frac{x}{2})^4 .....] ~~~~(1)$$ Using $${-p \choose k}=\frac{-p(-p-1)(-p-2)....(-p-k+1)}{k!},$$ we find that $${-3 \choose 1}=-3, {-3 \choose 2}=-3.-4/2=6, {-3 \choose 3}=-3.-4.-5/6 =-10, {-3\choose 4} =-3.-4.-5.-6/24 = 15....$$ So on and so forth. Inserting these co-efficients in (1) we gwt a serie which is valid for $|x|<2$.

When $|x|>1$, then $$f(x)=-(1-\frac{2}{x})^{-3}=-\left [1-{-3\choose 1}(\frac{2}{x})+{-3 \choose 2} (\frac{2}{x})^2-{-3 \choose 3} (\frac{2}{x})^3+..\right] $$

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