# Power Series Representation of $f(x) = \frac{x^3}{(2-x)^3}$

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.

• I think what you have done is fine. You can also write the final answer as $\sum\limits_{n=2}^{\infty} 2^{-n} \binom {n} {2} x^{n+1}$. – Kavi Rama Murthy Jan 24 '20 at 9:36
• 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? – swordlordswamplord Jan 24 '20 at 9:41
• Strart with $x=2t$ to make life easier. – Claude Leibovici Jan 24 '20 at 10:08

$$f(x)=\frac{x^3}{(2-x)^3}=\frac{x^3}{8}(1-\frac{x}{2})^{-3}=\frac{x^3}{8}[1-{-3 \choose 1}\frac{x}{2}+{-3 \choose 2} (\frac{x}{2})^2 -{-3 \choose 3}(\frac{x}{2})^3$$ $$+{-3 \choose k} (\frac{x}{2})^4 .....] ~~~~(1)$$ 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 $$f(x)=-(1-\frac{2}{x})^{-3}=-\left [1-{-3\choose 1}(\frac{2}{x})+{-3 \choose 2} (\frac{2}{x})^2-{-3 \choose 3} (\frac{2}{x})^3+..\right]$$