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.

  • 1
    $\begingroup$ 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

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.