3
$\begingroup$

I was just hoping to confirm that the following manipulations make sense:

Say I begin with $\frac{1}{(1-x)^n}$. Then we have $(1-x)^{-n} = $$\sum$ $-n\choose k$ $(-x)^k$ = $\sum$ $(-1)^k$ $n+k-1 \choose k$ $(-x)^k$, where our sum runs over k = $0$ to $\infty$. Does this make sense?

ADDED:

Verifying my example: $1/(1-3x)^2$ gives $\sum (n+1)3^nx^n$ is that correct?

$\endgroup$
  • $\begingroup$ The sum runs over $k$ from zero to infinity. $\endgroup$ – Gerry Myerson Apr 24 '13 at 12:58
  • $\begingroup$ Right, I'll make that edit. $\endgroup$ – user73041 Apr 24 '13 at 12:59
  • $\begingroup$ You should have $(-x)^k$ in the sum instead of $(x)^k$, otherwise it looks good. $\endgroup$ – vadim123 Apr 24 '13 at 13:07
  • $\begingroup$ Got it, in which case we get $(-1)^k$ $(-1)^k$ in front, which is why we needn't worry about them at all in this case, but in the case of $\frac {1}{(1+x)^n}$ we do have to, right? $\endgroup$ – user73041 Apr 24 '13 at 13:10
  • 1
    $\begingroup$ The example looks OK to me. $\endgroup$ – Gerry Myerson Apr 24 '13 at 22:49
1
$\begingroup$

The easiest way to establish the power series expansion is to use that differentation and taking powers just differ by a factor: $$\delta_{x} (1-x)^{-n} = n! (1-x)^{-(n+1)}$$

For example $\left(\frac{1}{1-x}\right)^{\prime} = 2 \left(\frac{1}{1-x}\right)^{2}$ and so on. So $$\left(\frac{1}{1-x}\right)^{n} = \delta_{x}^{n-1}\left(\frac{1}{1-x}\right)= \frac{1}{(n-1)!}\sum_{k=0}^{\infty} \delta_{x}^{n-1}(x^{k}) = \sum_{m=0}^{\infty} \frac{(m+n-1)!}{(n-1)! m!} x^{m} =\sum_{m=0}^{\infty} \binom{m+n-1}{n-1} x^{m} $$

$\endgroup$

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.