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?


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

  • $\begingroup$ 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
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    $\begingroup$ The example looks OK to me. $\endgroup$ – Gerry Myerson Apr 24 '13 at 22:49

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 $$\left(\frac{1}{1-x}\right)^{n} = \delta_{x}^{n-1}\left(\frac{1}{1-x}\right)= \frac{1}{(n-1)!}\sum_{k=0}^{\infty} \delta_{x}^{n-1}(x^{k}) = \sum_{m=0}^{\infty} \frac{(m+n-1)!}{(n-1)! m!} x^{m} =\sum_{m=0}^{\infty} \binom{m+n-1}{n-1} x^{m} $$


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