This is a curiosity I when looking at the binomial theorem. Say you have an ordinary generating function $\displaystyle\sum_{n,m\geq 0}\binom{n}{m}x^ny^m$. This looks kind of like $\displaystyle\sum_{i=0}^n\binom{n}{i}x^iy^{n-i}=(x+y)^n$.

Is there likewise a nice expression for the first generating function?

  • $\begingroup$ Yep. Write it as an outer sum over $n$ together with an inner sum over $m$. $\endgroup$ – Qiaochu Yuan Dec 14 '11 at 23:41
  • $\begingroup$ How are you defining ${n\choose m}$ for $m>n$? $\endgroup$ – Chris Taylor Dec 14 '11 at 23:41
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    $\begingroup$ @Chris: the only reasonable definition is that it be equal to $0$. $\endgroup$ – Qiaochu Yuan Dec 14 '11 at 23:42

Of course, $$ \sum_{n,m \ge 0} \binom{n}{m} x^n y^m = \sum_{n=0}^\infty x^n \sum_{m=0}^n \binom{n}{m} y^m = \sum_{n=0}^\infty x^n (1+y)^n = \frac{1}{1-x(1+y)} $$

  • $\begingroup$ Oh, that's very nice. Thank you. $\endgroup$ – VPL Dec 14 '11 at 23:52
  • $\begingroup$ doesnt this over $m$ too much? ie $m=0,m=0,1,m=0,1,2,$ etc $\endgroup$ – yoyo Dec 14 '11 at 23:53

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