I've recently come across the following identity: $$\frac{1}{\sqrt{n!m!}}\bigg(\frac{\mathrm{d}}{\mathrm{d}Z^{\ast}}\bigg)^{m}\big(Z^{\ast}\big)^{n}\bigg\vert_{Z^{\ast}\to 0}=\delta_{n,m}\;.$$ Here is a link to a book referencing it.

I would like to derive this identity, but I'm getting a bit stuck doing so (basically, I can't get the right product of factorials). Here's what I've got so far: $$\bigg(\frac{\mathrm{d}}{\mathrm{d}Z^{\ast}}\bigg)^{m}\big(Z^{\ast}\big)^{n}= n(n-1)(n-2)\cdots (n-m+1)\big(Z^{\ast}\big)^{n-m} \\ = \frac{n!}{(n-m)!}\big(Z^{\ast}\big)^{n-m}\;.\quad\;\,\,\,$$ Maybe I'm being stupid and there's is something trivial I'm missing here? Any help would be much appreciated.

Edit: I understand how the Kronecker delta arises, i.e., the right-hand side will only be non-zero in the case where $n=m$, that is, when there are no powers of $Z^{\ast}$ remaining. I just don't understand how the product of factorials arises?

  • $\begingroup$ You can just do it by cases if you like: consider what happens if 1) $m < n$, 2) $m = n$, 3) $m > n$. Remember that we are substituting $Z^* = 0$ ultimately! (Basically, in case 1), you still have powers of $Z^*$ remaining, and in case 3), you have differentiated so much that the function has become $0$ already, so in either of these cases we should get $0$ for the derivative when $Z^* = 0$.) $\endgroup$ – Minus One-Twelfth Mar 1 at 14:53
  • $\begingroup$ @MinusOne-Twelfth Thanks. I see how the Kronecker delta arises - the right-hand side will only be non-zero in the case where there are no powers of $Z^{\ast}$ remaining. I just don't see where the product of factorials comes from? $\endgroup$ – Will Mar 1 at 15:11
  • $\begingroup$ @kimchilover Ah yes, you're right. I'll fix that now. $\endgroup$ – Will Mar 1 at 15:56

If $m>n$, you've differentiated so often the function is identically $0$. If $m<n$, the function is proportional to a positive power of $Z$, so vanishes at $Z=0$. If $m=n$, the claim is trivial.

  • $\begingroup$ Thanks for your answer. I get the kronecker delta part of the claim, I just don't see where the factorials arise from? $\endgroup$ – Will Mar 1 at 16:19
  • $\begingroup$ @Will You can rearrange the $m=n$ case as the claim that $\frac{d}{dZ^n}Z^n$ is $n!$ at $Z=0$. In fact, it is at any $Z$. You can prove it by induction, since $\frac{d}{dZ^k}Z^k=k!\implies\frac{d}{dZ^{k+1}}Z^{k+1}=(k+1)\frac{d}{dZ^k}Z^k=(k+1)!$. $\endgroup$ – J.G. Mar 1 at 17:48
  • $\begingroup$ Why is it split up into a product of to factorials though, i.e. why is it $\sqrt{n!m!}$ rather than just $n!$? $\endgroup$ – Will Mar 1 at 18:27
  • $\begingroup$ @Will Either answer is right because we only get a non-zero answer when $m=n$, but whoever stated the theorem probably just wanted to use an $m\leftrightarrow n$-symmetric function as the coefficient. $\endgroup$ – J.G. Mar 1 at 18:36
  • $\begingroup$ Ah, fair enough. So it’s just for convenience then. $\endgroup$ – Will Mar 1 at 19:57

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