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From Pirani's lectures on General Relativity I got the following identity which he asks the reader to prove by induction which is easy

$$\partial_{\alpha_1} \partial_{\alpha_2} ... \partial_{\alpha_n} \frac{1}{r}= (-1)^n \frac{(2n)!}{2^n n!} \frac{x^{\alpha_1} x^{\alpha_2} ... x^{\alpha_n}}{r^{2n+1}}+ \text{kronecker delta terms}$$

However, I am not entirely satisfied by the proof by induction since any induction proof assumes knowledge of the given expression. The expression in question is often found out by trial and error method, which seems like a formidable task for this particular identity given its complexity (or I am just dumb).

Is there a systematic proof of this identity which does not use induction or if not how can we guess the expression after computing it for a few small values of $n$ or via some ansatz. I am guessing there must be an elegant method to arrive at this formula "without guessing" as the coefficient $\frac{(2n)!}{2^n n!}$ seems familiar from somewhere: probably Rodrigues formula for Legendre polynomials.

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  • $\begingroup$ I have not tried to write out details, but assuming $r$ is Euclidean distance to the origin, it looks promising to write $1/r$ as the composition of the one-variable function $f(u) = 1/\sqrt{u}$ and the quadratic polynomial $u = (x^1)^2 + \cdots + (x^n)^2$ (whose mixed partials all vanish), and repeatedly apply the chain and product rules. $\endgroup$
    – Andrew D. Hwang
    Commented May 13 at 14:42
  • $\begingroup$ This becomes very tedious soon enough, and the general form of the coefficients is very hard to glean even after doing 4 derivatives. $\endgroup$
    – Sanjana
    Commented May 13 at 15:28
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    $\begingroup$ For reference, the exercise is at the bottom of page 285 and is written as shown. That said, it would be helpful to include the explicit version given on the previous page: $$\partial_{\alpha_1} \partial_{\alpha_2} \ldots \partial_{\alpha_N} \frac{1}{r}=\left[\frac{\partial^N}{\partial x^{\alpha_1}\partial x^{\alpha_2}\cdots \partial x^{\alpha_N}}\frac{1}{|\vec{x}-\vec{x}'|}\right]_{\vec{x}'=0}$$ $\endgroup$
    – Semiclassical
    Commented May 13 at 15:47
  • $\begingroup$ Also, the coefficient $(2n)!/(2^n n!)$ is known as the (odd) double factorial $(2n-1)!!$. $\endgroup$
    – Semiclassical
    Commented May 13 at 16:01

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Wikipedia provides the following multivariate version of Faà di Bruno 's formula:

$${\partial^{n} \over \partial x_{1}\cdots \partial x_{n}} f(y)=\sum_{\pi \in \Pi }f^{(\left|\pi \right|)}(y)\cdot \prod_{B\in \pi } {\partial^{\left|B\right|}y \over \prod _{j\in B}\partial x_{j}}$$

where

  • $\pi$ runs through the set $\Pi$ of all partitions of $\{1,\ldots,n\}$,

  • $B\in \pi$ runs through the list of all "blocks" of the partition $\pi$, and

  • $|A|$ denotes the cardinality of the relevant set

In this case, we have $f(y)=1/r=y^{-1/2}$ and $y=r^2=x_1^2+\cdots +x_n^2$. If I interpret the formula correctly, the term of interest here is when $\pi$ consists of $n$ blocks each containing one element. (The Kronecker delta terms occur when some of these blocks aref merged.) This yields

$$f^{(n)}(y)\cdot \prod_{k=1}^n \frac{\partial y}{\partial x_k}$$

The second term immediately simplifies to $2^n x_1\cdots x_n$. For the first term, we have \begin{align} \frac{\partial^n}{\partial y^n}y^{-1/2} &=-\frac{1}{2}\frac{\partial^{n-1}}{\partial y^{n-1}}y^{-3/2}\\ &=\cdots\\ &=-\frac{1}{2}\cdot -\frac{3}{2}\ldots\cdot -\left(n-\frac12\right) y^{-1/2-n}\\ &=\left(-\frac{1}{2}\right)^n \frac{(2n-1)!!}{r^{2n+1}} \end{align} Hence together this yields $$(-1)^n (2n-1)!!\frac{x_1\cdots x_n}{r^{2n+1}}.$$ But $$\frac{(2n)!}{2^n n!}=\frac{(2n)(2n-1)(2n-2)\cdots(2)(1)}{(2n)(2n-2)\cdots 2}=(2n-1)(2n-3)\cdots (1)=(2n-1)!!$$ so this form is equivalent to the one given.

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