2
$\begingroup$

There are two equations showed in Gel'fand and Shilov's book (Generalized Functions I Properties and Operations) on page 183 and 185:

$${\delta}^{(k-1)}(1-x^2)=\frac{(-1)^{k-1}}{2^kx^{k-1}}[{\delta}^{(k-1)}(x-1)-{\delta}^{(k-1)}(x+1)];\tag{1}$$

$${\delta}^{(k)}(f(x))=\sum_n \frac{1}{|f'(x_n)|}(\frac{1}{f'(x)}\frac{d}{dx})^k{\delta}(x-x_n),\tag{2}$$ where $x_n$ are $n$ simple roots of $f(x).$

I think the first equation should satisfy the second equation. But it seems not. Why? And I would like to know the right equation.

$\endgroup$
0
$\begingroup$

TL;DR: Formula (2) is correct, which can easily be checked$^1$ by using test functions. However, formula (1) is incorrect.

We calculate$^2$ $$ u_k(x)~:=~(-1)^k\delta^{(k)}(1\!-\!x^2) ~=~\delta^{(k)}(x^2\!-\!1)~=~\left.\delta^{(k)}(y) \right|_{y=x^2-1}$$ $$~=~\left.\left(\frac{d}{dy}\right)^k\delta(y)\right|_{y=x^2-1} ~=~\left(\frac{1}{2x}\frac{d}{dx}\right)^k\delta(x^2\!-\!1) ~=~\left(\frac{1}{2x}\frac{d}{dx}\right)^k\frac{1}{2}\sum_{\pm}\delta(x\!\mp\!1). \tag{A}$$

By anti-normal-ordering (=ordering derivatives to the left) of eq. (A) we get for the first few $k\in\mathbb{N}_0$:

$$\begin{align}u_{k=0}(x)&~=~\frac{1}{2}\sum_{\pm}\delta(x\!\mp\!1),\cr u_{k=1}(x)&~=~\frac{1}{2x}\frac{d}{dx}\frac{1}{2}\sum_{\pm}\delta(x\!\mp\!1)\cr &~=~\left(\frac{d}{dx}\frac{1}{2x}+\frac{1}{2x^2}\right)\frac{1}{2}\sum_{\pm}\delta(x\!\mp\!1)\cr &~=~\frac{1}{4}\sum_{\pm}\pm \delta^{\prime}(x\!\mp\!1)+\frac{1}{4}\sum_{\pm}\delta(x\!\mp\!1),\cr u_{k=2}(x)&~=~\left(\frac{1}{2x}\frac{d}{dx}\right)^2\frac{1}{2}\sum_{\pm}\delta(x\!\mp\!1)\cr &~=~\left(\frac{d^2}{dx^2}\frac{1}{4x^2}+\frac{d}{dx}\frac{3}{4x^3}+\frac{3}{4x^4}\right)\frac{1}{2}\sum_{\pm}\delta(x\!\mp\!1)\cr &~=~\frac{1}{8}\sum_{\pm}\delta^{\prime\prime}(x\!\mp\!1)+\frac{3}{8}\sum_{\pm}\pm \delta^{\prime}(x\!\mp\!1)+\frac{3}{8}\sum_{\pm}\delta(x\!\mp\!1),\cr &~~~\vdots\end{align}\tag{B}$$

By normal-ordering (=ordering derivatives to the right) of eq. (A) we get for the first few $k\in\mathbb{N}_0$:

$$\begin{align}u_{k=0}(x)&~=~\frac{1}{2}\sum_{\pm}\delta(x\!\mp\!1),\cr u_{k=1}(x)&~=~\frac{1}{2x}\frac{d}{dx}\frac{1}{2}\sum_{\pm}\delta(x\!\mp\!1)\cr &~=~\frac{1}{4x}\sum_{\pm}\delta^{\prime}(x\!\mp\!1),\cr u_{k=2}(x)&~=~\left(\frac{1}{2x}\frac{d}{dx}\right)^2\frac{1}{2}\sum_{\pm}\delta(x\!\mp\!1)\cr &~=~\left(\frac{1}{4x^2}\frac{d^2}{dx^2}-\frac{1}{4x^3}\frac{d}{dx}\right)\frac{1}{2}\sum_{\pm}\delta(x\!\mp\!1)\cr &~=~\frac{1}{8x^2}\sum_{\pm}\delta^{\prime\prime}(x\!\mp\!1) -\frac{1}{8x^3}\sum_{\pm} \delta^{\prime}(x\!\mp\!1),\cr &~~~\vdots\end{align}\tag{C}$$ which is different from eq. (1). It is interesting how anti-normal-ordering (B) and normal-ordering (C) generate very different-looking expressions for the same underlying mathematical distributions.

--

$^1$ In this answer, we make repeated use of the following distribution identities: $$\delta(f(x))~=~\sum_{i}^{f(x_i)=0}\frac{1}{|f'(x_i)|}\delta(x\!-\!x_i), \tag{D}$$ $$ \{f(x)-f(y)\}\delta(x\!-\!y)~=~0, \tag{E}$$ and derivatives thereof: $$ \left(\frac{d}{dx}\right)^k[\{f(x)-f(y)\}\delta(x\!-\!y)]~=~0. \tag{F}$$

$^2$ We have for convenience shifted OP's definition $k-1\to k$ and removed an over-all sign factor $(-1)^k$. This can of course easily be reinstalled.

$\endgroup$

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.