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In a physics problem I'm currently considering generating functions containing the term $1/\sqrt{t^2}$, where earlier in the derivation I have restricted my attention to the cases $\lvert xt\rvert<\frac12$ (by expanding a geometric series).

Suppose I have produced \begin{equation} G(x,t)=\frac{\sqrt{(1-2xt)^2}}{2(1-2xt)\sqrt{t^2}}\exp\left(\frac{(1-2xt)^2}{4t^2}\right)\Gamma\left(\frac12,\frac{(1-2xt)^2}{4t^2}\right) \end{equation} Finding the $n$'th coefficient requires taking the $n$'th derivative with respect to $t$, dividing by $n!$, and setting $t=0$.

What are the considerations for treating these derivatives of the absolute value in a generating function in the distributional sense, and what does the existence of $\sqrt{t^2}$ in $G$ mean for what the coefficients encode? For what purpose may I argue that its derivative is a sign function, it's second derivative a delta distribution etc.? And if we may argue that it is "legal" to call this a $\delta$ distribution: what is the meaning of that, since I will not be considering any integrals?

Derivation:

\begin{align} \exp\left(-\frac{\partial_x^2}4\right)\left(\frac1{1-2xt}\right) &= \exp\left(-\partial_x^2/4\right)\left(1-2xt\right)^{-1}\\ &= \sum_{m\geq0}\left(-\frac14\right)^m \frac{(2t)^{2m}}{\left(1-2xt\right)^{1+2m}}\frac{(2m)!}{m!}\\ &= \frac1{1-2xt}\sum_{m\geq0}(-)^m(2m-1)!! \left(\frac{\left(1-2xt\right)^{2}}{2t^{2}}\right)^{-m}\\ &= \frac1{1-2xt}\frac{\exp\left(\frac{(1-2xt)^2}{4t^2}\right)\Gamma\left(\frac12,\frac{(1-2xt)^2}{4t^2}\right)}{\sqrt{2\frac{2t^2}{(1-2xt)^2}}} \end{align} but also \begin{align} \exp\left(-\frac{\partial_x^2}4\right)\sum_{n\geq0} (2xt)^n &= \sum_{n\geq0}2^n t^n\sum_{m\leq n/2} \left(-1\right)^m\left(\frac12\right)^{2m}\frac{\Gamma(1+n)}{m!\Gamma(1+n-2m)}(x)^{n-2m}\\ &=\sum_{n\geq0}t^nH_n(x) \end{align}

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    $\begingroup$ There is too much soup of words. You better pose the original problem, instead of asking a "modelled" version, when you don't understand how it should be modelled. Taking coefficients of a generating function doesn't necessarily require taking derivatives. Algebraically this is the same as $\sum_{k=0}^{\infty}2^kx^kt^{k-1}$. See XY question. $\endgroup$ Commented Oct 9 at 17:30
  • $\begingroup$ You should include the exact function and its derivation (and in a hurry, before your question gets closed). $\endgroup$ Commented Oct 9 at 17:34
  • $\begingroup$ Thank you -- equation is updated now. $\endgroup$ Commented Oct 9 at 17:37
  • $\begingroup$ I must be missing something, but isnt ${\sqrt t^ 2}=|t|$? $\endgroup$ Commented Oct 9 at 21:24
  • $\begingroup$ @user619894 Yes precisely, which is not infinitely differentiable at $t=0$. So what are my options to extract a generic coefficient $[t^n]$ from the series without fully computing? $\endgroup$ Commented Oct 10 at 10:12

1 Answer 1

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For

$$G(x, t)=\frac{\sqrt{(1-2 t x)^2}}{2 \sqrt{t^2} (1-2 t x)}\, \exp\left(\frac{(1-2 t x)^2}{4 t^2}\right)\, \Gamma\left(\frac{1}{2},\frac{(1-2 x t)^2}{4 t^2}\right)\tag{1}$$

assuming $x\in\mathbb{R}$, then $\lim\limits_{t\to 0}\, G(x,t)=1$, and I believe you can compute $\underset{t\to 0}{\text{lim}}\frac{\partial ^nG(x,t)}{\partial t^n}$ for all $n$.

Using this approach Mathematica produces the following table of values for $a(n)$ where $G(x,t)=\sum\limits_{n=0}^\infty a(n)\, t^n$.

$$\begin{array}{cc} n & a(n) \\ 0 & 1 \\ 1 & 2 x \\ 2 & 4 x^2-2 \\ 3 & 4 x \left(2 x^2-3\right) \\ 4 & 4 \left(4 x^4-12 x^2+3\right) \\ 5 & 8 x \left(4 x^4-20 x^2+15\right) \\ 6 & 8 \left(8 x^6-60 x^4+90 x^2-15\right) \\ 7 & 16 x \left(8 x^6-84 x^4+210 x^2-105\right) \\ 8 & 16 \left(16 x^8-224 x^6+840 x^4-840 x^2+105\right) \\ 9 & 32 x \left(16 x^8-288 x^6+1512 x^4-2520 x^2+945\right) \\ 10 & 32 \left(32 x^{10}-720 x^8+5040 x^6-12600 x^4+9450 x^2-945\right) \\ \end{array}$$

As noted in a comment below, the values of $a(n)$ in the table above correspond to the Hermite Polynomials $H_n(x)$, so in fact one has

$$G(x, t)=\lim\limits_{N\to\infty} \left(\sum\limits_{n=0}^N H_n(x)\, t^n\right)\tag{2}.$$

Figure (1) below illustrates formula (2) for $G(x, t)$ above in orange overlaid on the reference function $G(x, t)$ defined in formula (1) above in blue where formula (2) is evaluated at $N=20$ and both formulas are evaluated at $x=\frac{1}{4}$. Note Mathematica exhibits an anomaly near $t=0$ in the evaluation of formula (1) for $G(x, t)$ above illustrated in blue, but Mathematica's evaluation of formula (2) for $G(x,t)$ above illustrated in orange eliminates this anomaly.

Illustration of formulas (1) and (2) for G(1/4,t)

Figure (1): Illustration of formulas (1) and (2) for $G\left(\frac{1}{4},t\right)$ in blue and orange respectively

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  • $\begingroup$ Thank you for the values. They are Hermite polynomials, but the concern remains with regards to the higher derivatives of $1/\sqrt{x^2}$. For the final expression I will consider $G^k$ so I hope to find a way to still find the $n$'th coefficients by using derivatives (hence my original question about the higher derivatives of expressions like $1/\sqrt{x^2}$ and possibly also the "sign function"esque expression in 1-2xt. $\endgroup$ Commented Oct 9 at 18:59

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