# Evaluation of $\exp\left(a\frac{d^2}{dx^2}\right)f(x)$

I know that

\begin{align*} \exp\left(a\frac{d}{dx}\right)f(x)=f(x+a)\,, \end{align*}

by comparing the Taylor expansions of both sides ($f(x)$ is an arbitrary function). However, if I have, where $f(x)$ is still arbitrary,

\begin{align*} \exp\left(a\frac{d^2}{dx^2}\right)f(x)\,, \end{align*} is there also a simple form for this expression?

Expanding the exponential only results in the even derivatives of $f(x)$, so I am reasonably sure that this is not equal to something as simple as, for example, $f(x+a)^2$. I then realized that

\begin{align*} \exp\left(a\frac{d^2}{dx^2}\right)f(x)=\exp\left(a\frac{d}{dx}\right)^{\frac{d}{dx}}f(x)\,, \end{align*}

If I expand the exponential, I get

\begin{align*} \left(1+a\frac{d}{dx}+\frac{a^2}{2}\frac{d^2}{dx^2}+\cdots\right)^{\frac{d}{dx}}f(x)\,. \end{align*}

However, now I am not sure how to treat this. I was thinking that this might require fractional calculus, but I am unsure how to proceed.

• It's possible it doesn't also have a simple form.
– anon
Jul 26, 2016 at 22:46
• What does it even mean to raise an expression to the power of an operator? Jul 26, 2016 at 22:52
• @ZachEffman If you're talking about the original ones with $\exp()$, those are defined with power series. If you're talking about the one at the end, maybe binomial series?
– anon
Jul 26, 2016 at 22:56
• @arctictern Some links of that would be of interest. Jul 26, 2016 at 22:58
• As for the need of fractional calculus, fractional calculus deals with $\frac{d^{1/2}}{dx^{1/2}}$ and similar things, which I don't see here. Jul 26, 2016 at 23:03

Set $D=d/dx$. Whenever the series defining $\exp\left(a D^2 \right) f(x)$ converges, we have the identity

$$\exp\left(a D^2 \right) f(x) =\int_{-\infty}^\infty \frac{1}{\sqrt{4\pi a}}e^{-(x-y)^2/4a} \:f(y) \:dy$$

This integral transform is a scaled version of the Weierstrass transform.

Similarly we have

$$\exp\left(a D^3 \right) f(x) =\int_{-\infty}^\infty \frac{1}{\sqrt{3a}}\mathrm{Ai}(\frac{y-x}{\sqrt{3a}}) \:f(y) \:dy$$

where $\mathrm{Ai}$ is the Airy function, and in general

$$\exp\left(a D^n \right) f(x) =\int_{-\infty}^\infty \mathcal{K}_n(x-y) \:f(y) \:dy$$

where the convolution kernel $\mathcal{K}_n(x)$ is the inverse Fourier transform (non-unitary, angular frequency) of $e^{a(-ip)^n}$, which can be expressed in terms of generalized hypergeometric functions ${}_0F_{n-2}$. This also works when $n$ is not a natural number. See this related question for how to find these expressions in the general case of $\exp(f(x, D))$, borrowing a "trick" from quantum mechanics.

Note that the functions $\mathcal{K}_n$ are solutions of the differential equation

$$an\frac{d^{n-1}\mathcal{K}_n(x)}{dx^{n-1}} + x\mathcal{K}_n(x) = 0$$

Indeed for $n=3$ this is a scaled Airy differential equation, for $n=2$ this gives a Gaussian, and even for $n=1$, the delta function $\delta(x+a)$ can be considered a "solution" of $xK_1(x)=-aK_1(x)$. This can be proven pretty straightforwardly from the expression of $\mathcal{K}_n$ as a Fourier transform.

• Great answer! How did you get your equation for $\exp(aD^n)f(x)$? Are there alternative methods to finding out what $\exp(aD^n)f(x)$ does beyond solving a Schrodinger-like equation? Jan 22, 2017 at 20:40
• @user85503 Thank you! In the case of $\exp(aD^n)$ we actually don't need the full Schrödinger-like formalism since only $D$ appears in the exponential. If we Fourier transform the starting expression, since $D$ turns into $-ip$ we get the product $\exp(a(-ip)^n) \tilde{f}(p)$, and then doing the inverse Fourier transform we get the convolution $\mathcal{F}^{-1}[\exp(a(-ip)^n)] \ast f(x)$, which is the integral above. I don't think there's a simpler method in the general case, but the Wikipedia article for Weierstrass transform has an alternative derivation for $n=2$ under the section I linked. Jan 22, 2017 at 21:04