5

Given that

$$\exp(aD)f(x)=f(x+a)$$

where $\exp(D)$ is the exponential of the differential operator $D$, is there a similar closed-form, general expression for $\exp(g(D))f(x)$, where $g(D)$ is a polynomial function of $D$?

user76284
  • 5,967
  • 4
    Note that all operators of the form $h(D)$ commute with one another. In particular, all $\exp(a_m D^m)$ are mutually commuting. So it should suffice to find $\exp(a_m D^m)f(x)$. – Semiclassical Aug 19 '14 at 01:14
  • Let me attempt a guess here. The differential operator $aD$ acts on $x$ by $$ x \mapsto x+aDx = x+a.$$ Infinitesimally, $aD$ generates a translation. What does $g(D)$ generate infinitesimally? $$ x \mapsto x+g(D)x $$ To be explicit, let $g(D)=g_o+g_1D+ \cdots g_{n-1}D^{n-1}+ D^n$. Note $D(Dx)=D(1)=0$ hence $g(D)x$ truncates to $g(D)x = g_ox+g_1$. Thus, $$ x \mapsto x+g_ox+g_1 =(1+g_o)x+g_1. $$ So, I would conjecture (wildly) that $exp(g(D))f(x) = f((1+g_o)x+g_1)$ which of course reduces to Taylor's theorem in the case $g_o=0$. – James S. Cook Aug 19 '14 at 03:33
  • @JamesS.Cook: unless I'm misinterpreting the meaning of $\exp$ here it seems that $g(D) = D^2$, $f(x) = x^2$ is a counterexample to your conjecture: $e^{t D^2} f(x) = f(x) + t f''(x) = x^2 + 2t \ne x^2$. – Anthony Carapetis Aug 19 '14 at 05:16
  • @AnthonyCarapetis Thanks, guess my conjecture is dead. We'll just have to calculate it as Semiclassical indicated. – James S. Cook Aug 19 '14 at 05:42

1 Answers1

1

In general if $a$ is a very nice function, not necessarily a pollinomial, then $$ a(D)f(x) = \int _{-\infty} ^\infty a(\xi) \hat f (\xi) e ^{2 \pi i \xi x } d\xi$$ Sadly there is no closed form for most $a$, that formula is a consequence of the fact that the Fourier transform satisfies $FD=FQ$ where $Q f(x)= x f(x)$.

There was a post with some nice examples I'm going to look for it

k76u4vkweek547v7
  • 1,314
  • Some examples http://math.stackexchange.com/questions/116633/is-there-a-formula-similar-to-fxa-ea-fracddxfx-to-express-f-al/126744#126744 – k76u4vkweek547v7 Jan 30 '16 at 22:23