What does $e^{\cos(D)}$ do?

Question

We know for example, $$e^D f(x) = f(x+1)$$ and for example, $$e^{-D^2} x^n = H_n(x/2)$$ But what if the exponent is also an exponential, like $$e^{-\cos(D)}$$ or $$e^{-e^D}\ \ \ ?$$ Do we at least know of any solutions?

Answer 1

In general $f(D)$ creates a linear operator of this type:
$\displaystyle f(D)=\sum_{n=0}^{\infty}\dfrac{f^{(n)}(0)}{n!}D^n$
Since $e^{-\cos(x)}$ is an even function I tried to use it with trigonometric function (since their derivatives are "periodic") and I saw that:
$e^{-\cos(D)}\sin(\alpha x)=e^{-\cosh(\alpha)}\sin(\alpha x)$
and
$e^{-\cos(D)}\cos(\alpha x)=e^{-\cosh(\alpha)}\cos(\alpha x)$
So technically the operator $e^{-\cos(D)}$ reduces the amplitude of the waves.