I have a homework question that is quite confusing of a concept to me. I am instructed to evaluate:
$$g(t) = e^{3\frac{d}{dt}}f(t)$$
$$f(t) = t^2$$
about t = 1. The way I would apply the taylor expansion is:
$$g(t) = \sum_{n=0}^\inf \frac{g^{(n)}(a)}{n!}(t-a)^n$$
where I would let $a = 1$, and then I would end up with some strange looking zeroth order term:
$$g^{(0)}(t) = e^{3\frac{d}{dt}}f(t)$$
and ever-stranger 1st order term...
$$g^{(1)}(t) = 3\frac{d[\frac{d}{dt}]}{dt}e^{3\frac{d}{dt}}f(t) + 2te^{3\frac{d}{dt}}$$
and so on. As I have no idea how the differential operator $\frac{d}{dt}$ behaves wrt a, how do I evaluate these? I know for a fact I'm doing something wrong, because Somehow the solutions guide indicates this evaluates to expand to:
$$\sum_{n=0}^{\inf}\frac{a^n}{n!}(\frac{d}{dt})^nf(t) = f(t+a)$$
I have no clue what happens here, so if anybody could clarify, I would be super appreciative (I have never seen a linear differential operator before, and the instructor did not really explain how they work, so any reading material you could recommend that takes me from a "thinking about differentials as infintesimally small change in one top quantity with respect to an infintesimally small change in another quantity (ie, $\frac{df(t)}{dt}$, which I can rationalize in my head as $\frac{\Delta f(t)}{\Delta t}$) to thinking about it as a linear operator (ie, $\frac{d}{dt}$ that can stand by itself without needing a function on top), I would be much appreciated! All the help I could find either assumed you knew it as the former or assumed you knew it as the latter, I couldn't find anything to bridge that gap, if that makes sense).
Taking the $0$th power of an operator to be the identity, we have $(3d/dt)^0f(t)=f(t).$ And $(3d/dt)^1f(t)=3df(t)/dt=3f'(t).$ And $(3d/dt)^2f(t)=(3d/dt)^1(3f'(t))=9f''(t).$ And $(3d/dt)^3f(t)=27f'''(t).$ Et cetera. So $$exp(3d/dt)f(t)=\sum_{j=0}^{\infty}j!^{-1}3(d/dt)^jf(t)=\sum_{j=0}^{\infty}j!^{-1}3^jf^{(j)}(t)$$ where $f^{(0)}(t)=f(t)$ and $f^{(j)}(t) $ is the $j$th derivative of $f(t)$ for $j>0.$
You should recognize this as the power series for $f(t+3).$
The symbol $3d/dt$ is an operator that sends functions to functions. It is customary, for an operator $J, $ to interpret $J^2(f)$ as $J(J(f)), $ and $(kJ)(f)$ as $k\cdot J(f)$ for constant $k$.