Raising a differential operator to a power

Question

I've been reading about the shift operator $E=e^{\frac{\mathrm{d}}{\mathrm{d}x}}$, which can be represented as

$$e^{\frac{\mathrm{d}}{\mathrm{d}x}} = I + \frac{\mathrm{d}}{\mathrm{d}x} + \frac{1}{2!} \frac{\mathrm{d}^2}{\mathrm{d}x^2} + \frac{1}{3!} \frac{\mathrm{d}^3}{\mathrm{d}x^3} + \ldots$$

My understanding of this Taylor series is that the latter differential operators in this series come from the $x^k$ factors of the summation terms, $\left( \frac{\mathrm{d}}{\mathrm{d}x} \right)^0, \left( \frac{\mathrm{d}}{\mathrm{d}x} \right)^1, \left( \frac{\mathrm{d}}{\mathrm{d}x} \right)^2,\ldots$. If, generally, $\left(\frac{\mathrm{d}f}{\mathrm{d}x}\right)^2 \ne \frac{\mathrm{d}^2f}{\mathrm{d}x^2}$, how does raising the differential operator to some power, $k$ yield the $k$th differential operator, as asserted here? That is, why does $\left(\frac{\mathrm{d}}{\mathrm{d}x}\right)^k = \frac{\mathrm{d}^k}{\mathrm{d}x^k}$

Answer 1

$\left(\frac d{dx}\right)^k = \frac{d^k}{dx^k}$ is the definition of $\left(\frac d{dx}\right)^k$. In other words, the mathematical community at large has, collectively, come to the conclusion that among all possible interpretations of $\left(\frac d{dx}\right)^k$, $\frac{d^k}{dx^k}$ makes the most sense. So that is what we have decided that it means.

And to me it doesn't seem strange at all that $\left(\frac{d}{dx}\right)^kf^k\neq \left(\frac{df}{dx}\right)^k$. Multiplication isn't always commutative, and $\frac{d}{dx}$ and $f$ is just another example of things that do not commute.

Answer 2

The key here is that you're exponentiating the operator itself as opposed to the operator applied to some given function.

Answer 3

The "multiplication" in the definition of $e^{\frac{d}{dx}}$ is composition, not pointwise multiplication. This is more or less standard notation.

One possible source of intuition for this is to look at the advection equation $\frac{\partial u}{\partial t}-\frac{\partial u}{\partial x}=0$ on the whole line, with initial condition $u(0,x)=u_0(x)$. You can write this as $\frac{\partial u}{\partial t}-\frac{\partial}{\partial x} u=0$. Now if you treat the operator $\frac{\partial}{\partial x}$ like a constant with respect to $t$, then this suggests applying $e^{-t \frac{\partial}{\partial x}}$ (whatever that means) to both sides to obtain

$$\frac{\partial}{\partial t} \left ( e^{-t \frac{\partial}{\partial x}} u \right ) = 0.$$

Thus for any fixed $x$, $\left ( e^{-t \frac{\partial}{\partial x}} u \right )(t,x)$ is constant. On the other hand we also know how to solve the advection equation without this formalism, obtaining $u(t,x)=u_0(x+t)$. Thus if we identify these two, we get that $e^{t \frac{\partial}{\partial x}} u_0(x) = u_0(x+t)$, which is what you're hoping to get out of Taylor expansion.

The exact same notation can be used to formally write down an asymptotic expansion of the solution of the heat equation.