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 $\left(\frac{\mathrm{d}}{\mathrm{d}x}\right)^k = \frac{\mathrm{d}^k}{\mathrm{d}x^k}$

  • $\begingroup$ Are you asking about why $\left(\frac d{dx}\right)^k = \frac{d^k}{dx^k}$? $\endgroup$ – Arthur Mar 20 at 14:11
  • $\begingroup$ Yes. Will edit to clarify. $\endgroup$ – mnosefish Mar 20 at 14:13

$\left(\frac d{dx}\right)^k = \frac{d^k}{dx^k}$ 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.

  • $\begingroup$ How does the Taylor series hold up if the operations behind $x^k$ and $\mathrm{d}^k/\mathrm{d}x^k$ aren't mathematically equivalent? This just seems like abuse of notation. Is it because $x^k$ is a composition of the multiplication operator? $\endgroup$ – mnosefish Mar 20 at 14:22
  • $\begingroup$ What do you mean "aren't mathematically equivalent"? If we write $x^3f(x)$, we mean $xxxf(x)$ and if we write $\left(\frac d{dx}\right)^3f(x)$ we mean $\frac{d}{dx}\frac{d}{dx}\frac{d}{dx}f(x)$. Seems perfectly equivalent to me. Note that the real notation abuse is the "exponents" in $\frac{d^k}{dx^k}$. One could even argue that in my answer, which term is defined in terms of which is actually reversed from what I say. But it is so much more common to see $\frac{d^k}{dx^k}$. $\endgroup$ – Arthur Mar 20 at 14:24
  • $\begingroup$ Not mathematically equivalent in that multiplying something by another something isn't the same as differentiating something. Would we expect the Taylor series to work out if we substituted $x + x + x + f(x)$ for $x^3 f(x)$. Is the above definition of the shift operator simply a matter of how we've defined the differential operator? $\endgroup$ – mnosefish Mar 20 at 14:48
  • $\begingroup$ Not mathematically equivalent in that sense, no. But mathematically equivalent in the sense that an exponent of $k$ means "take $k$ copies of it, place them next to one another, and perform whatever implicit operation is expected to be between them". $\endgroup$ – Arthur Mar 20 at 14:49

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.


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

  • $\begingroup$ Is there a proof for, or even an intuition behind why, raising the differential operator itself to a power is the same as composition? $\endgroup$ – mnosefish Mar 20 at 14:10
  • 1
    $\begingroup$ @mnosefish, "product" of operators = composition is a definition. $\endgroup$ – Martín-Blas Pérez Pinilla Mar 20 at 14:15
  • $\begingroup$ It's just by definition. $\endgroup$ – Gary Moon Mar 20 at 14:15
  • $\begingroup$ When we plug $\frac{d}{dx}$ into a power series we get a bunch of terms like $\left(\frac{d}{dx}\right)^k$. In order to make sense of the power series, we must define what is meant by $\left(\frac{d}{dx}\right)^k$. Since the product of operators is generally defined to be composition, we define $\left(\frac{d}{dx}\right)^k = \frac{d^k}{dx^k}.$ What do you mean by not mathematically equivalent? $\endgroup$ – Gary Moon Mar 20 at 14:27

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