# What is $e^{\frac d{dx}}$?

At the end of this video, 3blue1brown suggests it is possible to take $$e^{\frac d{dx}}$$.

So, what does $$e^{\frac d{dx}}$$ equal?

• It's a translation operator on the space of smooth functions. Apr 16, 2021 at 12:18
• Related question I asked some time ago: Differential operators with arbitrary functions? Due to the association with 3b1b, I added the popular-math tag; feel free to remove if you don't think it fits Apr 16, 2021 at 12:28

(Note on notation: we will be counting rows and columns from $$0$$, i.e. the first row will be row $$0$$, and the cell on row $$i$$ and column $$j$$ of a matrix $$A$$ will be denoted $$a_{ij}$$. Also, instead of writing $$\frac d{dx}$$ we will write $$D$$.)

In this answer we will be using the monomial basis as it is the basis used by 3blue1brown in an earlier video and also the simplest basis.

$$e^M=\sum^\infty_{n=0}\frac{M^n}{n!}$$

If we want to find $$e^D$$, we need to find $$D^n$$ for natural $$D$$.

Suppose

$$D^n=\left(\begin{matrix}\vec{v_0} & \vec{v_1} & \vec{v_2} & \vec{v_3} & \cdots\end{matrix}\right)$$

Then $$D^n$$ maps $$x^k$$ to the polynomial represented by $$\vec{v_k}$$.

Now, for natural $$n,k$$, we claim $$D^n x^k=\frac{k!}{(k-n)!}x^{k-n}$$.

The proof goes as follows:

When $$n=0$$, $$D^nx^k=x^k=\frac{k!}{(k-0)!}x^{k-0}=\frac{k!}{(k-n)!}x^{k-n}$$.

Suppose the formula is correct for $$n=m$$. Now consider the case $$n=m+1$$.

$$D^{m+1}x^k=D(D^mx^k)=D\left(\frac{k!}{(k-m)!}x^{k-m}\right)=\frac{k!}{(k-m)!}D\left(x^{k-m}\right)=\frac{k!}{(k-m)!}D\left(x^{k-m}\right)\frac{k!}{(k-m)!}(k-m)x^{k-m-1}=\frac{k!}{(k-m-1)!}x^{k-m-1}=\frac{k!}{(k-n)!}x^{k-n}$$

Hence, the column vector $$\vec{v_k}$$ of $$D^n$$ has all of its entries equal to $$0$$, except for when $$n-k\geq0$$, in which case the $$n-k$$th row entry is equal to $$\frac{k!}{(k-n)!}$$ instead.

Thus, if an entry of $$D^n$$ is at the $$j$$th column, it equals $$\frac{j!}{(j-n)!}$$ if $$i=j-n$$ and $$0$$ otherwise, i.e. $$D^n_{ij}=\delta_{i+n,j}\frac{j!}{(j-n)!}$$

$$\frac{D^n_{ij}}{n!}=\delta_{i+n,j}\frac{j!}{(j-n)!n!}=\delta_{i+n,j}\binom{j}{j-n}=\delta_{i+n,j}\binom{j}{i}$$

Where $$\delta$$ is the Kronecker Delta.

$$\left(e^D\right)_{ij}$$ (i.e. the cell on row $$i$$ and $$j$$ of $$e^D$$) is the sum of all $$\frac{D^n_{ij}}{n!}$$ from $$n=0$$ to $$n=\infty$$

$$\left(e^D\right)_{ij}=\sum^\infty_{n=0}\delta_{i+n,j}\binom{j}{i}$$

When $$i\leq j$$ then there exists exactly one natural $$n$$ such that $$i+n=j$$, hence $$\left(e^D\right)_{ij}=\binom{j}{i}$$ for $$i\leq j$$.

When $$i>j$$ then there does not exist such an $$n$$, hence $$\left(e^D\right)_{ij}=\sum^\infty_{n=0}0=\binom{j}{i}$$.

Regardless,

$$\left(e^D\right)_{ij}=\binom{j}{i}$$

This implies

$$e^D=\left(\begin{matrix} 1 & 1 & 1 & 1 & 1 & \cdots \\ 0 & 1 & 2 & 3 & 4 & \cdots \\ 0 & 0 & 1 & 3 & 6 & \cdots \\ 0 & 0 & 0 & 1 & 4 & \cdots \\ 0 & 0 & 0 & 0 & 1 & \cdots \\ \vdots & \vdots & \vdots & \vdots & \vdots & \ddots \end{matrix}\right)$$

• $$e^D$$ looks rather similar to a transposed version of Pascal's triangle
• $$e^D$$ maps $$x^j$$ to $$\sum^j_{i=0}\binom{j}{i}x^i$$, which is the binomial expansion of $$(x+1)^j$$. This means $$e^D[P(x)]=P(x+1)$$ for a polynomial $$P$$. This property has been noticed in the past.

Edit: Grey has pointed out that for any analytical function $$f$$, we note $$e^{aD}[f(x)]=f(x+a)$$.

• Note that for any real or complex $a$, $$e^{a\frac{d}{{dx}}} = \sum\limits_{n = 0}^\infty {\frac{{a^n }}{{n!}}\frac{{d^n }}{{dx^n }}} ,$$ thus by the Taylor formula $e^{a\frac{d}{{dx}}} f(x) = f(x + a)$ for functions analytic at $a$.
– Gary
Apr 16, 2021 at 8:18
• @Gary The Taylor series does not necessarily converge for all continuous functions, so that method doesn't give a result that is quite as general.
– Kyky
Apr 16, 2021 at 8:24
• Yes, but note that you are interchanging the limit operation and the $e^D$ operator, when extending your result to continuous functions. Is that justified?
– Gary
Apr 16, 2021 at 8:35
• Sorry, can you clarify what you mean by "limit operation"? Not quite sure what you mean by that.
– Kyky
Apr 16, 2021 at 8:38
• If $f(x)$ is continuous on a compact interval then there is a sequence of polynomials $p_n(x)$ converging uniformly to $f(x)$. Then $$e^D f(x) = e^D \mathop {\lim }\limits_{n \to + \infty } p_n (x) = \mathop {\lim }\limits_{n \to + \infty } e^D p_n (x) = \mathop {\lim }\limits_{n \to + \infty } p_n (x + 1) = f(x + 1)$$ provided $p_n$ converges to $f$ also at $x+1$ and that the limit and $e^D$ operators are interchangeable. The fact that the latter can be done is not clear to me.
– Gary
Apr 16, 2021 at 8:43