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?

  • $\begingroup$ It's a translation operator on the space of smooth functions. $\endgroup$ Commented Apr 16, 2021 at 12:18
  • $\begingroup$ 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 $\endgroup$ Commented Apr 16, 2021 at 12:28

1 Answer 1


(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.


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


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

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 $n=m+1$.


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


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 $n=\infty$


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



This implies

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

I'd like to make a few comments about this matrix:

  • $e^D$ 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 $e^{aD}[f(x)]=f(x+a)$.

  • 8
    $\begingroup$ 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 $a$. $\endgroup$
    – Gary
    Commented Apr 16, 2021 at 8:18
  • $\begingroup$ @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. $\endgroup$ Commented Apr 16, 2021 at 8:24
  • $\begingroup$ 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? $\endgroup$
    – Gary
    Commented Apr 16, 2021 at 8:35
  • $\begingroup$ Sorry, can you clarify what you mean by "limit operation"? Not quite sure what you mean by that. $\endgroup$ Commented Apr 16, 2021 at 8:38
  • $\begingroup$ 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. $\endgroup$
    – Gary
    Commented Apr 16, 2021 at 8:43

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