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?
 A: (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
$$e^D=\left(\begin{matrix}
1 & 1 & 1 & 1 & 1 & \cdots \\
0 & 1 & 2 & 3 & 4 & \cdots \\
0 & 0 & 1 & 3 & 6 & \cdots \\
0 & 0 & 0 & 1 & 4 & \cdots \\
0 & 0 & 0 & 0 & 1 & \cdots \\
\vdots & \vdots & \vdots & \vdots & \vdots & \ddots
\end{matrix}\right)$$
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 $e^{aD}[f(x)]=f(x+a)$.
