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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?

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  • $\begingroup$ 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

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(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 $e^{aD}[f(x)]=f(x+a)$.

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    $\begingroup$ 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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