Sum of Bell Polynomials of the Second Kind

Question

A problem of interest that has come up for me recently is solving the following.

$$\frac{d^{n}}{dt^{n}}e^{g(t)}$$

There is a formula for a general $n$-th order derivative of a composition as shown above: http://mathworld.wolfram.com/FaadiBrunosFormula.html

In terms of the Bell Polynomials, we can write

$$\frac{d^{n}}{dt^{n}}e^{g(t)}=e^{g(t)}\sum_{k=0}^{n}B_{n,k}(g'(t),g''(t),\cdots)$$

And the Bell polynomials of the second kind are shown in the Wolfram link above. I am wondering if there is a closed-form solution for the sum of the series of Bell Polynomials.

Answer 1

$B_{n}$ is the complete exponential Bell polynomial.

$$\sum_{k=1}^n B_{n,k}(g'(t),g''(t),...)=B_{n}(g'(t),g''(t),...)$$

$$B_{0,0}=1,\ B_0=1$$

$$\frac{d^{n}}{dt^{n}}e^{g(t)}=e^{g(t)}B_{n}(g'(t),g''(t),...)$$

For certain $g$, there are general closed-form expressions for $B_{n,k}$ and $B_{n}$ and therefore for $\frac{d^n}{dt^n}e^{g(t)}$ - often with constants that are combinatorial numbers, e.g. Bell numbers, Stirling numbers, partition numbers.

Examples:

$$B_{n,k}(a^t)=S_{n,k}\ln(a)^n(a^t)^k,\ \ B_n(a^t)=\sum_{k=1}^nS_{n,k}\ln(a)^n(a^t)^k$$

$$\frac{d^n}{dt^n}e^{a^t}=e^{a^t}\sum_{k=1}^nS_{n,k}\ln(a)^n(a^t)^k$$

$S_{n,k}$: Stirling numbers of the second kind

$$B_{n,k}(\log_a(t))=s_{n,k}\ln(a)^{-k}t^{-n},\ \ B_n(\log_a(t))=\sum_{k=1}^ns_{n,k}\ln(a)^{-k}t^{-n}$$

$$\frac{d^n}{dt^n}e^{\log_a(t)}=e^{\log_a(t)}\sum_{k=1}^ns_{n,k}\ln(a)^{-k}t^{-n}$$

$s_{n,k}$: Stirling numbers of the first kind

In 2012, I wrote a still unpublished article "On partial Bell polynomials for the higher derivatives of composed functions".