Strategy for expanding the function $e^{e^x-1}$ into Maclaurin series I am trying to find a way to expand the function $e^{e^x-1}$ into Maclaurin series but I don't know how to. I don't want to use the cumbersome Maclaurin formula of taking successive higher derivative.
Do you know how to nicely expand this function into Taylor series. I know that this is the generating function of Bell numbers.
Wolfram gives the expansion as:
$1-x+x^2+\dfrac{5x^3}{6}+\dfrac{5x^4}{8}...$
 A: The exponent is
\begin{eqnarray*}
e^x-1 = x+\frac{x^2}{2}+ \frac{x^3}{6}+\cdots.
\end{eqnarray*}
Now just plug this in the power series for the exponential function
\begin{eqnarray*}
\operatorname{exp}(e^x-1) = 1&+& \left(x+ \frac{x^2}{2}+ \frac{x^3}{6}+\cdots \right) \\
&+& \frac{1}{2} \left(x+ \frac{x^2}{2}+ \frac{x^3}{6}+\cdots \right)^2 \\
&+& \frac{1}{6} \left(x+ \frac{x^2}{2}+ \frac{x^3}{6}+\cdots \right)^3 \\
&+& \cdots  \\
\end{eqnarray*}
Expand to whatever order you need ...
\begin{eqnarray*}
\operatorname{exp}(e^x-1) = \color{red}{1}+\color{red}{1}x+ \frac{\color{red}{2}x^2}{2}+ \frac{\color{red}{5}x^3}{3!} +\cdots
\end{eqnarray*}
But of course ... as Ethan Bolker says in the comments the Bell numbers can be calculated much more easily using
\begin{eqnarray*}
B_{n+1} = \sum_{k=0}^{n}  \binom{n}{k} B_k.
\end{eqnarray*}
A: The composition of Taylor series for function such as $e^{f(x)}$ or $\log(f(x))$ requires patience first.
The trick I use is to first expand $f(x)$
$$f(x)=f(0)+x f'(0)+\frac{1}{2} x^2 f''(0)+O\left(x^3\right)$$ and continue to get
$$e^{f(x)}=e^{f(x_0)}\left(1+x f'(0)+\frac{1}{2} x^2 \left(f''(0)+f'(0)^2\right)+O\left(x^3\right)\right)$$
$$\log(f(x))=\log (f(0))+x\frac{ f'(0)}{f(0)}+\frac{1}{2} x^2
   \left(\frac{f''(0)}{f(0)}-\frac{f'(0)^2}{f(0)^2}\right)+O\left(x^3\right)$$
A: For $n\in\mathbb{N}$, the $n$th derivative of the function $e^{e^x}$ can be computed by
\begin{equation}\label{Bell-funct-derivative-eq}
\frac{\textrm{d}^ne^{e^x}}{\textrm{d} x^n}=e^{e^x}\sum_{k=1}^{n}S(n,k)e^{kx},
\end{equation}
where
\begin{equation*}
S(n,k)=\frac1{k!}\sum_{\ell=1}^{k}(-1)^{k-\ell}\binom{k}{\ell}\ell^n
\end{equation*}
for $n\ge k\ge1$ are the Stirling numbers of the second kind.
There are several proofs for this conclusion in the following paper:

*

*Feng Qi, Some inequalities for the Bell numbers, Proceedings of the Indian Academy of Sciences--Mathematical Sciences 127 (2017), no. 4, 551--564; available online at https://doi.org/10.1007/s12044-017-0355-2.

This result has been generalized and applied in the following papers:


*Feng Qi, Some inequalities and an application of exponential polynomials, Mathematical Inequalities & Applications 23 (2020), no. 1, 123--135; available online at https://doi.org/10.7153/mia-2020-23-10.

*Feng Qi, Da-Wei Niu, Dongkyu Lim, and Bai-Ni Guo, Some properties and an application of multivariate exponential polynomials, Mathematical Methods in the Applied Sciences 43 (2020), no. 6, 2967--2983; available online at https://doi.org/10.1002/mma.6095.

*Feng Qi, Integral representations for multivariate logarithmic polynomials, Journal of Computational and Applied Mathematics 336 (2018), 54--62; available online at https://doi.org/10.1016/j.cam.2017.11.047.

*Feng Qi, On multivariate logarithmic polynomials and their properties, Indagationes Mathematicae 29 (2018), no. 5, 1179--1192; available online at https://doi.org/10.1016/j.indag.2018.04.002.

