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:
$1-x+x^2+\dfrac{5x^3}{6}+\dfrac{5x^4}{8}...$
On
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 $$e^{f(x)}=e^{f(x_0)}\left(1+x f'(0)+\frac{1}{2} x^2 \left(f''(0)+f'(0)^2\right)+O\left(x^3\right)\right)$$ $$\log(f(x))=\log (f(0))+x\frac{ f'(0)}{f(0)}+\frac{1}{2} x^2 \left(\frac{f''(0)}{f(0)}-\frac{f'(0)^2}{f(0)^2}\right)+O\left(x^3\right)$$
On
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:
This result has been generalized and applied in the following papers:
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*}