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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:

$1-x+x^2+\dfrac{5x^3}{6}+\dfrac{5x^4}{8}...$

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    $\begingroup$ Any nice way to find the Taylor series would be a nice way to find the Bell numbers, and conversely. There are several that don't call for computing the derivatives. So the best you can do is look at the known ways to do that and hope to find one that suits your need. Wikipedia is pretty good on Bell numbers. $\endgroup$ – Ethan Bolker Nov 15 '19 at 2:53
  • $\begingroup$ @ Ethan Bolker: I wish to find a way to expand this functions. I have tried the derivative way and there is $e$ involved. I don't want that. $\endgroup$ – James Warthington Nov 15 '19 at 3:01
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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*}

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    $\begingroup$ +1 I was suggesting this in a comment when your answer appeared. $\endgroup$ – Ethan Bolker Nov 15 '19 at 3:07
  • $\begingroup$ So you substitute the whole expression into $e^x-1$, how do you simplify them to get to the final answer? $\endgroup$ – James Warthington Nov 15 '19 at 3:10
  • $\begingroup$ It is \begin{eqnarray*} e^X = 1+X+\frac{X^2}{2}+ \frac{X^3}{6}+\cdots. \end{eqnarray*} with $X$ replaced with $x+x^2/2+ \cdots$. $\endgroup$ – Donald Splutterwit Nov 15 '19 at 3:13
  • $\begingroup$ @Donald Splutterwit: How do you simply them to get to the final answer $1+1x+\frac{2x^2}{2}+\frac{5x^3}{3!}...$ $\endgroup$ – James Warthington Nov 15 '19 at 3:25
  • $\begingroup$ expand all the brackets and collect the $x^k$ terms. $\endgroup$ – Donald Splutterwit Nov 15 '19 at 3:31
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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 $$e^{f(x)}=e^{f(x_0)}\left(1+x f'(0)+\frac{1}{2} x^2 \left(f''(0)+f'(0)^2\right)+O\left(x^3\right)\right)$$ $$\log(f(x))=\log (f(0))+x\frac{ f'(0)}{f(0)}+\frac{1}{2} x^2 \left(\frac{f''(0)}{f(0)}-\frac{f'(0)^2}{f(0)^2}\right)+O\left(x^3\right)$$

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