How to find the first four terms of the maclaurin series of $f(x) = e^{e^x}$ I tried substituting the maclaurin series for $e^x$ into the equation, but that doesn’t seem to give the right answer.
I’m not sure what do besides just taking the derivatives.
 A: We know
$$
e^x = 1+x+\frac{x^2}{2}+\frac{x^3}{6}+\dots
$$
so then
$$\begin{align}
\exp(e^x) &= \exp\left(1+x+\frac{x^2}{2}+\frac{x^3}{6}+\dots\right)
\\ &= e^1 e^x e^{x^2/2} e^{x^3/6}\dots
\\
&= e \left(1+x+\frac{x^2}{2}+\frac{x^3}{6}+\dots\right)
\left(1+\frac{x^2}{2}+\frac{x^4}{8}+\dots\right)
\left(1+\frac{x^3}{6}+\dots\right)\cdots
\\ &=
e\left(1+x+x^2+\frac{5}{6} x^3 + \dots\right)
\end{align}$$
A: By Taylor with the substitution $x\leftrightarrow e^x$,
$$e^{e^x}=\sum_{n=0}^\infty \frac{(e^{x})^n}{n!}=\sum_{n=0}^\infty \frac{e^{nx}}{n!}=\sum_{n=0}^\infty \frac{\displaystyle\sum_{m=0}^\infty \dfrac{(nx)^m}{m!}}{n!}=\sum_{n=0}^\infty\sum_{m=0}^\infty\dfrac{n^mx^m}{n!m!}=\sum_{m=0}^\infty\left(\sum_{n=0}^\infty\dfrac{n^m}{n!}\right)\frac{x^m}{m!}.$$
Then we can evaluate the inner sums as follows:
$$\sum_{n=0}^\infty\dfrac{n^0}{n!}=\sum_{n=0}^\infty\dfrac1{n!}=e,$$
$$\sum_{n=0}^\infty\dfrac{n^1}{n!}=\sum_{n=0}^\infty\dfrac1{(n-1)!}=e,$$
$$\sum_{n=0}^\infty\dfrac{n^2}{n!}=\sum_{n=0}^\infty\dfrac{n-1+1}{(n-1)!}=\sum_{n=0}^\infty\dfrac1{(n-2)!}+\sum_{n=0}^\infty\dfrac1{(n-1)!}=2e,$$
$$\sum_{n=0}^\infty\dfrac{n^3}{n!}=\sum_{n=0}^\infty\dfrac{(n-2)(n-1)+3(n-1)+1}{(n-1)!}
\\=\sum_{n=0}^\infty\dfrac1{(n-3)!}+3\sum_{n=0}^\infty\dfrac1{(n-2)!}+\sum_{n=0}^\infty\dfrac1{(n-1)!}=5e.$$
Note that the coefficients so computed are the Bell numbers https://en.wikipedia.org/wiki/Bell_number and
$$e^{e^x}=e\sum_{m=0}^\infty B_m\frac{x^m}{m!}.$$
A: Let $f(x)=e^x-1$. Consider the first $4$ terms of the MacLaurin series of $f$:$$x+\frac12x^2+\frac16x^3.\tag1$$Now replace each $x$ in $(1)$ by $(1)$ itself:$$x+\frac12x^2+\frac16x^3+\frac12\left(x+\frac12x^2+\frac16x^3\right)^2+\frac16\left(x+\frac12x^2+\frac16x^3\right)^3.\tag2$$Now, in $(2)$ take only those terms whose degree is smaller than $4$:$$x+x^2+\frac56x^3.$$These are the first terms of the MacLaurin series of $f$. So, since\begin{align}e^{e^x}&=e\times e^{-1+e^x}\\&=e\times e^{f(x)}\\&=e\times f\bigl(f(x)\bigr)+e,\end{align}we have that the first terms of the MacLaurin series of $e^{e^x}$ are$$e+ex+ex^2+\frac56ex^3.$$
A: Hint:
$f(x)=e^{e^x}\\f'(x)=f(x)e^x\\f''(x)=(f'(x)+f(x))e^x=f(x)(e^x+e^{2x})\\f'''(x)=(f(x)+2f'(x)+f''(x))e^x=f(x)(e^x+3e^{2x}+e^{3x})$
