Find $f^{(100)}(0) $ and $f^{(101)}(0) $ if $f(x)=xe^{\arctan{x}}$ $$f(x)=xe^{\arctan{x}}$$
Part of my solution
$$f^{(n)}(x)=\sum_{k=0}^{n}\binom{n}{k}x^{(k)}(e^{\arctan{x}})^{(n-k)}=x(e^{\arctan{x}})^{(n)}+(e^{\arctan{x}})^{(n-1)}$$
First term probably disappears because $x=0$ but i don't know what to do with second term.  
 A: See that
$$e^{\arctan(x)}=\sum_{n=0}^\infty\frac{(\arctan(x))^n}{n!}$$
And
$$\arctan(x)=\sum_{n=0}^\infty(-1)^n\frac{x^{2n+1}}{2n+1}$$
which gives
$$e^{\arctan(x)}=\sum_{n=0}^\infty\frac{\left(\sum_{k=0}^\infty(-1)^k\frac{x^{2k+1}}{2k+1}\right)^n}{n!}$$
Note that you only want the coefficient of $x^{100,101}$ (by Taylor's theorem), so we can safely ignore all things that result in larger exponents. (and the case $n=0$)
$$\sum_{n=1}^{101}\frac{\left(\sum_{k=0}^{50}(-1)^k\frac{x^{2k+1}}{2k+1}\right)^n}{n!}$$
I'm not exactly sure what you'd do from here, but it is somewhat hand calculate-able from here.
Note that since $101$ is prime, very few cases result in a $x^{101}$.  More cases arrive for $x^{100}$, but it is still within reasonable to calculate.
For $x^{101}$:
$$\left(\frac1{101}-\frac1{101!}\right)x^{101}$$
which is clear since no other combination of $n,k$ will produce $x^{101}$.
For $x^{100}$, the problem is much harder but not impossible.  The multinomial theorem provides light on the problem and lets it reduce down to fewer scenarios of $n\in\lbrace2,4,10,100\rbrace$ (using factorizations and the fact that $n$ can't be odd)

For $n=2$ (I only considered the coefficients of the $x^{100}$ term):
$$a_2=\frac{\frac1{1(99)}+\frac1{3(97)}+\frac1{5(95)}+\dots+\frac1{97(3)}+\frac1{99(1)}}{-2}=-\left(\sum_{p=0}^{24}\frac1{(2p+1)(100-(2p+1))}\right)$$
$$=\sum_{p=0}^{24}\frac1{4p^2-196p-101}$$

After some work, I arrive at the conclusion that for $n\in\lbrace4,10,100\rbrace$, the problem is too tedious.
A: $\frac{f^{(n)}(0)}{n!}$ is the nth coefficient of the series expansion of your function.
We have
$$f(x) = x\sum_{n=0}^{\infty}{\frac{\left(\arctan{x}\right)^n}{n!}}$$
and
$$\arctan{x} = \sum_{n=0}^{\infty}{\frac{(-1)^nx^{2n+1}}{2n+1}}$$
so
$$f(x) = x\sum_{n=0}^{\infty}{\frac{\left(\sum_{n=0}^{\infty}{\frac{(-1)^nx^{2n+1}}{2n+1}}\right)^n}{n!}}$$
to get the 100th coefficient we can look at the odd compositions of 99 (since we multiplied by $x$).
$$f^{(100)}(0) = 100! \left( \frac{(-1)^{49}}{1!} \cdot \frac{1}{99} 
 + \frac{(-1)^{48}}{3!}\sum_{a+b+c=48}{\frac{1}{(2a+1)(2b+1)(2c+1)}}
 \\+ \frac{(-1)^{47}}{5!}\sum_{a+b+c+d+e=47}{\frac{1}{(2a+1)(2b+1)(2c+1)(2d+1)(2e+1)}} + \dots + \frac{(-1)^1}{99!}\right)$$
I'm stuck here though. I hope someone knows how to simplify these sums.
A: Partial answer:
As you pointed out,
$$\begin{align}f^{[n]}(x)&=\sum_{i=0}^n{n\choose i}x^{[i]}\left(e^{\tan^{-1}(x)}\right)^{[n-i]}
\\
&=x\left(e^{\tan^{-1}(x)}\right)^{[n]}+n\left(e^{\tan^{-1}(x)}\right)^{[n-1]}
\end{align}$$
We set $g(x)=\left(e^{\tan^{-1}(x)}\right)$, $h(x)=\tan^{-1}(x)$ and consider $g^{[n]}(x)$.
$\begin{align}g'(x)&=\left(e^{h(x)}\right)'
\\
&=h'(x)g(x)
\\
g''&=h''g+h'g'
\\
&=h''g+h'h'g
\\
&=(h''+(h')^2)g
\\
g'''&=(h'''+2h'h'')g+(h''+(h')^2)h'g
\\
&=(h'''+3h'h''+(h')^3)g
\\
g^{[4]}&=(h^{[4]}+3(h''h''+h'h''')+3(h')^2h'')g+(h'''h'+3(h')^2h''+(h')^4)g
\\
&=(h^{[4]}+4(h')^2h''+3(h'')^2+4h'h'''+(h')^4)g
\\
g^{[5]}&=(h^{[5]}+4(2(h')h''h''+(h')^2h''')+6(h'')h'''+4(h''h'''+h'h^{[4]})+4(h')^3h'')g\dots
\\
&\quad\dots+(h'h^{[4]}+4(h')^3h''+3h'(h'')^2+4(h')^2h'''+(h')^5)g
\\
&=(h^{[5]}+8(h')^3h''+11h'(h'')^2+8(h')^2h'''+10h''h'''+5h'h^{[4]}+(h')^5)g
\end{align}$
Note that each bracketed term is a sum of the form:
$$\sum_{i=0}^{n}C_i\prod\left(h^{[m]}\right)^p$$
Where $C_i$ are integer constants and for each $i$, $\sum{pm}=n$. In fact, all sums of $pm$ equal to $n$ are represented as terms. For example, $3=(3\cdot1)=(1+2)=(1\cdot3)$ so there are terms with $(h')^3$, $h'h''$ and $h'''$. This is strikingly similar to the binomial series but I've yet to determine the constants. We would probably be able to use Faa di Bruno's formula for this.
We know that $h'(x)=\frac{1}{x^2+1}$, which we equate to a geometric series in $-x^2$ and generalise.
$$\begin{align}
h'(x)&=\sum_{i=0}^\infty(-x^2)^i
\\
h''(x)&=\sum_{i=1}^\infty(-1)^i\cdot2ix^{2i-1}
\\
h^{[n]}(x)&=\sum_{i=n-1}^\infty(-1)^i(2i)_{n-1}x^{2i-(n-1)}
\end{align}$$
Where $(x)_n=\prod_{i=0}^{n-1}(x-i)$ and $x^0=1$ when $x=0$.
Hence, $h'(0)=1, h''(0)=0, h'''(0)=-2, h^{[5]}=4!,h^{[7]}=-6!\dots$ and in general:
$$h^{[n]}(0)=\begin{cases}0&n\:\mathrm{even}
\\
(-1)^{\frac{n-1}{2}}(n-1)!&n\:\mathrm{odd}
\end{cases}$$
We may substitute this into the multi-chain-rule expansion of $g^{[n]}$(x) at $x=0$ using that $g(0)=1$.
$\begin{align}g'(0)&=1
\\
g''(0)&=1
\\
g'''(0)&=-2+1
\\
g^{[4]}(0)&=4(-2)+1
\\
g^{[5]}(0)&=4!+8(-2)+1
\dots
\end{align}$
The problem becomes finding the general coefficients $C_i$ of the odd terms in the expansion of $g^{[n]}$ but I'm unsure how to proceed.
