TITLE: Derive taylor series of $e^{\sin(x)}$ in two different ways
QUESTION [0 upvotes]: I need to find the Taylor series of $e^{\sin(x)}$ up to $x^4$ in two different ways. First I derived it by calculating the derivatives of the function, and I found the answer $P_4(x) = 1+x+ \frac{x^2}{2} - \frac{x^4}{8}$. Now I need to use the Taylor series of $e^y$ and plug in the Taylor series of $\sin(x)$ to find an answer. After that I need to draw a conclusion. So I know the Taylor series of $e^y$ up to $y^4$ looks like $P_4(y) = 1 + y + \frac{y^2}{2} + \frac{y^3}{6} + \frac{y^4}{24}$ and the Taylor series of $\sin(x)$ up to $x^4$ looks like $P_4(x) = x - \frac{x^3}{6}$. I substituted the Taylor series of $\sin(x)$ into the $y$ variable of the Taylor series of $e^y$, but it doesn't give me the same answer as the answer I got by using the first four derivates of $e^{\sin(x)}$. Am I doing anything wrong?

REPLY [0 votes]: The Faa di Bruno formula can be described in terms of partial Bell polynomials $B_{n,k}(x_1,x_2,\dotsc,x_{n-k+1})$ by
\begin{equation}\label{Bruno-Bell-Polynomial}
\frac{\textrm{d}^n}{\textrm{d} x^n}f\circ h(x)=\sum_{k=0}^nf^{(k)}(h(x)) B_{n,k}\bigl(h'(x),h''(x),\dotsc,h^{(n-k+1)}(x)\bigr).
\end{equation}

The partial Bell polynomials $B_{n,k}$ satisfy
\begin{multline}\label{bell-sin-eq}
B_{n,k}\biggl(-\sin x,-\cos x,\sin x,\cos x,\dotsc, \cos\biggl[x+\frac{(n-k+1)\pi}{2}\biggr]\biggr)\\
=\frac{(-1)^k\cos^kx}{k!}\sum_{\ell=0}^k\binom{k}{\ell}\frac{(-1)^\ell}{(2\cos x)^\ell}
\sum_{q=0}^\ell\binom{\ell}{q}(2q-\ell)^n \cos\biggl[(2q-\ell)x+\frac{n\pi}2\biggr]
\end{multline}
and
\begin{multline}\label{bell-sin=ans}
B_{n,k}\biggl(\cos x,-\sin x,-\cos x,\sin x,\dotsc, \sin\biggl[x+\frac{(n-k+1)\pi}{2}\biggr]\biggr)\\
=\frac{(-1)^k\sin^{k}x}{k!}\sum_{\ell=0}^k\binom{k}{\ell}\frac1{(2\sin x)^{\ell}}
\sum_{q=0}^\ell(-1)^q\binom{\ell}{q}(2q-\ell)^n \cos\biggl[(2q-\ell)x+\frac{(n-\ell)\pi}2\biggr].
\end{multline}
Taking $x\to0$ leads to
\begin{multline}\label{bell-sin-eq=0}
B_{n,k}\biggl(0,-1,0,1,\dotsc, \cos\frac{(n-k+1)\pi}{2}\biggr)\\
=\frac{(-1)^k}{k!}\biggl(\cos\frac{n\pi}2\biggr) \sum_{\ell=0}^k\frac{(-1)^\ell}{2^\ell}\binom{k}{\ell}
\sum_{q=0}^\ell\binom{\ell}{q}(2q-\ell)^n
\end{multline}
and
\begin{multline}\label{bell-sin=ans=0}
B_{n,k}\biggl(1,0,-1,0,\dotsc, \sin\frac{(n-k+1)\pi}{2}\biggr)\\
=\frac{(-1)^k}{k!2^k} \biggl[\cos\frac{(n-k)\pi}2\biggr] \sum_{q=0}^k(-1)^q\binom{k}{q}(2q-k)^n
=\biggl[\cos\frac{(n-k)\pi}2\biggr]2^{n-k}S_{-k/2}(n,k),
\end{multline}
where
\begin{equation*}%\label{S(n,k,x)-satisfy-eq}
S_r(n,k)=\frac1{k!}\sum_{j=0}^k(-1)^{k-j}\binom{k}{j}(r+j)^n, \quad n\ge k\ge0.
\end{equation*}

By virtue of some formulas above-mentioned, we have
\begin{align}
\frac{\textrm{d}^n\exp(\sin x)}{\textrm{d} x^n}
&=\sum_{k=0}^n\exp(\sin x) B_{n,k}(\cos x, -\sin x, -\cos x,\sin x,\dotsc)\\
&\to\sum_{k=0}^n B_{n,k}(1, 0, -1,0,\dotsc), \quad x\to0\\
&=\sum_{k=0}^n \biggl[\cos\frac{(n-k)\pi}2\biggr]2^{n-k}S_{-k/2}(n,k).
\end{align}
Consequently, we find
\begin{equation}
\exp(\sin x)=\sum_{n=0}^\infty\Biggl(\sum_{k=0}^n \biggl[\cos\frac{(n-k)\pi}2\biggr]2^{n-k}S_{-k/2}(n,k)\Biggr)\frac{x^n}{n!}.
\end{equation}


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