Finding the Maclaurin series of $e^{\sin x}$ by comparing coefficients

Question

I believe I have found a nice way to find the Maclaurin series of $e^{\sin x}$. Please check if there are any mistakes with my working. Is this method well known?

Answer 1

(Working finds up to $x^5$ and assumes knowledge that the Maclaurin series of $\cos x\approx1-\frac{x^2}{2}+\frac{x^4}{24}$.)

Let $f(x)= e^{\sin x}$, therefore $f'(x)=\cos x*e^{\sin x}=\cos x*f(x)$ by chain rule.

Assume that $f(x)$ has a Maclaurin series and let that series $f(x)=a+bx+cx^2+dx^3+ex^4+fx^5$, therefore $f^{'}(x)=b+2cx+3dx^2+4ex^3+5fx^4$.

Sub in known expressions into $f^{'}(x)=\cos x*f(x)$ to get: $b+2cx+3dx^2+4ex^3+5fx^4=(1-\frac{x^2}{2}+\frac{x^4}{24})(a+bx+cx^2+dx^3+ex^4+fx^5)$.

Expand the right hand side to get: $RHS=a+bx+(c-\frac{1}{2})x^2+(d-\frac{b}{2})x^3+(e-\frac{c}{2}+\frac{a}{24})x^4+(f-\frac{d}{2}+\frac{b}{24})x^5$.

Comparing coefficients of $LHS$ and $RHS$ yields:

$b=a$

$2c=b$

$3d=c-\frac{1}{2}$

$4e=d-\frac{b}{2}$

$5f=e-\frac{c}{2}+\frac{a}{24}$.

When $x=0$, $f(x)=1$ and $f(x)=a$, therefore $a=1$. Which can be used to find:

$b=1$

$c=\frac{1}{2}$

$d=0$

$e=-\frac{1}{8}$

$f=-\frac{1}{15}$.

This means the Maclaurin series of $e^{\sin x}=1+x+\frac{1}{2}x^2-\frac{1}{8}x^4-\frac{1}{15}x^5$ which is indeed to correct series.

Answer 2

At the site https://math.stackexchange.com/a/4658587, it was obtained that \begin{equation*} \operatorname{e}^{\sin x} =1+\sum_{n=1}^\infty\Biggl\{\sum_{k=1}^n \biggl[\cos\frac{(n-k)\pi}2\biggr] \frac{(-1)^k}{(2k)!!} \sum_{q=0}^k(-1)^q\binom{k}{q}(2q-k)^n\Biggr\}\frac{x^n}{n!}. \end{equation*}