Series expansion for $e^{\sin x}$ and coefficient comparison

Question

I am looking for a way to expand the function $e^{\sin x}$ into Taylor series without employing the cumbersome Taylor formula and come across this thread: Finding the Maclaurin series of $e^{\sin x}$ by comparing coefficients

I understand almost everything of Bernard's answer except the red underlied part. How do you obtain $(c-\dfrac{1}{2})x^2+(d-\frac{b}{2})x^3$ and so on?

Answer 1

Looks like they are simply combining coefficients by exponent. That is when you multiply out the polynomial you'll have $c$ a coefficient of $x^2$ and $-\frac{a}{2}$ a coefficient. So $$cx^2 -\frac{a}{2}x^2 =(c-\frac{a}{2})x^2.$$ And so on of course.

Answer 2

For the $x^2$ term it seems there is a typo indeed we should have

$$cx^2-\frac12 ax^2 =\left(c-\frac a2\right)x^2$$

then

$$ dx^3-\frac12b x^3=\left(d-\frac b2\right)x^3$$

and

$$1\cdot ex^4+\left(-\frac {x^2}2\right) \cdot cx^2+\frac {x^4}{24} \cdot a=\left(e-\frac c 2+\frac a{24}\right)x^4$$

and finally

$$1\cdot fx^5+\left(-\frac {x^2}2\right) \cdot dx^3+\frac {x^4}{24} \cdot bx=\left(f-\frac d 2+\frac b{24}\right)x^5$$

Answer 3

Note: this is one of my nothing-original-here answers.

If $g(x) =e^{f(x)} $, where $f(x) =\sum_{k=1}^{\infty} a_kx^k $, then $g'(x) =f'(x)e^{f(x)} =f'(x)g(x) $ and $f'(x) =\sum_{k=1}^{\infty} ka_kx^{k-1} =\sum_{k=0}^{\infty} (k+1)a_{k+1}x^{k} $.

Assuming $g(x) =\sum_{n=0}^{\infty} b_nx^n $, then $g'(x) =\sum_{n=1}^{\infty} nb_nx^{n-1} =\sum_{n=0}^{\infty} (n+1)b_{n+1}x^{n} $ so that

$\begin{array}\\ \sum_{n=0}^{\infty} (n+1)b_{n+1}x^{n} &=\sum_{k=0}^{\infty} (k+1)a_{k+1}x^{k}\sum_{n=0}^{\infty} b_nx^n\\ &=\sum_{m=0}^{\infty}x^m\sum_{k=0}^m (k+1)a_{k+1}b_{m-k}\\ &=\sum_{m=0}^{\infty}x^m\sum_{k=0}^m (m-k+1)a_{m-k+1}b_{k}\\ \end{array} $

so that $(n+1)b_{n+1} =\sum_{k=0}^n (n-k+1)a_{n-k+1}b_{k} $.

This gives the recursion $b_{n+1} =\frac1{n+1}\sum_{k=0}^n (n-k+1)a_{n-k+1}b_{k} $.

Answer 4

For $n\ge1$, by the Faa di Bruno formula and some properties of the partial Bell polynomials, we have \begin{align*} \bigl(\operatorname{e}^{\sin x}\bigr)^{(n)} &=\sum_{k=1}^n \operatorname{e}^{\sin x} B_{n,k}\biggl(\cos x,-\sin x,-\cos x,\sin x,\dotsc, \sin\biggl[x+\frac{(n-k+1)\pi}{2}\biggr]\biggr)\\ &\to \sum_{k=1}^n B_{n,k}\biggl(1,0,-1,0,\dotsc, \sin\frac{(n-k+1)\pi}{2}\biggr), \quad x\to0\\ &=\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. \end{align*} Consequently, we obtain \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*}

References

1. F. Qi, Derivatives of tangent function and tangent numbers, Appl. Math. Comput. 268 (2015), 844--858; available online at https://doi.org/10.1016/j.amc.2015.06.123.
2. Feng Qi, Da-Wei Niu, Dongkyu Lim, and Yong-Hong Yao, Special values of the Bell polynomials of the second kind for some sequences and functions, Journal of Mathematical Analysis and Applications 491 (2020), no. 2, Paper No. 124382, 31 pages; available online at https://doi.org/10.1016/j.jmaa.2020.124382.