Derive taylor series of $e^{\sin(x)}$ in two different ways
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?
$\endgroup$ 04 Answers
$\begingroup$After computing$$1+\left(x-\frac{x^3}6\right)+\frac12\left(x-\frac{x^3}6\right)^2+\frac16\left(x-\frac{x^3}6\right)^3+\frac1{24}\left(x-\frac{x^3}6\right)^4$$you must eliminate the monomials whose degree is greater than $4$. And then you will get the same answer as before.
$\endgroup$ 5 $\begingroup$Let $e^{\sin x}=\sum_{r=0}^\infty a_rx^r$
$\sin x=\ln(\sum_{r=0}^\infty a_rx^r)$
Differentiate both sides with respect to $x$
$\cos x(\sum_{r=0}^\infty a_rx^r)=\sum_{r=1}^\infty a_rrx^{r-1}$
Expand $\cos x$ and compare the constant and the coefficients of $x,x^2,x^3,x^4$ to find $a_r,0\le r\le4$
$\endgroup$ $\begingroup$Let\begin{equation*} D(x)=\sum_{k=0}^\infty d_kx^k \end{equation*}be a power series expansion. Then the function $E(x)=e^{D(x)}$ has the power series expansion\begin{equation*} E(x)=\sum_{k=0}^\infty e_kx^k, \end{equation*}where the coefficients $e_k$ for $k\in\{0\}\cup\mathbb{N}$ satisfy\begin{align} e_0&=e^{d_0},\\ e_k&=\frac1k\sum_{\ell=1}^k\ell d_\ell e_{k-\ell} =\frac1k\sum_{\ell=0}^{k-1}(k-\ell)d_{k-\ell}e_{\ell},\quad k\in\mathbb{N},\\ e_n&=e^{d_0}\left(d_n+\sum_{j=1}^{n-1}\sum_{\substack{\sum_{i=0}^jm_i=n,\\ m_i\ge1, 0\le i\le j}} \prod_{i=0}^j\frac{m_id_{m_i}}{n-\sum_{q=0}^{i-1}m_q}\right), \quad n\in\mathbb{N}, \end{align}and\begin{equation}\label{alpha-k-power-eq} e_k= e^{d_0}\sum_{j=1}^k\frac1{j!} \sum_{\substack{\sum_{\ell=1}^ji_\ell=k,\\ i_\ell\ge1,1\le\ell\le j}} \prod_{\ell=1}^jd_{i_\ell}, \quad k\in\mathbb{N}. \end{equation}These conclusions can be found in the paper [1] below.
[1] Feng Qi, Xiao-Ting Shi, and Fang-Fang Liu, Expansions of the exponential and the logarithm of power series and applications, Arabian Journal of Mathematics 6 (2017), no. 2, 95--108; available online at .
Consequently, we finally can obtain$$ \exp(\sin x)=1+x+\frac{x^2}{2}-\frac{x^4}{8}-\frac{x^5}{15}-\frac{x^6}{240}+\frac{x^7}{90}+\frac{31 x^8}{5760}+\frac{x^9}{5670}-\frac{2951 x^{10}}{3628800}-\frac{x^{11}}{3150}+\dotsm. $$
$\endgroup$ $\begingroup$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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