Taylor series for $\sin^5 x$ at zero

I'm trying to obtain the taylor series for $f=\sin^5 x$ at the point $\alpha = 0$, by using the known Taylor series of $\sin x$, then taking its $5$th power, as such $$\Bigl(x-\frac{x^3}{3!}+\frac{x^5}{5!}+ \cdots \Bigl)^5 $$

Now all that needs to be done, is to calculate the coefficient of each term individually. This doesn't seems like a good solution, and I feel like I'm really misunderstanding the point of Taylor series at this point. Anyway, a bigger confusion I have, is that if you where to start calculating the derivatives of $\sin^5(x)$, it would not really yield anything meaningful. For example ( the first derivative ) $$f^{(1)} = 5\sin^4x\cdot \cos x$$

So $f^{(1)}(\alpha) = 0$ when $\alpha = 0$, and so the first term is just $\frac{0}{1!}\cdot(x-\alpha)^1=0$. The 2nd and third terms are $0$, as the corresponding derivatives are $0$ at $\alpha = 0$. This is not reflected in the actual series, which has non-zero terms in their place. If we want to find the $nth$ power of a known Taylor series, what are the practical steps involved?

Solution 1:

Your method will work but is rather tedious. I'd much rather turn multiplication into addition using Chebyshev's formula: $$ \sin^5(x) =\frac{1}{16}(10 \sin(x) - 5 \sin(3 x) + \sin(5 x)); $$then knowing the Maclaurin series for $\sin(x)$, you can get the others by direct substitution and then add up the coefficients.

Solution 2:

An accurate answer is as follows.

1. 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}

2. 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*}

3. These formulas can be applied to establish general formulas of the $n$th derivatives for functions of the types $f(\sin x)$ and $f(\cos x)$, such as $\sin^\alpha x$, $\cos^\alpha x$, $\sec^\alpha x$, $\csc^\alpha x$, $e^{\pm\sin x}$, $e^{\pm\cos x}$, $\ln\cos x$, $\ln\sin x$, $\ln\sec x$, $\ln\csc x$, $\sin\sin x$, $\cos\sin x$, $\sin\cos x$, $\cos\cos x$, $\tan x$, and $\cot x$.

4. By virtue of some formulas above-mentioned, we have \begin{align} \frac{\textrm{d}^n\sin^5x}{\textrm{d} x^n} &=\sum_{k=0}^n\langle5\rangle_k \sin^{5-k}x B_{n,k}(\cos x, -\sin x, -\cos x,\sin x,\dotsc)\\ &\to5!B_{n,5}(1, 0, -1,0,\dotsc), \quad x\to0\\ &=5! \biggl[\cos\frac{(n-5)\pi}2\biggr]2^{n-5}S_{-5/2}(n,5). \end{align} Consequently, we find \begin{equation} \sin^5x=5!\sum_{n=0}^\infty\biggl[\cos\frac{(n-5)\pi}2\biggr]\frac{2^{n-5}S_{-5/2}(n,5)}{n!}x^n =5!\sum_{k=0}^\infty(-1)^k\frac{2^{2k}S_{-5/2}(2k+5,5)}{(2k+5)!}x^{2k+5}. \end{equation}

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