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 were 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 $n$th power of a known Taylor series, what are the practical steps involved?

  • $\begingroup$ I'm not sure I follow, $f(x) = \sin^5 x$ does not have nonzero term until $x^5$. Both methods will agree. If you differentiate and obtain $f^{(5)}(0)/5!$ you should obtain that it equals 1, in line with the first approach. Note in the first approach that the first nonzero term in power series will be $x^5$. $\endgroup$
    – Gregory
    May 13, 2020 at 15:36
  • 2
    $\begingroup$ The problem looks messy, but I think the least-messy solution would be to use the fact that $\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$. $\endgroup$
    – Botond
    May 13, 2020 at 15:36
  • $\begingroup$ The simplest (comparatively) would probably be to linearise $\sin^5x$. $\endgroup$
    – Bernard
    May 13, 2020 at 15:41

4 Answers 4


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

  • 1
    $\begingroup$ You beat me to it :) Anyway, I found that this yields $\sin^5(x) = \dfrac{5}{16} \displaystyle\sum_{n=0}^{\infty} \dfrac{(-1)^nx^{2n+1}}{(2n+1)!}(5^{2n} -3^{2n+1} + 2)$. The $n=0$ and $n=1$ terms vanish which means the series starts from an $x^5$ term, as expected... so hopefully I haven't made any algebraic error. $\endgroup$
    – peek-a-boo
    May 13, 2020 at 15:47
  • $\begingroup$ Thanks @peek-a-boo, I was too lazy to actually compute the terms :) $\endgroup$
    – Integrand
    May 13, 2020 at 15:48

For $m,k\in\mathbb{N}=\{1,2,\dotsc\}$, we have \begin{equation}\label{sin-ell-d} \frac{\textrm{d}^m\sin^kx}{\textrm{d} x^m} =\frac{(-1)^k}{2^k} \sum_{q=0}^k(-1)^q\binom{k}{q}(2q-k)^m \cos\biggl[(m-k)\frac\pi2+(2q-k)x\biggr]. \end{equation} For $\ell\in\mathbb{N}$, we have \begin{equation}\label{sin-poer-exp} \sin^\ell x=\frac{(-1)^\ell}{2^\ell}\sum_{q=0}^\ell(-1)^q\binom{\ell}{q} \cos\biggl[(2q-\ell)x-\frac{\ell}2\pi\biggr]. \end{equation} Employing anyone of these two formulas, one can derive Maclarurin's series expansion of the function $\sin^5x$ and, generally, the function $\sin^nx$ for $n\in\mathbb{N}$.

These two formulas can be found in the following paper:

  1. Bai-Ni Guo and Feng Qi, On the Wallis formula, International Journal of Analysis and Applications 8 (2015), no. 1, 30--38.

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}


  1. B.-N. Guo, D. Lim, and F. Qi, Maclaurin's series expansions for positive integer powers of inverse (hyperbolic) sine and tangent functions, closed-form formula of specific partial Bell polynomials, and series representation of generalized logsine function, Appl. Anal. Discrete Math. 16 (2022), in press; available online at https://doi.org/10.2298/AADM210401017G.
  2. 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.
  3. F. Qi and J. Gelinas, Revisiting Bouvier's paper on tangent numbers, Adv. Appl. Math. Sci. 16 (2017), no. 8, 275--281.
  4. F. Qi and B.-N. Guo, An explicit formula for derivative polynomials of the tangent function, Acta Univ. Sapientiae Math. 9 (2017), no. 2, 348--359; available online at https://doi.org/10.1515/ausm-2017-0026.
  5. F. Qi and B.-N. Guo, Explicit formulas for special values of the Bell polynomials of the second kind and for the Euler numbers and polynomials, Mediterr. J. Math. 14 (2017), no. 3, Art. 140, 14 pages; available online at https://doi.org/10.1007/s00009-017-0939-1.
  6. 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.
  7. F. Qi, G.-S. Wu, and B.-N. Guo, An alternative proof of a closed formula for central factorial numbers of the second kind, Turkish J. Anal. Number Theory 7 (2019), no. 2, 56--58; available online at https://doi.org/10.12691/tjant-7-2-5.
  8. C.-F. Wei and F. Qi, Several closed expressions for the Euler numbers, J. Inequal. Appl. 2015, Paper No. 219, 8 pages; available online at https://doi.org/10.1186/s13660-015-0738-9.
  9. A.-M. Xu and G.-D. Cen, Closed formulas for computing higher-order derivatives of functions involving exponential functions, Appl. Math. Comput. 270 (2015), 136--141; available online at https://doi.org/10.1016/j.amc.2015.08.051.
  10. J.-L. Zhao, Q.-M. Luo, B.-N. Guo, and F. Qi, Remarks on inequalities for the tangent function, Hacet. J. Math. Stat. 41 (2012), no. 4, 499--506.
  • $\begingroup$ Alternative answers to Maclaurin's series expansions or to the $n$th derivatives of the function $\sin^nx$ can be found in the paper [Yu. A. Brychkov, Power expansions of powers of trigonometric functions and series containing Bernoulli and Euler polynomials, Integral Transforms Spec. Funct. vol. 20 (2009), no. 11-12, 797--804; available online at doi.org/10.1080/10652460902867718 ] and in the handbook [A. P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev, Integrals and Series, Elementary Functions,Vol. 1, Gordon and Breach, NewYork, London, 1986]. $\endgroup$
    – qifeng618
    Dec 14, 2021 at 10:52

A general answer to this question is the following result.

Theorem. For $\ell\in\mathbb{N}$ and $x\in\mathbb{R}$, we have \begin{equation}\label{sine-power-ser-expan-eq} \biggl(\frac{\sin x}{x}\biggr)^{\ell} =1+\sum_{j=1}^{\infty} (-1)^{j}\frac{T(\ell+2j,\ell)}{\binom{\ell+2j}{\ell}} \frac{(2x)^{2j}}{(2j)!}, \end{equation} where \begin{equation}\label{T(n-k)-EF-Eq(3.1)} T(n,k)=\frac1{k!} \sum_{\ell=0}^{k}(-1)^{\ell}\binom{k}{\ell}\biggl(\frac{k}2-\ell\biggr)^n. \end{equation} denotes central factorial numbers of the second kind.

For details of the proof for the above result, please read the paper below.


  1. Feng Qi and Peter Taylor, Series expansions for powers of sinc function and closed-form expressions for specific partial Bell polynomials, Applicable Analysis and Discrete Mathematics 18 (2024), no. 1, 92–115; available online at https://doi.org/10.2298/AADM230902020Q.

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .