# Use the Taylor series for $\sin(x)$ to find a Taylor series for $f(x) = p(\sin x)$ around point $x=0$, where $p(x) = 4x^3 - 3x$

Consider the polynomial $$p(x) = 4x^3 - 3x.$$ Use the Taylor series for $$\sin(x)$$ to find a Taylor series for $$f(x) = p(\sin x)$$ around point $$x=0$$.
Don't just calculate this by differentiating right away. Instead, start with the first few terms of the Taylor series for $$\sin(x)$$, and substitute those into the given polynomial to figure out the first few terms.

I am not sure what exactly they want me to do here. The first few terms of $$\sin x$$ are: $$x - \frac{x^3}{6} + \frac{x^5}{120} -+\cdots.$$

Are they just looking for me to plug this expression into the original polynomial?

The answer that the site linked below shows is $$-3x + \frac{9}{2}x^3+ \frac{-81}{40}\,x^5 . . .$$

Were they actually cubing trinomials to find the answer, or am I missing an easier method?

https://ximera.osu.edu/course/kisonecat/multivariable-calculus/master/taylor-series/remainders

• We have $\sin x = x-\dfrac{x^3}{6}+\dfrac{x^5}{120}+O\left(x^6\right)$ Now, $$4 \left(\dfrac{x^5}{120}-\dfrac{x^3}{6}+x\right)^3-3 \left(\dfrac{x^5}{120}-\dfrac{x^3}{6}+x\right)$$ Look at the expanded form using Wolfram Alpha
– Moo
Commented May 5, 2017 at 23:37
• @Moo Try this instead: proofwiki.org/wiki/Triple_Angle_Formulas/Sine Commented May 6, 2017 at 0:05
• @Moo In my opinion, expanding a series raised to a power is more error prone. Commented May 6, 2017 at 0:06
• I do not disagree that there is a clean trig approach, but this may not always be the case.
– Moo
Commented May 6, 2017 at 0:07
• @Moo I personally just recognize $\sin(kx)$ and $\cos(kx)$ expanded for $k=1,2,3,4$ off the top of my head. Commented May 6, 2017 at 0:09

Hint:

First show that

$$4\sin^3(x)-3\sin(x)=-\sin(3x)$$

Then show that

$$f(x)=-\sin(3x)=\sum_{n=0}^\infty\frac{(-1)^{n+1}(3x)^{2n+1}}{(2n+1)!}$$

• Commented Dec 14, 2021 at 12:41

It is known that $$\sin x=\sum_{k=0}^\infty(-1)^k\frac{x^{2k+1}}{(2k+1)!}, \quad x\in\mathbb{R}.$$ On page 43 in the handbook [1] below, there is the series expansion $$\sin^3x=\frac14\sum_{k=1}^\infty(-1)^{k+1}\bigl(3^{2k+1}-3\bigr)\frac{x^{2k+1}}{(2k+1)!}, \quad x\in\mathbb{R}.$$ Consequently, we obtain $$p(\sin x)=4\sin^3x-3\sin x=\sum_{k=1}^\infty(-1)^{k+1}\frac{(3x)^{2k+1}}{(2k+1)!}, \quad x\in\mathbb{R}.$$

Reference

1. I. S. Gradshteyn and I. M. Ryzhik, Table of Integrals, Series, and Products, Translated from the Russian, Translation edited and with a preface by Daniel Zwillinger and Victor Moll, Eighth edition, Revised from the seventh edition, Elsevier/Academic Press, Amsterdam, 2015; available online at https://doi.org/10.1016/B978-0-12-384933-5.00013-8.

A general and alternative answer to this kind of questions is the following theorem obtained in Lemma 2 of the paper [1] below.

For $$\ell\in\mathbb{N}$$ and $$x\in\mathbb{R}$$, assume that the value of the function $$\bigl(\frac{\sin x}{x}\bigr)^{\ell}$$ at the point $$x=0$$ is $$1$$. Then $$$$\label{sine-power-ser-expan-eq} \biggl(\frac{\sin x}{x}\biggr)^{\ell} =\sum_{j=0}^{\infty} (-1)^{j}\frac{R\bigl(\ell+2j,\ell,-\frac{\ell}{2}\bigr)}{\binom{\ell+2j}{\ell}} \frac{(2x)^{2j}}{(2j)!},$$$$ where $$R\bigl(\ell+2j,\ell,-\frac{\ell}{2}\bigr)$$ is given by$$$$\label{S(n,k,x)-satisfy-eq} R(n,k,r)=\frac1{k!}\sum_{j=0}^k(-1)^{k-j}\binom{k}{j}(r+j)^n$$$$ for $$r\in\mathbb{R}$$ and $$n\ge k\ge0$$.

The notation $$R(n,k,r)$$ stands for weighted Stirling numbers of the second kind, which was first introduced in [2] below.

References

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.
2. L. Carlitz, Weighted Stirling numbers of the first and second kind, I, Fibonacci Quart. 18 (1980), no. 2, 147--162.