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?
Best Answer
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.