Generalized nested sine function [closed]

Solution 1:

There may be other extensions of $\sin^n$, but this one is fairly straightforward to compute. Essentially, given integer $n$, construct a Maclaurin polynomial for $\sin^n(x)$ centered at $x=0$. To do this, we need to compute $M_p(n)=(-1)^p\frac{d^{2p+1}}{dx^{2p+1}}\sin^n(x)\mid_{x=0}$. It turns out that given some $p$, $M_p(n)$ forms a polynomial in $n$ of degree $p$ (although I haven't proven this, it works for $p\leq 3$). This can be easily extended to any rational (or real) number by simply just plugging those numbers into the polynomial in $n$ which describes $M_p(n)$. Therefore, all you need to do is calculate $M_p(0),M_p(1),\ldots,M_p(p-1),M_p(p)$ for a total of $p+1$ points, which will determine the polynomial. Then, $\sin^n(x)=\sum_{p=0}^{\infty} (-1)^pM_p(n)\frac{x^{2p+1}}{(2p+1)!}$.

I have found $M_p(n)$ for $p\leq 3$. So far, the approximation for $\sin^n(x)$ around $x=0$ (which works for any real $n$) looks like: $$ \sin^n(x\approx 0)=x-n\left(\frac{x^{3}}{3!}\right)+n\left(5n-4\right)\left(\frac{x^{5}}{5!}\right)-\frac{n\left(175n^{2}-336n+164\right)}{3}\left(\frac{x^{7}}{7!}\right)+O(x^9) $$

Edit: I just realized that because computing high derivatives of nested $\sin$ functions is computationally expensive, you can also compute that high derivative of a smaller composition of $\arcsin$ functions instead. So calculate $M_p\left(-\left\lfloor \frac{p}{2}\right\rfloor\right),\ldots, M_p\left(\left\lceil \frac{p}{2}\right\rceil\right)$, again for a total of $p+1$ points, where $\sin^{-n}(x)=\arcsin^{n}(x)$.