I was wondering if there exists a power series for the sin of a power series, in other words: which is the formula for the coefficients $\xi_{\lambda}$ in terms of the $f_{\lambda}$ in the expansion:
$\sin \left(\sum_{\lambda=0}^{\infty} f_{\lambda}\epsilon^{\lambda} \right) = \sum_{\lambda=0}^{\infty} \xi_{\lambda}\epsilon^{\lambda}$
assuming that $\sin(f_{0})=0$? Thanks a lot in advance!
On
The problem will be simpler if summing to $n$ instead of $\infty$. $$\sin \left(\sum_{\lambda=0}^{n} f_{\lambda}\epsilon^{\lambda} \right) = \sum_{\lambda=0}^{n} \xi_{\lambda}\epsilon^{\lambda}$$ In such a case (this is the tricky part), we could consider that $$\sum_{\lambda=0}^{n} f_{\lambda}\epsilon^{\lambda}=\sum_{\lambda=0}^{n} f_{\lambda}\epsilon^{\lambda}+O(\epsilon^{n+1})$$ and $$\sum_{\lambda=0}^{n} \xi_{\lambda}\epsilon^{\lambda}=\sum_{\lambda=0}^{n} \xi_{\lambda}\epsilon^{\lambda}+O(\epsilon^{n+1})$$
and compose Taylor series.
For a try, use $n=4$. Without any restriction, this would give $$\sin (f_0)+f_1 \epsilon \cos (f_0)+\epsilon ^2 \left(f_2 \cos (f_0)-\frac{1}{2} f_1^2 \sin (f_0)\right)+\epsilon ^3 \left(-f_2 f_1 \sin (f_0)-\frac{1}{6} f_1^3 \cos (f_0)+f_3 \cos (f_0)\right)+\epsilon ^4 \left(\frac{1}{24} f_1^4 \sin (f_0)-\frac{1}{2} \left(f_2^2+2 f_1 f_3\right) \sin (f_0)-\frac{1}{2} f_2 f_1^2 \cos (f_0)+f_4 \cos (f_0)\right)+O\left(\epsilon ^5\right)$$ For sure, if $\sin(f_0)=0$ this would simplify a lot to $$f_1 \epsilon \cos (f_0)+f_2 \epsilon ^2 \cos (f_0)-\frac{1}{6} \epsilon ^3 \left(\left(f_1^3-6 f_3\right) \cos (f_0)\right)-\frac{1}{2} \epsilon ^4 \left(\left(f_1^2 f_2-2 f_4\right) \cos (f_0)\right)+O\left(\epsilon ^5\right)$$ which could again simplify since $\cos(f_0)=\pm 1$. So, we should have $$\pm\left(f_1 \epsilon +f_2 \epsilon ^2 -\frac{1}{6} \left(f_1^3-6 f_3\right)\epsilon ^3 -\frac{1}{2} \left(f_1^2 f_2-2 f_4\right)\epsilon ^4 +O\left(\epsilon ^5\right) \right)$$
I think that the generalization could be difficult (not to say improssible if we make $n\to \infty$).
There is a series, but it’s not so easy to calculate the coefficients. The most natural thing is to write $F(x)=\sum_{\lambda=1}^\infty f_\lambda x^\lambda$, and then use the series expansion of $\bigl(F(x)\bigr)^n$ for as many odd values of $n$ as you need. Note that if you tell yourself that you’re willing to settle for, say, the first $7$ terms of the resulting series, you can throw away all monomials with exponents $>7$, and thus have no more work than to calculate the first, third, fifth, and seventh powers of $F$.
There are fancier ways of doing this, and you can read about “Formal power series” in Wikipedia, where about half-way through the topic of composition of power series is broached.