Expansion of $x$ in powers of $u$

Question

Given: $$\sin(x) = u \sin(x+a),\qquad {u<1}$$

How do I expand $x$ in powers of $u$?

I tried using Taylor series but it failed to proceed.

Answer 1

Here is what I did. I started with $\, x_0 := O(u). \,$ Define by recursion $\, x_{n+1} := \sin^{-1}(u \sin(x_n + a)). \,$ The first few values are $\, x_1 = \sin(a)\, u + O(u^2), \, x_2 = \sin(a)\, u + \sin(2a)\,u^2/2 + O(u^3). \,$ Using the pattern and taking it to the limit, I found that $\, x = \log( (1 - u \exp(-ia)) / (1 - u\exp(ia)) )/(2i). \,$ In the limit $\, x = \sum_{n=1}^\infty \sin(na)\,u^n/n . \,$

Another method uses exponentials. Let $\, X := \exp(i x), \, A := \exp(i a) \,$ and substitute them in equation $\, \sin(x) = u \sin(x+a) \,$ to get $\, (X - 1/X) = u (X A - 1/(X A)) \,$ and solving for $\, X \,$ gives $ X^2 = (1 - u/A) / (1 - A u). \,$ Using $\, \log(1-x) = - \sum_{n>0} x^n/n \,$ now gives the power series in $\, u.$