How do you solve this functional equation?

Question

$$f(x)f(y)=f(x+y)(f(x)+f(y)-2\cos\gamma)+1.$$

I want to obtain the solution $f(x)=\sin(x+\gamma)/\sin x$, instead of the one which is well defined at 0, which I have posted a question before. Is there any way to do that instead of simple substitution, and can one show the uniqueness of this solution?

This functional equation is actually obtained from the Yang-Baxter equation, with the representation $R_i(x)=f(x)+e_i$, where $e_i$ is recognized in the Temperley-Lieb algebra.

Answer 1

For meromorphic functions you can use the substitution $f(x)=g(x)+e^{i\gamma}$ to get $$g(x)g(y)+e^{i\gamma}g(x)+e^{i\gamma}g(y)+e^{2i\gamma}=(g(x+y)+e^{i\gamma})(g(x)+g(y)+2e^{i\gamma}-2\cos\gamma)+1$$ which is $$g(x)g(y)=g(x+y)(g(x)+g(y)+e^{i\gamma}-e^{-i\gamma})$$ Substitute $g(x)=h(x)(e^{i\gamma}-e^{-i\gamma})$: $$h(x)h(y)=h(x+y)(h(x)+h(y)+1)$$ Substitute $h(x)=1/j(x)$: $$j(x+y)=j(x)+j(y)+j(x)j(y)$$ Substitute $j(x)=k(x)-1$: $$k(x+y)=k(x)k(y)$$ which is a form of Cauchy's functional equation with meromorphic solutions $k(x)=e^{2i\lambda x}.$ The factor of $2i$ is just for convenience. Unwinding all this gives

\begin{align*} f(x) &=(e^{i\gamma}-e^{-i\gamma})\frac{1}{e^{2i\lambda x}-1}+e^{i\gamma}\\ &=\frac{e^{i\gamma}-e^{-i\gamma}}2\frac{e^{2i\lambda x}+1}{e^{2i\lambda x}-1}+\frac{e^{i\gamma}+e^{i\gamma}}2\\ &=\sin(\gamma)\frac{\cos(\lambda x)}{\sin(\lambda x)}+\cos(\gamma)\\ &=\frac{\sin(\lambda x+\gamma)}{\sin(\lambda x)} \end{align*}

Answer 2

The general solution with a simple pole at $\;x=0\;$ is $\;f(x):=\sin(xk+\gamma)/\sin(x k)\;$ where $k\ne 0$ is a constant. This can be verified by assuming a Laurent series expansion for $f(x)$ and, assuming the functional equation, solving for the series coefficients one by one.