Recurrence involving derivative

Question

I would like to get a closed form of $A_n(x)$ if verifies the following recurrence relation

$$A_n(x)=\frac{d}{dx}\left(\frac{A_{n-1}(x)}{a-\cos x}\right)\,\,\,\text{and}\,\,\,A_0(x)=1.$$

Really I need to know the general term of $A_n(0)$.

Any ideas or suggestions are welcome.

Answer 1

This is not a complete solution.

Setting $n=\infty$, and the recurrence equation becomes $$A_{\infty}(x)=\frac{d}{dx}\left(\frac{A_{\infty}(x)}{a-\cos x}\right)$$

The solution to it is:

$$A_{\infty}(x) = C(a - \cos x)\exp(a x - \sin x) $$ where $C$ is a constant of integration.

Answer 2

The first few terms:

• $A_0(x)=1$
• $\displaystyle A_1(x)=\frac{-\sin x}{(a-\cos x)^2}$
• $\displaystyle A_2(x)=3A_1^2(x)+\frac{-\cos x}{(a-\cos x)^3}$
• $\displaystyle A_3(x)=-15A_1^3(x)+10A_1(x)A_2(x)+\frac{\sin x}{(a-\cos x)^4}$
• $\displaystyle A_4(x)=105A_1^4(x)-105A_1^2(x)A_2(x)+15A_1(x)A_3(x)+10A_2^2(x)+\frac{\cos x}{(a-\cos x)^5}$