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I'm trying to solve the following recurrence equation, for $\alpha$ and $\beta$ in $\mathbb{R}$.

$$\forall n\in \mathbb{Z}, \; (\alpha+\beta n^2)\, f(n)=\frac{f(n+1)+f(n-1)}{2}$$

I know that if I had an $n$ instead of a $n^2$, I could introduce Bessel functions of the first kind using the relation $J_{n+1}(x)+J_{n-1}(x)={2n \over x} J_n(x)$, but this is not the case here...

Is there by any chance a special function defined by my recurrence relation ?

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First, do the Fourier transform $$\beta\frac{d^{2}}{d\xi^{2}}f(\xi)+(\cos(\xi)-\alpha){f(\xi)}=0$$ Then do the rescaling $$\xi=2\eta$$ and you get the Mathieu Equation $$\frac{d^{2}f(\eta)}{d\eta^{2}}+\Big(\frac{4}{\beta}\cos(2\eta)-\frac{4\alpha}{\beta}\Big)f(\eta)=0$$ It is solved by Mathieu functions. A good reference is e. kamke differentialgleichungen!

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    I understand ! By any chance, if we want to compute the inverse Fourier transform, is there any closed expression for the solution ? Is there at least a generating function for Mathieu functions ? – 2017-02-20