<s>[INST] Solving a functional equation and general theory of functional equations $f(x-1)$ + $f(x+1)$ = $kf(x)$
given that $k>1$ then I need to find $f(x)$. 
I am clueless about solving functional equations please help me this . Also give some general advice about how to solve functional equations.
 [/INST] If we write $\displaystyle F(x) = \sum_{n=0}^\infty f(n)x^n$, then we have
$$f(n-1) + f(n+1) = kf(n)$$
$$\sum_{n=0}^\infty f(n-1)x^n + \sum_{n=0}^\infty f(n+1)x^n  = kF(n)$$
$$\sum_{n=-1}^\infty f(n)x^{n+1} + \sum_{n=1}^\infty f(n)x^{n-1}  = kF(x)$$
$$f(-1)+\sum_{n=0}^\infty f(n)x^{n+1} -f(0)x^{-1}+\sum_{n=0}^\infty f(n)x^{n-1}  = kF(x)$$
$$f(-1)+xF(x) -f(0)x^{-1}+ x^{-1}F(x)  = kF(x)$$
$$f(-1) - f(0)x^{-1}  = F(x)(k - x-x^{-1})$$
$$F(x) = \frac{f(0) - xf(-1)}{x^2 - kx +1}$$
If we use $U_n$ for Chebyshev polynomials of the second kind, we find that  $\displaystyle\sum_{n=0}^\infty U_n(y/2)x^n = \frac{1}{x^2 -yx +1}$, and
$$F(x)= \Big(f(0) - xf(-1))\Big) \sum_{n=0}^\infty U_n(k/2)x^n$$
$$\sum_{n=0}^\infty f(n)x^n = \sum_{n=0}^\infty \Big [f(0)U_n(k/2) - f(-1) U_{n-1}(k/2)\Big ]x^n$$
$$f(n) = f(0)U_n(k/2) - f(-1) U_{n-1}(k/2)$$
Now with
$$U_n(x) = \frac{(x + \sqrt{x^2-1})^{n+1}- (x - \sqrt{x^2-1})^{n+1}}{2\sqrt{x^2-1}}$$
You have a (messy) closed form.
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