<s>[INST] Solving the functional equation $x[f(x+1)-f(x-1)]=1$ 
Possible Duplicate:
Solving the functional equation $f(x+1) - f(x-1) = g(x)$ 

How do I approach this problem $x[f(x+1)-f(x-1)]=1$.
 [/INST] $x(f(x+1)-f(x-1))=1$
$f(x+1)-f(x-1)=\dfrac{1}{x}$
$x\to x+1$:
$f(x+2)-f(x)=\dfrac{1}{x+1}$
$x\to2x$:
$f(2x+2)-f(2x)=\dfrac{1}{2x+1}$
$f(2(x+1))-f(2x)=\dfrac{1}{2x+1}$
$f(2x)=\sum_x\dfrac{1}{2x+1}+\Theta_1(x)$, where $\Theta_1(x)$ is an arbitrary periodic function with unit period
Since $\lim_{x\to+\infty}\dfrac{1}{2x+1}=0$, so according to http://en.wikipedia.org/wiki/Indefinite_sum#Mueller.27s_formula, the result can be further simplified to
$f(2x)=\sum_{n=0}^\infty\left(\dfrac{1}{2n+1}-\dfrac{1}{2x+2n+1}\right)+\Theta_1(x)$, where $\Theta_1(x)$ is an arbitrary periodic function with unit period
$f(x)=\sum_{n=0}^\infty\left(\dfrac{1}{2n+1}-\dfrac{1}{x+2n+1}\right)+\Theta(x)$, where $\Theta(x)$ is an arbitrary periodic function with period $2$
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