Solving the functional equation $x[f(x+1)-f(x-1)]=1$ [duplicate]

Possible Duplicate:
Solving the functional equation $f(x+1) - f(x-1) = g(x)$

How do I approach this problem $x[f(x+1)-f(x-1)]=1$.

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marked as duplicate by Marvis, Pedro Tamaroff♦, lhf, Argon, t.b.Jul 3 '12 at 4:10

Over what domain? It can't be true at x=0 – Ross Millikan Jan 19 '11 at 16:56
I'm looking for any solution; any connected domain will work. – Benji Jan 19 '11 at 17:01
For large enough $x$, try $\log x$ =) – Pedro Tamaroff Jul 3 '12 at 0:14
The question for which this one was closed answers part of this question: that this equation can be solved and any solution is unique up to a $2$-periodic function. However, this question can be solved explicitly, whereas the other question is too general. I think that this might be a reason for reopening this question. – robjohn Jul 3 '12 at 12:45

Pick any function $f(x)$ on $(0,2]$. Then $f(x)$ on $(2,\infty )$ is determined-just step downward by 2's until you get into $(0,2]$. A similar technique works for $x<0$ but you step up.

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Let $$f(x)=\sum_{k=1}^\infty\left(\frac{1}{2k-1}-\frac{1}{2k+x-1}\right)\tag{1}$$ then \begin{align} f(x+1)-f(x-1) &=\sum_{k=1}^\infty\left(\frac{1}{2k+x-2}-\frac{1}{2k+x}\right)\\ &=\lim_{N\to\infty}\left(\frac1x-\frac{1}{2N+x}\right)\\ &=\frac1x\tag{2} \end{align} and \begin{align} f(x) &=\frac12\sum_{k=1}^\infty\left(\frac{1}{k-\frac12}-\frac{1}{k+\frac{x-1}{2}}\right)\\[6pt] &=\frac12\psi\left(\frac{x+1}{2}\right)-\frac12\psi\left(\frac{1}{2}\right)\\ &=\frac12\psi\left(\frac{x+1}{2}\right)+\log(2)+\frac{\gamma}{2}\tag{3} \end{align} where $\psi$ is the Digamma function.

Two functions defined by $f(x+1)-f(x-1)$ differby a function which is $2$-periodic. Thus, a function that is defined by $(2)$, would differ from $(3)$ by a $2$-periodic function.

Therefore, $$f(x)=\frac12\psi\left(\frac{x+1}{2}\right)+\varphi(x)\tag{4}$$ where $\varphi(x)$ is any $2$-periodic function.

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$x(f(x+1)-f(x-1))=1$

$f(x+1)-f(x-1)=\dfrac{1}{x}$

$x\to x+1$:

$f(x+2)-f(x)=\dfrac{1}{x+1}$

$x\to2x$:

$f(2x+2)-f(2x)=\dfrac{1}{2x+1}$

$f(2(x+1))-f(2x)=\dfrac{1}{2x+1}$

$f(2x)=\sum_x\dfrac{1}{2x+1}+\Theta_1(x)$, 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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