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

Can you help me evaluate this function? I have tried to substitute values into $x$ to find the function, use Maclaurin series expansion and... And how do I solve these type of questions?

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  • Sorry i am new to stackexchange –  Aug 03 '16 at 10:53
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    http://math.stackexchange.com/questions/1860910/difficult-functions-evaluation-problem – juantheron Aug 03 '16 at 10:56
  • Do you want to evaluate it at a specific value of x or find an analytic form for $f(x)$ ? – ITA Aug 03 '16 at 11:02
  • @juantheron That solution is fine but is there anyway for solving f(g(x))+f(h(x))=j(x) for arbitary functions g(x),h(x) and j(x) –  Aug 03 '16 at 11:05
  • @ivan I want to find a closed form of f(x) –  Aug 03 '16 at 11:07
  • Do you have any more information about $f$? for example, what it's domain is? – Morgan Rogers Aug 03 '16 at 11:07
  • @Morgan Rogers No. But i believe that f(x) is continous for all x except for x=1 and that that f(x) might be the sum of the inverses of x^n. –  Aug 03 '16 at 11:11
  • We might need some more information about $f(x)$. Replacing $x \to 1/x$ we can generate

    $$ f\left( {\frac{1}{x}} \right) + f\left( {\frac{x}{{1 - x}}} \right) = \frac{1}{{1 - x}} $$

    which has a very similar form unto sign difference. So if its an odd function it might have a different form than if its even.

     – ITA Aug 03 '16 at 11:19
  • @Ivan Abraham can you elaborate more? –  Aug 03 '16 at 11:20

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