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The question is : Determine all functions f satisfying the functional relation $f(x)+f\left(\frac{1}{1-x}\right)=\frac{2(1-2x)}{x(1-x)}$ where $x$ is a real number . Also $x$ is not equal to $0$ or $1$. Here $f:\mathbb{R}-\{0,1\} \to \mathbb{R}$.

I try the following in many ways. But I can't proceed.I put many things at the place of $x$ to get the answer but I failed again and again.

Stefan4024
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Sufaid Saleel
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1 Answers1

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Let $P(x)$ be the assertation of the function equation. Then:

$$P(x) \implies f(x)+f\left(\frac{1}{1-x}\right)=\frac{2(1-2x)}{x(1-x)}$$

$$P\left(\frac{1}{1-x}\right) \implies f\left(\frac{1}{1-x}\right) + f\left(\frac{x-1}{x}\right) = \frac{2(x+1)(1-x)}{x}$$

$$P\left(\frac{x-1}{x}\right) \implies f\left(\frac{x-1}{x}\right) + f(x) = \frac{2x(2-x)}{x-1}$$

Adding the first and third equation and subtracting the second one will yield $2f(x)$ on the LHS and you should be able to manipulate the RHS.

Stefan4024
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    +1 Nice solution! What was the motivation behind solving it this way? – Ovi Sep 23 '16 at 13:41
  • @Ovi I was trying to get some cancelations on the LHS. I initially tried to make $f(x) = f\left(\frac{1}{1-x}\right)$, but it didn't work. Then I tried this cycle thing and it happened to work. When I was younger I came across a similar problem, where I had to find the $2000$ or so element in a recursive sequence. The sequence was quite similar to $x,\frac{1}{1-x} \dots$, but I know it was 8 terms long. Luckily this was just $3$. – Stefan4024 Sep 23 '16 at 13:57