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Let $F(x)$ be the real-valued function defined for all real $x$ except for $x = 0$ and $x = 1$ and satisfying the functional equation $$F(x) + F\left(\frac{x-1}x\right) = 1+x$$ Find $F(x)$ satisfying these conditions.

Write $F(x)$ as a rational function with expanded polynomials in the numerator and denominator.

I substituted $x = \frac{x-1}x, -\frac{1}{x-1}$. This gave me the system of equations

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

However, solving this system of equations gave me $F(x) = \frac{x^3-x^2-1}{2(x^2-2)}$, which is incorrect. Where did I go wrong?

I. Kan
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1 Answers1

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It's just a simplification error. Let $g(x)=\frac{x-1}{x}$. Then, note $$ g(x)=1-\frac{1}{x},\quad g(g(x))=\frac{1}{1-x},\quad g(g(g(x)))=x. $$ And so you have the system: \begin{align*} 1+x&=F(x)+F(g(x)),\\ 1+g(x)&=F(g(x))+F(g(g(x))),\\ 1+g(g(x))&=F(g(g(x)))+F(x) \end{align*} from which \begin{align*} F(x)&=[F(x)+F(g(x))+F(g(g(x)))]-[F(g(x))+F(g(g(x)))]\\ &=\frac{1}{2}[1+x+1+g(x)+1+g(g(x))]-(1+g(x))\\ &=\frac{x^3-x^2-1}{2(x^2-\color{red}{x})}. \end{align*}

yurnero
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  • Sorry, that's what I meant to put in. It was a typo - when I solved it that was also what I got. But that answer was incorrect. Edit: when I expanded it to $2x^2-x$, it accepted the answer. Thanks for the help! – I. Kan Apr 24 '18 at 01:12
  • @I.Kan Our solutions differ slightly. Are you saying that is due to a typo and in fact our solutions are the same? In any case, $\frac{x^3-x^2-1}{2(x^2-\color{red}{x})}$ is correct. – yurnero Apr 24 '18 at 01:15
  • When I put in the $\frac{x^3-x^2-1}{2x^2-2x}$, the system accepted the answer. – I. Kan Apr 24 '18 at 01:26