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I'd like to ask how to find all solutions to the functional equation $f(x)\cdot f(f(x)+\frac{1}{x})=1$, where $f: (0, +\infty)\to\mathbb{R}$ is strictly increasing?

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  • $\begingroup$ Related math.stackexchange.com/questions/1051745/… ? $\endgroup$
    – Sil
    Mar 31, 2019 at 18:21
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    $\begingroup$ $f(x) = 1$ is a solution $\endgroup$
    – fGDu94
    Mar 31, 2019 at 18:31
  • $\begingroup$ Over what domains? $\mathbb{R}\to \mathbb{R}$ with $x\neq 0$? Please specify! $\endgroup$
    – user574848
    Mar 31, 2019 at 23:27
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    $\begingroup$ "I'd like to ask how to find all solutions to the functional equation..." please include a context and your attempts/thoughts to solve this. Otherwise will not be probably considered a good question and will attract downvotes. $\endgroup$
    – Sil
    Apr 7, 2019 at 9:19

2 Answers 2

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I'll answer the problem by myself. For any $x$, let $f(x)=a$. Then, $f(a+\frac{1}{x})=\frac{1}{a}$. Furthermore, we have $f(a+\frac{1}{x})f(f(a+\frac{1}{x})+\frac{1}{a+\frac{1}{x}})=1$. Therefore, $f(\frac{1}{a}+\frac{1}{a+\frac{1}{x}})=a$. Since $f$ is strictly increasing, we have $\frac{1}{a}+\frac{1}{a+\frac{1}{x}}=x$. Thus, the value of $a$ can be obtained by solving the quadratic equation.

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  • $\begingroup$ Nice. Maybe the $a$ by itself obfuscates the fact that it depends on $x$, more standard way to write it would be to keep it as $f(x)$ and derive same result, but good work. $\endgroup$
    – Sil
    Apr 7, 2019 at 9:12
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Hint:

Let $g(x)=f(x)+\dfrac{1}{x}$ ,

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

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

$\therefore\left(g(x)-\dfrac{1}{x}\right)\left(g(g(x))-\dfrac{1}{g(x)}\right)=1$

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    $\begingroup$ Sorry, but I still cannot see how to derive the formula of $g(x)$. $\endgroup$
    – Hang Wu
    Apr 2, 2019 at 15:31

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