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Consider this functional equation: $$f(x+\frac{1}{x})=x, x\ne0$$ We can solve it easily by setting $y = x+\frac{1}{x}$ and solving for $x$. It's just a quadratic equation in $x$: $x=\frac{y\pm\sqrt{y^2-4}}{2}$ so finally we have $f(y)=\frac{y\pm\sqrt{y^2-4}}{2}$ - or if we want to, we can change $y$ back to $x$: $$f(x)=\frac{x\pm\sqrt{x^2-4}}{2}$$ It can be easily checked that this solution satisfies the original equation.

However the problem is that $x+\frac{1}{x}$ is symmetric about the multiplicative inverse for $x$. That is, if we replace $x$ with $\frac{1}{x}$ it doesn't change. Thus we can do it with the argument on the left hand side of the equation - but the right hand side will then change to $\frac{1}{x}$ and our solution will no longer work.

Furthermore, if we do the change, we will get that $$\forall x\ne0, \frac{1}{x} = f(x+\frac{1}{x})=x$$ thus $\forall x\ne0,x = \frac{1}{x}$ which means we have a contradiction and thus no such function $f(x)$ exist.

Which solution method is correct? Am I missing something here?

Alma Do
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    $f(y)=\frac{y\pm\sqrt{y^2-4}}2$ is not a function. It is two functions. You can't solve this functional equation for all $x$ for precisely that reason. $g(x)=x+1/x$ has a left inverse, $g\circ f(y)=y,$ but $f\circ g(x)\neq x.$ – Thomas Andrews Oct 21 '22 at 18:59
  • In general, if $f(g(x))$ is one-to-one, then $g(x)$ is one-to-one. You want $f(g(x))=x!$ which is one-to-one, but $g(x)=x+1/x$ is not one-to-one. – Thomas Andrews Oct 21 '22 at 20:20

1 Answers1

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How do you want to interpret the $\pm$ in the first attempt? $f$ is a function, so it cannot have more than one output for any given input. As it stands, you have two outputs for any given $|x| \geq 2$.

This agrees with the second method, which ultimately says your $f$ has two outputs $x, 1/x$ for any given $x+\frac1x$. It also implies that no matter which choice of $+$ or $-$ you make in your first attempt, your result will fail to satisfy the functional equation for some [infinitely many] $x$.

Brian Moehring
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  • I meant the $\pm$ as two solutions? If I chose only one sign, say $+$ and then check it on the original equation, it works. So can't we have two solutions with different signs? – Alma Do Oct 21 '22 at 19:12
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    @AlmaDo If you choose $+$, then you get, for $0 < x < 1$, $1/x$, not $x$. You only get $x$ on $0 < x < 1$ if you had chosen the $-$, but if you choose the $-$, you get $1/x$ for $x > 1$. You cannot choose it consistently to have $f(x + 1/x) = x$ for all $x \neq 0$. – Brian Moehring Oct 21 '22 at 19:19