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What are the general solutions of the functional equations?

$$ f(x,y)+f(y,z)=\frac{1}{f(x,z)} $$

$$ f(x,y)f(y,z)f(x,z)=1 $$

Umar
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3 Answers3

2

I am assuming the variables are real.

For the first, setting $x=y=z$ gives $2f(x,x)=\frac 1{f(x,x)}$, so $f(x,x)=\pm\frac {\sqrt 2}2$

As Ivan Loh points out, the sign choice must be the same for all $x$:
$\quad$$y=z$ gives $f(x,z)+f(z,z)=\frac1f{(x,z)}$
$\quad$$x=y$ gives $f(x,x)+f(x,z)=\frac 1{f(x,z)}$
$\quad$so $f(x,x)=f(z,z).$

Assume $f(x,x)=\frac {\sqrt 2}2$:
$\quad$Setting $x=z$ gives $f(x,y)+f(y,x)=\sqrt 2$
$\quad$Setting $y=z$ gives $f(x,z)+\frac {\sqrt 2}2 = \frac 1{f(x,z)}$
$\quad$This has solutions $f(x,z)=\frac {\sqrt 2}2, -\sqrt 2$
$\quad$But only the first satisfies the original equation

Now if $f(x,x)=-\frac{\sqrt 2}2$
$\quad$Setting $x=z$ gives $f(x,y)+f(y,x)=-\sqrt 2$
$\quad$Setting $y=z$ gives $f(x,z)-\frac {\sqrt 2}2 = \frac 1{f(x,z)}$
$\quad$This has solutions $f(x,z)=-\frac {\sqrt 2}2, \sqrt 2$
$\quad$But only the first solves the original equation

So $f(x,y)$ is the same for all $x,y$ and is $\pm \frac {\sqrt 2}2$

For the second, setting $x=y=z$ gives $f(x,x)=1$
Setting $x=z$ we get $f(x,y)f(y,x)=1$ or $f(x,y)=\frac 1{f(y,x)}$
Setting $x=y$ we have $f(x,z)^2=1$
So $f(x,y)=\pm 1$ for all $x,y$

As Hagen von Eitzen points out, we can divide the reals into two equivalence classes in any way we want (including one of them empty). Let $a,b$ be representatives of the two classes. We have $f(a,a)=f(b,b)=1, f(a,b)=f(b,a)=-1$

Ross Millikan
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  • Why can't $f(x,z)=-1$? – vadim123 May 08 '13 at 14:13
  • @vadim123: for the second? because then $f(x,x)^3$ would equal $-1$ – Ross Millikan May 08 '13 at 14:18
  • You've proved that $f(x,z)^2=1$ for all $x,z$. Perhaps this is sometimes 1, sometimes -1. In those cases where $x=z$, it's 1, but maybe for other cases -1. – vadim123 May 08 '13 at 14:19
  • @vadim123: good point. It applies to the first as well. I'll think on it. – Ross Millikan May 08 '13 at 14:20
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    Write $x\sim y$ if $f(x,y)=1$. Then you've shown that $\sim$ is reflexive and symmetric. As it is also transitive (by the original equation), $\sim$ is an equivalence relation. It cannot have more than two equivalence classes, but any equivalence relation with (at most) two classes solves the original equation. – Hagen von Eitzen May 08 '13 at 14:25
  • @vadim123: in the first, the $x=z$ line proves the claim that $f(x,y)=\frac {\sqrt 2}2$, the root $-\sqrt 2$ is not allowed. – Ross Millikan May 08 '13 at 14:29
  • @RossMillikan Yes, a sum of two elements $\in{\frac{\sqrt 2}2,-\sqrt 2}$ that produces $\sqrt 2$ msut be $\frac{\sqrt 2}2+\frac{\sqrt 2}2$. – Hagen von Eitzen May 08 '13 at 14:31
  • @HagenvonEitzen: I realized that immediately and deleted. I have incorporated your comment in your answer. Thanks. – Ross Millikan May 08 '13 at 14:36
  • For the first, $f(x, y)=-\frac{\sqrt{2}}{2} , \forall x, y \in \mathbb{R}$ is also a solution. This solution was missed because in your second line, $2f(x, x)=\frac{1}{f(x, x)}$ gives $f(x, x)=\pm \frac{\sqrt{2}}{2}$, instead of just $\frac{\sqrt{2}}{2}$. – Ivan Loh May 08 '13 at 15:20
  • @IvanLoh: correct. That ruins a lot. I'll work on it – Ross Millikan May 08 '13 at 15:25
  • Fortunately, this can easily be worked around: $y=z$ gives $f(x, z)+f(z, z)=\frac{1}{f(x, z)}$, and $x=y$ gives $f(x, x)+f(x, z)=\frac{1}{f(x, z)}$ so $f(x, x)=f(z, z)$. – Ivan Loh May 08 '13 at 15:26
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Under the false assumption that they need to both hold,

From the first equation, $f$ is never zero. From the second equation, $f(x,y)f(y,z)=\frac{1}{f(x,z)}$; combining with the first we get $$f(x,y)+f(y,z)=f(x,y)f(y,z)$$

Taking $x=y=z$ we get $2f(x,x)=f(x,x)^2$; since $f(x,x)\neq 0$ we have $f(x,x)=2$ for all $x$. Taking $x=y$ we get $2+f(y,z)=2f(y,z)$ so $f(y,z)=2$ for all $y,z$. But this doesn't satisfy the second equation, so there are no solutions.

vadim123
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-1

There are infinitely many(whose cardinality is $2^\Bbb R$) solutions $f:\Bbb R\to \Bbb R$ of the simple functional equation $$ f(x)^2=1. $$ Let $A$ be an arbitrary subset of $\Bbb R$. Then, $f(x)=1$ for $x\in A$, $f(x)=-1$ for $x\in A^c$ are the solutions of the equation.

Therefore, the above proof(by Millikan) makes no sense.

Chung. J
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  • I would suggest that the (version of the) proof (that you responded to) made sense, at least to two readers. It was incorrect. A better approach would be to leave this as a comment to that effect, or to write up the correct proof fully. – Ross Millikan May 08 '13 at 15:17
  • Sorry! I didn't know the approach other ways. I agree that his proof will give some idea for a complete proof. – Chung. J May 08 '13 at 15:56