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How many function satisfy $f(x)=f^{-1}(x)$ if the $f(x)$ have only have one subtraction operation and one division operation?

Please help enlighten me if you could. By the way, it looks like no way could start the proof!

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Please check out the post 66093, and comments therein. –  Sasha Oct 12 '11 at 1:08
I don't understand what "have only one subtraction and one division operation" means. Are you literally restricting attention to only functions of the form $(a-b)/c$, $c/(a-b)$ and $a-(b/c)$ where zero, one or two of the letters are $x$ and the other(s) are constants? –  anon Oct 12 '11 at 1:22
@anon - it could also be (b/c)-a –  Victor Oct 12 '11 at 1:28
Technically anything in the form $b/c-a$ can be put into the form $a'-b'/c'$ so they're the same, but more importantly: these are the only functions you're asking the question about? If so, then it's just a special case of linear fractional transformations associated to matrices $A\in GL_2(F)$, $F=\mathbb{R}$ such that $A^2=I_2$... –  anon Oct 12 '11 at 1:36
@Victor: Well, there are intuitive ways to understand what LFT's are in and of themselves, but that's a bit beyond the scope of the question. There is a very simple way, though: go through all three cases and solve for the restrictions on $a,b,c$ that make $f(f(x))=x$ (and also make $f$ invertible). Also, I don't have a college degree :) –  anon Oct 12 '11 at 1:45

1 Answer 1

up vote 3 down vote accepted

Suppose $f(x)=(x-a)/b$. Then $f(f(x))=((x-a)/b-a)/b=(x/b^2)-(a/b^2)-(a/b)$. To have $f(f(x))$ identically equal $x$, we need $b^2=1$ and $(a/b^2)+(a/b)=0$. So $b=1$, $a=0$, and $f(x)=x$, or else $b=-1$, $a$ is arbitrary, $f(x)=c-x$ for any $c$.

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You didn't take into account functions like $x/(x-1)$. –  anon Oct 12 '11 at 2:50
@anon, you are right. I made an interpretation of what OP meant by "one subtraction and one division." I took it to mean, start with $x$, subtract some number, and divide by some number. Other interpretations will lead to other functions, such as the one you suggest. Only OP can tell us what the question really means. –  Gerry Myerson Oct 12 '11 at 3:33
Gerry: I made my interpretation based on this question @Victor asked a few hours ago. –  anon Oct 12 '11 at 3:37

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