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Are linear functions the only functions that retain their form when inverted -i.e. an exponential function becomes a log function when inverted, a square function becomes a square root when inverted, but a linear function remains linear when inverted?

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  • $\begingroup$ The inverse of $x\mapsto 1/x$ coincides with itself. $\endgroup$ – egreg Jan 18 '14 at 16:54
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The functions in question are those whose implicit equations are symmetrical in x and y. E.g., if $xy=a=$ constant, then $x(y)=\dfrac ay$ , and $y(x)=\dfrac ax$ . Or if $x^n\pm y^n=r^n$, then $x(y)=\sqrt[n]{1\mp y^n}$, and $y(x)=\sqrt[n]{1\mp x^n}$. Etc.

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  • $\begingroup$ Hi Lucian, this looks like a reasonable criterion (that the implicit form should contain some kind of symmetry), but how should “symmetrical” be interpreted? If I have an expression like $y=ax+b$, the implicit form $y-ax=b$ looks far from symmetric and yet the inverse is of the same form. $\endgroup$ – pppqqq Jan 18 '14 at 17:44
  • $\begingroup$ @pppqqq: I guess that would depend on how you interpret the expression "the same form". I took to mean "identical". If you take it to mean "similar", then I guess that provides you with more leeway, since its meaning is looser, and more subjective. $\endgroup$ – Lucian Jan 18 '14 at 17:52
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Möbius transforms have inverses of the same form. The inverse of the function $$f(x) = \frac{ax+b}{cx+d}$$ is another function of the same type: $$f^{-1}(x) = \frac{dx-b}{-cx+a}.$$

This subsumes both the linear functions $x\mapsto ax+b$ (take $c=0, d=1$) and the reciprocal function $x\mapsto \frac bx$ (take $a=d=0, c=1$) as special cases.

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