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I have a simple, continuous, real-valued function $\sigma$, whose functional form I know, and I know that two other invertible functions, $f$ and $g$, satisfy the following relationships:

  • $f\circ\sigma\circ\sigma\circ f^{-1}=1$,
  • $g\circ\sigma\circ\sigma\circ g^{-1}=1$,
  • $g^{-1}\circ\sigma\circ f\circ f\circ \sigma \circ g^{-1}=1$,
  • $f^{-1}\circ\sigma\circ g\circ g\circ \sigma \circ f^{-1}=1$.

where $(a\circ b)(x)=a(b(x))$ is function composition, $a^{-1}$ is the inverse of function $a$, and $1$ is the identity function.

How do I go about finding $f$ and $g$, or any further relationships between them? If $\sigma=1$ then $f=g=h$ for any invertible function $h$ seems to work, but how about more generally? Any suggestions of topics to investigate very warmly received.

Klangen
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AnotherShruggingPhysicist
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1 Answers1

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We have that

$$f\circ\sigma\circ\sigma\circ f^{-1}=1\iff \sigma\circ\sigma=1$$

$$g\circ\sigma\circ\sigma\circ g^{-1}=1\iff \sigma\circ\sigma=1$$

$$g^{-1}\circ\sigma\circ f\circ f\circ \sigma \circ g^{-1}=1\iff \sigma\circ f\circ f\circ \sigma=g\circ g$$

$$f^{-1}\circ\sigma\circ g\circ g\circ \sigma \circ f^{-1}=1\iff \sigma\circ g\circ g\circ \sigma=f\circ f$$

If we assume $\sigma =1$ then we have

$$f\circ f=g\circ g.$$ But we can get $f,g$ from the relation above.

Some solutions:

  • If $f=g$ we are done.
  • Assuming the domain is $\mathbb{R}$ and that $f\circ f=g\circ g=1$ we can have $f(x)=x, g(x)=k-x$ or $f(x)=a-x, g(x)=b-x.$
  • Assuming the domain is $\mathbb{R}\setminus\{0\}$ and that $f\circ f=g\circ g=1$ we can have $f(x)=\dfrac 1x, g(x)=k-x$ or $f(x)=a-x, g(x)=b-x.$
mfl
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  • Thanks @mfl - I was hoping that there'd be a nice solution where $\sigma \neq 1$, but by your first line it looks like that might not be possible? I hadn't realised that $f\circ\sigma\circ\sigma\circ f^{-1}=1$ necessitated $\sigma\circ\sigma=1$. – AnotherShruggingPhysicist Aug 22 '18 at 13:07
  • Note that $f\circ\sigma\circ\sigma\circ f^{-1}=1\iff f\circ\sigma\circ\sigma\circ f^{-1}\circ f=1\circ f\iff f\circ\sigma\circ\sigma=f.$ And thus $f\circ\sigma\circ\sigma=f\iff f^{-1}\circ f\circ\sigma\circ\sigma=f^{-1}\circ f\iff \sigma\circ\sigma=1 .$ – mfl Aug 22 '18 at 13:20
  • Assume $\sigma(x)=1-x.$ Then $f(x)=a-x,g(x)=b-x,$ and $f(x)=x,b(x)=b-x$ are solutions. – mfl Aug 22 '18 at 13:29
  • Are there any nonlinear functions $\sigma$ such that $\sigma\circ\sigma=1$? – AnotherShruggingPhysicist Aug 22 '18 at 13:59
  • For example $1/x$ and $\sqrt{1-x^2}$ – mfl Aug 22 '18 at 14:01
  • That's really helpful, thanks: is there a general way to go from the four equations you derived in the answer to the functional forms in your bullet points, that I could use for, say, $\sigma(x)=1/x$? – AnotherShruggingPhysicist Aug 22 '18 at 14:03
  • If $\sigma(x)=1/x$ then $f(x)=1/x, g(x)=x$ and $f(x)=1/x, g(x)=k-x$ are solutions. – mfl Aug 22 '18 at 14:08