I was looking at the following problem: "If a graph of a function rotated 90 degrees about the origin, then it is not changed. Is there such a function?" The only one I can think of if is $f(x)=0$ defined on $x=0$. Is it possible to prove that this is the only such function? I understand that due to the rotational symmetry $f(-f(x))=x$ but how do I go from here?
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If I'm understanding correctly, $f$ would have to be one-to-one, otherwise it would fail the vertical line test after rotation. – Randall Aug 28 '17 at 19:38
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A necessary condition: $y=f(x)\iff f(-y)=x$ – ajotatxe Aug 28 '17 at 20:08
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ajotatxe observation. Will allow for a method of picking functions point by point. $f(x) = \sqrt{1-x^2}; -1\le x < -1/2;0\le x < 12; f(x) = -\sqrt{1-x^2}$ otherwise, works. – fleablood Aug 28 '17 at 20:25
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One example defined on all of $\mathbb R$ would be
$ \raise 5em \hbox{$ f(x) = \begin{cases}
0 & x=0 \\
(-1)^{\lceil x\rceil}x -1 & x>0 \\
(-1)^{\lfloor x\rfloor}x+1 & x<0 \end{cases} $} $
Bonus question: Is there a solution with finitely many discontinuities? I can't find one.
hmakholm left over Monica
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@Henning Makholm. Thank you! The second part of the problem was "prove that $f(b)=b$ has a unique solution" but I think I can take care of that one now. – Vasili Aug 29 '17 at 12:27
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To answer hmakholm left over Monica's question there is one example of such a function with a finite number of discontinuities:
$ \raise 3em \hbox{$ f(x) = \begin{cases} 0 & x=0 \\ \tan{x}+\frac{\pi}{2} & 0<x<\frac{\pi}{2}\\ \tan(x)-\frac{\pi}{2} & -\frac{\pi}{2} <x<0 \\ -\arctan(x-\frac{\pi}{2}) & \frac{\pi}{2} \leq x \\ -\arctan(x+\frac{\pi}{2}) & x \leq -\frac{\pi}{2} \\ \end{cases} $} $
LIR
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Rotating that by 90 degrees doesn't even give a function (and so in particular not itself) because it is not injective: $f(-\frac\pi2)=f(0)=f(\frac\pi2)=0$. – hmakholm left over Monica Feb 29 '20 at 03:29
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@hmakholmleftoverMonica Could $x=\frac{\pi}{2}$ and $x=-\frac{\pi}{2}$ be excluded from the domain? – nickgard Feb 29 '20 at 11:48