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On k-regular functions

Mathematics Asked by Mathematician 42 on December 25, 2021

My father asked me the following question:

A real function $fcolon mathbb{R}to mathbb{R}$ is called $k$-regular if for each $yin mathbb{R}$, the equation $f(x)=y$ has precisely $k$ distinct solutions.

  1. Show that there are no continuous $k$-regular functions for $k$ even.
  2. For each $k$ odd, provide an example of a continuous $k$-regular function.

Solving this question was fairly easy but led me to the following problem:

For each $ain mathbb{R}^+$ (including $0$), consider the function $f_acolon mathbb{R}to mathbb{R}:xmapsto x+asin(x)$. Clearly $f_0(x)=x$ yields a $1$-regular function. Plotting these functions for $a=4$ and $a=5$, it becomes fairly obvious that there should be a value of $4leq aleq 5$ such that $f_a$ is $3$-regular. (The value should be ‘close’ to 4.5 as can be seen on a plot).

How can we prove that such a value exists and is it feasable to accurately find this value? More generally, given $k=2l+1$, for which value of $a$ is $f_a$ a $k$-regular function?

I tried the following wishful thinking: I want that $f_a'(b)=0=1+acos(b)$ and that $f(b)=0=b+asin(b)$. From this one finds that $b-1=tan(b)$. According to wolframalpha, -4.53360… is a solution to this equation and thus $a=frac{-b}{sin(b)}$ yields a value close to 4.6 which seems like a decent candidate.

Based on the above equations, there are many more candidates but only two solutions two $b-1=tan(b)$ seem to yield a $3$-regular function $f_a$. I’m not really seeing why these particular values should work and the others fail.

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