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I need to find a continuous function which takes every real value exactly 2n+1 times, for any $n \in \mathbb{N} $

Thank you in advance

3 Answers 3

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Hint:

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    This is great! +12012-10-09
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I will show you one way to do the case $n=1$. You can generalize it to all $n$.

Start with a function $\phi:[0,1]\to[0,1]$ whose graph is

enter image description here

$\phi$ takes every $y\in(0,1)$ exactly three times, but $0$ and $1$ only two. Now define $f$ on $[k,k+1]$ as $\phi(x-k)+k$, $k\in\mathbb{Z}$.

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    If you insist on a formula, here is one: $f(x)=\sin^2(3\,\pi(x-\lfloor x\rfloor)/2)-\lfloor x\rfloor$, where $\lfloor x\rfloor$ is the **floor function** (greatest integer less that $x$.)2012-10-10
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Same idea:

enter image description here

$f(x) = \sin(x)+ax$, where $a \approx 0.21723$ is the solution of $\sqrt{1-a^2} - a\pi-a\arccos(a) = 0$, so that one local maximum value equals a subsequent local minimum value.