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Investigating the $$\cosh (\pi \sin (\pi r))$$ Looked at the graph and seems it is somehow related to some combination of $sin$ and some unknwon constants $$1.68 \pi (\sin (2 \pi r-1.68)+2.4)-2 \pi$$

The below is the graph of both functions, where blue is the first one and yellow is the second function. Is there any idea if the constants can be tuned enough so we get an exact relation?

enter image description here

Adding more details based on @Oliver Oloa comment: $$\cosh (\pi \sin (\pi r)) = 2 \sum _{n=0}^{\infty} (-1)^n I_{2 n}(\pi ) \cos (2 \pi n r)-I_0(\pi )$$

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    $\begingroup$ it has period $1$ and is even; it has a Fourier cosine series adapted to that period. $\endgroup$ Commented Sep 16 at 21:35
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    $\begingroup$ You can stop the summation at $n=3$ $\endgroup$ Commented Sep 17 at 6:56

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Hint. One may use the Fourier series of $e^{\alpha\sin (\pi x)}$:

$$ e^{\alpha\sin (\pi x)}= I_0(\alpha) + 2\sum_{n= 0}^\infty(-1)^n I_{2n+1}(\alpha)\sin((2n+1) \pi x)+2\sum_{n= 1}^\infty(-1)^n I_{2n}(\alpha)\cos(2n\pi x) $$ where $I_n$ is the modified Bessel function of the first kind.

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    $\begingroup$ Thank you modified the original post based on this comment. $\endgroup$ Commented Sep 17 at 5:13
  • $\begingroup$ Hi Olivier ! Very long time no see. Cheers :-) $\endgroup$ Commented Sep 17 at 6:09
  • $\begingroup$ Hi Claude! Hope you are OK :-) $\endgroup$ Commented Sep 21 at 20:49
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$$\cosh(\pi\sin(\pi r))$$ oscillates between $\cosh(0)=1$ and $\cosh(\pi)\approx 11.59$, hence around $6.30$ with an amplitude $5.30$.

You might hope to approximate your function with a shifted and scaled sine of $2\pi r$ because the hyperbolic cosine is approximately parabolic ($\cosh(x)\approx 1+\frac12x^2$) and $\sin(2\theta+\frac\pi2)=\cos(2\theta)=2\sin^2(\theta)$ is also quadratic.

But for the mapping to be exact, you would need $\cosh(x)=1+\frac12x^2$ exactly, or $\sqrt{2(\cosh(x)-1)}$ to be linear, which it isn't.

enter image description here

If you want to perfectly match the extrema, use

$$\frac{\cosh(1)+1}2-\frac{\cosh(1)-1}2\cos(2\pi r)$$

enter image description here

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For $0\leq r \leq 1$, you can use the series expansion of $\cosh(x)$ and make $x=\pi\sin(\pi r)$.

Using $$\cosh(x)=\sum_{n=0}^p \frac {x^{2n}}{(2n)!}+O(x^{2p+2})$$ consider the norm $$\Phi_p=\int_0^1\Bigg(\cosh (\pi \sin (\pi r))- \sum_{n=0}^p \Bigg)^2\,dr$$

$$\left( \begin{array}{cc} p & \Phi_p \\ 3 & 1.34327\times 10^{-02} \\ 4 & 1.35670\times 10^{-04} \\ 5 & 6.67033\times 10^{-07} \\ 6 & 1.77333\times 10^{-09} \\ 7 & 2.76048\times 10^{-12} \\ \end{array} \right)$$

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  • $\begingroup$ thank you, very interesting. $\endgroup$ Commented Sep 19 at 4:27

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