MHB Interpolating Points with Continuous Modular Functions?

[SatyaDas]

Define a continuous function $$F(x;n)$$ that interpolates points (x, x mod n) for a given integer n and all integer x. For example $$F(x;2)=\frac{1}{2}-\frac{1}{2}\cos\left(\pi x\right)$$ interpolates all points (x, x mod 2) when x is an integer. Similarly $$F(x;3)$$ should interpolate points (0,0), (1,1), (2,2), (3,0), (4,1), and so on and so forth.

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[Opalg]

As a first shot, here's $F(x;3) = 1 - \cos\left(\frac{2\pi x}3\right) - \frac1{\sqrt3}\sin\left(\frac{2\pi x}3\right)$:

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[SatyaDas]

As a first shot, here's $F(x;3) = 1 - \cos\left(\frac{2\pi x}3\right) - \frac1{\sqrt3}\sin\left(\frac{2\pi x}3\right)$:

Nice attempt. Below is my output for F(x;3).

 

[Opalg]

Of course, if you only require $F(x;n)$ to be continuous then you can use a sawtooth function consisting of straight line segments from $(kn,0)$ to $(kn+n-1,n-1)$ and from $(kn+n-1,n-1)$ to $((k+1)n,0)$ (for all $k\in\Bbb{Z}$). But I am assuming that you want $F(x;n)$ to be a smooth function. So it presumably needs to be a trigonometric function.

For $n=4$ I'm getting $F(x;4) = \frac32 - \cos\bigl(\frac{\pi x}2\bigr) - \sin\bigl(\frac{\pi x}2\bigr) - \frac12\cos(\pi x)$. I don't yet see what the general formula should be, but it seems that the constant term in $F(x;n)$ must be $\frac{n-1}2$.

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[SatyaDas]

Of course, if you only require $F(x;n)$ to be continuous then you can use a sawtooth function consisting of straight line segments from $(kn,0)$ to $(kn+n-1,n-1)$ and from $(kn+n-1,n-1)$ to $((k+1)n,0)$ (for all $k\in\Bbb{Z}$). But I am assuming that you want $F(x;n)$ to be a smooth function. So it presumably needs to be a trigonometric function.

For $n=4$ I'm getting $F(x;4) = \frac32 - \cos\bigl(\frac{\pi x}2\bigr) - \sin\bigl(\frac{\pi x}2\bigr) - \frac12\cos(\pi x)$. I don't yet see what the general formula should be, but it seems that the constant term in $F(x;n)$ must be $\frac{n-1}2$.

Indeed, I meant smooth function and your finding of constant term matches with mine. So, high five. Below is my output for F(x;4). I guess it will help to get the idea that there is a pattern.

 

[SatyaDas]

Here is my solution.
Since $F(x;n)$ is periodic with period $n$, we can assume that the function can be expressed as
$$
F(x;n)=\sum_{i=0}^{n-1}\left(a_i\cos\left(\frac{2\pi i x}{n}\right)+b_i\sin\left(\frac{2\pi i x}{n}\right)\right).
$$
There are total of $2n$ unknowns.
We get $n$ equations by using the fact
$$
F(j;n)=j\text{ for all }j\in Z\text{ and }0\le j\le n-1.
$$
We need $n$ more equations so that we can find all the unknowns. For that purpose we can impose more restrictions on the properties of the function $F(x;n)$. If we assume $F'(x;n)=0$ for all $x\in Z$ then we get $n$ more equations. So, now we have a system of $2n$ equations with $2n$ unknowns. If we solve them we get:
$$
a_0=\frac{n-1}{2},\\
a_i=-\frac{n-i}{n} \text{ for all } 0<i<n,\\
b_0=0,\text{ and}\\
b_i=-\frac{n-i}{n}\cot\left(\frac{i\pi}{n}\right)\text{ for all } 0<i<n.
$$
If we simplify things we get
$$
F\left(x;n\right)=\frac{n-1}{2}-\sum_{i=1}^{n-1}\left(1-\frac{i}{n}\right)\csc\left(\pi\frac{i}{n}\right)\sin\left(\pi\frac{i}{n}\left(2x+1\right)\right).
$$
This graph can be visualized interactively at desmos:
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