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Today in math class we hade a discussion about linear regression, which is all about finding the best (though not perfect) linear equation that passes through a countable set of points, and people then started wondering about having an arbitary polynomial $p(x)$ reaching every point "perfectly", i.e. $p(x_k) = y_k$ for any list $((x_0,y_0), \cdots (x_n,y_n))$. This usually requires a polynomial of degree $n-1$, and is rather hard to find by hand.

But I remember reading this post a while ago, and started to figure, whether an equation of the form $$y= \alpha \sin(ax) + \beta \sin(bx + \xi), \alpha, \beta,a,b \in \mathbb{R} , \xi \in \mathbb{R} \backslash \mathbb{Q}$$ (i.e. $\xi$ being irrational) would be able to pass through any countable list of points in $\mathbb{R}^2$, my motivation is the following

  • Since the function is nonperiodic we can scale it down such that it will be able to pass through points close to one another

  • Since the individual $\sin$-functions are periodic, we have an extra $2 \pi n$ when we take the $\arcsin$, which will give us a countable amount of equations which we can use to solve using the conditions in any given list.

My motivation may be a tad vague, but I hope you'll atleast be able to get the idea of what I'm after, a trigonometric function that passes through any point of our choise.

Thanks in advance

Periodic Sqare well
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  • For those who may not be aware: The Lagrange Interpolation Formula is a straightforward way to generate the degree $n-1$ polynomial that hits $n$ points (with distinct $x$ coordinates). (I don't know that I'd consider this "hard to find", although it is tedious to compute by hand. That said, I wouldn't expect by-hand-computablility when trigonometrically interpolating infinitely many points.) – Blue Apr 17 '18 at 17:58
  • Thanks, at this point I'm mostly just interested in wheteher my argument hold or not, but I'll investegate the formula. – Periodic Sqare well Apr 17 '18 at 18:01
  • This is not a new concept. if points are equally spaced, coefficient of trigonometric polynomial passing thru a set of points can be found using discrete Fourier transform (DFT). Look up trigonometric interpolation. – Vasili Apr 17 '18 at 18:15
  • If you have $n$ parameters, you can model a curve that will intersect $n$ points. – Doug M Apr 17 '18 at 19:24

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