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Consider a sine function parameterized by amplitude, frequency, phase, and constant offset. In other words,

$h(x) = a\sin(fx + p) + c\tag*{}$

where $a$, $f$, $p$, and $o$ are arbitrary real numbers. Suppose we are also given a set of points $\{\,(x_1, y_1),$ $(x_2, y_2),$ $\ldots,$ $(x_n, y_n)\,\}.$ The only constraint on the points is that the $x_i$ are distinct. That is, $i\ne j \implies x_i \ne x_j.$ What is the maximum value $N$ for $n$ such that there always exist choices for $a$, $f$, $p$, and $c$ so that $h(x_i) = y_i$ for all integers $i \in [1, n]\text ?$

A degrees-of-freedom argument suggests that $N \le 4.$ Is $N=4\text ?$ If not, what is it?

An analogous problem for the exponential function suggests that degrees of freedom isn't always an achievable bound. For instance, if $g(x) = ae^{bx} + c,$ then a degree-of-freedom argument suggests that $g(x)$ can always choose $a,$ $b,$ and $c$ such that $g(x)$ interpolates three given points. However, the exponential function is monotonic, so we could never fit a point set that has a local extremum (say, $\{\,(1,0),\ (2,1),\ (3,0)\,\}).$ I haven't been able to find an analogous argument for the sine function, but I'm unconvinced that $N=4.$

Ted Hopp
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  • I think a simple periodicity argument can be used to rule out even three points: $(0,0)$, $(\sqrt 2,0)$ and $(2,0)$. – Henrik supports the community Feb 23 '24 at 16:34
  • Check out the Points $(0,1)$ , $(1,e)$ , $(2,e^2)$ , $(3,e^3)$ : I think $N=3$ is the Maximum. – Prem Feb 23 '24 at 16:35
  • Your argument will work when we have no $C$ , @Henriksupportsthecommunity , With $C$ , we may "handle" those Points !! – Prem Feb 23 '24 at 16:42
  • $c$ doesn't change the period, but nothing says all three $0$'s (in my example) have to be achieved at the same point in the period, so you're right. – Henrik supports the community Feb 23 '24 at 17:03
  • When $C$ is not available , then the 2 $0$ values will fix the Period somehow or the other & then the $3^{rd}$ $0$ may not fit on that Period. When $C$ is available , then the 2 $0$ values may not fix the Period & then the $3^{rd}$ $0$ may fit somewhere or the other. The $C$ is the Issue here , @Henriksupportsthecommunity , to make $N=3$ the Maximum, – Prem Feb 23 '24 at 17:49
  • @Henriksupportsthecommunity - You need some variability in the $y_i.$ If $y_1 = y_2 = \cdots = y_n,$ then setting $a=0$ and $c=y_1$ works (with $f$ and $p$ irrelevant) for any number of points. – Ted Hopp Feb 23 '24 at 19:43
  • @TedHopp: Right, I hadn't thought about that the possibility that $\sin$ could just vanish. – Henrik supports the community Feb 23 '24 at 22:35

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