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 Quote by disregardthat The problem with polynomials of higher degree is that they cannot be periodic, so I think it would be hard to accomplish that which you describe.
Actually that would work. We know that
$f(x) = \sin(2 \pi x / L)$
has period L, so in our example L=6.

Then all we need to do is find the Newton polynomial solving
$f(0) \mapsto 0$
$f(1) \mapsto 1$
$f(2) \mapsto 1$
and so on. Call this p(y).

Then $p \circ f (x)$ gives what we want.