1

If I want to Fourier transform a function $$t \in [-1,1] \to f $$ but this function and it's first $n$ differentials are not equal at the edges : $$f^{(k)}(-1) \neq f^{(k)}(1) , k \in \{0,1,\cdots,n\}$$, which function $g$ can I additively adjust $f$ with so that the resulting function $h(t) = f(t)+g(t)$ has as sparse a Fourier transform as possible? Let us for simplicitys sake assume that f is awfully nicely behaving on the inside of the interval $]-1,1[$ having bounded $0$:th to $k$:th derivatives there.

Now the trivial solution is to select $g(t)=-f(t)$. This one I am not interested in as it does not help me in any sense. See I still need to store the information about $g$ as this is for data compression purposes.

Own work One simple approach I have thought about is to use a linear compensatory function $$2g(t) = -f(-1)-f(1) + t\cdot(f(-1)-f(1))$$ This will remove the step discontinuity as $h(t)$ should be 0 at the edges, but probably leave us having discontinuities in all the derivatives.

Having started with a polynomial... for the purpose of expanding into being able to attack and remove higher order discontinuities it could be tempting to continue into the realm of polynomial splines or Bezier curves or something like that?

Joako
  • 1,957
mathreadler
  • 26,534
  • A linear $g(t)$ was exactly my first thought. Why do you think it would introduce discontinuities in the derivatives? – aschepler Jun 29 '23 at 23:02
  • 1
    Not introduce new ones. I mean it may not remove any discontinuities already present in the derivatives. Derivative from the right $f'(-1)$ may differ from derivative from the left $f'(1)$ and this may still be the case for $h$ in those points if choosing such a linear $g$. But allowing $g$ to be a higher order polynomial maybe we can smooth those out as well. – mathreadler Jun 29 '23 at 23:42
  • 1
    Your idea of a compensatory function seems like a good one. Your linear function has the right impact on the function values. You want a function that also compensates the derivatives. You just need a polynomial with suitable derivatives at the two ends of the interval. Hermite interpolation will give you this. – bubba Jun 30 '23 at 08:10

0 Answers0