1
$\begingroup$

I have a smooth function :

$$\frac{1}{2\pi} \sin(2\pi x )~,$$

that I would like to approximate on an interval $[-a, b]$. However I already know that all the inputs are guarantied to be close to an integer :

$x = i + \epsilon$, where $-a \leq i \leq b \in \mathbb{Z}$ and $-2^{-e} \leq \epsilon \leq 2^{-e}$, for some given $e$.

So I do not need to approximate the function over the whole interval, but only near the area where my inputs will fall.

Is there a way to compute the Chebyshev nodes in such a way that I can concentrate them near the expected input values in order to reduce the approximation degree for an equivalent precision? If yes, how would that be done?

$\endgroup$
2
  • $\begingroup$ Note that the function you want to approximate is periodic and the value of the approximation at $x = i + \epsilon$ is independent of $i$. So you essentially don't care about the approximation outside of $[-2^e, 2^e]$. This means it suffices to approximate the function accurately on $[-2^e, 2^e]$. So you can use rescaled Chebyshev nodes on this interval. Then you can evaluate the approximation at $x + j$, where $j$ is an integer such that $x + j \in [-2^e, 2^e]$. $\endgroup$ – Ruben Nov 26 '19 at 8:17
  • $\begingroup$ Thing is, I'm doing this evaluation homomorphically, so I don't have access to $x$, I really need the approximation to be in the full specified range. But since I posted this question, I found a way to do it : compute individually the rescaled nodes around each integers (as if those were individual approximations of small degree) and interpolate the resulting Chebyshev polynomial using all the nodes together. $\endgroup$ – Pro7ech Nov 27 '19 at 10:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.