4
$\begingroup$

I have a function

$$f(x) = \left\{ \begin{array}{lr} 0 & 0 \leq x < 1/3\\ q(x) & 1/3 \leq x < 2/3 \\ 1 & 2/3\leq x \leq 1 \end{array} \right.\\ $$ where $q(x)$ is some analytic function that interpolates between the points $x=1/3, 2/3$ and matches the first derivative at these points (say a spline fit). I want to approximate this with a polynomial (as Weierstrauss' theorem says I can) and want to understand how the error falls off as I increase the degree of the approximating polynomial, $p_n(x)$, where the error is defined as $\epsilon_n = \sup|f(x)-p_n(x)|$ .

My initial idea was to choose set of orthogonal polynomials (say Legendre polynomials, $l_m(x)$) and then try express $f(x) \approx \sum_{m=1}^n a_m l_m(x)$. The error can then be bounded by the part of the series that has been cut off, and we should be able to bound this. However, $f(x)$ is not analytic and hence there I have no idea if extracting the $a_m$ is possible in the conventional way of

$$a_m = \int_0^1 f(x)l_m(x) dx $$.

is this is a valid method of extracting the coefficients $a_m$ given $f(x)$ is not analytic? If not, is there a way to go about it?

More generally, is there a better way of find out how the error scales with the degree of the approximating polynomial?

$\endgroup$

2 Answers 2

3
$\begingroup$

You can use Chebyshev series. Or if you want the best uniform approximation to a given degree, try the Remez algorithm.

$\endgroup$
6
  • $\begingroup$ Is there a particular advantage of using Chebyshev polynomials over other orthogonal sets? $\endgroup$ Commented Jul 6, 2017 at 16:34
  • 1
    $\begingroup$ The Chebyshev series for $f(x)$ on $[-1,1]$ is essentially the Fourier cosine series of $f(\cos(t))$ on $[0,2\pi]$, which is quite well-behaved if $f$ is continuous and piecewise smooth. $\endgroup$ Commented Jul 6, 2017 at 17:44
  • 1
    $\begingroup$ I think the easiest to understand is the convolution with $n e^{-\pi n^2 x^2}$ which is analytic $\endgroup$
    – reuns
    Commented Jul 6, 2017 at 17:57
  • $\begingroup$ Would an expansion with Legendre polynomials, theoretically, work just as well (although potentially take longer to converge)? I ask as what I'm really interested in is getting some sort of analytic form for the coefficients, which seems to be harder with Chebyshev polynomials as they're orthogonal with respect to the 1/sqrt(1-x^2) weighting. $\endgroup$ Commented Jul 6, 2017 at 20:00
  • $\begingroup$ To add to that, by the identity theorem we know that f(x) doesn't have a Taylor expansion, so what is our series of Chebyshev/Legendre polynomials actually converging to? $\endgroup$ Commented Jul 6, 2017 at 21:58
1
$\begingroup$

I suggest you take a look at the Chebfun system, and the documentation that goes along with it, especially this page. In my view, it's the state-of-the-art in polynomial approximation. The web site includes some chapters from a book by Nick Trefethen that explain how Chebfun's approximation algorithms operate. Briefly, they work by interpolating the given function at the extrema of Chebyshev polynomials. There are theorems guaranteeing that this approach will give convergence as the polynomial degree increases, provided the given function is somewhat smooth (a Lipschitz condition); the function certainly doesn't need to be analytic.

If you really want a best uniform approximation, then you need the Remez algorithm, which is also implemented in Chebfun. But Remez is a much more complex algorithm and its results are usually only a bit better than you get from interpolation, so, if you're writing the code yourself, it may not be worth the trouble.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .