# Exact Chebyshev expansion coefficients of a function

Mark Richardson, 13th June 2012

## Contents

(Chebfun example approx/ExactChebCoeffs.m)

[Tags: #Chebyshevcoefficients, #residue]

## 1. Introduction

In this example, we shall compare the results of the Chebfun construction process to a known closed-form formula for the Chebyshev expansion coefficients of a function with a pole.

This gives us an excellent excuse to visit some interesting approximation theory from the 1960s!

## 2. The residue method of Elliott

For certain functions, explicit formulas for the coefficients in the Chebsyhev series expansion may be obtained using a contour-integral technique described by Elliott [1]. Here is how it works:

The Chebsyhev coefficients of a Lipschitz-continuous function f can be determined by the integral $$ a_n = \frac{2}{\pi} \int_{-1}^{1} \frac{f(x)T_n(x)}{\sqrt{1-x^2}} {\rm d} x.$$

If f(x) is analytic within a particular contour C in the complex-plane, then by Cauchy's integral formula, we can also write $$ f(x) = \frac{1}{2 \pi i} \int_{C} \frac{f(z)}{z-x} {\rm d} z.$$

If C is large enough to enclose the unit interval, then the second of these two formulas can be subsituted into the first and the orders of integration interchanged to give

$$ a_n = \frac{1}{\pi^2 i} \int_C f(z) \int_{-1}^{1} \frac{T_n(x) {\rm d} x}{(z-x)\sqrt{1-x^2}} {\rm d} z.$$

The integral with respect to x can be computed exactly so that we end up with $$a_n = \frac{1}{\pi i} \int_{C} \frac{f(z)}{\sqrt{z^2-1}(z \pm \sqrt{z^2-1})^n} {\rm d} z.$$

Here, we note that $\rho = |z \pm \sqrt{z^2-1}|$ is the parameter of the usual Bernstein Ellipse $E_\rho$ with the point $z = x + iy$ on its boundary.

So, if $f$ is a function with a pole at $z_0$ whose integral around $E_\rho$ tends to zero as $\rho \to \infty$, then by the Residue theorem we have $$ a_n = \frac{-2r_0}{\sqrt{z_0^2-1}(z_0 \pm \sqrt{z_0^2-1})^n},$$ where $r_0$ is the residue of the pole at $z_0$.

## 3. A function with pole

As an example, consider the function $$ f(x) = \frac{1}{5 + x} .$$

This function can be represented in Chebfun by an interpolant in 17 points.

f = @(x) 1./(5+x); fc = chebfun(f); k = 1:length(fc);

The function has a pole at -5 and corresponding residue 1. Substituing these values into the above formula then gives the us following exact expression for the Chebyshev expansion coefficients $$ a_n = \frac{1}{\sqrt{6}} \frac{(-1)^n}{(5+\sqrt{24})^n} $$

The theoretical coefficients match the ones computed by Chebfun, apart from floating point representation and aliasing effects. The $a_0$ coefficient is out by the usual factor of 2.

exact_coeffs = flipud( (1/sqrt(6)*(-1).^(k-1)./(5+sqrt(24)).^(k-1))' ); cheb_coeffs = chebpoly(fc)'; display([exact_coeffs cheb_coeffs exact_coeffs-cheb_coeffs]) chebpolyplot(fc) title('Chebyshev coefficients of 1/(5+x)') xlabel('n'), ylabel('log(|a_n|)'), grid on

ans = 0.000000000000000 0.000000000000000 -0.000000000000000 -0.000000000000000 -0.000000000000000 0.000000000000000 0.000000000000005 0.000000000000005 0.000000000000000 -0.000000000000047 -0.000000000000047 -0.000000000000000 0.000000000000461 0.000000000000461 0.000000000000000 -0.000000000004565 -0.000000000004565 -0.000000000000000 0.000000000045188 0.000000000045188 0.000000000000000 -0.000000000447312 -0.000000000447312 -0.000000000000000 0.000000004427932 0.000000004427932 0.000000000000000 -0.000000043832012 -0.000000043832012 -0.000000000000000 0.000000433892186 0.000000433892186 0.000000000000000 -0.000004295089853 -0.000004295089853 -0.000000000000000 0.000042517006342 0.000042517006342 0.000000000000000 -0.000420874973563 -0.000420874973563 -0.000000000000000 0.004166232729288 0.004166232729288 -0.000000000000000 -0.041241452319315 -0.041241452319315 0 0.408248290463863 0.204124145231932 0.204124145231932

References:

[1] D. Elliott, The evaluation and estimation of the coefficients in the Chebyshev series expansion of a function, Math. Comp. 18 (1964).