Chebyshev Density and Potential/Runge Phenomenon

[brru25]

Homework Statement

Show that the integral from -1 to 1 of p(x)*log|z-x| dx equals log|z - sqrt(z^2 -1)| / 2, where p(x) = 1 / (pi*sqrt(1-x^2))

2. Other information

This topic comes from Chebyshev interpolation. p(x) is the Chebyshev density.

The Attempt at a Solution

The best idea I could come up with was to use z = x + iy and substitute that into log|z-x| to get log|x+iy-x| = log|iy| = log(y) since |iy| = sqrt(0^2 + y^2) = sqrt(y^2) = y. That left me with just one term with an x to integrate. Then I used trig substitution of x = sin(theta) to have the integral become:

integral from -pi/2 to pi/2 of log(y)/pi d(theta). That left me with an answer of just log(y) which is clearly not right, or at least not in the form that should be. I think my approach to the integration is completely off.

 

[bemer42]

Hello,

Is this question from the exercises of chapter 5 of the text 'Spectral Methods' by Trefethen? If so, I believe equation (5.9) of that text is wrong. The potential should have the form

phi(z) = log(|z+sqrt(z^2 - 1)/2|)

so that the minus sign is actually a plus. I'm working on the first exercise. Your question seems to be pertaining to the fifth exercise. Not sure if this helps.