Chebyshev series

Remarks

Example: for $f(x) = x^2$, one finds $a_0 = 1$, $a_2 = 1/2$ and $a_k = 0$ otherwise. Check: $\frac{2}{\pi} \int_{-1}^{1} x^2 (1 - x^2)^{-1/2} \, dx = \frac{2}{\pi} \cdot \frac{\pi}{2} = 1$, and $\frac{1}{2} + \frac{1}{2}(2x^2 - 1) = x^2$.

Non-example: $f(x) = (1 - x^2)^{-1/2}$ has no Chebyshev series, since $\int_{-1}^{1} (1 - x^2)^{-1} \, dx$ diverges.

With $x = \cos \theta$ and $g(\theta) = f(\cos \theta)$, one gets $a_k = \frac{2}{\pi} \int_0^{\pi} g(\theta) \cos(k\theta) \, d\theta$. So the Chebyshev series of $f$ is the Fourier series of the even function $g$, and convergence results transfer. For example, if $f$ is Lipschitz on $[-1, 1]$, the series converges uniformly to $f$. The $T_k$ are orthogonal polynomials for the weight $(1 - x^2)^{-1/2}$, and, when $\int_{-1}^{1} f(x)^2 (1 - x^2)^{-1/2} \, dx < \infty$, the partial sums are the least squares approximations of $f$ for this weight.