Chebyshev series for sine

In last week’s post on polynomial approximations for sine, I showed that the polynomial based on Chebyshev series was much better than a couple alternatives. I calculated a few terms of the Chebyshev series for sin(πx) but didn’t include the calculations in that blog post. I calculated the series coefficients numerically, but this post will show how to calculate the coefficients analytically.


The Chebyshev series for a function f(x) on [−1, 1] is given by

f(x) = \frac{a_0}{2} + \sum_{n=1}^\infty a_n T_n(x)

where Tn(x) is the nth Chebyshev polynomial of the first kind. The coefficients are given by

a_n = \frac{2}{\pi} \int_{-1}^1 \frac{f(x) \,T_n(x)}{\sqrt{1-x^2}} \, dx

One way of defining the polynomials Tn(x) is

T_n(\cos\theta) = \cos(n \theta)

and so the change of variables x = cos θ lets us conclude

a_n = \frac{2}{\pi} \int_0^\pi f(\cos \theta) \cos n\theta \, d\theta

Series for sin(πx)

Now for our particular function, f(x) = sin(πx), we know by symmetry that the coefficients with even subscripts will be zero. This is because sine is an odd function, and Tn is an even function when n is even,

Using equation 10.9.2 here we can prove that if n = 2k + 1 then

a_n = (-1)^k 2 J_n(\pi)

where Jn is the nth Bessel function of the first kind.

(The preceding sentence was the conclusion to a fair amount of fumbling around on my part. As is often the case in mathematics, the length of the write-up is unrelated to the length of the discovery process.)

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