Abstract
The usual way to determine the asymptotic behavior of the Chebyshev coefficients for a function is to apply the method of steepest descent to the integral representation of the coefficients. However, the procedure is usually laborious. We prove an asymptotic upper bound on the Chebyshev coefficients for the k t h k^{th} integral of a function. The tightness of this upper bound is then analyzed for the case k = 1 k=1 , the first integral of a function. It is shown that for geometrically converging Chebyshev series the theorem gives the tightest upper bound possible as n → ∞ n\rightarrow \infty . For functions that are singular at the endpoints of the Chebyshev interval, x = ± 1 x=\pm 1 , the theorem is weakened. Two examples are given. In the first example, we apply the method of steepest descent to directly determine (laboriously!) the asymptotic Chebyshev coefficients for a function whose asymptotics have not been given previously in the literature: a Gaussian with a maximum at an endpoint of the expansion interval. We then easily obtain the asymptotic behavior of its first integral, the error function, through the application of the theorem. The second example shows the theorem is weakened for functions that are regular except at x = ± 1 x=\pm 1 . We conjecture that it is only for this class of functions that the theorem gives a poor upper bound.