Chebyshev polynomials/Proofs

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[edit] Second derivative at end points

The second derivative of the Chebyshev polynomial of the first kind is

T''_n = n \frac{n T_n - x U_{n - 1}}{x^2 - 1}

which, if evaluated as shown above, poses a problem because it is indeterminate at \scriptstyle x = \pm 1. Since the function is a polynomial, (all of) the derivatives must exist for all real numbers, so the taking to limit on the expression above should yield the desired value:

T''_n(1) = \lim_{x \to 1} n \frac{n T_n - x U_{n - 1}}{x^2 - 1}

where only x = 1 is considered for now. Factoring the denominator:

T''_n(1) = \lim_{x \to 1} n \frac{n T_n - x U_{n - 1}}{(x + 1)(x - 1)} = \lim_{x \to 1} n \frac{\frac{n T_n - x U_{n - 1}}{x - 1}}{x + 1}.

Since the limit as a whole must exist, the limit of the numerator and denominator must independently exist, and

T''_n(1) = n \frac{\lim_{x \to 1} \frac{n T_n - x U_{n - 1}}{x - 1}}{\lim_{x \to 1} (x + 1)} = \frac{n}{2} \lim_{x \to 1} \frac{n T_n - x U_{n - 1}}{x - 1} .

The denominator (still) limits to zero, which implies that the numerator must be limiting to zero, ie Un − 1(1) = nTn(1) = n which will be useful later on. Since the numerator and denominator are both limiting to zero, L'Hôpital's rule applies:

\begin{align} T''_n(1) & = \frac{n}{2} \lim_{x \to 1} \frac{\frac{d}{dx}(n T_n - x U_{n - 1})}{\frac{d}{dx}(x - 1)} \\ & = \frac{n}{2} \lim_{x \to 1} \frac{d}{dx}(n T_n - x U_{n - 1}) \\ & = \frac{n}{2} \lim_{x \to 1} \left(n^2 U_{n - 1} - U_{n - 1} - x \frac{d}{dx}(U_{n - 1})\right) \\ & = \frac{n}{2} \left(n^2 U_{n - 1}(1) - U_{n - 1}(1) - \lim_{x \to 1} x \frac{d}{dx}(U_{n - 1})\right) \\ & = \frac{n^4}{2} - \frac{n^2}{2} - \frac{1}{2} \lim_{x \to 1} \frac{d}{dx}(n U_{n - 1}) \\ & = \frac{n^4}{2} - \frac{n^2}{2} - \frac{T''_n(1)}{2} \\ T''_n(1) & = \frac{n^4 - n^2}{3}. \\ \end{align}

The proof for x = − 1 is similar, with the fact that Tn( − 1) = ( − 1)n being important.