Chebyshev polynomials of the second kind

The Chebyshev polynomials of the second kind are defined recursively by $\begin{align*} U_0(x) &= 1, \\ U_1(x) &= 2x, \\ U_{n+1}(x) &= 2xU_n(x) - U_{n-1}(x), \end{align*}$ or equivalently by $U_n(x) = \frac{\sin ((n + 1) \arccos x)}{\sqrt{1-x^2}}.$

Proof of equivalence of the two definitions

In the proof below, $U_n(x)$ will refer to the recursive definition. Let $y = \arccos x$ , so that $\cos y = x$ .

For the $n = 0$ base case, $\frac{\sin y}{\sqrt{1-x^2}} = \frac{\sqrt{1-x^2}}{\sqrt{1-x^2}} = 1 = U_0(x),$ using the Pythagorean identity $\sin^2 y + \cos^2 y = 1$ and the fact that the range of $\arccos x$ is $[0,\pi]$ , on which $\sin y$ is nonnegative.

For the $n = 1$ base case, we apply the sine double angle formula to get $\frac{\sin 2y}{\sqrt{1-x^2}} = \frac{2\cos y \sin y}{\sqrt{1-x^2}} = \frac{2x\sqrt{1-x^2}}{\sqrt{1-x^2}} = 2x = U_1(x).$

Now for the inductive step, we assume the trigonometric definition holds for $n - 1$ and $n$ and prove that it also holds for $n + 1$ . From the sine sum and difference identities we have $\sin((n+2)y) = \sin((n+1)y + y) = \sin((n+1)y)\cos y + \cos((n+1)y)\sin y$ and $\sin ny = \sin((n+1)y - y) = \sin((n+1)y)\cos y - \cos((n+1)y)\sin y.$ The sum of these equations is $\sin((n+2)y) + \sin ny = 2\sin((n+1)y)\cos y;$ rearranging and dividing by $\sqrt{1-x^2}$ , $\frac{\sin((n+2)y)}{\sqrt{1-x^2}} = \frac{2\sin((n+1)y)\cos y}{\sqrt{1-x^2}} - \frac{\sin ny}{\sqrt{1-x^2}}.$ Substituting our original assumptions yields $\frac{\sin((n+2)y)}{\sqrt{1-x^2}} = 2xU_n(x) - U_{n-1}(x) = U_{n+1}(x),$ as desired.

Note that division by $\sqrt{1-x^2}$ implies that the trigonometric definition of Chebyshev polynomials of the second kind is only valid on $(-1,1)$ , in contrast to the interval of $[-1,1]$ on which the trigonometric definition of Chebyshev polynomials of the first kind holds.

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Chebyshev polynomials of the second kind

Proof of equivalence of the two definitions