View source for Chebyshev polynomials of the second kind

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

==Proof of equivalence of the two definitions==
In the proof below, <math>U_n(x)</math> will refer to the recursive definition. Let <math>y = \arccos x</math>, so that <math>\cos y = x</math>.

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

For the <math>n = 1</math> base case, we apply the [[Trigonometric identities#Double-angle identities|sine double angle formula]] to get <cmath>\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).</cmath>

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

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