3.5 Chebyshev Polynomials of the Second Kind

Definition 3.5.1. The Chebyshev polynomials of the second kind are defined in terms of Chebyshev polynomials of the first kind as follows. Differentiating \[T_n(x) = \cos n\theta \] with respect to \(\, x = \cos \theta \), get \[\frac {d}{dx} T_n(x) = -n\, \sin \theta \, \frac {d\theta }{dx} = \frac {n\, \sin n\theta }{\sin \theta }\] we then define Chebyshev polynomials of the second kind by \[U_n(x) = \frac {1}{n + 1}\, \frac {d}{dx}T_{n+1}(x) = \frac {\sin [(n + 1)\theta ]}{\sin \theta }\, , \hspace {0.3cm} n = 0, \, 1, \, 2, \, \cdots \cdots \] We observe that in other context Chebyshev polynomials of the second kind have been defined by \[U_n(x) = \sin \left (n\, \cos ^{-1}(x)\right ) = \sin n\theta , \, \hspace {0.3cm}\text {where}\hspace {0.3cm} x = \cos \theta .\]

Theorem 3.5.2 (Recurrence Relation).

(a).: \(U_n(x) = x U_{n-1}(x) + T_n(x), \hspace {0.4cm} n = 1, \, 2, \, \cdots \cdots \)
(b).: \(U_n(x) = 2x U_{n - 1}(x) - U_{n-2}(x), \hspace {0.4cm} n = 2, \, 3, \cdots \cdots \)
(c).: \((1 - x^2)U_n'(x) = nx U_n(x) + nU_{n-1}(x).\)

Proof.

(a).: \begin {align*} U_n(x) & = \frac {\sin [(n + 1)\theta ]}{\sin \theta }\\ & = \frac {\sin n\theta \cos \theta + \sin \theta \cos n\theta }{\sin \theta } = \cos \theta \left (\frac {\sin n\theta }{\sin \theta }\right ) + \cos n\theta \\ & = xU_{n-1}(x) + T_n(x). \end {align*}
(b).: Use the identity \[\sin [(n+1)\theta ] + \sin [(n - 1)\theta ] = 2\sin n\theta \sin \theta .\]
(c).: Exercise

□

Theorem 3.5.3 (Orthogonality). \[\int _{-1}^1U_n(x)\, U_m(x)\, (1 - x^2)^{\frac {1}{2}}\, dx = \begin {cases} 0, & m\neq n\\ \frac {\pi }{2}, & m=n\neq 0\\ 0, & m = n= 0\\ \end {cases}\]

Proof. Let \(x = \cos \theta , \hspace {0.3cm} dx = -\sin \theta \, d\theta .\) Then \[\int _{-1}^1U_n(x)\, U_m(x)\, (1 - x^2)^{\frac {1}{2}}\, dx = \int _0^{\pi }\sin [(n+1)\theta ]\,\sin [(m+1)\theta ]\, d\theta = 0, \hspace {0.3cm} m\neq n.\] □

Exercise 3.5.4. Find the values of

(a).: \(U_n(1)\)
(b).: \(U_n(-1)\)
(c).: \(U_{2n}(0)\)
(d).: \(U_{2n + 1}(0).\)

Show that \(U_n(x)\) is the solution of the Chebyshev equation \[(1 - x^2)\, \frac {d^2y}{dx^2} - x\, \frac {dy}{dx} + n^2 y = 0.\]

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