Chebyshev polynomials of the first kind

The Chebyshev polynomials of the first kind are defined recursively by $T_0(x) = 1,$ $T_1(x) = x,$ $T_{n+1}(x) = 2xT_n(x) - T_{n-1}(x),$ or equivalently by $T_n(x) = \cos (n \arccos x).$

Proof of equivalence of the two definitions

In the proof below, $T_n(x)$ will refer to the recursive definition.

For the $n = 0$ base case, $\cos(0 \arccos x) = \cos 0 = 1 = T_0(x);$ for the $n = 1$ base case, $\cos(1 \arccos (x)) = \cos(\arccos (x)) = x = T_1(x).$

Now for the inductive step, let $y = \arccos x$ , so that $x = \cos y$ . We then assume that $\cos ((n-1)y) = T_{n-1}(x)$ and $\cos ny = T_n(x)$ , and we wish to prove that $\cos ((n+1)y) = T_{n+1}(x)$ .

From the cosine sum and difference identities we have $\cos ((n+1)y) = \cos (ny+y) = \cos ny \cos y - \sin ny \sin y$ and $\cos ((n-1)y )= \cos (ny-y) = \cos ny \cos y + \sin ny \sin y.$ The sum of these equations is $\cos ((n+1)y) + \cos ((n-1)y) = 2 \cos ny \cos y;$ rearranging, $\cos ((n+1)y) = 2 \cos y \cos ny - \cos ((n-1)y).$ Substituting our assumptions yields $\cos ((n+1)y) = 2xT_n(x) - T_{n-1}(x) = T_{n+1}(x),$ as desired.

Composition identity

For nonnegative integers $m$ and $n$ , the identity $T_{mn} = T_m(T_n(x))$ holds.

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Chebyshev polynomials of the first kind

Contents

Proof of equivalence of the two definitions

Composition identity

First proof

Second proof (Induction)