Power Series Expansion for Chebyshev Polynomial of the Second Kind

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Theorem

The $n$th Chebyshev polynomial of the second kind can be expressed as a power series expansion in the form:

\(\ds \map {U_n} x\) \(=\) \(\ds \sum_{k \mathop = 0}^\infty \paren {-1}^k \dbinom {n + 1} {2 k + 1} x^{n - 2 k} \paren {1 - x^2}^k\)
\(\ds \) \(=\) \(\ds \dbinom {n + 1} 1 x^n - \dbinom {n + 1} 3 x^{n - 2} \paren {1 - x^2} + \dbinom {n + 1} 5 x^{n - 4} \paren {1 - x^2}^2 - \dbinom {n + 1} 7 x^{n - 6} \paren {1 - x^2}^3 + \cdots\)

where $n \in \N$.


Proof

From the Definition of Chebyshev Polynomial of the Second Kind, we have:

$\map {U_n} {\cos \theta} \sin \theta = \map \sin {\paren {n + 1} \theta}$

From De Moivre's Formula, we have:

$\map \cos {\paren {n + 1} \theta} + i \map \sin {\paren {n + 1} \theta} = \paren {\cos \theta + i \sin \theta}^{n + 1}$

As $n \in \Z_{>0}$, we use the Binomial Theorem on the right hand side, resulting in:

$\ds \map \cos {\paren {n + 1} \theta} + i \map \sin {\paren {n + 1} \theta} = \sum_{k \mathop \ge 0} \binom {n + 1} k \paren {\cos^{n + 1 - k} \theta} \paren {i \sin \theta}^k$

When $k$ is odd, the expression being summed is imaginary.

Equating the imaginary parts of both sides of the equation, replacing $k$ with $2 k + 1$ to make $k$ odd, gives:

\(\ds \map \sin {\paren {n + 1} \theta}\) \(=\) \(\ds \sum_{k \mathop \ge 0} \paren {-1}^k \dbinom {n + 1} {2 k + 1} \paren {\cos^{n + 1 - \paren {2 k + 1} } \theta} \paren {\sin^{2 k + 1} \theta}\)
\(\ds \) \(=\) \(\ds \sin \theta \sum_{k \mathop \ge 0} \paren {-1}^k \dbinom {n + 1} {2 k + 1} \paren {\cos^{n - \paren {2 k} } \theta} \paren {\sin^2 \theta}^k\)
\(\ds \) \(=\) \(\ds \sin \theta \sum_{k \mathop \ge 0} \paren {-1}^k \dbinom {n + 1} {2 k + 1} \paren {\cos^{n - \paren {2 k} } \theta} \paren {1 - \cos^2 \theta}^k\) Sum of Squares of Sine and Cosine
\(\ds \) \(=\) \(\ds \sin \theta \sum_{k \mathop = 0}^\infty \paren {-1}^k \dbinom {n + 1} {2 k + 1} x^{n - 2 k} \paren {1 - x^2}^k\) $\cos \theta \to x$
\(\ds \leadsto \ \ \) \(\ds \map {U_n} x\) \(=\) \(\ds \sum_{k \mathop = 0}^\infty \paren {-1}^k \dbinom {n + 1} {2 k + 1} x^{n - 2 k} \paren {1 - x^2}^k\)

$\blacksquare$


Also see


Sources

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