Relationship between Chebyshev Polynomial of the First and Second Kind/Formulation 3

Theorem

Let $\map {T_n} x$ denote the Chebyshev polynomial of the first kind of order $n$.

Let $\map {U_n} x$ denote the Chebyshev polynomial of the second kind of order $n$.

Then:

$\ds \map {U_n} x = \dfrac 1 \pi \int_{-1}^1 \dfrac {\map {T_{n + 1} } v \rd v} {\paren {v - x} \sqrt {1 - v^2} }$

Proof

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Sources

• 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 30$: Chebyshev Polynomials: Relationships Between $\map {T_n} x$ and $\map {U_n} x$: $30.44$
• 2009: Murray R. Spiegel, Seymour Lipschutz and John Liu: Mathematical Handbook of Formulas and Tables (3rd ed.) ... (previous) ... (next): $\S 31$: Chebyshev Polynomials: Relationships Between $\map {T_n} x$ and $\map {U_n} x$: $31.44.$