Relationship between Chebyshev Polynomial of the First and Second Kind/Formulation 4
Jump to navigation
Jump to search
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:
- $\ds \map {U_n} x = \dfrac 1 \pi \int_{-1}^1 \dfrac {\sqrt {1 - v^2} \map {U_{n - 1} } v \rd v} {x - v}$
Proof
![]() | This theorem requires a proof. You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by crafting such a proof. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{ProofWanted}} from the code.If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page. |
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 30$: Chebyshev Polynomials: Relationships Between $\map {T_n} x$ and $\map {U_n} x$: $30.45$
- 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 $\map {U_n} x$: $31.45.$