Difference between revisions of "Orthogonality of Chebyshev T on (-1,1)"
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==Theorem== | ==Theorem== | ||
The following formula holds for $m,n \in \{0,1,2,\ldots\}$: | The following formula holds for $m,n \in \{0,1,2,\ldots\}$: | ||
− | $$\int_{-1}^1 \dfrac{T_m(x)T_n(x)}{\sqrt{1-x^2}} | + | $$\int_{-1}^1 \dfrac{T_m(x)T_n(x)}{\sqrt{1-x^2}} \mathrm{d}x = \left\{ \begin{array}{ll} |
0 &; m \neq n \\ | 0 &; m \neq n \\ | ||
\dfrac{\pi}{2} &; m=n\neq 0 \\ | \dfrac{\pi}{2} &; m=n\neq 0 \\ | ||
Line 10: | Line 10: | ||
==Proof== | ==Proof== | ||
− | ==References= | + | ==References== |
+ | * {{BookReference|An Introduction to Orthogonal Polynomials|1978|T.S. Chihara|prev=Orthogonality relation for cosine on (0,pi)|next=Chebyshev T}} $(1.3)$ (<i>note: only mentions the $m \neq n$ case</i>) | ||
[[Category:Theorem]] | [[Category:Theorem]] | ||
[[Category:Unproven]] | [[Category:Unproven]] |
Latest revision as of 22:36, 19 December 2017
Theorem
The following formula holds for $m,n \in \{0,1,2,\ldots\}$: $$\int_{-1}^1 \dfrac{T_m(x)T_n(x)}{\sqrt{1-x^2}} \mathrm{d}x = \left\{ \begin{array}{ll} 0 &; m \neq n \\ \dfrac{\pi}{2} &; m=n\neq 0 \\ \pi &; m=n=0, \end{array} \right.$$ where $T_m$ denotes Chebyshev polynomials of the first kind.
Proof
References
- 1978: T.S. Chihara: An Introduction to Orthogonal Polynomials ... (previous) ... (next) $(1.3)$ (note: only mentions the $m \neq n$ case)