Difference between revisions of "Orthogonality of Chebyshev T on (-1,1)"
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(Created page with "==Theorem== The following formula holds for $m,n \in \{0,1,2,\ldots\}$: $$\int_{-1}^1 \dfrac{T_m(x)T_n(x)}{\sqrt{1-x^2}} dx = \left\{ \begin{array}{ll} 0 &; m \neq n \\ \dfrac...") |
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==Theorem== | ==Theorem== | ||
The following formula holds for $m,n \in \{0,1,2,\ldots\}$: | The following formula holds for $m,n \in \{0,1,2,\ldots\}$: | ||
| − | $$\int_{-1}^1 \dfrac{T_m(x)T_n(x)}{\sqrt{1-x^2}} | + | $$\int_{-1}^1 \dfrac{T_m(x)T_n(x)}{\sqrt{1-x^2}} \mathrm{d}x = \left\{ \begin{array}{ll} |
0 &; m \neq n \\ | 0 &; m \neq n \\ | ||
\dfrac{\pi}{2} &; m=n\neq 0 \\ | \dfrac{\pi}{2} &; m=n\neq 0 \\ | ||
Revision as of 22:30, 19 December 2017
Theorem
The following formula holds for $m,n \in \{0,1,2,\ldots\}$: $$\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.
