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 $m,n \in \{0,1,2,\ldots\}$:
$$\int_{-1}^1 \dfrac{T_m(x)T_n(x)}{\sqrt{1-x^2}} dx = \left\{ \begin{array}{ll}
+
$$\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 \\
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==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