Difference between revisions of "Chebyshev T"
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| − | <strong>Theorem (Orthogonality):</strong> The following | + | <strong>Theorem (Orthogonality):</strong> The following formula holds: |
$$\int_{-1}^1 \dfrac{T_m(x)T_n(x)}{\sqrt{1-x^2}} dx = \left\{ \begin{array}{ll} | $$\int_{-1}^1 \dfrac{T_m(x)T_n(x)}{\sqrt{1-x^2}} dx = \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 \\ | ||
| − | \pi | + | \pi &; m=n=0. |
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\end{array} \right.$$ | \end{array} \right.$$ | ||
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Revision as of 10:37, 23 March 2015
Chebyshev polynomials of the first kind are orthogonal polynomials defined by $$T_n(x) = \cos(n \mathrm{arccos}(x)).$$
Properties
Theorem: The polynomials $T_n(x)$ and $U_n(x)$ are two independent solutions of the following equation, called Chebyshev's equation: $$(1-x^2)\dfrac{d^2y}{dx^2}-x\dfrac{dy}{dx}+n^2y=0.$$
Proof: █
Theorem: The following formula holds: $$T_{n+1}(x)-2xT_n(x)+T_{n-1}(x)=0.$$
Proof: █
Theorem (Orthogonality): The following formula holds: $$\int_{-1}^1 \dfrac{T_m(x)T_n(x)}{\sqrt{1-x^2}} dx = \left\{ \begin{array}{ll} 0 &; m \neq n \\ \dfrac{\pi}{2} &; m=n\neq 0 \\ \pi &; m=n=0. \end{array} \right.$$
Proof: █
