Difference between revisions of "Chebyshev T"

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(Created page with "Chebyshev polynomials of the first kind are orthogonal polynomials defined by $$T_n(x) = \cos(n \mathrm{arccos}(x))$$")
 
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Chebyshev polynomials of the first kind are [[orthogonal polynomials]] defined by
 
Chebyshev polynomials of the first kind are [[orthogonal polynomials]] defined by
$$T_n(x) = \cos(n \mathrm{arccos}(x))$$
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$$T_n(x) = \cos(n \mathrm{arccos}(x)).$$
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=Properties=
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<div class="toccolours mw-collapsible mw-collapsed" style="width:800px">
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<strong>Theorem:</strong> The polynomials $T_n(x)$ and $U_n(x)$ are two independent solutions of the following equation, called Chebyshev's equation:
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$$(1-x^2)\dfrac{d^2y}{dx^2}-x\dfrac{dy}{dx}+n^2y=0.$$
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<div class="mw-collapsible-content">
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<strong>Proof:</strong> █
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</div>
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</div>
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<div class="toccolours mw-collapsible mw-collapsed" style="width:800px">
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<strong>Theorem:</strong> The following formula holds:
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$$T_{n+1}(x)-2xT_n(x)+T_{n-1}(x)=0.$$
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<div class="mw-collapsible-content">
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<strong>Proof:</strong> █
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</div>
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</div>
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<div class="toccolours mw-collapsible mw-collapsed" style="width:800px">
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<strong>Theorem (Orthogonality):</strong> The following formulas hold:
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$$\int_{-1}^1 \dfrac{T_m(x)T_n(x)}{\sqrt{1-x^2}} dx = \left\{ \begin{array}{ll}
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0 &; m \neq n \\
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\dfrac{\pi}{2} &; m=n\neq 0 \\
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\pi &; m=n=0
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\end{array} \right.$$
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and
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$$\int_{-1}^1 \dfrac{U_m(x)U_n(x)}{\sqrt{1-x^2}} dx = \left\{ \begin{array}{ll}
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0 &; m \neq n \\
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\dfrac{\pi}{2} &; m=n\neq 0\\
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0 &; m=n=0.
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\end{array} \right.$$
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<div class="mw-collapsible-content">
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<strong>Proof:</strong> █
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</div>
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</div>

Revision as of 10:27, 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 formulas hold: $$\int_{-1}^1 \dfrac{T_m(x)T_n(x)}{\sqrt{1-x^2}} dx = \left\{ \begin{array}{ll} 0 &; m \neq n \\ \dfrac{\pi}{2} &; m=n\neq 0 \\ \pi &; m=n=0 \end{array} \right.$$ and $$\int_{-1}^1 \dfrac{U_m(x)U_n(x)}{\sqrt{1-x^2}} dx = \left\{ \begin{array}{ll} 0 &; m \neq n \\ \dfrac{\pi}{2} &; m=n\neq 0\\ 0 &; m=n=0. \end{array} \right.$$

Proof: