Associated Legendre function/Catalogs: Difference between revisions

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imported>Paul Wormer
(New page: {{subpages}} The associated Legendre functions through ''l'' = 6 are: :<math> \begin{align} P_0^0(x) &= 1 \\ \\ P_1^0(x) &= x \\ P_1^1(x) & = (1-x^2)^{1/2} \\ \\ P_2^0(x) &= \tfrac{1}{2}...)
 
imported>Paul Wormer
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P_4^0(x) &= \tfrac{1}{8}(35x^4- 30x^2 + 3)\\
P_4^0(x) &= \tfrac{1}{8}(35x^4- 30x^2 + 3)\\
P_4^1(x) &= \tfrac{1}{2}(1-x^2)^{1/2} (35x^3 - 5x) \\
P_4^1(x) &= \tfrac{1}{2}(1-x^2)^{1/2} (35x^3 - 15x) \\
P_4^2(x) &= \tfrac{1}{2}(1-x^2)(105x^2 -15)  \\
P_4^2(x) &= \tfrac{1}{2}(1-x^2)(105x^2 -15)  \\
P_4^3(x) &=  105 (1-x^2)^{3/2} x  \\
P_4^3(x) &=  105 (1-x^2)^{3/2} x  \\
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P_6^0(x) &= \tfrac{1}{16}(231x^6- 315x^4 + 105x^2 -5)\\
P_6^0(x) &= \tfrac{1}{16}(231x^6- 315x^4 + 105x^2 -5)\\
P_6^1(x) &= \tfrac{1}{8}(1-x^2)^{1/2} (693x^5 - 630x^3 + 105)    \\
P_6^1(x) &= \tfrac{1}{8}(1-x^2)^{1/2} (693x^5 - 630x^3 + 105x)    \\
P_6^2(x) &= \tfrac{1}{8}(1-x^2)(3465 x^4 - 1890 x^2 +105)  \\
P_6^2(x) &= \tfrac{1}{8}(1-x^2)(3465 x^4 - 1890 x^2 +105)  \\
P_6^3(x) &= \tfrac{1}{2} (1-x^2)^{3/2} (3465x^3-945x)  \\
P_6^3(x) &= \tfrac{1}{2} (1-x^2)^{3/2} (3465x^3-945x)  \\

Latest revision as of 07:56, 10 September 2009

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An informational catalog, or several catalogs, about Associated Legendre function.

The associated Legendre functions through l = 6 are: