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Legendre Polynomials as an Orthogonal Basis
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[QUOTE="jambaugh, post: 6413165, member: 76054"] Your (source's) wording is confusing. Specifically: "show that for any polynomial with p a set of... less than ##n##". What is p here? Does it mean: "with p a set of [I]polynomials in ##L^2([-1,1])## [/I]with degree less than n"? Does it mean" "with p [I]in[/I] a set of [I]Legendre polynomials in[/I] ..."? Legendre polynomials are mutually orthogonal under the ##L^2([-1,1])## norm and indeed you can derive them fairly easily up to scalar multipliers by the Gram-Schmidt procedure applied to the power basis ##\{ 1,x,x^2,x^3, \ldots\}##. Starting with ##P_0(x)=1## and ##P_1(x)=x## (already orthogonal), then ##p_2(x)=x^2## an ## \langle p_2,P_1\rangle = 0## but: [tex]\langle p_2,P_0\rangle = \int_{-1}^1 1\cdot x^2 dx = 2/3[/tex] So we take: ## P_2 = p_2 -\frac{\langle P_0,p_2\rangle}{\langle P_0,P_0\rangle}\cdot P_0## and that linear combination will be orthogonal to ##P_0##. In this case with ##\langle P_0,P_0\rangle = 2## we get: [tex]P_2 = p_2 - \frac{1}{3}P_0,\quad P_2(x) = x^2 -\frac{1}{3}[/tex] However with standard normalization the actual degree 2 Legrange polynomial is 3/2 times this one, namely ##P_2(x)=\frac{1}{2}(3x^2-1)##. (Standard normalization assures ##P_n(1)=1, P_n(-1)=(-1)^n##.) That's all I can say without better context to understand the ?mis-worded? question. [/QUOTE]
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Legendre Polynomials as an Orthogonal Basis
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