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Physics LibreTexts

5.11: Legendre Polynomials

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In this section we cover just enough about Legendre polynomials to be useful in the following section. Before starting, I want you to expand the following expression, by the binomial theorem, for |x|<1, up to x4 :

1(12xcosθ+x2)1/2.

Please do go ahead and do it. Well, you probably won’t, so I’d better do it myself:

I’ll start with

(1X)1/2=1+12X+38X2+516X3+35128X4...

and therefore

[1x(2cosθx)]1/2=1+12x(2cosθx)+38x2(2cosθx)2+516x3(2cosθx)3+35128x4(2cosθx)4...

=1+xcosθ12x2+38x2(4cos2θ4xcosθ+x2)+516x3(8cos3θ12xcos2θ+6x2cosθx3)+35128x(16cos4θ32xcos3θ+24x2cos2θ8x3cosθ+x4)...

=1+xcosθ+x2(12+32cos2θ)+x3(32cosθ+52cos3θ)+x4(38154cos2θ+358cos4θ)...

The coefficients of the powers of x are the Legendre polynomials Pl(cosθ), so that

1(12xcosθ+x2)1/2=1+xP1(cosθ)+x2P2(cosθ)+x3P3(cosθ)+x4P4(cosθ)+...

The Legendre polynomials with argument cosθ can be written as series of terms in powers of cosθ by substitution of cosθ for x in Equations 1.12.5 in Section 1.12 of Chapter 1. Note that x in Section 1 is not the same as x in the present section. Alternatively they can be written as series of cosines of multiples of θ as follows.

P0=1P1=cosθP2=14(3cos2θ+1)P3=18(5cos3θ+3cosθ)P4=164(35cos4θ+20cos2θ+9)P5=1128(63cos5θ+35cos3θ+30cosθ)P6=1512(231cos6θ+126cos4θ+105cos2θ+50)P7=11024(429cos7θ+231cos5θ+189cos3θ+175cosθ)P8=(6435cos8θ+3432cos6θ+2772cos4θ+2520cos2θ+1225)/214

For example, P6(cosθ) can be written either as given by Equation 5.11.7, or as given by Equation 1, namely

P6=116(231c6315c4+105c25), where c=cosθ.

The former may look neater, and the latter may look “awkward” because of all the powers. However, the latter is far faster to compute, particularly when written as nested parentheses:

P6=(5+C(105+C(315+231C)))/16, where C=cos2θ.


This page titled 5.11: Legendre Polynomials is shared under a CC BY-NC 4.0 license and was authored, remixed, and/or curated by Jeremy Tatum via source content that was edited to the style and standards of the LibreTexts platform.

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