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Mathematical Methods for Physicists: A concise introduction - Site Map

Mathematical Methods for Physicists: A concise introduction - Site Map

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SPECIAL FUNCTIONS OF MATHEMATICAL PHYSICS<br />

where s n …x† is given by<br />

s n …x† ˆXn<br />

kˆ0<br />

u k …x†v nk …x† ˆu 0 …x†v n …x†‡‡u n …x†v 0 …x†:<br />

Now the Cauchy product <strong>for</strong> our two series is given by<br />

X 1<br />

X n<br />

nˆ0 kˆ0<br />

<br />

a nk z nk e …nk†i<br />

<br />

a k z k e ki ˆ X1<br />

nˆ0<br />

z n Xn<br />

kˆ0<br />

a k a nk e …n2k†i : …7:18†<br />

In the inner sum, which is the sum of interest to us, it is straight<strong>for</strong>ward to prove<br />

that, <strong>for</strong> n 1, the terms corresponding to k ˆ j and k ˆ n j are identical except<br />

that the exponents on e are of opposite sign. Hence these terms can be paired, and<br />

we have <strong>for</strong> the coecient of z n ,<br />

P n …cos † ˆa 0 a n …e ni ‡ e ni †‡a 1 a n1 …e …n2†i ‡ e …n2†i †‡<br />

ˆ 2‰ a 0 a n cos n ‡ a 1 a n1 cos…n 2† ‡Š:<br />

…7:19†<br />

If n is odd, the number of terms is even and each has a place in one of the pairs. In<br />

this case, the last term in the sum is<br />

a …n1†=2 a …n‡1†=2 cos :<br />

If n is even, the number of terms is odd and the middle term is unpaired. In this<br />

case, the series (7.19) <strong>for</strong> P n …cos † ends with the constant term<br />

a n=2 a n=2 :<br />

Using Eq. (7.17) to compute values of the a n , we ®nd from the unit coecient of z 0<br />

in Eqs. (7.18) and (7.19), whether n is odd or even, the speci®c expressions<br />

9<br />

P 0 …cos † ˆ1; P 1 …cos † ˆcos ; P 2 …cos † ˆ…3 cos 2 ‡ 1†=4<br />

P 3 …cos † ˆ…5 cos 3 ‡ 3 cos †=8<br />

P 4 …cos † ˆ…35 cos 4 ‡ 20 cos 2 ‡ 9†=64<br />

P 5 …cos † ˆ…63 cos 5 ‡ 35 cos 3 ‡ 30 cos †=128<br />

P 6 …cos † ˆ…231 cos 6 ‡ 126 cos 4 ‡ 105 cos 2 ‡ 50†=512<br />

>=<br />

: …7:20†<br />

>;<br />

Orthogonality of Legendre polynomials<br />

The set of Legendre polynomials fP n …x†g is orthogonal <strong>for</strong> 1 x ‡1. In<br />

particular we can show that<br />

Z ‡1<br />

2=…2n ‡ 1† if m ˆ n<br />

P n …x†P m …x†dx ˆ<br />

0 if m 6ˆ n : …7:21†<br />

1<br />

304

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