SPECIAL FUNCTIONS
which on collecting terms gives
∑∞n=0{(n+2)(n+1)an+2−[n(n+1)−(+1)]an}xn=0.The recurrence relation is therefore
an+2=[n(n+1)−(+1)]
(n+1)(n+2)an, (18.2)forn=0, 1 , 2 ,.... If we choosea 0 = 1 anda 1 = 0 then we obtain the solution
y 1 (x)=1−(+1)x^2
2!+(−2)(+1)(+3)x^4
4!−···, (18.3)whereas on choosinga 0 = 0 anda 1 = 1 we find a second solution
y 2 (x)=x−(−1)(+2)x^3
3!+(−3)(−1)(+2)(+4)x^5
5!−···. (18.4)By applying the ratio test to these series (see subsection 4.3.2), we find that both
series converge for|x|<1, and so their radius of convergence is unity, which
(as expected) is the distance to the nearest singular point of the equation. Since
(18.3) contains only even powers ofxand (18.4) contains only odd powers, these
two solutions cannot be proportional to one another, and are therefore linearly
independent. Hence, the general solution to (18.1) for|x|<1is
y(x)=c 1 y 1 (x)+c 2 y 2 (x).18.1.1 Legendre functions for integerIn many physical applications the parameterin Legendre’s equation (18.1) is
an integer, i.e.=0, 1 , 2 ,.... In this case, the recurrence relation (18.2) gives
a+2=[(+1)−(+1)]
(+1)(+2)a=0,i.e. the series terminates and we obtain a polynomial solution of order.In
particular, ifis even, theny 1 (x) in (18.3) reduces to a polynomial, whereas ifis
oddthesameistrueofy 2 (x) in (18.4). These solutions (suitably normalised) are
called theLegendre polynomialsof order; they are writtenP(x) and are valid
for all finitex. It is conventional to normaliseP(x) in such a way thatP(1) = 1,
and as a consequenceP(−1) = (−1). The first few Legendre polynomials are
easily constructed and are given by
P 0 (x)=1, P 1 (x)=x,P 2 (x)=^12 (3x^2 −1), P 3 (x)=^12 (5x^3 − 3 x),P 4 (x)=^18 (35x^4 − 30 x^2 +3), P 5 (x)=^18 (63x^5 − 70 x^3 +15x).