512 Higher Engineering Mathematics
from which,ar+ 2 =ar[
(c+r− 1 )(c+r)+ 2 (c+r)−k^2 −k](c+r+ 1 )(c+r+ 2 )=ar[(c+r)(c+r+ 1 )−k(k+ 1 )]
(c+r+ 1 )(c+r+ 2 ) (46)
Whenc= 0 ,ar+ 2 =ar[r(r+ 1 )−k(k+ 1 )]
(r+ 1 )(r+ 2 )
Forr= 0 ,a 2 =a 0 [−k(k+ 1 )]
( 1 )( 2 )
Forr= 1 ,a 3 =a 1 [( 1 )( 2 )−k(k+ 1 )]
( 2 )( 3 )=−a 1 [k^2 +k−2]
3!=−a 1 (k− 1 )(k+ 2 )
3!
Forr=2,a 4 =a 2 [( 2 )( 3 )−k(k+ 1 )]
( 3 )( 4 )=−a 2[
k^2 +k− 6]( 3 )( 4 )=−a 2 (k+ 3 )(k− 2 )
( 3 )( 4 )=−(k+ 3 )(k− 2 )
( 3 )( 4 ).a 0 [−k(k+ 1 )]
( 1 )( 2 )=a 0 k(k+ 1 )(k+ 3 )(k− 2 )
4!
Forr=3,a 5 =a 3 [( 3 )( 4 )−k(k+ 1 )]
( 4 )( 5 )=−a 3 [k^2 +k−12]
( 4 )( 5 )=−a 3 (k+ 4 )(k− 3 )
( 4 )( 5 )=−(k+ 4 )(k− 3 )
( 4 )( 5 ).−a 1 (k− 1 )(k+ 2 )
( 2 )( 3 )=a 1 (k− 1 )(k− 3 )(k+ 2 )(k+ 4 )
5!and so on.Substituting values into equation (43) gives:y=x^0{
a 0 +a 1 x−a 0 k(k+ 1 )
2!x^2−a 1 (k− 1 )(k+ 2 )
3!x^3+a 0 k(k+ 1 )(k− 2 )(k+ 3 )
4!x^4+a 1 (k− 1 )(k− 3 )(k+ 2 )(k+ 4 )
5!x^5+ ···}i.e.y=a 0{
1 −k(k+1)
2!x^2+k(k+1)(k−2)(k+3)
4!x^4 −···}+a 1{
x−(k−1)(k+2)
3!x^3+(k−1)(k−3)(k+2)(k+4)
5!x^5 −···}
(47)From page 506, it was stated that if two solutions of
the indicial equation differ by an integer, as in this case,
wherec=0 and 1, and if one coefficient is indetermi-
nate, as with whenc=0, then the complete solution is
always given by using this value ofc. Using the second
value ofc,i.e.c=1 in this problem, will give a series
which is one of the series in the first solution.(This may
be checked forc=1andwherea 1 =0; the result will be
the first part of equation (47) above).Legendre’s polynomials
(A polynomial is an expression of the form:
f(x)=a+bx+cx^2 +dx^3 + ···).Whenkin equation
(47) above is an integer, say,n, one of the solutionseries
terminates after a finite number of terms. For example,
ifk=2, then the first series terminates after the term in
x^2. The resulting polynomial inx, denoted byPn(x),is
called aLegendre polynomial.Constantsa 0 anda 1 are
chosen so thaty=1whenx=1. This is demonstrated
in the following worked problems.Problem 13. Determine the Legendre polynomial
P 2 (x).Since in P 2 (x),n=k=2, then from the first part of
equation (47), i.e. the even powers ofx:y=a 0{
1 −2 ( 3 )
2!x^2 + 0}
=a 0 { 1 − 3 x^2 }a 0 is chosen to makey=1whenx= 1i.e. 1=a 0 { 1 − 3 ( 1 )^2 }=− 2 a 0 , from which,a 0 =−1
2