Power Series Solutions to the Legendre Equation

1 Power Series Solutions to the Legendre EquationPower Series Solutions to the Legendre EquationDepartment of MathematicsIIT GuwahatiRA/RKSMA-102 (2016) Power Series Solutions to the Legendre EquationThe Legendre equationThe Equation (1 x2)y 2xy + ( + 1)y= 0,(1)where is any real constant, is called Legendre s Z+, the Equation has polynomial Solutions calledLegendre polynomials. In fact, these are the same polynomialthat encountered earlier in connection with the Eqn. (1) can be rewritten as[(x2 1)y ] = ( + 1)y,which has the formT(y) = y, whereT(f) = (pf ) , withp(x) =x2 1 and = ( + 1).

2 RA/RKSMA-102 (2016) Power Series Solutions to the Legendre EquationNote that the nonzero Solutions of (1) are eigenfunctions ofTcorresponding to the eigenvalue ( + 1).Sincep(1) =p( 1) = 0,Tis symmetric with respect to theinner product(f,g) = 1 1f(x)g(x) , eigenfunctions belonging to distinct eigenvalues (2016) Power Series Solutions to the Legendre EquationPower Series solution for the Legendre equationThe Legendre Equation can be put in the formy +p(x)y +q(x)y= 0,wherep(x) = 2x1 x2andq(x) = ( + 1)1 x2,ifx26= (1 x2)= n=0x2nfor|x|<1, bothp(x) andq(x) havepower Series expansions in the open interval ( 1,1).

3 Thus, seek a Power Series solution of the formy(x) = n=0anxn,x ( 1,1).RA/RKSMA-102 (2016) Power Series Solutions to the Legendre EquationDifferentiating term by term, we obtainy (x) = n=1nanxn 1andy = n=2n(n 1)anxn ,2xy = n=12nanxn= n=02nanxn,and(1 x2)y = n=2n(n 1)anxn 2 n=2n(n 1)anxn= n=0(n+ 2)(n+ 1)an+2xn n=0n(n 1)anxn= n=0[(n+ 2)(n+ 1)an+2 n(n 1)an] (2016) Power Series Solutions to the Legendre EquationSubstituting in (1), we obtain(n+ 2)(n+ 1)an+2 n(n 1)an 2nan+ ( + 1)an= 0,n 0,which leads to a recurrence relationan+2= ( n)( +n+ 1)(n+ 1)(n+ 2)

4 , we obtaina2= ( + 1)1 2a0,a4= ( 2)( + 3)3 4a2= ( 1)2 ( 2)( + 1)( + 3)4!a0,..a2n= ( 1)n ( 2) ( 2n+ 2) ( + 1)( + 3) ( + 2n 1)(2n)! (2016) Power Series Solutions to the Legendre EquationSimilarly, we can computea3,a5,a7,..,in terms ofa1and obtaina3= ( 1)( + 2)2 3a1a5= ( 3)( + 4)4 5a3= ( 1)2( 1)( 3)( + 2)( + 4)5! +1= ( 1)n( 1)( 3) ( 2n+ 1)( + 2)( + 4) ( + 2n)(2n+ 1)!a1 Therefore, the Series fory(x) can be written asy(x) =a0y1(x) +a1y2(x),wherey1(x) = 1 + n=1( 1)n ( 2) ( 2n+2) ( +1)( +3) ( +2n 1)(2n)!

5 X2n,andy2(x) =x+ n=1( 1)n( 1)( 3) ( 2n+1) ( +2)( +4) ( +2n)(2n+1)!x2n+ (2016) Power Series Solutions to the Legendre EquationNote: The ratio test shows thaty1(x) andy2(x) converges for|x|<1. These solutionsy1(x) andy2(x) satisfy the initialconditionsy1(0) = 1,y 1(0) = 0,y2(0) = 0,y 2(0) = (x) andy2(x) are independent, the general solution ofthe Legendre Equation over ( 1,1) isy(x) =a0y1(x) +a1y2(x)with arbitrary (2016) Power Series Solutions to the Legendre EquationObservationsCase I. When = 0 or = 2m, we note that ( 2) ( 2n+ 2) = 2m(2m 2) (2m 2n+ 2) =2nm!

6 (m n)!and( + 1)( + 3) ( + 2n 1) = (2m+ 1)(2m+ 3) (2m+ 2n 1)=(2m+ 2n)!m!2n(2m)!(m+n)!.Then, in this case,y1(x) becomesy1(x) = 1 +(m!)2(2m)!m k=1( 1)k(2m+ 2k)!(m k)!(m+k)!(2k)!x2k,which is a polynomial of degree 2m. In particular, for = 0,2,4(m= 0,1,2), the corresponding polynomials arey1(x) = 1,1 3x2,1 10x2+ (2016) Power Series Solutions to the Legendre EquationNote that the seriesy2(x) is not a polynomial when is evenbecause the coefficients ofx2n+1is never II. When = 2m+ 1,y2(x) becomes a polynomial andy1(x) is not a this case,y2(x) =x+(m!)

7 2(2m+ 1)!m k=1( 1)k(2m+ 2k+ 1)!(m k)!(m+k)!(2k+ 1)!x2k+ example, when = 1,3,5 (m= 0,1,2), the correspondingpolynomials arey2(x) =x,x 53x3,x 143x3+ (2016) Power Series Solutions to the Legendre EquationThe Legendre polynomialLetPn(x) =12n[n/2] r=0( 1)r(2n 2r)!r!(n r)!(n 2r)!xn 2r,where [n/2] denotes the greatest integer n/2. Whennis even, it is a constant multiple of the polynomialy1(x). Whennis odd, it is a constant multiple of the polynomialy2(x).The first five Legendre polynomials areP0(x) = 1,P1(x) =x,P2(x) =12(3x2 1)P4(x) =18(35x4 30x2+ 3),P5(x) =18(63x5 70x3+ 15x).

8 RA/RKSMA-102 (2016) Power Series Solutions to the Legendre EquationFigure : Legendre polynomial over the interval [ 1,1]RA/RKSMA-102 (2016) Power Series Solutions to the Legendre EquationRodrigues s formula for the Legendre polynomialsNote that(2n 2r)!(n 2r)!xn 2r=dndxnx2n 2rand1r!(n r)!=1n!(nr).Thus,Pn(x) in (2) can be expressed asPn(x) =12nn!dndxn[n/2] r=0( 1)r(nr)x2n [n/2]<r n, the termx2n 2rhas degree less thann,so itsnth derivative is zero. This givesPn(x) =12nn!dndxnn r=0( 1)r(nr)x2n 2r=12nn!dndxn(x2 1)n,which is known as Rodrigues (2016) Power Series Solutions to the Legendre EquationProperties of the Legendre polynomialsPn(x) For eachn 0,Pn(1) = 1.

9 Moreover,Pn(x) is the onlypolynomial which satisfies the Legendre Equation (1 x2)y 2xy +n(n+ 1)y= 0andPn(1) = 1. For eachn 0,Pn( x) = ( 1)nPn(x). 1 1Pn(x)Pm(x)dx={0ifm6=n,22n+1ifm= (2016) Power Series Solutions to the Legendre Equation Iff(x) is a polynomial of degreen, we havef(x) =n k=0ckPk(x),whereck=2k+ 12 1 1f(x)Pk(x)dx. It follows from the orthogonality relation that 1 1g(x)Pn(x)dx= 0for every polynomialg(x) with deg(g(x))<n.** End **RA/RKSMA-102 (2016)}