Transcription of Legendre Polynomials: Rodriques’ Formula and Recursion ...
1 Legendre Polynomials: Rodriques Formula and Recursion RelationsJackson says By manipulation of the power series solutions it is possible to obtain a compactrepresentation of the Legendre polynomials known as Rodrigues Formula . Here is a proofthat Rodrigues Formula indeed produces a solution to Legendre s differential equation. Fromthe differential equation, assuming a series solutionPn=Pajxj( = 0) we obtained therelationaj+2=j(j+ 1) n(n+ 1)(j+ 1)(j+ 2)aj[JDJ ( ), with = 0]. Withj=n 2k, this is satisfied byan 2k= ( 1)k(2n 2k)!2nk! (n k)! (n 2k)!,where 1/2nis conventional. So, we can writePn(x) =[n/2]Xk=0( 1)k(2n 2k)!
2 2nk! (n k)! (n 2k)!xn 2k,where [n/2] denotes the greatest integer or the integer part. For integern 2k, this isPn(x) =[n/2]Xk=0( 1)k2nk!(n k)! ddx nx2n 2k=12nn! ddx nnXk=0n!k! (n k)!( 1)k(x2)n k,where the extra terms introduced by extending the upper limit of the sum from [n/2] tonhave zero derivative. By the binomial theorem, this expression isPn(x) =12nn! ddx n(x2 1)n.( )Jackson next says, From Rodrigues Formula it is a straightforward matter to obtain re-cursion relations for thePn. To this end, first prove some relations that are useful in manyapplications. LetD=d/dx. Then, for any functionf(x),Dl(xf) =x(Dlf) +l(Dl 1f).
3 This can be proved by induction: it clearly holds forl= 0, for which it readsxf=xf, andforl= 1 by the product rule for derivatives,D(xf) =x(Df) +f(Dx). Suppose it holds forl 1; thenDl(xf) =D[Dl 1(xf)] =D[x(Dl 1f+ (l 1)Dl 2f]= [x(Dlf) +Dl 1f] + (l 1)Dl 1f=x(Dlf) +l(Dl 1f).2 Apply forg(x) =xf(x):Dl(x2f) =Dl(x fx) =xDl(fx) +lDl 1(fx)=x[x(Dlf) +lDl 1f] +l[xDl 1f+ (l 1)Dl 2f]=x2(Dlf) + 2lx(Dl 1f) +l(l 1) (Dl 2f).This procedure iterated leads to the general and perhaps well known resultDn(fg) =nXk=0n!k! (n k)!(Dkf)(Dn kg),and so in particularDl[(x2 1)f] = (x2 1) (Dlf) + 2lx(Dl 1f) +l(l 1) (Dl 2f).( )Now, use ( ) to prove the middle of Jackson s ( ).)
4 ApplyDto Rodrigues Formula forPl+1, first commutingDlas above and then applying the product rule for derivatives anumber of times:DPl+1=D 12l+1(l+ 1)!Dl+1(x2 1)l+1 =12(l+ 1)D2 Dl12ll!(x2 1)(x2 1)l =12(l+ 1)12ll!D2h(x2 1)Dl(x2 1)l+ 2lxDl 1(x2 1)l+l(l 1)Dl 2(x2 1)li=12(l+ 1)hD((x2 1)DPl) + (2Pl+ 2xDPl) + (2lxDPl+ 4l Pl) +l(l 1) Legendre s equation, the first term isl(l+ 1)Pl. Gathering terms inxDPlandPl,DPl+1=12(l+ 1)h2(l+ 1)2Pl+ 2(l+ 1) , finally, the desired result,DPl+1= (l+ 1)Pl+xDPl.( )It seems likely that there is an easier way to get here, but this is another one:DPl+1=Dh12l+1(l+ 1)!Dl+1(x2 1)l+1i=12(l+ 1)12ll!
5 DlhD2(x2 1)l+1i=12(l+ 1)12ll!Dlh4l(l+ 1)x2(x2 1)l 1+ 2(l+ 1)(x2 1)li=12ll!Dlh2l(x2 1 + 1)(x2 1)l 1+ (x2 1)li=12ll!Dlh2l(x2 1)l 1+ (2l+ 1)(x2 1)li=DPl 1+ (2l+ 1) beginning to end, this saysDPl+1 DPl 1= (2l+ 1)Pl.( )
