ORTHOGONAL FUNCTIONS: THE LEGENDRE, - LSUMath

1 ORTHOGONAL functions : THE LEGENDRE, LAGUERRE, AND HERMITE POLYNOMIALSTHOMAS COVERSON, SAVARNIK DIXIT, ALYSHA HARBOUR,AND TYLER legendre , Laguerre, and Hermite equations areall homogeneous second order sturm -Liouville equations. Usingthe sturm -Liouville Theory we will be able to show that polynomialsolutions to these equations are ORTHOGONAL . In a more generalcontext, finding that these solutions are ORTHOGONAL allows us towrite a function as a Fourier series with respect to these legendre , Laguerre, and Hermite equations have many realworld practical uses which we will not discuss here. We will only focuson the methods of solution and use in a mathematical sense. In solvingthese equations explicit solutions cannot be found. That is solutionsin in terms of elementary functions cannot be found. In many cases itis easier to find a numerical or series is a generalized Fourier series theory which allows one to writea functionf(x) as a linear combination of an ORTHOGONAL system offunctions 1(x), 2(x).

2 , n(x),.. on [a,b]. The series produced is calledthe Fourier series with respect to the ORTHOGONAL system. While thecoefficients ,which can be determined by the formulacn= baf(x) n(x)dx ba 2n(x)dx,are called the Fourier coefficients with respect to the ORTHOGONAL are concerned only with showing that the legendre , Laguerre, andHermite polynomial solutions are ORTHOGONAL and can thus be used toform a Fourier series. In order to proceed we must define an innerproduct and define what it means for a linear operator to be define an inner product y1|y2 = bay1(x)y2(x)dxwherey1,y2 C2[a,b]and two functions aresaid to be ORTHOGONAL if(y1|y2) = bay1(x)y2(x) = linear operator L is self-adjoint if Ly1|y2 = y1|Ly2 for ally1, COVERSON, SAVARNIK DIXIT, ALYSHA HARBOUR, AND TYLER sturm -Liouville TheoryA sturm -Liouville equation is a homogeneous second order differen-tial equation of the form( )[p(x)y ] +q(x)y+ r(x)y= 0wherep(x),r(x)>0 on the interval [a,b] and where the functionq(x)is real-valued.

3 In order to make the problem simpler to solve we assumep(x),p (x),r(x),q(x) C[a,b]. We rewrite the equation in the formof an eigenvalue equation by defining a linear operatorLonC2[a,b] as( )Ly= [p(x)y ] +q(x) defined, the sturm -Liouville equation can be written in theform( )Ly+ r(x)y= 0 Now we impose boundary conditions such thaty C2[a,b] so thatLwill be self-adjoint with respect to the inner product defined abovewhich allows us to rewrite differential equations of the same form toshow that its solutionsy1,y2 C2[a,b] form an ORTHOGONAL basis. Itis also necessary to note that ify6= 0 andy BC2[a,b] is a solutiontoLy+ ry= 0 thenyis an eigenfunction and is an (y, ) is an want to know the boundary conditions necessary for L tobe self adjoint. We want Ly1|y2 y1|Ly2 = 0. Note that [p(x)y 1] +q(x)y1|y2 y1|[p(x)y 2] +q(x)y2 = ba(p y 1y2+py 1y2+qy1y2 y1p y 2 y1py 2 y1q1y2)dx= ba(p y 1y2+py 1y2 y1p y 2 y1py 2)dx= ba[p(y 1y2 y 2y1)] dx=p(b)(y 1(b)y2(b) y 2(b)y1(b)) p(a)(y1(a)y2(a) y 2(a)y1(a))In order for the equality to hold we wish to impose the boundary con-ditionsy(a) =y(b) = 0y (a) =y (b) = 0.

4 With these conditions wesay thaty BC2[a,b].Lemma eigenvalues of a sturm -Liouville problem are functions : THE legendre , LAGUERRE, AND HERMITE 0,y BC2[a,b] is a solution and satisfiesLy+ ry= 0 and compute Ly|y = y|Ly .Ly= ry ry|y = y| ry ry|y = y| ry ry|y = y| ry ry|y = y|ry bayyr(x)dx= bayyr(x)dx ba|y(x)|2r(x)dx= ba|y(x)|2r(x)dx = R With this equality, we have a new inner product called the weightedinner product( ) y1|y2 r= bay1(x)y2(x)r(x)dxwherey1,y2 C2[a,b] and (y|y)>0 wheny6= (y1, 1),(y2, 2)are eigenpairs where 16= 2theny1andy2are COVERSON, SAVARNIK DIXIT, ALYSHA HARBOUR, AND TYLER know thatLis self-adjoint becasuey BC2[a,b],Ly= ry, and R. Ly1|y2 = y1|Ly2 1ry1|y2 = y1| 2ry2 1ry1|y2 = y1| 2ry2 1ry1|y2 = y1| 2ry2 1 ry|y = 2 y1|ry2 1 bay1y2r(x)dx= 2 bay1y2r(x)dx 1 y1|y2 r= 2 y1|y2 r( 1 2) y1|y2 r= 0 y1,y2 r= 0 Thereforey1andy2are legendre PolynomialsThe legendre Differential Equation is( )(1 x2)y 2xy +n(n+ 1)y= 0,n R,x ( 1,1)We know thatx= 0 is an ordinary point of equation ( ).

5 We see thatwhen we divide by the coefficient (1 x2) thatx ( 1,1). We will seelater that the property of orthogonality falls out on the interval [ 1,1]by the sturm -Liouville Theory. In order to find the series solution tothis differential equation we will use the power series (x) = k=0akxky (x) = k=1akkxk 1y (x) = k=2akk(k 1)xk 2 Insert these terms into the original equation ( ) to obtainORTHOGONAL functions : THE legendre , LAGUERRE, AND HERMITE POLYNOMIALS5(1 x2) k=2akk(k 1)xk 2 2x k=1akkxk 1+n(n+ 1) k=0akxk= gives k=2akk(k 1)xk 2 k=2akk(k 1)xk 2 k=1akkxk+n(n+ 1) k=0akxk= 0making powers and indicies equal k=0ak+2(k+ 2)(k+ 1)xk k=0akk(k 1)xk 2 k=0ak(k)xk+ k=0n(n+ 1)akxk= 0simplify k=0[(k+ 2)(k+ 1)ak+2 (k)(k 1)ak 2kak+n(n+ 1)ak]xk= 0equating coefficients(k+ 2)(k+ 1)ak+2 (k)(k 1)ak 2kak+n(n+ 1)ak= 0solving forak+2gives us a recurrence relationak+2=k(k+ 1) n(n+ 1)(k+ 2)(k+ 1)ak.( )6 THOMAS COVERSON, SAVARNIK DIXIT, ALYSHA HARBOUR, AND TYLER are looking for polynomial solutions.

6 If we assume oursolution has degreeLthenL k=0akxk=a0+a1x+a2x2+ +aL+ 0xL+1+ 0xL+2+ Where all the terms followingaLwill be zero, whileaL6= 0. So weknow,aL+2=aLL(L+ 1) n(n 1)(L+ 2)(L+ 1)= 0L(L+ 1) n(n 1)(L+ 2)(L+ 1)= 0L(L+ 1) n(n+ 1) = 0L(L+ 1) =n(n+ 1)L=norL= (n+ 1)SoL=nis our solution because all terms aftern+1 are zero. Thereforethe degree of our polynomial solution is n where n is an get two linearly independent series solutions from the recurrencerelation ( ). The first solutions comes from the even values the second solution comes from the odds values ofk. We assumea06= 0 anda16= (x) =a0[1 n(n+ 1)2x2+(n 2)n(n+ 1)(n+ 3)4!x4 (n 4)(n 2)n(n+ 1)(n+ 3)(n+ 5)6!x6+ ]y2(x) =a1[x (n 1)(n+ 2)3!x3+(n 3)(n 1)(n+ 2)(n+ 4)5!x5+ ]Where both solutions are valid forx ( 1,1). Finding the Le-gendre polynomials can be very long and difficult. There are manymethods including Rodrigue s Formula that are useful in finding theseORTHOGONAL functions : THE legendre , LAGUERRE, AND HERMITE POLYNOMIALS7polynomials.

7 The first five legendre Polynomials turn out to beP0(x) = 1P1(x) =xP2(x) =12(3x2 1)P3(x) =12x(5x2 3)P4(x) =18(35x4 30x2+ 3) By rewriting the legendre Polynomial as a sturm -Liouville problem,we can prove its orthgonality. We find that the operator can be writtenasLy= [(1 x2)y ] .wheref,g C[ 1,1]. After imposing the conditions that anyf,g BC2[ 1,1] wheneverf,gmeet the conditions. We want Lf|g = f|Lg . That is we wantLto be self-adjoint. Lf|g f|Lg = 0 1 1Lf(x)g(x) f(x)Lg(x)dx= 1 1((1 x2)f ) g(x) f(x)((1 x2)g ) dx= 1 1( 2xf + (1 x2)f )g f( 2xg (1 x2)g )dx= 1 1(1 x2)f g 2xf g+ 2xfg (1 x2)fg dx= 1 1[(1 x2)(f g g f)] dx= [(1 x2)(f g g f)]1 1= self-adjoint with no imposed COVERSON, SAVARNIK DIXIT, ALYSHA HARBOUR, AND TYLER OTTOL etynandym,wheren6=m, be polynomial solutions to the differ-ential equation,Lyn= n(n+ 1)y. n(n+ 1) yn|ym = Lyn|ym = yn|Lym = yn| m(m+ 1)ym = m(m+ 1) yn|ym So, n(n+ 1) yn|ym = m(m+ 1) yn|ym , sincen6=m, yn|ym = could have also used Lemma ( ) to say that the legendre polyno-mials are ORTHOGONAL due to the sturm -Liouville theory.

8 The Legendrepolynomaials are ORTHOGONAL on the interval [ 1,1] with respect to thethe weight functionr(x) = Laguerre PolynomialsThe Laguerre differential equation is( )xy + (1 x)y +ny= 0, n R, x [0, ).We know thatx= 0 is a regular singular point of equation ( ). Inorder to find the series solution to this differential equation we mustuse the Frobenius method which is useful for solving equations of theform( )x2y +xp(x)y +q(x)y= will use the Frobenius method to find a series solution to equation( ) of the form( )y1(x) =xr k=0akxk, a06= 0whereris the root of the indicial equation( )r(r 1) +p0r+ compare the Laguerre equation( ) to our standard form equation( ). We multiply ( ) byxto obtain the equationORTHOGONAL functions : THE legendre , LAGUERRE, AND HERMITE POLYNOMIALS9( )x2+ (1 x)xy +nxyWe see thatp(x) = (1 x) andq(x) =nxp(0) = 1 andq(0) = our indicial equation isr(r 1) +r= 0orr2= the roots of the indcial equation arer1=r2= 0.]

9 Both roots areequal so we will have a second linearly independent equation, which wewill not use, of the the form( )y2(x) =y1(x) lnx+xr1 k= will now find the recurrence realtion for the coefficients fory1(x)by direct substitution ofy1(x) into equation ( ).y(x) =xr k=0akxk= k=0akx(k+r)y (x) = k=1(k+r)akx(k+r 1)y (x) = k=2(k+r)(k+r 1)akx(k+r 2)Plug above equations into ( ) to get10 THOMAS COVERSON, SAVARNIK DIXIT, ALYSHA HARBOUR, AND TYLER OTTOx k=2(k+r)(k+r 1)akx(k+r 2)+ (1 x) k=1(k+r)akx(k+r 1)+n k=0akx(k+r)= 0which gives us k=2(k+r)(k+r 1)akx(k+r 1)+ k=1(k+r)akx(k+r 1) k=1(k+r)akx(k+r)+ k=0nakx(k+r)= 0making all powers equal k=1(k+r+ 1)(k+r)ak+1x(k+r)+ k=0(k+r+ 1)ak+1x(k+r) k=1(k+r)akx(k+r)+ k=0nakx(k+r)= 0set r=0 to getORTHOGONAL functions : THE legendre , LAGUERRE, AND HERMITE POLYNOMIALS11 k=1(k+ 1)(k)ak+1xk+ k=0(k+ 1)ak+1xk k=1kakxk+ k=0nakxk= the terms withk= 1 we see that thek= 0 terms add nothing so k=0[(k+ 1)(k)ak+1+ (k+ 1)ak+1 kak+nak]xk= to the uniqueness of a power series we set the coefficients equal tozero and solve for the recurrence relation.

10 [(k+ 1)(k)ak+1+ (k+ 1)ak+1 kak+nak] = 0[(k+ 1)(k) + (k+ 1)]ak+1+ ( k+n)ak= 0ak+1=(k n)ak[(k+ 1)(k) + (k+ 1)]12 THOMAS COVERSON, SAVARNIK DIXIT, ALYSHA HARBOUR, AND TYLER OTTOThe coefficients area1= na0a2=(1 n)a14=n(n 1)a022a3=(2 n)a29= n(n 1)(n 2)a022 32a4=(3 n)a316=n(n 1)(n 2)(n 3)a022 32 [(k 1) n]ak 1k2=( 1)kn(n 1)(n 2)..(n k+ 1)22 32 42 k2=( 1)kn!a0(k!)2(n k)!Therefore by substituting these coefficients into equation ( ) we ob-tain the series solution for equation ( ).yn(x) =a0(1 nx+n(n 1)22x2 n(n 1)(n 2)22 32x3( )+n(n 1)(n 2)(n 3)a022 32 42x4+ +( 1)kn!(k!)2(n k)!xk+ )Which is( )yn(x) = k=0( 1)kn!a0(k!)2(n k)!xk, k= 0,1,2,..Whenn= 0,1,2,3,..the series ( ) ends. That is the terms after thenth term are zero. If we takea0to bek! then we get polynomials. Wecall these polynomials Laguerre polyomials. The Laguerre PolynomialsORTHOGONAL functions : THE legendre , LAGUERRE, AND HERMITE POLYNOMIALS13areL0(x) = 1( )L1(x) = x+ 1L2(x) =x2 4x+ 2L3(x) = x3+ 9x2 18x+ 6L4(x) =x4 16x3+ 72x2 96x+ 24 Ln(x) =n k=0( 1)k(n!)