An Orthogonality Property of Legendre Polynomials

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1 A Orthogoality Property of Legedre Polyomials L. Bos, A. Naraya, N. Leveberg 3 ad F. Piazzo 4 April 7, 05 Abstract We give a remarable secod othogoality property of the classical Legedre polyomials o the real iterval [, ]: Polyomials up to degree from this family are mutually orthogoal uder the arcsie measure weighted by the degree- ormalized hristoffel fuctio. Keywords: Legedre polyomials, hristoffel fuctio, equilbrium measure Let P (x) deote the classical Legedre polyomial of degree ad + P(x) := P (x) its orthoormalized versio. Thus, with δ i,j the Kroecer delta, the family P satisfies P i (x)p j (x) dx = δ i,j, i, j 0 We cosider the ormalized (reciprocal of) the associated hristoffel fuctio K (x) := + (P(x)). () As is well ow, K (x)dx teds wea- to the equilibrium measure of complex potetial theory for the iterval [, ], ad more precisely, =0 lim K (x) =, x (, ) π x Dept. of omputer Sciece, Uiversity of Veroa, Italy leoardpeter.bos@uivr.it Mathematics Dept., Uiversity of Massachusetts Dartmouth, North Dartmouth, Massachusetts, USA ail.araya@umassd.edu 3 Dept. of Mathematics, Idiaa Uiversity, Bloomigto, Idiaa, USA levebe@idiaa.edu 4 Dept. of Mathematics, Uiversity of Padua, Italy fpiazzo@math.uipd.it

2 locally uiformly. I other words, lim K (x) π =, x (, ) x locally uiformly, ad it would ot be uexpected that, at least asymptotically, Pi (x)pj (x) K (x) π dx δ ij, 0 i, j. x The purpose of this ote is to show that the above is actually a idetity. Theorem With the above otatio Pi (x)pj (x) K (x) π dx = δ ij, 0 i, j. () x We expect this result to have use i applied approximatio problems. For example, oe applicatio lies i polyomial approximatio of fuctios { from poitevaluatios. Our result idicates that the fuctios {Q j (x)} j=0 } P K(x) j (x) j=0 are a orthoormal set o (, ) uder the Lebesgue desity π. If we x geerate Mote arlo samples from this desity, evaluate a uow fuctio at these samples, ad perform least-squares regressio usig Q j as a basis, the a stability factor for this problem is give by max x [,] j=0 Q j (x) = + []. I fact, this is the smallest attaiable stability factor, ad therefore this approximatio strategy has optimal stability. The remaider of this documet is devoted to the proof of (). Proof of Theorem. We chage variables lettig x = cos(θ) to arrive at the equivalet expressio π π 0 Pi (cos(θ))p j (cos(θ)) dθ = δ ij, K (cos(θ)) 0 i, j which by symmetry is equivalet to π π 0 Pi (cos(θ))p j (cos(θ)) dθ = δ ij, 0 i, j. (3) K (cos(θ)) Now, for z, let J(z) := (z + /z)/. The for z = e iθ i the itegral (3) we obtai, dθ = iz dz, cos(θ) = J(z), ad the equatio becomes πi z Pi (J(z))P j (J(z)) dz = δ ij, 0 i, j (4) K (J(z)) where is the uit circle, orieted i the couter-clocwise directio. The proof is a direct calculatio of (4) based o the followig lemmas. First ote that K (cos(θ)) is a positive trigoometric polyomial (of degree ). By the Féjer-Riesz Factorizatio Theorem there exists a trigoometric polyomial, T (θ) say, such that K (cos(θ)) = T (θ).

3 I geeral the coefficiets of the factor polyomial, T (θ) i this case, are algebraic fuctios of the coefficiets of the origial polyomial. However i our case we have the explict (essetially) ratioal factorizatio. Propositio (Féjer-Riesz Factorizatio of K (J(z))) Let F (z) := d dz The ( z + P (J(z)) ) = (+)z P (J(z))+ z (z ) P (J(z)). (5) K (J(z)) = Proof. To begi, oe may easily verify that ( + ) F (z)f (/z). (6) Hece F (/z) = z {( + )z P (J(z)) z (z ) P (J(z))}. (7) ( z F (z)f (/z) = z {( + ) z (P (J(z))) z ( ) ) (P (J(z)) } ( z = ( + ) (P (J(z))) z ) (P (J(z)). Now otice that ( z z ) = ( z ) 4 z (z + + z ) 4 = 4 = J (z) so that F (z)f (/z) = ( + ) (P (J(z))) (J(z) )(P (J(z))). The result follows the from Lemma, below. Lemma For all x, we have K (x) = ( ( + ) (P (x)) (x )(P ( + ) (x)) ). Proof. First, we collect the followig ow idetities cocerig Legedre polyomials []: (hristoffel-darboux formula) P (x) = + [ P + (x)p (x) P + (x)p (x) ] (8a) =0 3

4 (Three-term recurrece) ( + )P + (x) = ( + )xp (x) P (x) (8b) (Differetiated three-term recurrece) ( + )P +(x) = ( + ) (P (x) + xp (x)) P (x) (8c) (x )P (x) = (xp (x) P (x)) (8d) ( + )P (x) = P +(x) P (x) (8e) We easily see from the hristoffel-darboux formula that K (x) = ( P + (x)p (x) P + (x)p (x) ). Hece the result holds iff ( + )P +(x)p (x) ( + )P + (x)p (x) = ( + ) (P (x)) (x )(P (x)) (8c),(8b),(8d) { ( + ) (P (x) + xp (x)) P (x) } P (x) {( + )xp (x) P (x)} P (x) = ( + ) (P (x)) (xp (x) P (x)) P (x) ( + ) (P (x)) P (x)p (x) + P (x)p (x) = ( + ) (P (x)) xp (x)p (x) + P (x)p (x) (P (x)) P (x)p (x) = xp (x)p (x) xp (x) = P (x) + P (x) (8c) ( ( + )P + + (x) + P (x) ) P (x) = P (x) + P (x) ( + )P +(x) = ( + )P (x) + ( + )( + )P (x), ad this last relatio is the same as the relatio (8e). There is somewhat more that ca be said about F (z). 4

5 Lemma Let F (z) be the polyomial of degree defied i (5). The all of its zeros are simple ad lie i the iterior of the uit dis. Proof. The polyomial Q (z) := z + P (J(z)) = z {z P (J(z))} has a zero at z = 0 ad its other zeros are those of P (J(z)) which are those z for which J(z) = r (, ), a zero of P (x). But J(z) = r (, ) (z + /z)/ = r z rz + = 0 z = r ± i r. I particular z = for the zeros of z P (J(z)). It follows the from the Gauss- Lucas Theorem that the zeros of F (z) are i the covex hull of z = 0 ad certai poits o the uit circle, i.e., are all i the closed uit dis. Suppose a zero of F (z) satisfies z =. By Propositio, K (J(z)) = ( + ) F (z)f (/z), so that K (J(z)) also vaishes. But z = implies that J(z) [, ], ad K (J(z)) thus caot vaish. Therefore, o zeros of F lie o the uit circle. To see that the zeros are all simple, a elemeary calculatio ad the ODE for Legedre polyomials gives us F (z) = ( + )z P (J(z)) + {z ( + )z }P (J(z)). Hece F (z) = F (z) = 0 if ad oly if ( ) z + ( z P (J(z)) (+) z z + z P z (J(z)) ) ( 0 = 0 ). But the determiat of this matrix is ( + )(z ( + )/z) ( + )(z )/z = ( + )/z 0. Hece F (z) = F (z) = 0 if ad oly if z P (J(z)) = z P (J(z)) = 0 if ad oly if P (J(z)) = P (J(z)) = 0, which is ot possible as P (x) has oly simple zeros. 5

6 The itegral (4) ca therefore be expressed as πi z P i (J(z))P j (J(z)) K (J(z)) dz = πi = πi = πi where we defie the polyomial of degree, ( + )z Pi (J(z))P j (J(z)) dz F (z)f (/z) ( + )z Pi (J(z))P j (J(z)) F (z)z dz F (/z) ( + )z Pi (J(z))P j (J(z)) dz F (z)g (z) G (z) := z F (/z). (9) As all the zeros of F (z) are i the iterior of the uit dis, the zeros of G (z) are all exterior to the (closed) uit dis. The followig formulas for F (z) ad G (z) will be useful. Lemma 3 We have ad F (z) = G (z) = z z {(( + )z )P (J(z)) zp (J(z))} z z {(z ( + ))P (J(z)) + zp (J(z))}. Proof. From the formula (5) we have ad from (7), F (z) = ( + )z P (J(z)) + z z P (J(z)) G (z) = ( + )z P (J(z)) z z P (J(z)). From the Legedre Polyomial idetity (8d) with x = J(z), we obtai z z P (J(z)) = z+ z J(z)P (J(z)) z+ z P (J(z)). ombiig these gives the result. It is also iterestig to ote that F (z) is a certai Hypergeometric fuctio. Lemma 4 We have. The polyomial y = F (z) is a solutio of the ODE ( z )y + ( )z y + 6y = 0. z 6

7 . If F (z) =: f (z ) the y = f (z) is a solutio of the Hypergeometric ODE z( z)y + (c (a + b + )z)y aby = 0 with a =, b = 3/ ad c = /. ( ) 3. f (z) = F (a, b; c; z). ( ) 4. F (z) = F (a, b; c; z ). 5. ( ) F (z) = ( + ) ( ( ) z =0 ( )( ) = ( + ) z. =0 6. If we write F (z) = =0 c z, the G (z) = = c z =0 =0 ) ( ) + c z. + Proof. Equatio () is easily verified usig the ODE for P (x) ad the defiitio of F (z), (5). () follows by chagig variables z = z. (3) follows as F (a, b; c; z) is the oly polyomial solutio of the Hypergeometric ODE. The costat of proportioality is calculated by otig that the leadig coefficiet of F (a, b; c; z) is + whereas that of f (z) is ( + )/ times the leadig coefficiet of P (x), i.e., ( + )/ ( ) /. (4) is trivial from the defitio f (z ) := F (z). (5) follows from the fact that F (a, b; c; z) = =0 with (a) the risig factorial, ad calculatig (a) = ( )! ( )!, (b) = so that ( + )!! (a) (b) =! (c) (a) (b) (c) z!,, (c) = ( ) ()!( )! ( )!! ( + ) ( ) ). (6) follows easy from the fact that G (z) := z F (/z) = =0 c z ad that c is easily computed from the formula for c give i (5). ( 7

8 Returig the the proof of the Theorem, we will actually show that ( + )z P (J(z)) (a) dz = 0,, πi F (z)g (z) (b) πi ( + )z P 0 (J(z)) dz =. F (z)g (z) The Theorem follows directly as, for i, j we may expad P i (x)p j (x) = =0 a P (x) for certai coefficiets a. From (a) ad (b) we the coclude that But for i j, πi ( + )z Pi (J(z))P j (J(z)) dz = a 0. F (z)g (z) a 0 = P i (x)p j (x)p 0 (x)dx = 0 as P 0 (x) = ad Pi (x) ad P j (x) are orthogoal. While for i = j, we have a 0 = (P i (x)) P 0 (x)dx =. We will ow compute the partial fractio decompositio of the itegrads i (a) ad (b), ( + )z P (J(z)), F (z)g (z) which will ivolve the followig two pairs of families of fuctios. Defiitio We defie A () 0 (z) := z, A () (z) := z, with B () 0 (z) := z, B () (z) := z, ( + + )A () + ad ( + + )B () + Further, we let := (( + ) + )J(z)A() := (( + ) + )J(z)B() (z) ( + )A() (z), =,, (z) ( + )B() (z), =,,. () 0 (z) := z, () (z) := z ( + )z, D () 0 (z) := z, D () ( + ) z (z) := z, 8

9 with ( ) () + (z) := (( )+)J(z)() (z) ( +)() (z), ad ( )D () + (z) := (( )+)J(z)D() (z) ( +)D() (z),. Propositio We have ( + )z P + (z) = A () (z)g (z) + B () (z)f (z), = 0,,, (0) so that ad so that ( + )z P + (J(z)) F (z)g (z) = A() (z) F (z) + B() (z) G (z), = 0,,, ( + )z P (z) = () (z)g (z) + D () (z)f (z), 0 () ( + )z P (J(z)) F (z)g (z) = () (z) F (z) + D() (z) G (z), 0. Proof (by iductio). As A () 0 = () 0 = z ad B () 0 = D () 0 = z, the = 0 case is i commo ad we calculate as desired. A () 0 (z)g (z) + B () 0 (z)f (z) = z (F (z) + G (z)) (Lemma 3) = z z z {(( + )z ) + (z ( + ))}P (J(z)) = z z {( + )z ( + )}P (J(z)) = ( + )z P (J(z)), 9

10 We ow prove (0) for the case =. We calculate, usig Lemma 3, A () (z)g (z) + B () (z)f (z) = z G (z) + z F (z) = z (G (z) + z F (z)) = z z z {[(z ( + ))P (J(z)) + zp (J(z))] + z [(( + )z )P (J(z)) zp (J(z))]} = z z {( + )(z4 )P (J(z)) z(z )P (J(z))} = z {( + )(z + )P (J(z)) zp (J(z))} = z {( + )(z + /z)p (J(z)) P (J(z))} = z {( + )J(z)P (J(z)) P (J(z))} (8b) = ( + )z P + (J(z)). Now for () for =. We calculate, agai usig Lemma 3, () (z)g (z) + D () (z)f (z) { ( + )z = z } ( + ) z G (z) + F (z) = z z { ( + )z z [(z ( + ))P (J(z)) + zp (J(z))] } ( + ) z + [(( + )z )P (J(z)) zp (J(z))] { = z ( + )z } z z ( + ) z P (J(z)) = z z ((( + )z ( + ))P (J(z)) = ( + )z P (J(z)). The rest of the proof proceeds by iductio. Assumig that (0) ad () hold from 0 up to a certai, we prove that they also hold for +. To this 0

11 ed we calculate A () + (z)g (z) + B () + (z)f (z) = = (( + ) + )J(z)A() (z) ( + )A() (( + ) + )J(z)B() (z) (z) ( + )B() {(( + ) + )J(z)[A() (z) G (z) F (z) (z)g (z) + B () (z)f (J(z))] ( + )[A () (z)g (z) + B () F (J(z))] = ( + )z + + {(( + ) + )J(z)P +(J(z)) ( + )P + (J(z))} (by the iductio hypothesis) = ( + )z P ++ (J(z)), by the three-term recursio formula for Legedre polyomials (8b) with degree m = +. Similarly, () + (z)g (z) + D () + (z)f (z) = (( ) + )J(z)() (z) ( + )() + = (( ) + )J(z)D() (z) (z) ( + )D() {(( ) + )J(z)[() (z) G (z) F (z) (z)g (z) + D () (z)f (J(z))] ( + )[ () (z)g (z) + D () F (J(z))] = ( + )z {(( ) + )J(z)P (J(z)) ( + )P ( ) (J(z))} (by the iductio hypothesis) = ( + )z P (+) (J(z)), usig the reverse three-term recursio, with m =. mp m (x) = (m + )xp m (x) (m + )P m+ (x) Due to the J(z) factor i the recursive defiitios of Defiitio, the fuctios A () (z), B() (z), () (z) ad D() (z) are all Lauret polyomials. It is easy to verify that they have the forms: +( 3) A () (z) = j= (+) a z,

12 +( ) B () (z) = j= ( ) +( ) () (z) = j= (+) +( ) D () (z) = j= (+) b z, c z, d z,. I particular, for 0 the fuctios A () polyomials of degree at most. Hece, for, ( + )z P (J(z)) dz = { πi F (z)g (z) πi (z), B() (z), () p(z) F (z) dz + (z) are all } q(z) G (z) dz for certai polyomials p(z) ad q(z) of degree at most. q(z) Now, G (z) dz = 0 as all the zeros of G (z) lie outside the (closed) uit dis. Further, if we let z j j, be the (simple) zeros of F (z), we may write p(z) F (z) = p(z)/c R j = F (z)/c z z j j= where c is the leadig coefficiet of F (z). Hece πi A (z) F (z) dz = πi πi But j= R j is the leadig coefficiet (of z ) of p(z)/c, i.e., 0, as p(z) is of degree at most. It follows that πi j= R j. ( + )z P (J(z)) dz = 0,. F (z)g (z) The cases P 0 (J(z)) ad P (J(z)) are special. First cosider the case P (J(z)). We have from Propositio, with =, ( + )z P (J(z)) = A () j= (z)g (z) + B () (z)f (z). j=+ 3 However, A () (z) = a j z j has a z while B () (z) = b j z j is still a poly- omial, of degree at most. Therefore it is still the case that B () (z) G (z) dz = 0 (the zeros of G (z) beig all oustide the uit dis). We eed to show that A () (z) dz = 0. Write A() (z) = q(z) + c/z where q(z) is a polyomial of F (z)

13 degree 3. The A () (z) F (z) dz = q(z) F (z) dz + c zf (z) dz. The first itegral o the right is zero as the coefficiet i q(z) of z is 0. For the secod itegral, decompose { c zf (z) dz = c F (0) z (F } (z) F (0))/z dz F (z) = { } F (0) z dz (F (z) F (0))/z dz. F (z) The first itegral is trivially πi, while the secod is πi times the coefficiet of z i (F (z) F (0))/z divided by the leadig coefficiet (of z ) i F (z), i.e., πi. Hece, ideed Lastly we calculate πi A () (z) dz = 0. F (z) ( + )z P 0 (J(z)) dz = ( + )z F (z)g (z) πi F (z)g (z) dz. We still have () from Propositio, However, () ( + )z P (z) = () (z) ad D () (z) have the form (z)g (z) + D () (z)f (z) () (z) = c z, j= D () (z) = d z, j= i.e., are both of the form p(z) + c/z where p(z) is a polyomial of degree. Specifically, write () (z) = p(z) + c/z ad D () (z) = q(z) + d/z. We thus have the followig expressio: πi ( + )z P 0 (z) dz = F (z)g (z) πi = πi [ + () (z) F (z) dz + πi p(z) F (z) dz + q(z) G (z) dz + c zf (z) dz D () (z) G (z) dz d zg (z) dz We eed to show that this expressio has value. Now we have already show q(z) that dz = 0 ad also remared that dz = 0. Hece we eed zf (z) G (z) ] 3

14 to calculate p(z) d dz ad dz. But the first of these is just πi F (z) πi zg (z) the (leadig) coefficiet of z i p(z), i.e., i () (z) divided by the leadig coefficiet of F (z). From Lemma 4 we have that ( ) F (z) = ( + ) z + ( ) ad it is easy to verify by iductio that also () (z) = ( + ) z +. Hece p(z) dz =. πi F (z) For the secod itegral, decompose as before d πi zg (z) dz = d πi G (0) { z (G (z) G (0))/z G (z) } dz. The secod itegral above is zero as all the zeros of G (z) are outside the closed uit dis. The first itegral is trivially d/g (0). From Lemma 4 we have that ( ) G (0) = ( + ) whereas the coefficiet of z i B () the same value. Hece πi (z) is easily verified by iductio to have d zg (z) dz = ad we have show that ( + )z P 0 (J(z)) dz = + =, πi F (z)g (z) as claimed. Refereces [] Albert ohe, Mar A. Daveport, ad Day Leviata, O the Stability ad Accuracy of Least Squares Approximatios, Foudatios of omputatioal Mathematics 3 (03), o. 5, [] Gábor Szegö, Orthogoal Polyomials, 4th ed., America Mathematical Soc.,