This page has been left blank deliberately. . . Chemnitz, CMS2013, - PowerPoint PPT Presentation

This page has been left blank deliberately. . . Chemnitz, CMS2013, September of 2013 – p. 1

Testing the remote control. . . Chemnitz, CMS2013, September of 2013 – p. 2

The “ ( x n ) ” sequence Paul Nevai paul@nevai.us Chemnitz, CMS2013, September of 2013 – p. 3

Introduction Linear difference equations are the discrete version of linear differential equations, and have a theory that parallels the latter. Even here, the problems can be quite difficult. E.g., going from Fibonacci’s F n +1 = F n + F n − 1 to, say, � � 1 + sin 1 F n +1 = F n + F n − 1 n is highly nontrivial, c.f. Poincaré’s theorem on linear difference equations. Destroying linearity by, say, F n +1 = F 2 n + F n − 1 can lead to very interesting problems. Chemnitz, CMS2013, September of 2013 – p. 4

def = Orthogonal Polynomials OPs Let α be a positive Borel measure with infinite support in R and let p n ( α, x ) = γ n ( α ) x n + l.d.t. , def n ∈ N 0 = { 0 , 1 , 2 , . . . } , denote the OPs w.r.t. α . (l.d.t. def = lower degree terms) Orthogonality: � p m p n dα = δ m,n . R The three-term recurrence: xp n = a n +1 p n +1 + b n p n + a n p n − 1 where a n = γ n − 1 /γ n > 0 and b n ∈ R “describe” the symmetry of the measure w.r.t. a vertical line. Chemnitz, CMS2013, September of 2013 – p. 5

Favard’s theorem THEOREM. (Favard, 1935) Given ( a n > 0) and ( b n ∈ R ) , if ( p n ) satisfy the three-term recurrence xp n = a n +1 p n +1 + b n p n + a n p n − 1 then they are OPs w.r.t. some, not necessarily unique, α in R . Chemnitz, CMS2013, September of 2013 – p. 6

Hermite polynomials The Hermite polynomials, well, the Hermite functions, are the eigenfunctions of the Fourier transform. So there can be no doubt that they are worthy of study. Chemnitz, CMS2013, September of 2013 – p. 7

Hermite polynomials The Hermite polynomials, well, the Hermite functions, are the eigenfunctions of the Fourier transform. So there can be no doubt that they are worthy of study. PUZZLE. What is the ugliest word in mathematical English? Chemnitz, CMS2013, September of 2013 – p. 7

Hermite polynomials The Hermite polynomials, well, the Hermite functions, are the eigenfunctions of the Fourier transform. So there can be no doubt that they are worthy of study. PUZZLE. What is the ugliest word in mathematical English? ANSWER. IMHO, it’s a tie between eigenfunction and eigen- value. Note. IMHO means “in my humble opinion”. Chemnitz, CMS2013, September of 2013 – p. 7

Hermite polynomials The Hermite polynomials, well, the Hermite functions, are the eigenfunctions of the Fourier transform. So there can be no doubt that they are worthy of study. PUZZLE. What is the ugliest word in mathematical English? ANSWER. IMHO, it’s a tie between eigenfunction and eigen- value. Note. IMHO means “in my humble opinion”. In the next few slides I will show how to obtain certain proper- ties of Hermite polynomials using “nothing” but orthogonality. The very same ideas can be applied to much more “sophis- ticated” weight functions. Chemnitz, CMS2013, September of 2013 – p. 7

Hermite polynomials ( h n ) w ( x ) def − x 2 � � dα ( x ) = w ( x ) dx = exp , x ∈ R . where Orthogonality: � h m h n w = δ m,n . R Chemnitz, CMS2013, September of 2013 – p. 8

Hermite polynomials ( h n ) w ( x ) def − x 2 � � dα ( x ) = w ( x ) dx = exp , x ∈ R . where Orthogonality: � h m h n w = δ m,n . R The three-term recurrence: � � n + 1 n xh n = h n +1 + 2 h n − 1 , n = 0 , 1 , 2 , . . . , 2 that is, � n a n = ( n = 2 , 3 , . . . ) & b n = 0 ( n = 0 , 1 , 2 , . . . ) 2 and R x 2 exp − x 2 � � � dx = Γ (3 / 2) Γ (1 / 2) = 1 a 2 a 2 0 = 0 & 1 = 2 . � R exp ( − x 2 ) dx Chemnitz, CMS2013, September of 2013 – p. 8

a n for h n The following tricky computation of ( a n ) probably goes back to Shohat and then Freud, probably innocently, repeated the same argument; it is based on the the property of w that w ( x ) ′ = − 2 x w ( x ) y ′ = − 2 x y. i.e. Chemnitz, CMS2013, September of 2013 – p. 9

a n for h n The following tricky computation of ( a n ) probably goes back to Shohat and then Freud, probably innocently, repeated the same argument; it is based on the the property of w that w ( x ) ′ = − 2 x w ( x ) y ′ = − 2 x y. i.e. On one hand, � � � ( h n h n − 1 ) ′ w = w orth � � h ′ n h n − 1 + h n h ′ h ′ = n h n − 1 w = n − 1 R R R Chemnitz, CMS2013, September of 2013 – p. 9

a n for h n The following tricky computation of ( a n ) probably goes back to Shohat and then Freud, probably innocently, repeated the same argument; it is based on the the property of w that w ( x ) ′ = − 2 x w ( x ) y ′ = − 2 x y. i.e. On one hand, � � � ( h n h n − 1 ) ′ w = w orth � � h ′ n h n − 1 + h n h ′ h ′ = n h n − 1 w = n − 1 R R R � � n γ n x n − 1 + l.d.t. h n − 1 w orth x n − 1 h n − 1 w orth � � = n γ n = R R Chemnitz, CMS2013, September of 2013 – p. 9

a n for h n The following tricky computation of ( a n ) probably goes back to Shohat and then Freud, probably innocently, repeated the same argument; it is based on the the property of w that w ( x ) ′ = − 2 x w ( x ) y ′ = − 2 x y. i.e. On one hand, � � � ( h n h n − 1 ) ′ w = w orth � � h ′ n h n − 1 + h n h ′ h ′ = n h n − 1 w = n − 1 R R R � � n γ n x n − 1 + l.d.t. h n − 1 w orth x n − 1 h n − 1 w orth � � = n γ n = R R n γ n � = n γ n � n − 1 w = n γ n − 1 x n − 1 + l.d.t. h n − 1 w orth h 2 � � . γ n − 1 γ n − 1 a n R R Chemnitz, CMS2013, September of 2013 – p. 9

a n for h n On the other hand, by integration by parts, � � � ( h n h n − 1 ) ′ w h n h n − 1 w ′ = 2 ibp h n h n − 1 x w orth = − = 2 a n R R R Chemnitz, CMS2013, September of 2013 – p. 10

a n for h n On the other hand, by integration by parts, � � � ( h n h n − 1 ) ′ w h n h n − 1 w ′ = 2 ibp h n h n − 1 x w orth = − = 2 a n R R R because, in general, � � γ n − 1 x n − 1 + l.d.t. x p n p n − 1 dα orth dα orth � � = x p n = R R γ n − 1 � p n ( γ n x n + l.d.t. ) dα = γ n − 1 � n dα = γ n − 1 p 2 = a n . γ n γ n γ n R R Chemnitz, CMS2013, September of 2013 – p. 10

a n for h n On the other hand, by integration by parts, � � � ( h n h n − 1 ) ′ w h n h n − 1 w ′ = 2 ibp h n h n − 1 x w orth = − = 2 a n R R R because, in general, � � γ n − 1 x n − 1 + l.d.t. x p n p n − 1 dα orth dα orth � � = x p n = R R γ n − 1 � p n ( γ n x n + l.d.t. ) dα = γ n − 1 � n dα = γ n − 1 p 2 = a n . γ n γ n γ n R R So � n ( h n h n − 1 ) ′ w = 2 a n . = a n R Chemnitz, CMS2013, September of 2013 – p. 10

b n for h n The coefficient b n = 0 because w is even. In general, from the recurrence formula, � x p 2 b n = n dα. R From the so-to-speak Hermitian point of view, � � � � n w = − 1 = 1 � ′ w = n w ′ ibp n w orth x h 2 h 2 h 2 � h n h ′ b n = = 0 . n 2 2 R R R R Chemnitz, CMS2013, September of 2013 – p. 11

b n for h n The coefficient b n = 0 because w is even. In general, from the recurrence formula, � x p 2 b n = n dα. R From the so-to-speak Hermitian point of view, � � � � n w = − 1 = 1 � ′ w = n w ′ ibp n w orth x h 2 h 2 h 2 � h n h ′ b n = = 0 . n 2 2 R R R R HOMEWORK. Find the a n ’s and the b n ’s for the weight function w ( x ) def − β 2 x 2 + β 1 x + β 0 � � with x ∈ R where β 2 > 0 . = exp Chemnitz, CMS2013, September of 2013 – p. 11

b n for h n The coefficient b n = 0 because w is even. In general, from the recurrence formula, � x p 2 b n = n dα. R From the so-to-speak Hermitian point of view, � � � � n w = − 1 = 1 � ′ w = n w ′ ibp n w orth x h 2 h 2 h 2 � h n h ′ b n = = 0 . n 2 2 R R R R HOMEWORK. Find the a n ’s and the b n ’s for the weight function w ( x ) def − β 2 x 2 + β 1 x + β 0 � � with x ∈ R where β 2 > 0 . = exp HOMEWORK. What if x ∈ R + for w above instead of x ∈ R ? Chemnitz, CMS2013, September of 2013 – p. 11

DE for h n Let k < n − 1 & let r k be a polynomial of degree at most k . Then � � � ( h n r k ) ′ w ibp n r k w orth h ′ = = − 2 ( h n r k ) x w = R R R � h n ( r k x ) w orth − 2 = 0 R Chemnitz, CMS2013, September of 2013 – p. 12

DE for h n Let k < n − 1 & let r k be a polynomial of degree at most k . Then � � � ( h n r k ) ′ w ibp n r k w orth h ′ = = − 2 ( h n r k ) x w = R R R � h n ( r k x ) w orth − 2 = 0 R so that h ′ n = const × h n − 1 . Thus, we have an Appell sequence, and by comparing leading coefficients √ h ′ n = 2 n h n − 1 and, therefore, √ √ � � h ′′ 2 n h ′ n = n − 1 = 2 n 2 ( n − 1) h n − 2 = 2 n ( n − 1) h n − 2 . Chemnitz, CMS2013, September of 2013 – p. 12