Hankel determinants, continued fractions, orthgonal polynomials, and hypergeometric series Ira M. Gessel with Jiang Zeng and Guoce Xin LaBRI June 8, 2007
Continued fractions and Hankel determinants There is a close relationship between continued fractions, Hankel determinants, and orthogonal polynomials.
Continued fractions and Hankel determinants There is a close relationship between continued fractions, Hankel determinants, and orthogonal polynomials. Why are we interested in these things?
Continued fractions and Hankel determinants There is a close relationship between continued fractions, Hankel determinants, and orthogonal polynomials. Why are we interested in these things? Continued fractions count certain “weighted Motzkin paths” which encode objects of enumerative interest such as partitions and permutations. (Flajolet)
Continued fractions and Hankel determinants There is a close relationship between continued fractions, Hankel determinants, and orthogonal polynomials. Why are we interested in these things? Continued fractions count certain “weighted Motzkin paths” which encode objects of enumerative interest such as partitions and permutations. (Flajolet) Hankel determinants arise in some enumeration problems, for example, counting certain kinds of tilings or alternating sign matrices.
Orthogonal polynomials � � A sequence of polynomials p n ( x ) n ≥ 0 , where p n ( x ) has degree n , is orthogonal if there is a linear functional L on polynomials � � � p m ( x ) 2 � such that L p m ( x ) p n ( x ) = 0 for m � = n , but L � = 0. We will assume that L ( 1 ) = 1.
Orthogonal polynomials � � A sequence of polynomials p n ( x ) n ≥ 0 , where p n ( x ) has degree n , is orthogonal if there is a linear functional L on polynomials � � � p m ( x ) 2 � such that L p m ( x ) p n ( x ) = 0 for m � = n , but L � = 0. We will assume that L ( 1 ) = 1. � � are µ n = L ( x n ) . p n ( x ) The moments of the sequence
Orthogonal polynomials � � A sequence of polynomials p n ( x ) n ≥ 0 , where p n ( x ) has degree n , is orthogonal if there is a linear functional L on polynomials � � � p m ( x ) 2 � such that L p m ( x ) p n ( x ) = 0 for m � = n , but L � = 0. We will assume that L ( 1 ) = 1. � � are µ n = L ( x n ) . So p n ( x ) The moments of the sequence µ 0 = 1.
Orthogonal polynomials � � A sequence of polynomials p n ( x ) n ≥ 0 , where p n ( x ) has degree n , is orthogonal if there is a linear functional L on polynomials � � � p m ( x ) 2 � such that L p m ( x ) p n ( x ) = 0 for m � = n , but L � = 0. We will assume that L ( 1 ) = 1. � � are µ n = L ( x n ) . So p n ( x ) The moments of the sequence µ 0 = 1. � � Theorem. The sequence p n ( x ) of monic polynomials is � � orthogonal if and only if there exist numbers a n n ≥ 0 and � � b n n ≥ 1 , with b n � = 0 for all n ≥ 1 such that xp n ( x ) = p n + 1 ( x ) + a n p n ( x ) + b n p n − 1 ( x ) , n ≥ 1 with p 0 ( x ) = 1 and p 1 ( x ) = x − a 0 .
The connection We then have the continued fraction ∞ 1 µ k x k = � b 1 x 2 k = 0 1 − a 0 x − b 2 x 2 1 − a 1 x − 1 − a 2 x − · · · and the Hankel determinant det ( µ i + j ) 0 ≤ i , j < n is equal to b n − 1 b n − 2 · · · b 2 n − 2 b 1 . 1 2
If we’re interested in Hankel determinants or continued fractions, what good are the orthogonal polynomials?
If we’re interested in Hankel determinants or continued fractions, what good are the orthogonal polynomials? There is a systematic treatment of all the “classical” orthogonal polynomials, that have explicit expressions as hypergeometric series, called the Askey scheme. In a comprehensive paper (or web site) by Koekoek and Swarttouw, you can look up all of these orthogonal polynomials and find out everything you might want to know about them. . .
If we’re interested in Hankel determinants or continued fractions, what good are the orthogonal polynomials? There is a systematic treatment of all the “classical” orthogonal polynomials, that have explicit expressions as hypergeometric series, called the Askey scheme. In a comprehensive paper (or web site) by Koekoek and Swarttouw, you can look up all of these orthogonal polynomials and find out everything you might want to know about them. . . except the moments.
If we’re interested in Hankel determinants or continued fractions, what good are the orthogonal polynomials? There is a systematic treatment of all the “classical” orthogonal polynomials, that have explicit expressions as hypergeometric series, called the Askey scheme. In a comprehensive paper (or web site) by Koekoek and Swarttouw, you can look up all of these orthogonal polynomials and find out everything you might want to know about them. . . except the moments. What are the moments of the classical orthogonal polynomials?
If we’re interested in Hankel determinants or continued fractions, what good are the orthogonal polynomials? There is a systematic treatment of all the “classical” orthogonal polynomials, that have explicit expressions as hypergeometric series, called the Askey scheme. In a comprehensive paper (or web site) by Koekoek and Swarttouw, you can look up all of these orthogonal polynomials and find out everything you might want to know about them. . . except the moments. What are the moments of the classical orthogonal polynomials? What are their generating functions?
If we’re interested in Hankel determinants or continued fractions, what good are the orthogonal polynomials? There is a systematic treatment of all the “classical” orthogonal polynomials, that have explicit expressions as hypergeometric series, called the Askey scheme. In a comprehensive paper (or web site) by Koekoek and Swarttouw, you can look up all of these orthogonal polynomials and find out everything you might want to know about them. . . except the moments. What are the moments of the classical orthogonal polynomials? What are their generating functions? Why do some, but not all of them, have nice exponential generating functions?
If we’re interested in Hankel determinants or continued fractions, what good are the orthogonal polynomials? There is a systematic treatment of all the “classical” orthogonal polynomials, that have explicit expressions as hypergeometric series, called the Askey scheme. In a comprehensive paper (or web site) by Koekoek and Swarttouw, you can look up all of these orthogonal polynomials and find out everything you might want to know about them. . . except the moments. What are the moments of the classical orthogonal polynomials? What are their generating functions? Why do some, but not all of them, have nice exponential generating functions? What are the orthogonal polynomials whose moments are the Genocchi numbers?
How do we find the moments of a sequence of orthogonal polynomials? The key is to find numbers α 0 , α 1 , α 2 , . . . such � � that we can evaluate L ( x + α 0 )( x + α 1 ) · · · ( x + α k ) for each k . (Usually this isn’t too hard to do.)
How do we find the moments of a sequence of orthogonal polynomials? The key is to find numbers α 0 , α 1 , α 2 , . . . such � � that we can evaluate L ( x + α 0 )( x + α 1 ) · · · ( x + α k ) for each k . (Usually this isn’t too hard to do.) We then apply the following Lemma: ∞ t n 1 � ( x + α 0 )( x + α 1 ) · · · ( x + α n − 1 ) ( 1 + α 0 t ) · · · ( 1 + α n t ) = 1 − xt . n = 0 Proof. We have the indefinite sum m t n � ( x + α 0 ) · · · ( x + α n − 1 ) ( 1 + α 0 t ) · · · ( 1 + α n t ) n = 0 t m + 1 1 � � = 1 − ( x + α 0 ) · · · ( x + α n ) , 1 − xt ( 1 + α 0 t ) · · · ( 1 + α n t ) which is easily proved by induction. The lemma follows by taking m → ∞ .
Now suppose that � � L ( x + α 0 )( x + α 1 ) · · · ( x + α n − 1 ) = M n . We apply L to ∞ t n 1 � ( x + α 0 )( x + α 1 ) · · · ( x + α n − 1 ) ( 1 + α 0 t ) · · · ( 1 + α n t ) = 1 − xt . n = 0 to get ∞ t n � 1 � � M n ( 1 + α 0 t ) · · · ( 1 + α n t ) = L 1 − xt n = 0 ∞ ∞ L ( x k ) t k = � � µ k t k . = k = 0 k = 0
Let’s look at an example. First, the standard notation: ( α ) n = α ( α + 1 ) · · · ( α + n − 1 ) � a 1 , . . . , a p ∞ � � ( a 1 ) n · · · ( a p ) n � z n � p F q � z = � b 1 , . . . , b p n ! ( b 1 ) n · · · ( b q ) n n = 0 The Hahn polynomials are defined by � − n , n + α + β + 1 , − x � � � Q n ( x ; α, β, N ) = 3 F 2 � 1 � α + 1 , − N They are orthogonal with respect to the linear functional L given by N � α + x �� β + N − x � � � � L p ( x ) = p ( x ) , x N − x x = 0 By applying Vandermonde’s theorem we find that = ( α + β + 2 ) N ( α + 1 ) k ( α + β + 2 + N ) k � � L ( x + α + 1 ) k . N ! ( α + β + 2 ) k
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