Is Analysis Necessary? Ira M. Gessel Brandeis University Waltham, - PDF document

Is Analysis Necessary? Ira M. Gessel Brandeis University Waltham, MA gessel@brandeis.edu Special Session on Algebraic and Analytic Combinatorics AMS Fall Eastern Meeting University of Connecticut, Storrs CT October 29, 2006

An example from Bak and Newman’s Complex Analysis n � 2 � n � 2 n � � Prove that = . k n k =0 n � 2 � n � n � 1 + 1 � = constant term in (1 + z ) n k z k =0 � n dz � 1 � 1 + 1 (1 + z ) n = 2 πi z z C (1 + z ) 2 n 1 � = dz z n +1 2 πi C = coefficient of z n in (1 + z ) 2 n � 2 n � = . n

A more interesting example The Bernoulli numbers of the second kind are defined by ∞ x n x � b n n ! = log(1 + x ) n =0 x 2 x 3 x 4 = 1 + x 2! − 1 2! + 1 3! − 19 6 4 30 4! x 5 x 6 x 7 + 9 5! − 863 6! + 1375 7! + · · · 4 84 24 Prove that they alternate in sign.

Since � 1 � 1 x (1+ x ) t dt = � � exp t log(1+ x ) dt = log(1 + x ) , 0 0 and ∞ � t � (1 + x ) t = � x n , n n =0 we have � 1 � 1 � t � b n = n ! dt = t ( t − 1) · · · ( t − n + 1) dt. n 0 0 Making the change of variables t = 1 − u , we have for n ≥ 1 � 1 b n = ( − 1) n − 1 (1 − u ) u ( u + 1) · · · ( u + n − 2) du. 0 Since the integrand is positive, ( − 1) n − 1 b n > 0 . Can we do this without analysis?

Expanding u ( u + 1) · · · ( u + n − 2) in terms of unsigned Stirling numbers of the first kind as n − 1 � c ( n − 1 , i ) u i i =0 and using the integral � 1 1 (1 − u ) u i du = ( i + 1)( i + 2) , 0 we find the explicit formula n − 1 1 � ( − 1) n − 1 b n = ( i + 1)( i + 2) c ( n − 1 , i ) . i =0 We could eliminate the analysis by defining a linear functional L on polynomials by L ( t n ) = 1 /n , and proving various properties of this linear functional. Alternatively, the explicit formula gives us a hint to finding a simple direct proof.

We have ∞ x n x � ( − 1) n − 1 b n n ! = log(1 − x ) , n =0 so ∞ x n n ! = d x � ( − 1) n b n +1 dx log(1 − x ) n =0 = e P − P − 1 , P 2 where P = − log(1 − x ) = � ∞ n =1 x n /n has positive coefficients. So ∞ ∞ x n P j − 2 � � ( − 1) n b n +1 n ! = j ! n =0 j =2 and the right side clearly has positive coefficients.

Residues A useful tool in some computations is a change of variables for residues: By Cauchy’s theorem, the coefficient of x − 1 in a i a i x i is 1 � Laurent series A ( x ) = � C A ( x ) dx . 2 πi Using the change of variables for integrals, we get a change of variables formula for the coefficient of x − 1 in a Laurent series. 1 1 � � C ′ A ( f ( x )) f ′ ( x ) dx A ( x ) dx = 2 πi 2 πi C so the coefficient of x − 1 in A ( x ) is the same as that of A ( f ( x )) f ′ ( x ) This change of variables can be done using only formal power series: We define the residue i a i x i to be the coefficient of a formal Laurent series � of x − 1 .

Theorem (Jacobi) If f ( x ) = f 1 x + f 2 x 2 + · · · , where f 1 � = 0 , then for any formal Laurent series A ( x ) , res A ( x ) = res A ( f ( x )) f ′ ( x ) . Proof. By linearity we may assume that A ( x ) = x k . If k � = − 1 then f k f ′ = ( f k +1 ) ′ / ( k + 1) so res f k f ′ = res x k = 0 since a derivative has residue 0. For k = − 1 we have res f ′ f = res f 1 + 2 f 2 x + · · · f 1 x + f 2 x 2 + · · · = res 1 x + · · · = 1 . A similar change of variables theorem holds for residues for Laurent series in several variables. An easy consequence is the Lagrange inversion formula, which expresses the coefficients of powers of the compositional inverse f �− 1 � in terms of the coefficients of powers of f : if g = f �− 1 � then � n � x [ x n ] g k = k n [ x n − k ] f

A different application of residues was given by D. Zagier in proving a reciprocity theorem for higher- order Dedekind sums: Define λ a + 1 λ b + 1 � d ( p ; a, b ) = λ a − 1 λ b − 1 λ p =1 λ � =1 (and similarly for d ( p ; a, b, c, . . . ) ) By applying the fact that the sum of the residues of a rational function is 0 to the rational function ( x a + 1)( x b + 1)( x c + 1) 1 ( x a − 1)( x b − 1)( x c − 1) , 2 x where a , b , and c are pairwise relatively prime, Zagier obtained the following reciprocity theorem for these Dedekind sums: cp ( c ; b, a ) = 1 − a 2 + b 2 + c 2 1 ap ( a ; b, c )+ 1 bp ( b ; a, c )+ 1 3 abc

This result can be proved by partial fraction expansion instead of residues. More generally, any result that can be proved by applying contour integration to rational functions (in one or more variables) can be proved by partial fractions. Examples include Stanley’s monster reciprocity theorem and the approach of Andrews et al. to MacMahon’s partition analysis, as shown by Guoce Xin.

D-finite series and P-recursive sequences A power series f ( x ) is D-finite if it satisfies a differential equation of the form m p i ( x ) d i f � dx i = 0 , i =0 where the p i ( x ) are polynomials. A sequence a 0 , a 1 , a 2 , . . . is P-recursive if it satisfies a recurrence of the form n � q j ( n ) a n + j = 0 , j =0 for all n ≥ 0 , where the q j ( n ) are polynomials. It is well known that a sequence a 0 , a 1 , . . . is P-recursive if and only if its generating function i =0 a i x i is D-finite. � ∞

Theorem. tan x is not D-finite. Analytic Proof. A solution of the differential equation i =0 p i ( x ) d i f � m dx i = 0 can have singularities only at the zeroes of p 0 ( x ) . In particular, it can have only finitely many singularities. But tan x has infinitely many singularities. Formal Power Series Proof. (Carlitz) Since dx tan x = 1 + tan 2 x , it follows easily by induction d that for each j , d j dx j tan x is a polynomial in tan x . So if tan x is D-finite, it is algebraic, and thus e ix , and therefore e x , is algebraic. But it is easy to prove that e x is not algebraic. Note: A formal power series g ( x ) is algebraic if there exist polynomials r 0 ( x ) , r 1 ( x ) , . . . , r m ( x ) , not all 0, such that m r j ( x ) g ( x ) j = 0 . � j =0

Non-algebraicity ∞ (3 n )! n ! 3 x n is not algebraic. � F ( x ) = Theorem. n =0 Analytic proof #1. (Stanley) By a theorem of n a n x n is algebraic and a n ≈ cn r α n , Jungen, if � as n → ∞ , where c ∈ C − { 0 } , r ∈ R , r < 0 , and α ∈ C − { 0 } , then r is 1 2 plus an integer. But by Stirling’s formula, √ (3 n )! 3 2 π n − 1 27 n , n ! 3 ≈ so F ( x ) cannot be algebraic. Analytic proof #2 (sketch): F ( x ) satisfies a differential equation (a special case of the hypergeometric differential equation) that allows us to determine the behavior of F at its branch points and thereby to show its monodromy group is infinite. Thus F has infinitely many branches, and is therefore not algebraic.

A Bernoulli Number Identity Let ∞ B k � k ( k − 1) x k − 1 , B ( x ) = k =2 where B k is the k th Bernoulli number, let 3 x 2 + 7 36 x 3 + · · · � 2 x − 2 log(1 + x ) = x − 1 u ( x ) = and let v ( x ) = � ∞ n =0 v n x n /n ! be the compositional inverse of u ( x ) ; i.e., u ( v ( x )) = x . Thus v 1 = 1 , v 2 = 2 / 3 , v 3 = 1 / 6 , v 4 = − 4 / 45 , and so on. Theorem. (de Bruijn) ∞ x k � 2 − k v 2 k +1 k ! = e B ( x ) . k =0 Note that this is an identity for formal power series only, as the series have radius of convergence 0.

Proof : We show that both sides are Analytic asymptotic expansions of ( e/n ) n √ n ! , 2 πn where n = 1 /x , as n → ∞ . By the Euler-Maclaurin summation formula we have n � log n ! = log i i =1 ∞ B k � ≈ n (log n − 1) + 1 k ( k − 1) n − ( k − 1) . 2 log 2 πn + k =2 For the second formula, we start with � ∞ e − u u n du n ! = 0 � ∞ � n dx, = e − n n n +1 e − x (1 + x ) � − 1 where we have made the substitution u = n (1 + x ) .

Note that − x 2 2 + x 3 e − x (1+ x ) = exp � � � � − x +log(1+ x ) = exp 3 + · · · . We want to apply a change of variables that makes this e − z 2 / 2 , so want − x + log(1 + x ) = − z 2 / 2 . This gives z = u ( x ) , so x = v ( z ) . So � K 2 z 2 n v ′ ( z ) dz e − 1 n ! ≈ e − n n n +1 − K � ∞ ∞ v j +1 2 z 2 n z j dz e − 1 � ≈ e − n n n +1 j ! −∞ j =0 ∞ (2 k )! · √ πn − k − 1 v 2 k +1 2 (2 k )! � ≈ e − n n n +1 2 k k ! . k =0