7 Asymptotics for Meromorphic Functions

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1 Lecture G jacques@ucsd.edu 7 Asymptotics for Meromorphic Functions Hadamard s Theorem gives a broad description of the exponential growth of coefficients in power series, but the notion of exponential growth is very crude in describing the coefficients. In this section we concentrate on asymptotics of rational and meromorphic generating functions. 7. Pringsheim s Theorem A dominant singularity of a function f(z) analytic at zero with radius of convergence ρ > 0 is a singularity on the circle z = ρ. The saddle point method gives a broad approach when the power series has a strong enough dominant singularity at z = ρ. In some cases, the saddle point method can be greatly simplified, in particular, in the case of meromorphic functions with finitely many poles. The first step towards a theorem for meromorphic functions is Pringsheim s Theorem: Theorem (Pringsheim s Theorem) Let A C[[X]] have non-negative real coefficients and finite radius of convergence ρ > 0. Then z = ρ is a dominant singularity of A. Proof We proved in Hadamard s Theorem that there is a singularity of A on z = ρ, otherwise a compactness argument shows that A converges on a disc z σ with σ > ρ. Suppose for a contradiction that A is analytic at z = ρ. Then there is an ε > 0 such that A converges on z ρ ε. In particular A converges at ρ ε and so 4 A(z) = n 0 b n (z ρ + ε 4 )n is convergent. If a n = [X n ]A(X) then the binomial theorem shows ( ) n b k = a n (ρ ε 4 k )n k. n=0 Also A(z) converges at ρ + ϵ and so performing an expansion around ρ ε with radius ε 4 2 we get A(ρ + ε) = ( ) n a 4 n (ρ ε 4 k )n k ( ε 2 )k. k=0 n=0 This is an absolutely convergent sum since a k 0 for all k, and so by the binomial theorem, A(ρ + ε) = a 4 n (ρ + ε 2 )n. n=0

2 It follows that lim n a n (ρ + ε 2 )n = 0 which contradicts that the radius of convergence of A is ρ. If A is analytic at zero and has non-negative real coefficients, then within the radius of convergence A(z) = A(z) and so dominant singularities come in complex conjugate pairs. Pringsheim s Theorem guarantees at least the dominant singularity z = ρ, although there may be others with equal modulus, such as for ( z 2 ) /2 which has z = ± as the dominant singularities. 7.2 Meromorphic Functions We are now ready to describe the asymptotics of power series of meromorphic functions, and in particular, asymptotics for rational functions. The main point is that the two are essentially the same, since we can via Laurent series expansion write a meromorphic function as a rational function plus an analytic term this is sometimes called the method of subtracted singularities. Theorem 2 (Meromorphic Asymptotics) Let A C[[X]] have finite radius of convergence ρ > 0 and suppose A is analytic on z = ρ and has poles z, z 2,..., z k in the disc z ρ of orders m, m 2,..., m k respectively. Then there exist polynomials P i (X) of degree m i such that k [X n ]A(X) P i (z i )z n = O(ρ n ). i= In particular, if A is a rational function, then i [X n ]A(X) = k i= P i (z i )z n i. Proof For the statement on rational functions, we may give a direct proof using partial fraction decomposition (similar to the linear recurrence equations theorem) and we leave it as an exercise. The first identity can be proved directly via Cauchy s Residue Theorem, which furthermore shows that the O(ρ n ) term is at most ρ n sup z =ρ A(z) ; here we give a direct proof based on the rational case. The key idea is that for some polynomials Q i (z) of degree m i, B(z) = A(z) k i= n= m i Q i (z) k (z z i ) = A(z) B n i (z) i= is analytic on the circle γ of radius ρ. This follows by collecting all negative powers of z z i in the Laurent series expansion of A at each pole z i. Now since B i (z) is a rational function we have from the rational case that [X n ]B i (X) = P i (z i )z n i for some polynomial 2

3 P i of degree m i. It remains to consider B(z), and we use a saddle point bound: let γ be the curve z = ρ and observe [X n ]B(X) = B(z) dz ρ sup B(z) ρ n = O(ρ n ) 2πi zn+ z =ρ since B is analytic on γ. This proves the theorem. γ A key fact here is that if there is a unique dominant singularity of A(z) then we can obtain from the theorem an asymptotic formula for [X n ]A(X), and even more, the more the singularities are spaced out, the better choice we can make for ρ to minimize the error term ρ n sup z =ρ A(z). 7.3 Four short applications Surjections. Recall that the number of surjections from [n] onto an initial segment of the integers is R n = r=0 r!s n,r where S n,r is the Stirling number of the second kind the number of partitions of [n] into r non-empty sets. As a power series R(X) with coefficients R n, we get using exponential generating functions: R(X) = n=0 r=0 r! S n,r n! Xn = r=0 (exp(x) )r r! r! = 2 exp(x). Now this is a meromorphic function with simple poles at log 2+2ikπ. The closest singularity to the origin is log 2 and all other singularties have modulus at least log 2 + 4π 2 > 6. Taking the circle z = 6 in the meromorphic asymptotics theorem, we get R n n! = P (log 2) (log 2) n + O(6 n ). To determine the constant we just compute the real limit lim z log 2 z log 2 2 exp(z) = 2 from which one obtains R n = n! 2 (log 2) n + O(n!6 n ). This is a striking example of how close we get to the actual number of surjections, with the error term 6 n explaining the precision. 3

4 Alignments. An alignment is an ordered sequence of cycles partitioning [n]. For instance, the permutation (2,, 3) gives rise to two alignments, namely (2, ), (3) and (3), (2, ). Let A n be the number of alignments of [n]. By the product lemma, the exponential generating function for alignments satisfies Ψ(X) = + log( X) since log( X) is the exponential generating function for cycles. A singularity lies at z = /e, coming from log( z) =. According to the meromorphic asymptotics theorem, A n n! e ( e ) n. A tiling example. On exercise sheet, it was required to find a recurrence relation for the number of domino tilings of a 3 n grid. If a n is the number of tilings, one finds a n = 4a n 2 a n 4. Note that a n = 0 if n is odd. One can solve this recurrence relation, but instead we move directly to the asymptotic behavior of a n. The characteristic equation is x 4 4x 2 + = 0. The two roots closest to the origin are ± 2 3 and each is a simple pole of the generating function. By the rational function asymptotics theorem, a n c (2 3) n/2. The generating function is (with initial conditions a 0 = 0 = a = a 3 and a 2 = 3: Φ(X) = 3X 4X 2 + X 4 Taking limits of (x 2 3)Φ(x) as x 2 3 we get a n 3 2 (2 3) n/2 = 3 2 (2 + 3) n/2. For the 4 n grid, it can be shown that the number a n of tilings satisfies a n = a n + 5a n 2 + a n 3 a n 4. 4

5 The root of the characteristic equation closest to the origin is α = ( (7 + ) 29) 4 and therefore for some constant β > 0, a n βα n. The value of β is found as above and is left as an exercise. Compositions into primes. Let P be the set of primes. The number a n of compositions of an integer n into primes is generated by Φ(X) = Φ P (X) = X 2 X 3 X An approximation to the smallest root α of the denominator is found by considering A(X) = X p and Then for any x [0, ), B(X) = p P :p r p P :p r X p + Xr+ X. A(x) Φ(x) B(x) and if α m is the smallest real root of A(x) = and β m is the smallest real root of B(x) =, then lim m α m = lim m β m = α. So we can find α = and then a n γα n where γ = A similar approach can be used for compositions into parts from other sequences of integers. For instance, one could let Q be the set of prime numbers of the form k 2 + with k N. Then we have Φ(X) = X 2 X 3 X 5 X 7 X We can approximate the smallest root α = of the denominator as above, and so obtain an approximate asymptotic formula for the number of compositions of an integer n into primes of the form k 2 + : k N. This is despite the notorious open question of determining whether there are infinitely many primes which are one more than an integer square. 5

6 7.4 Strings with forbidden substrings The autocorrelation polynomial for a string σ of letters from a finite alphabet Σ is the polynomial c(x) defined by c i = [X i ]c(x) = σ j = σ i+j for j =, 2,..., σ i where i = 0,,..., σ. So for instance, for binary strings, if σ = 00 then c(x) = + X 3, if σ = 000 then c(x) = + X + X 2 and if σ = 00 then c(x) = + X 2. Observe that if S is a set of strings avoiding σ and T is the set of strings with σ at the end and no prior appearance of σ, then S T = {ε} SΣ and if σ = σ σ 2... σ k are the letters of σ, then ( ) Sσ = T (σ k i+ σ k i+2... σ k ). c i = In the latter equality, we decompose strings in Sσ by their leftmost appearance of σ as a substring. Combining the formal equations gives Φ S + Φ T = + Σ XΦ S and Solving the equations we get Φ S (X) = X k Φ S = Φ T c(x). c(x) X k + ( Σ X)c(X). This is a rational function to which the rational function asymptotics theorem can be applied to get asymptotics. For instance, suppose we want to consider the number of strings of length n from the alphabet Σ = {, 2, 3} that do not contain 232 (contiguously). Then c(x) = + X 4 and Φ S (X) = + X 4 X 5 + ( 3X)( + X 4 ) = + X 4 3X + X 4 2X 5. The denominator has a unique root α = and this is a dominant singularity. The next nearest one to the origin has modulus about Taking ρ =.05 say, we get that the number of strings of length n with no 232 is asymptotic to a constant times α n and we find that constant in the usual way using limits. 6

7 7.5 A statistical theorem We recall that if Φ S (X) is the generating function for a finite set S of configurations, then Φ (X)/Φ(X) at X = is the expected weight or average weight of configurations in S. A power series A(z) is called aperiodic if A(z) is not identically equal to B(z d ) for any d 2 and B analytic at the origin. In this section, if T is a set of configurations with weight function ω, then T is implicitly assumed to have a weight function that is additive with respect to T, so that Φ T (X) = /( Φ T (X) by the product lemma. Theorem 3 Let T be a set of configurations with weight function ω and suppose that Φ T (z) has radius of convergence ρ > 0 and Φ T (ρ) > and Φ is aperiodic. Let α (0, ρ) be the unique root of Φ T (x) =. Then [X n ]Φ T (X) α n+ Φ T (α). Furthermore, the average number of parts in a string from T of weight n is asymptotic to n + Φ T + (α). (α) (α)2 αφ T Φ T Proof Since Φ T has non-negative coefficients, Φ T (z) > 0 for z (0, ρ) so α is well-defined. Furthermore Φ T (z) is analytic on (0, α) and has a simple pole at z = α since Φ T (z) > 0 for z (0, ρ). Pringsheim s Theorem shows that Φ T (z) has radius of convergence α and Φ T (z) (z α)φ T (α) as z α. Now one can show using the aperiodicity of Φ T that there exists r > α such that Φ T (z) has only one pole, α, in the disc z α r, and therefore Φ T (z) is meromorphic and the meromorphic functions theorem applies. One computes [X n ]Φ T (X) = ] αφ T (α)[xn z/α + O(r n ) α n+ Φ T (α) as required. The average number of parts is ] [X n ]Φ Y = T (X)[Xn Y Φ T (X) Y = Φ T (X) ] [X n ]Φ T (X)[Xn ( Φ T (X)) 2 where Y as usual marks parts. Now Φ T /( Φ T ) 2 as a pole of order two at α. By the meromorphic functions theorem, Φ T (z) ( Φ T (z)) 2 α 2 Φ T (α)2 ( z/α) 2 as z α. Computing the coefficients and using the first part of the theorem gives the result. 7

8 The reader should check what happens if A(z) is not aperiodic, and what the appropriate version of the above theorem would be in that case (and why it does not hold as stated if A(z) is not aperiodic). One also can compute higher moments: for example one can compute the variance of the number of parts in compositions, and so on. There are many other asymptotic statistical theorems that can be derived from generating functions, for now these are beyond the scope of this course. Average number of parts in compositions. If we consider compositions of n, then there are 2 n compositions in total, and the average number of parts by the last theorem is asymptotic to n/2. This is confirmed by the exact average which is n k= ( ) n k k 2 = n +. n 2 Suppose we consider now the average number of parts in a composition of n in which all parts have size at least two. If T = {2, 3, 4,... } then Φ T (X) = Φ T (X) =. X2 X Clearly Φ T (z) is aperiodic with radius of convergence ρ =. The unique root of Φ T (X) = in (0, ) is α = ( 5 )/2. Therefore the number of compositions is asymptotic to α n+ Φ T (α) = ( 5 + ) n. 5 2 The average number of parts in these compositions is computed from the theorem: it is asymptotically equal to ( )n 5( ) 0.276n. We recall we also computed statistics for binary strings on exercise sheet, for example, the average number of blocks in a binary string of length n, or the average number of occurrences of a particular substring. One can develop similar theorems to the one given here for various other statistics of T. 8