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

Prove that

n

• k=0

n k 2 = 2n n

• .

n

• k=0

n k 2 = constant term in (1 + z)n

• 1 + 1

z n = 1 2πi

• C

(1 + z)n

• 1 + 1

z n dz z = 1 2πi

• C

(1 + z)2n zn+1 dz = coefficient of zn in (1 + z)2n = 2n n

• .

A more interesting example

The Bernoulli numbers of the second kind are defined by

∞

• n=0

bn xn n! = x log(1 + x) = 1 + x 2! − 1 6 x2 2! + 1 4 x3 3! − 19 30 x4 4! + 9 4 x5 5! − 863 84 x6 6! + 1375 24 x7 7! + · · · Prove that they alternate in sign.

Since 1 (1+x)t dt = 1 exp

• t log(1+x)
• dt =

x log(1 + x), and (1 + x)t =

∞

• n=0

t n

• xn,

we have bn = n! 1 t n

• dt =

1 t(t − 1) · · · (t − n + 1) dt. Making the change of variables t = 1 − u, we have for n ≥ 1 bn = (−1)n−1 1 (1 − u)u(u + 1) · · · (u + n − 2) du. Since the integrand is positive, (−1)n−1bn > 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

• i=0

c(n − 1, i)ui and using the integral 1 (1 − u)ui du = 1 (i + 1)(i + 2), we find the explicit formula (−1)n−1bn =

n−1

• i=0

1 (i + 1)(i + 2)c(n − 1, i). We could eliminate the analysis by defining a linear functional L on polynomials by L(tn) = 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

∞

• n=0

(−1)n−1bn xn n! = x log(1 − x), so

∞

• n=0

(−1)nbn+1 xn n! = d dx x log(1 − x) = eP − P − 1 P 2 , where P = − log(1 − x) = ∞

n=1 xn/n has positive

• coefficients. So

∞

• n=0

(−1)nbn+1 xn n! =

∞

• j=2

P j−2 j! 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 Laurent series A(x) =

i aixi is 1 2πi

• C A(x) dx.

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 2πi

• C

A(x) dx = 1 2πi

• C′ A(f(x))f ′(x) dx

so the coefficient of x−1 in A(x) is the same as that

• f A(f(x))f ′(x) This change of variables can be done

using only formal power series: We define the residue

• f a formal Laurent series

i aixi to be the coefficient

• f x−1.

Theorem (Jacobi) If f(x) = f1x + f2x2 + · · · , where f1 = 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) = xk.

If k = −1 then f kf ′ = (f k+1)′/(k + 1) so res f kf ′ = res xk = 0 since a derivative has residue 0. For k = −1 we have res f ′ f = res f1 + 2f2x + · · · f1x + f2x2 + · · · = 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

• f powers of f: if g = f −1 then

[xn] gk = k n[xn−k] x f n

A different application of residues was given by

• D. Zagier in proving a reciprocity theorem for higher-
• rder Dedekind sums:

Define d(p; a, b) =

• λp=1

λ=1

λa + 1 λa − 1 λb + 1 λb − 1 (and similarly for d(p; a, b, c, . . . )) By applying the fact that the sum of the residues

• f a rational function is 0 to the rational function

1 2x (xa + 1)(xb + 1)(xc + 1) (xa − 1)(xb − 1)(xc − 1), where a, b, and c are pairwise relatively prime, Zagier

• btained the following reciprocity theorem for these

Dedekind sums: 1 ap(a; b, c)+ 1 bp(b; a, c)+ 1 cp(c; b, a) = 1− a2 + b2 + c2 3abc

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

• i=0

pi(x)dif dxi = 0, where the pi(x) are polynomials. A sequence a0, a1, a2, . . . is P-recursive if it satisfies a recurrence of the form

n

• j=0

qj(n)an+j = 0, for all n ≥ 0, where the qj(n) are polynomials. It is well known that a sequence a0, a1, . . . is P-recursive if and only if its generating function ∞

i=0 aixi is D-finite.

Theorem. tan x is not D-finite. Analytic Proof. A solution of the differential equation m

i=0 pi(x)dif dxi = 0 can have singularities only at the

zeroes of p0(x). In particular, it can have only finitely many singularities. But tan x has infinitely many singularities. Formal Power Series Proof. (Carlitz) Since

d dx tan x = 1 + tan2 x, it follows easily by induction

that for each j, dj

dxj tan x is a polynomial in tan x. So

if tan x is D-finite, it is algebraic, and thus eix, and therefore ex, is algebraic. But it is easy to prove that ex is not algebraic. Note: A formal power series g(x) is algebraic if there exist polynomials r0(x), r1(x), . . . , rm(x), not all 0, such that

m

• j=0

rj(x)g(x)j = 0.

Non-algebraicity

Theorem. F(x) =

∞

• n=0

(3n)! n!3 xn is not algebraic. Analytic proof #1. (Stanley) By a theorem of Jungen, if

n anxn is algebraic and an ≈ cnrαn,

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, (3n)! n!3 ≈ √ 3 2π n−127n, so F(x) cannot be algebraic. Analytic proof #2 (sketch): F(x) satisfies a differential equation (a special case

• f

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(x) =

∞

• k=2

Bk k(k − 1)xk−1, where Bk is the kth Bernoulli number, let u(x) =

• 2x − 2 log(1 + x) = x − 1

3x2 + 7 36x3 + · · ·

and let v(x) = ∞

n=0 vnxn/n! be the compositional

inverse of u(x); i.e., u(v(x)) = x. Thus v1 = 1, v2 = 2/3, v3 = 1/6, v4 = −4/45, and so on.

• Theorem. (de Bruijn)

∞

• k=0

2−kv2k+1 xk k! = eB(x). Note that this is an identity for formal power series

• nly, as the series have radius of convergence 0.

Analytic Proof: We show that both sides are asymptotic expansions of (e/n)n √ 2πn n!, where n = 1/x, as n → ∞. By the Euler-Maclaurin summation formula we have log n! =

n

• i=1

log i ≈ n(log n − 1) + 1

2 log 2πn + ∞

• k=2

Bk k(k − 1)n−(k−1). For the second formula, we start with n! = ∞ e−uun du = e−nnn+1 ∞

−1

• e−x(1 + x)

n dx, where we have made the substitution u = n(1 + x).

Note that e−x(1+x) = exp

• −x+log(1+x)
• = exp
• −x2

2 +x3 3 +· · ·

• .

We want to apply a change of variables that makes this e−z2/2, so want −x + log(1 + x) = −z2/2. This gives z = u(x), so x = v(z). So n! ≈ e−nnn+1 K

−K

e−1

2z2nv′(z) dz

≈ e−nnn+1

∞

• j=0

vj+1 j! ∞

−∞

e−1

2z2nzj dz

≈ e−nnn+1

∞

• k=0

v2k+1 (2k)! · √πn−k−1

2 (2k)!

2kk! .