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A2

Orthogonality and Irrationality

Overview

This appendix is an application of orthogonalization of particular Müntz systems to the proof of the irrationality of ζ(3) and some other familiar numbers. It reproduces Apéry’s remarkable proof of the irrationality of ζ(3) in the context of orthogonal systems.

Orthogonality and Irrationality

Apéry’s wonderful proof of the irrationality of ζ(3) amounts to showing that

0 < |d_{n}^{3}a_{n}ζ(3) − b_{n}| → 0 ,

where b_{n} is an integer,

a_{n} := ∑_{k=0}^{n} \binom{n + k}{k}^{2}\binom{n}{k}^{2} ,

and

d_{n} := lcm{1, 2, …, n} .

Here lcm denotes the least common multiple; see van der Poorten [79] and Beukers [79]. In [81] Beukers recast the proof using Padé approximations.

Orthogonality and Irrationality 373

Many, maybe most, irrationality proofs may be based on approximation by Padé approximants and related orthogonal polynomials; see, for example, Borwein [91a], [92] or Chudnovsky and Chudnovsky [84]. It is the intention of this appendix to try to put the proof of the irrationality of ζ(3) into the framework of orthogonality. From the general orthogonalization of the system

(x^{λ_{0}}, x^{λ_{1}}, x^{λ_{2}}, …)

on [0, 1], where the numbers λ_{j} are nonnegative and distinct, specializing to the case when

λ_{2j} = j  and  λ_{2j+1} = j + ϵ ,  j = 0, 1, …

where ϵ decreases to 0, is a very natural thing to do. This is how one should interpret orthogonalizing the system (x^{0}, x^{0}, x^{1}, x^{1}, …). This leads to orthogonal functions that generalize the Legendre polynomials and are of the form

p_{n}(x) log x + q_{n}(x)

with polynomials p_{n}, q_{n} ∈ P_{n} of degree n. Legendre polynomials are closely tied to irrationality questions concerning log (see Borwein and Borwein [87], Chapter 11), and higher-order analogs prove to be the basis of dealing with the irrationality of the trilog (∑_{n=1}^{∞} x^{n}/n^{3}) for some values of x. We think that the proof of the irrationality of ζ(3) flows quite naturally from this point of view. Although in the end (Lemma A.2.3) we get back to Beukers’ integral approach to the irrationality of ζ(3) (as indeed we must). What follows, up to one application of the prime number theorem, is a self-contained proof of the irrationality of ζ(3).

The orthogonalization in question is the content of the first theorem.

Theorem A.2.1. Let

G_{n}(x) := \frac{1}{2πi} ∫_{Γ} \frac{∏_{k=0}^{n−1} (t + k + 1)^{2}}{∏_{k=0}^{n}(t − k)^{2}} (t + n + 1)x^{t} dt ,

where Γ is any simple contour containing t = 0, 1, …, n. Then

G_{n}(x) = p_{n}(x) log x + q_{n}(x) ,

where

p_{n}(x) = ∑_{k=0}^{n} \binom{n}{k}^{2}\binom{n + k}{k}^{2}(n + k + 1) x^{k} .

Furthermore, we have the orthogonality relations

∫_{0}^{1} G_{n}(x)x^{k} dx = 0 ,  k = 0, 1, …, n;
∫_{0}^{1} G_{n}(x)(log x)x^{k} dx = 0 ,  k = 0, 1, …, n − 1 ;

374 A2. Orthogonality and Irrationality

and

∫_{0}^{1} G_{n}^{2}(x) dx = \frac{1}{2n + 1} .

Proof. As in Section 3.4, the representation of G_{n} is just the evaluation of the integral at the poles, t = 0, 1, …, n, by the residue theorem. The proof of the orthogonality conditions is a straightforward exercise on evaluating

∫_{0}^{1} x^{k}(log x)^{j}G_{n}(x) dx

by interchanging the order of integration as in the proof of Theorem 3.4.3. ⊓ ⊔

We need to modify these forms marginally to give a zero at 1, as in the next result.

Theorem A.2.2. Let G_{n} and Γ be defined as in Theorem A.2.1. Then

F_{n}(x) := \frac{1}{x^{n+1}} ∫_{0}^{x} y^{n}G_{n}(y) dy

= \frac{1}{2πi} ∫_{Γ} \frac{∏_{k=0}^{n−1}(t + k + 1)^{2}}{∏_{k=0}^{n}(t − k)^{2}} x^{t} dt = A_{n}(x) log x + B_{n}(x) ,

where

A_{n}(x) := ∑_{k=0}^{n} \binom{n + k}{k}^{2}\binom{n}{k}^{2}x^{k}

and

B_{n}(x) := ∑_{k=0}^{n} c_{k}x^{k}

with

c_{k} := \binom{n + k}{k}^{2}\binom{n}{k}^{2}\{∑_{i=0}^{n−1} \frac{2}{k + i + 1} − ∑_{{i=0 \atop i≠k}}^{n} \frac{2}{k − i}\} .

Furthermore,

\begin{aligned}∫_{0}^{1} F_{n}(x)F_{m}(x) dx & = \frac{2δ_{n,m}}{(2n + 1)^{3}} ,  δ_{n,m} := \{ \begin{matrix}0 & if n ≠ m \\ 1 & if n = m; \end{matrix} \\ ∫_{0}^{1} F_{n}(x)x^{k}(log x)^{j} dx & = 0 ,  k = 0, 1, …, n − 1 , j = 0, 1 ; \end{aligned}

and

F_{n}(1) = 0 ,  F′_{n}(1) = 1 ,  F′′_{n}(1) = 2n^{2} + 2n − 1 .

Proof. This follows much as in Theorem A.2.1. The fact that F_{n}(1) = 0 is just the statement that G_{n} is orthogonal to x^{n}. ⊓ ⊔

The fact that F_{n}(1) = 0 is critical in what follows. It immediately gives that B_{n}(x) has a zero at 1, so B_{n}(x)/(1 − x) is a polynomial, as in part c] of the following corollary. (This is the only part of the corollary we need, but the corollary is of some interest in its own right.)

Corollary A.2.3. Let F_{n} be defined as in Theorem A.2.2. Then

a] F_{n} has 2n + 1 zeros on (0, 1].

b] A_{n}(x) has all real negative zeros.

c] B_{n}(x)/(1 − x) is a polynomial with all real negative zeros that interlace the zeros of A_{n}(x).

Proof. The orthogonality conditions give 2n zeros of F_{n} on (0, 1) in a standard fashion, and there is one zero at 1. The real negative zeros of A_{n}(x) and B_{n}(x) and their interlacing follow from the fact that

log x + \frac{B_{n}(x)}{A_{n}(x)}

has 2n + 1 zeros on (0, 1], and known results on interpolating Stieltjes transforms by rational functions (see, for example, Baker and Graves Morris [81] or Borwein [83]). ⊓ ⊔

One can, from the integral representation, deduce the next corollary, which is also not actually needed in the proof of the irrationality of ζ(3).

Corollary A.2.4. Both F_{n} and A_{n} satisfy

x^{2}(y′′_{n} − y′′_{n−1}) − 2nx(y′_{n} + y′_{n−1}) + x(y′_{n} − y′_{n−1}) + n^{2}(y_{n} − y_{n−1}) = 0 .

From Theorem A.2.2 we obtain (see E.2 b]) the following:

Corollary A.2.5. Let d_{n} := lcm{1, 2, …, n}. Then the polynomial d_{n}B_{n}(x) has integer coefficients.

We get an approximation to ζ(3) by integration over the unit square.

Theorem A.2.6.

−\frac{1}{2} ∫_{0}^{1} ∫_{0}^{1} \frac{F_{n}(xy)}{1 − xy} dx dy = A_{n}(1)ζ(3) + R_{n} ,

where 2d_{n}^{3}R_{n} is an integer.

Proof. Recall that

−\frac{1}{2} ∫_{0}^{1} ∫_{0}^{1} \frac{log(xy)}{1 − xy} = ζ(3)

since

∫_{0}^{1} ∫_{0}^{1} (xy)^{n} log(xy) dx dy = \frac{−2}{(n + 1)^{3}} .

We have

\frac{F_{n}(xy)}{1 − xy} = \frac{A_{n}(xy) − A_{n}(1)}{1 − xy} log(xy) + \frac{A_{n}(1)}{1 − xy} log(xy) + \frac{B_{n}(xy)}{1 − xy} .

The first term of the last expression is a polynomial multiple of log(xy), and the last term is a polynomial. Both have degree n − 1 in xy. Here we use part c] of Corollary A.2.3 in an essential way. Integrating the above equation with respect to x on [0, 1] and with respect to y on [0, 1], we get the identity of the theorem. The fact that d_{n}^{3}R_{n} is an integer can be seen as follows. One d_{n} arises from each of the two integrations of a polynomial of degree n − 1 with integer coefficients and one d_{n} comes from Corollary A.2.5. ⊓ ⊔

Theorem A.2.7. The number ζ(3) is irrational.

Proof. It now suffices to show that there is an ϵ > 0 for which

0 < |∫_{0}^{1} ∫_{0}^{1} \frac{F_{n}(xy)}{1 − xy} dx dy| = O (\frac{1}{e^{(3+ϵ)n}} ) = o (\frac{1}{d_{n}^{3}} )

since by the prime number theorem, lcm{1, 2, …, n} = O(e^{(1+ϵ)n}) for every ϵ > 0; see Borwein and Borwein [87, p. 377]. (We use the notation b_{n} = o(a_{n}) if b_{n} = ϵ_{n}a_{n} with lim_{n→∞} ϵ_{n} = 0.) This can be proven in a number of ways, we chose to connect this proof via Padé approximants to the integral estimate due to Beukers. This is the content of the following results and, in particular, Lemma A.2.10. From Lemma A.2.10, the above estimates are easy since the integrand in the right-hand side in Lemma A.2.10 satisfies

0 < \frac{xyv(1 − x)(1 − y)(1 − v)}{1 − (1 − xy)v} ≤ (\sqrt{2} − 1)^{4}

on the open unit cube 0 < x, y, v < 1. ⊓ ⊔

The following lemma gives standard representations for the Padé approximants to log and can be checked by expanding the integrals; see E.3.

Lemma A.2.8 (Padé Approximants). We have

\frac{(n!)^{2}}{2πi} ∫_{Γ} \frac{x^{t} dt}{∏_{k=0}^{n}(t − k)^{2}} = (x − 1)^{2n+1} ∫_{0}^{1} \frac{v^{n}(1 − v)^{n} dv}{(1 − (1 − x)v)^{n+1}}
= p_{n}(x) log x + q_{n}(x) = O((x − 1)^{2n+1}) ,

where Γ is a simple contour containing t = 0, 1, …, n, and p_{n} and q_{n} are polynomials of degree n. (So p_{n}/q_{n} is the (n, n) Padé approximant to log x at 1.)

Lemma A.2.9 (Rodrigues-Type Formula). With Γ as in Theorem A.2.1,

\begin{aligned}F_{n}(xy) & = \frac{d^{n}}{dy^{n}} \frac{d^{n}}{dx^{n}} \frac{x^{n}y^{n}}{2πi} ∫_{Γ} \frac{(xy)^{t} dt}{∏_{k=0}^{n}(t − k)^{2}} \\ & = \frac{1}{(n!)^{2}} \frac{d^{n}}{dy^{n}} \frac{d^{n}}{dx^{n}} ∫_{0}^{1} \frac{(xy)^{n}v^{n}(1 − v)^{n}(xy − 1)^{2n+1}}{(1 − (1 − xy)v)^{n+1}} dv . \end{aligned}

Proof. This follows from differentiating the two representations derived from Lemma A.2.8 coupled with Theorem A.2.2. ⊓ ⊔

Lemma A.2.10. We have

\begin{aligned}∫_{0}^{1} ∫_{0}^{1} \frac{F_{n}(xy)}{1 − xy} & dx dy \\ & = − ∫_{0}^{1} ∫_{0}^{1} ∫_{0}^{1} \frac{[xyv(1 − x)(1 − y)(1 − v)]^{n} dx dy dv}{(1 − (1 − xy)v)^{n+1}} . \end{aligned}

Proof. For k, n nonnegative integers

\begin{aligned}∫{1 \atop 0} ∫{1 \atop 0} & (xy)^{n+k}(1 − x)^{n}(1 − y)^{n} dx dy \\ & = −\frac{1}{(n!)^{2}} ∫_{0}^{1} ∫_{0}^{1} \frac{1}{1 − xy} \frac{d^{n}}{dy^{n}} \frac{d^{n}}{dx^{n}} (xy)^{n+k}(xy − 1)^{2n+1} dx dy . \end{aligned}

(Both sides equal

(\frac{n!(n + k)!}{(2n + k + 1)!})^{2} ,

though this is not completely transparent; see E.4.) So

\frac{1}{(n!)^{2}} ∫_{0}^{1} ∫_{0}^{1} \frac{1}{1 − xy} \frac{d^{n}}{dy^{n}} \frac{d^{n}}{dx^{n}} x^{n}y^{n}(xy − 1)^{2n+1+k} dx dy

= − ∫_{0}^{1} ∫_{0}^{1} [xy(1 − x)(1 − y)]^{n} (xy − 1)^{k} dx dy .

Hence, on expanding (1 − (1 − xy)v)^{−n−1} in the following integrands,

\frac{1}{(n!)^{2}} ∫_{0}^{1} ∫_{0}^{1} \frac{1}{1 − xy} \frac{d^{n}}{dy^{n}} \frac{d^{n}}{dx^{n}} \frac{x^{n}y^{n}(xy − 1)^{2n+1}}{(1 − (1 − xy)v)^{n+1}} dx dy

= − ∫_{0}^{1} ∫_{0}^{1} \frac{[xy(1 − x)(1 − y)]^{n}}{(1 − (1 − xy)v)^{n+1}} dx dy ,

and with Lemma A.2.9 we are done. ⊓ ⊔

Comments, Exercises, and Examples.

The approach of this appendix follows Borwein, Dykshoorn, Erdélyi, and Zhang [to appear]. Beukers’ [79] very elegant recasting of Apéry’s proofs of the irrationality of ζ(2) and ζ(3) is also presented in Borwein and Borwein [87]. E.5 recasts the irrationality of ζ(2) = \frac{1}{6}π^{2} into a form similar to the proof of the irrationality of ζ(3). E.6 treats the irrationality of log 2. Mahler [31] casts transcendence results for exp and log in terms of general systems of Padé approximants.

E.1 Proof of Corollaries A.2.4 and A.2.5.

a] Prove Corollary A.2.4 from the explicit representations of Theorem A.2.2.

b] Prove Corollary A.2.5 from the formula for c_{k} in Theorem A.2.2.

Hint: Observe that if p is a prime and n < p^{α} ≤ n + k for some integer α, then p divides \binom{n+k}{k}. This is fairly straightforward from Euler’s formula for the largest power of a prime dividing a factorial. ⊓ ⊔

E.2 Formulas for ζ(n).

a] Show that

∫_{0}^{1} ⋯ ∫_{0}^{1} ∫_{0}^{1} \frac{dx_{1} dx_{2} ⋯ dx_{n}}{1 − x_{1}x_{2} ⋯ x_{n}} = ζ(n) := ∑_{k=1}^{∞} \frac{1}{k^{n}} .

b] Show that

∫_{0}^{1} \frac{log x}{1 − x} dx = −\frac{π^{2}}{6} .

c] Find a closed formula for

∫_{0}^{1} ⋯ ∫_{0}^{1} ∫_{0}^{1} \frac{log(x_{1}x_{2} ⋯ x_{n})}{1 − x_{1}x_{2} ⋯ x_{n}} dx_{1} dx_{2} ⋯ dx_{n}

in terms of ζ(n + 1).

E.3 Proof of Lemma A.2.8. Give a proof of Lemma A.2.8.

Orthogonality and Irrationality 379

E.4 The Identity in the Proof of Lemma A.2.10. Prove the identity

∫_{0}^{1} ∫_{0}^{1}(xy)^{n+k}(1 − x)^{n}(1 − y)^{n} dx dy

= −\frac{1}{(n!)^{2}} ∫_{0}^{1} ∫_{0}^{1} \frac{1}{1 − xy} \frac{d^{n}}{dy^{n}} \frac{d^{n}}{dx^{n}} (xy)^{n+k}(xy − 1)^{2n+1} dx dy

= (\frac{n!(n + k)!}{(2n + k + 1)!})^{2} ,

where k and n are nonnegative integers. Follow parts a] to c].

Let k and n be nonnegative integers. a] Show that

A_{n,k} := ∫_{0}^{1} ∫_{0}^{1} (xy)^{n+k}(1 − x)^{n}(1 − y)^{n} dx dy = (\frac{n!(n + k)!}{(2n + k + 1)!})^{2} .

b] Let

B_{n,k} := −\frac{1}{(n!)^{2}} ∫_{0}^{1} ∫_{0}^{1} \frac{1}{(1 − xy)} \frac{d^{n}}{dy^{n}} \frac{d^{n}}{dx^{n}} x^{n+k}(xy − 1)^{2n+1} dx dy .

Show that

n^{2}B_{n,k} =[(n + k)^{2} + (2n + 1)(2n + 2k + 1) + (2n + 1)(2n)]B_{n−1,k+2}
− [2(n + k)^{2} + (2n + 1)(2n + 2k + 1)]B_{n−1,k+1}
+ (n + k)^{2}B_{n−1,k} .

c] Show that A_{0,k} = B_{0,k}. Show that the values

A_{n,k} = (\frac{n!(n + k)!}{(2n + k + 1)!})^{2}

satisfy the recurrence relation established by part b] for the values B_{n,k}.

E.5 The Irrationality of π^{2}. Consider

F_{n}(x) := \frac{n!}{(2πi)} ∫_{Γ} \frac{∏_{k=1}^{n}(t + k)}{∏_{k=0}^{n}(t − k)^{2}} x^{t} dt ,

where Γ is any simple contour containing the poles at 0, 1, …, n.

380 A2. Orthogonality and Irrationality

a] Show that

∫_{0}^{1} \frac{F_{n}(x)}{1 − x} dx = ∑_{k=0}^{n} \binom{n + k}{k}\binom{n}{k}^{2}ζ(2) + R_{n} ,

where d_{n}^{2}R_{n} is an integer (d_{n} := lcm{1, 2, …, n} as before). Hint: Show that

F_{n}(x) = A_{n}(x) log x + B_{n}(x) ,

where

A_{n}(x) := ∑_{k=0}^{n} \binom{n + k}{k}^{2}\binom{n}{k}^{2}x^{k}

and

B_{n}(x) := ∑_{k=0}^{n} c_{k}x^{k}

with

c_{k} := \binom{n + k}{k}\binom{n}{k}^{2}\{∑_{i=1}^{n} \frac{1}{k + i} − ∑_{{i=0 \atop i≠k}}^{n} \frac{2}{k − i}\} .

Write

\frac{F_{n}(x)}{1 − x} = \frac{A_{n}(x) − A_{n}(1)}{1 − x} log x + \frac{A_{n}(1)}{1 − x} + \frac{B_{n}(x)}{1 − x} .

Show that F_{n}(1) = 0, hence B_{n}(1) = 0. Recall that

− ∫_{0}^{1} \frac{log x}{1 − x dx} = ζ(2) .

Now the conclusions follow, as in the proof of Theorem A.2.6. ⊓ ⊔ b] Show that

∫_{0}^{1} \frac{F_{n}(x)}{1 − x} dx = ∫_{0}^{1} ∫_{0}^{1} \frac{x^{n}y^{n}(1 − x)^{n}(1 − y)^{n}}{(1 − xy)^{n+1}} dx dy .

Hint: Use Lemma A.2.8. ⊓ ⊔ c] Show that there exists a constant c independent of n such that

0 < |∫_{0}^{1} \frac{F_{n}(x)}{1 − x} dx| ≤ c (\frac{\sqrt{5} − 1}{2})^{5n}

and deduce that ζ(2) = \frac{1}{6}π^{2} irrational. Hence π is irrational.

Orthogonality and Irrationality 381

E.6 The Irrationality of log 2. Let

p_{n}(x) := \frac{1}{n!} \frac{d^{n}}{dx^{n}} [x^{n}(1 − x)^{n}] = ∑_{k=0}^{n} \binom{n + k}{k}\binom{n}{k}(−1)^{k}x^{k}

be the nth Legendre polynomial on [0, 1]. a] Show that

∫_{0}^{1} \frac{p_{n}(x)}{1 + x} dx = ∫_{0}^{1} \frac{p_{n}(x) − p_{n}(−1)}{1 + x} dx + ∫_{0}^{1} \frac{p_{n}(−1)}{1 + x} dx
= p_{n}(−1) log 2 + R_{n} ,

where d_{n}R_{n} is an integer (d_{n} := lcm{1, 2, …, n} as before). b] Show that

|∫_{0}^{1} \frac{p_{n}(x)}{1 + x} dx| = ∫_{0}^{1} \frac{x^{n}(1 − x)^{n}}{(1 + x)^{n+1}} dx .

c] Use parts a] and b] to show that log 2 is irrational.