Transcendental Numbers

Basic Notions in Mathematics

Utrecht, March 31, 2005

1

Definition

A complex number, which is a zero of a non-trivial polynomial with integral coefficients iscalled algebraic. The set of algebraic numbersis denoted by Q.

Examples:

1. 3/5 zero of 5x− 3

2.√

2 zero of x2 − 2

3.√

1 + 3√5 zero of x6 − 3x4 + 3x2 − 6

4. . . .

Definition A complex number which is not al-gebraic is called transcendental.

2

First known transcendental numbers

Liouville numbers constructed by Liouville in

1844.

Example: a =∑∞

n=01

10n! is transcendental. In

decimals:

a = 1.1100010000000000000000010000 · · ·

3

A transcendence proof

Suppose a satisfies P (a) = 0 where P 6≡ 0 is

polynomial with integer coefficients. Denote

by aN the N-th aproximation aN =∑N

n=01

10n!.

Note, aN has denominator 10N ! and |a− aN | ≈10−(N+1)!.

Analytic upper bound:

|P (aN)| ≤ C(P )

10(N+1)!.

On the other hand: P (aN) is a non-zero ra-

tional number (if N sufficiently large) with de-

nominator dividing (10N !)d where d = degree(P ).

Hence (arithmetic lower bound):

1

10dN !≤ |P (aN)|.

Contradiction when N is sufficiently large. Hence

a transcendental.

4

Cantor’s ideas

Theorem The algebraic numbers form a count-able set.

Proof

• It suffices to show that the polynomialswith integer coeffcients form a countableset.

• For a polynomialP = anxn + an−1xn−1 + · · · a1x + a0with an 6= 0 we defineH(P ) = n + |a0|+ |a1|+ · · ·+ |an|.

• Enumerate the polynomials according tothe size of H(P ).

Cantor (1874): The set of (real) transcenden-tal numbers is not enumerable.

5

Famous transcendental numbers

1) π (Ludolphsche Zahl), proved transcenden-tal in 1882 by Lindemann. More generally:a ∈ Q and a 6= 0, then ea transcendental.

2) e, proved transcendental in 1873 by Her-mite.

3) 2√

2 (formerly Hilbert’s 7th problem) provedtranscendental by Gel’fond and Schneiderin 1934.

More generally, if a, b algebraic, a 6= 0,1and b irrational, then ab is transcendental.Example: i−2i = eπ.

4) 0.123456789101112131415161718 · · · (Cham-pernowne’s number) proven transcenden-tal by Mahler in 1936.

6

Famous transcendental numbers

5) Let a algebraic and 0 < |a| < 1. Then∑∞n=0 an2

is transcendental (Nesterenko, 1997).

6) Let {an}∞n=1 = 0,0,1,0,0,1,1,0, . . . be the

paper folding sequence and

f(x) = 1 +∑

n≥1

anxn.

Then f(a) is transcendental for any alge-

braic number a with 0 < |a| < 1 (K.Mahler).

Note:f(x2)− f(x) = x3/(x4 − 1).

7) Γ(1/3),Γ(1/4) shown transcendental by

G.Chudnovski (1980’s).

8) Zeros of the Bessel function J0 are tran-

scendental (Siegel, 1929).

7

Infamous transcendental numbers

1) γ (Euler’s constant), not known to be ir-

rational.

2) ζ(3), known to be irrational (Apery, 1978)

but not known to be transcendental.

3) e + π, not known to be irrational.

4) Γ(1/5) not known to be irrational.

5) log(2) log(3) not known to be irrational.

8

Main streams, I

Gel’fond-Schneider (1934):

a, b ∈ Q, a 6= 0,1, b 6∈ Q⇒ ab 6∈ QEquivalent: Let α1, α2, β1, β2 ∈ Q∗ and sup-pose that α1, α2 are mulitiplicatively indepen-dent. Then

β1 logα1 + β2 logα2 6= 0.

Theorem (A.Baker, 1967) Let α1, . . . , αn ∈ Q∗and suppose α1, . . . , αn are multiplicatively in-dependent . Then logα1, . . . , logαn are linearlyindependent over Q.

Quantitative version: Let α1, . . . , αn be as above.Then, there exists C > 0 such that for anyb1, . . . , bn ∈ Z, not all zero, we have

|b1 logα1 + · · ·+ bn logαn| > B−C

where B = maxi |bi|. Moreover, C can be com-puted explicitly in terms of the αi.

9

Mordell’s equation

Baker’s quantitative theorem can be used tosolve diophantine equations of various forms,e.g

Mordell’s equation y2 = x3 + k.

Theorem (Mordell) Let k ∈ Z and supposek 6= 0. The the equation y2 = x3 + k has atmost finitely many solutions in x, y ∈ Z.

Example, y2 = x3+17 has the integer solutions

32 = (−2)3 + 17

42 = (−1)3 + 17

52 = 23 + 17

92 = 83 + 17

232 = 433 + 17

3876612 = 52343 + 17

Baker’s theory provides us with an effectivemethod of solution.

10

Thue’s equation

Theorem (Thue,1909) Let F ∈ Z[x, y] homo-

geneous and k ∈ Z. If F has at least three

distinct (complex) zeros then the equation

F (x, y) = k

has at most finitely many solutions x, y ∈ Z.

Example, x3 + x2y − 2xy2 − y3 = 1 has the

solutions

(x, y) = (1,0), (0,−1), (−1,1)

(−1,−1)(2,−1)(−1,2)

(5,4), (4,−9), (−9,5)

Baker’s theory provides us with an effective

method of solution.

11

Catalan’s problem

Catalan (1844) Je vous prie, Monsieur, de vouloirbien enoncer, dans votre receuil, le theoremesuivant, que je crois vrai, bien que je n’aie pasencore reussi a le demontrer completement:d’autres seront peut-etre plus heureux:

Consider the infinite sequence of perfect pow-ers

1,4,8,9,16,25,27,32,32,36,49,64,

81,100,121,125,128, . . .

The only two consecutive numbers in this se-quence are 8 and 9.

R.Tijdeman (1976), using linear forms in log-arithms: There at most finitely many con-secutive perfect powers. They all lie belowexp exp exp exp(730).

P.Mihailescu (2002), using algebraic numbertheory: Catalan’s conjecture is true.

12

Main stream I, continued

Gel’fond-Schneider theory has been extended

to values of elliptic functions, period of elliptic

curves and abelian varieties, etc.

Theorem (Wustholz, 1982) Let G be a com-

mutative algebraic group defined over Q of di-

mension n. Let

expG : Cn → G(C)

be an exponential map defined over Q. Let

u = (u1, . . . , un) ∈ Cn be a point such that

u 6= 0 and expG(u) ∈ G(Q). Let Wu be the

smallest linear subspace of Cn, defined over

Q, which contains u. Then expG(Wu) is an

algebraic subgroup of G defined over Q.

13

Algebraic independence

Definition The numbers a1, . . . , an are called

algebraically independent (over Q) if for every

non-trivial polynomial P in n variables and co-

efficients in Z we have P (a1, . . . , an) 6= 0.

The transcendence degree of a set b1, . . . , bn

is the maximal number of algebraically inde-

pendent elements among b1, . . . , bn. Notation:

degtrQ(b1, . . . , bn).

Conjecture (Schanuel). For any a1, . . . , an that

are linearly independent over Q we have

degtrQ(a1, . . . , an, ea1, . . . , ean) ≥ n.

14

Exercises

Suppose Schanuel’s conjecture holds. Deduce

1. that π is transcendental

2. that if αi are Q-linear independent algebraic

numbers then eα1, . . . , eαn are algebraically

independent (Lindemann-Weierstrass the-

orem).

3. Gel’fond-Schneider theorem

4. Baker’s (qualitative) theorem on linear forms

in logarithms

15

Main stream II, E-functions

Lindemann-Weierstrass theorem: let α1, . . . , αn

be distinct algebraic numbers. Then eα1, . . . , eαn

are Q-linear independent.

Alternatively: consider the functions

f1(z) = eα1z, . . . , fn(z) = eαnz

Then their values at z = 1 are Q-linear inde-

pendent.

C.L.Siegel extended the functions fi(z) to a

more general class, the (Siegel) E-functions.

16

E-functions, definition

An entire function f(z) given by a powerseries

∞∑

n=0

ak

k!zk

is called a Siegel E-function if

1. a0, a1, a2, . . . ∈ Q

2. f(z) satisfies a linear differential equationwith coefficients in Q(z).

3. logH(a0, a1, . . . , aN) = O(N) for all N .

Here, H(α1, . . . , αn) denotes the logarithmic ab-solute height of the vector (α1, . . . , αn) ∈ Qn.

Example:

H(p1/q, p2/q, . . . , pn/q)

= max(|p1|, |p2|, . . . , |pn|, |q|)17

E-function examples

1. exp(αz) =∑∞

k=0(αz)k

k! where α ∈ Q∗.

2. J0(−z2) =∑∞

k=0z2k

k!k! =∑∞

k=0

(2kk

)z2k

(2k)!.

3. Confluent hypergeometric series qFp withrational parameters

∞∑

n=0

(µ1)n · · · (µp)n

(ν1)n · · · (νq)n

z(q+1−p)n

n!

Differential equations

y′ − αy = 0

zy′′ + y′ − 4zy = 0

Dq∏

j=1

(D + νj − 1)F = zp∏

i=1

(D + µi)F

where D = z ddz.

18

The main theorem

Let f1(z), . . . , fn(z) be E-functions satisfying a

system of n differential equations

d

dz

y1...

yn

= A

y1...

yn

where A is an n×n-matrix with entries in Q(z).

We assume that the common denominator of

the entries is T (z).

Theorem (Siegel-Shidlovskii, 1929, 1956).

Let α ∈ Q and αT (α) 6= 0. Then

degtrQQ(f1(α), f2(α), . . . , fn(α)) =

degtrQC(z)(f1(z), f2(z), . . . , fn(z))

19

A relation

f(z) =∞∑

k=0

((2k)!)2

(k!)2(6k)!(2916z)k

satisfies

FtQF = (z)

where

F =

f(z)

Df(z)

D2f(z)

D3f(z)

D4f(z)

, D = zd

dz

and

Q =

z − 324z2 −18z 198z −486z 324z

−18z −109

232 −28 18

198z 232 −120 297 −198

−486z −28 297 −729 486

324z 18 −198 486 −324

20

The last problem?

Theorem (Nesterenko-Shidlovskii, 1996). Let

f1(z), . . . , fn(z) be E-functions which satisfy a

system of n first order equations. Then there

is a finite set S such that for every ξ ∈ Q, ξ 6∈ S

the following statement holds. Any relation of

the form

α1f1(ξ) + · · ·+ αnfn(ξ)) = 0

is obtained by specialisation of a polynomial

linear relation

p1(z)f1(z) + · · ·+ pn(z)fn(z) = 0

with pi(z) ∈ Q(z) at z = ξ.

A similar statement holds for polynomial rela-

tions.

Conjecture S = {α|αT (α) = 0}.

21

Yves Andre’s work

Theorem (Y.Andre, 2000) Let f(z) be an E-function. Then f(z) satisfies a differential equa-tion of the form

zmy(m) +m−1∑

k=0

zkqk(z)y(k) = 0

where qk(z) ∈ Q[z] has degree ≤ m− k.

Corollary Let f(z) be an E-function with coef-ficients in Q and suppose that f(1) = 0. Then1 is an apparent singularity of the minimal dif-ferential equation satisfied by f .

Proof Consider f(z)/(1 − z). This is againE-function. So its minimal differential equa-tion has a basis of analytic solutions at z = 1.This means that the original minimal differen-tial equation for f(z) has a basis of analyticsolutions all vanishing at z = 1. So z = 1 isapparent singularity.

22

Andre’s proof of transcendence of π

Suppose 2πi algebraic. Then the E-functione2πiz − 1 vanishes at z = 1. The product overall conjugate E-functions is an E-function withrational coefficients vanishing at z = 1. Sothe above corollary applies. However linearforms in exponential functions satisfy differen-tial equations with constant coefficients, con-tradicting existence of a singularity at z = 1.

By a combination of Andre’s Theorem and dif-ferential galois theory one can show more.

Theorem (FB, 2004) Let f(z) be an E-functionand suppose that f(ξ) = 0 for some ξ ∈ Q∗.Then ξ is an apparent singularity of the mini-mal differential equation satisfied by f .

Additional use of a home-made zero lemmagives us:

Theorem (FB, 2004) The Nesterenko-Shidlovskiiconjecture holds with S = singularities ∪ 0.

23

Values at the singular points

Example f(z) = (z−1)ez. It satisfies (z−1)f ′ =zf and f(1) = 0.

More generally,

(z − ξ)k d

dz

f1(z)...

fn(z)

= A(z)

f1(z)...

fn(z)

where A(ξ) 6= O. Then,

A(ξ)

f1(ξ)...

fn(ξ)

= 0.

Theorem Let f(z) = (f1(z), . . . , fn(z)) be E-function solution of system of n first orderequations and suppose they are Q(z)-linear in-dependent. Then there exists an n×n- matrixB with entries in Q[z] and det(B) 6= 0 andE-functions e(z) = (e1(z), . . . , en(z) such thatf(z) = Be(z) and e(z) satisfies system of equa-tions with singularities in the set {0,∞}.

24