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Transcendental Numbers Basic Notions in Mathematics Utrecht, March 31, 2005 1

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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.

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First known transcendental numbers

Liouville numbers constructed by Liouville in

1844.

Example: a =∑∞

n=01

10n! is transcendental. In

decimals:

a = 1.1100010000000000000000010000 · · ·

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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.

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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.

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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.

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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).

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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.

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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.

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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.

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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.

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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.

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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.

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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.

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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

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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.

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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.

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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))

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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

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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}.

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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.

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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.

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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,∞}.

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