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Louville’s Theorem and Transcendence
Proof and its application for constructing a class of Transcendental Numbers
Swarnava Mukhopadhyay
[email protected]
Chennai Mathematical Institute
Plot No. H1 SIPCOT IT Park,
PO. Padur,Siruseri-603 103
Tamilnadu,India.
A Brief Introduction
Louville’s theorem basically says that any algebraic number cannot be approximated by a sequence of numbers convering to it after a certain degree
and thus this thoerem can be used to prove the existance of Transcendental
Number as well as produce a class of Transcendental Numbers.We will look
at a proof of Louville’s theorem and also constructed some class of transcendental numbers.
The theorem states that
If x is an algebraic number then there exists c ≥ 0 such that for all integer p and natural
q we have,
c
p
6 |x − |
d
q
q
where d is the degree of the irreducible polynomial which is satisfied by x. I will give
a proof of Louville’s Theorem and also an estimate of the value of the constant c given
in the statementand use the theorem for constructing a certain class of Transcendental
Numbers.
There are certain class of number known as the Louville’s number which can be constructed out of the Louville’s Theorem but all transcendental numbers are not Louville’s
number. Louville proved this theorem in 1844 using countinued fractions and then in
1854 he reformulated it removing the dependence of the continued fractions which enabled him to publish a class of transcendental numbers by which he proved the existance
of transcendental numbers both real and complex.
Claim 1. The inclusion of Q in R is strict
Proof. The statement means that ∃ x such that x ∈ R but x ∈
/ Q. So we have to find a
number which is in R but it is not in
Q.
√
√
So
√ we start by taking the number 2. We know that 2 ∈ R but we will show that
2∈
/ Q.
We consider the map from
R→R
given as follows
f :R→R
x → x2 − 2
It is a continous function from R to R. Now f (2) ≥ 0 and f (0) ≤ 0.
So by the Intermediate Value Theorem we can say that it will take the value 0 within
1
the interval [0,2]. If
√
2 is rational,then we can express
√
√
2 as
p
q
where p and q are mutually coprime. Squaring we get the following,
2=
2 = p2 q 2
⇒ p2 = 2q 2
⇒ p2 is even.
It follows that q is also even. Since q 2 is even we can say that q √
is also even. This
contradicts the fact that p and q are mutually co-prime. Therefore 2 ∈
/ Q. Thus the
inclusion of Q in R is strict.
2
Let z be a complex algebraic number. It is the root of a polynomial
with co-efficients in Q. So z is also a root of that polynomial.
If a polynomial p(x) ∈ R[x] where R is a integral domain and let F be its fraction field.
If the polynomial p(x) is reducible in its fraction field F [x] then by Gauss Lemma it is
also reducible over the ring R[x].
Let the polynomial be P (x).
P (x) = (x − z)(x − z)Q(x)
Suppose P is a product of irreducible polynomials. So by assumption we can decompose
P into a product of irreducible polynomials. say P = P1 .P2 .P3 . · · · .Pn
But P (z) = 0 ⇒∃ i such that Pi = 0. We can conclude that Pi is irreducible. After changing the notations and multiplying P by an integer we assume that P ∈ Z is irreducible.
Let d be its degree.We are going to prove the following.
Claim 2. Let
pn
qn
where n ∈ N and pn and qn belongs to N such that.
pn
=z
n→∞ qn
lim
We will prove that
pn
1
≤ 2| − z||P 0 (z)|
d
qn
qn
Proof. If A is a polynomial and B is a polynomial in Q[x]. Then let D bethe gcd(A, B).
So D divides both A and B. Now Q[x] is a Euclidean Domain. So by Bezout’s Theorem
we have can write D as
D = AM + BN
where M and N are also polynomials. Now since P is a irreducible polynomial which
leads us to conclude that P 0 and P are co-prime. So P (z) = 0 ⇒ P 0 (z) 6= 0
⇒ |P 0 (z)| ≥ 0
Otherwise it contradicts the irreducibility of P .Therefore P is the polynomial of degree
d such that P (z) = 0,where z is a complex algebric number.
Now let pqnn where n ∈ N be the sequence of numbers converging to z as defined earlier
and N is the set of natural numbers. By Taylor’s Expansion we have,
P (x) = (x − z)P 0 (z) +
(x − z)2 00
(x − z)d d
P (z) + · · · +
P (z)
2!
d!
2
Since the polynomial is of degree d and it is also a smooth function too. Its derivatives
higher than d is zero. Putting x = pqnn in the above expression we have that.
( pqn − z)2 00
( pqn − z)d d
pn
pn
) = ( − z)P 0 (z) + n
P (z) + · · · + n
P (z)
qn
qn
2!
d!
Taking modulus on both sides and applying triangle in-equality we get that the following
expression.
P(
Since
pn
qn
( pqnn − z)2 00
( pqnn − z)d d
pn
pn
0
|P ( )| ≤ |( − z)||P (z)| + |
P (z) + · · · +
P (z)|
qn
qn
2!
d!
is close enough to z and since z is algebraic we have
pn
P ( ) 6= 0
qn
Let the irreducible polynomial satisfied by z is P (x) = a0 xd + a1 xd−1 + · · · + ad .
Multiplying both sides of the above equation by qnd we have that
pn
d
qnd P ( ) = a0 pdn + a1 pd−1
n qn + · · · + ad qn
qn
wherea0 , a1 , a2 , a3 , · · · , ad are integers. So we have
pn
qnd P ( ) ∈ Z
qn
pn
|a0 pdn + a1 pnd−1 + · · · + an qnd |
)| =
qn
qnd
pn
1
|P ( )| ≥ d
qn
qn
d
d
. Since |a0 pn + · · · + an qd | is a positive integer. Now
pn
pn
→ z ⇒ P( ) → 0
qn
qn
|P (
d
X
d
X
(x − z)i
|(x − z)i |
i
|
P (z)
|≤
|P (z)|
i!
i!
n=2
n=2
Let
pn
qn
i
= x and it converges to z. Suppose |x − z| ≤ δ
d
X
d
|(x − z)i | X i
δi
|P (z)|
≤
|P (z)|
i!
i!
n=2
n=2
i
i
for δ ≤ 1 We choose δ such that
≤ (d − 1)max( |P i!(z)| )δ 2
|P i (z)| 2
).δ ≤ P 0 (z).δ
i!
Since P 0 (z) =
6 0 So we have a choice for δ and this is possible only because of the series
coverges.
|P 0 (z)|
δ≤
i
(d − 1)max( |P i!(z)| )
∀ i ≥ 2 and we preferably choose δ 6= 0.
|P 0 (z)|
δ ≤ min(1,
)
i
(d − 1)max( |P i!(z)| )
(d − 1)max(
3
pn
qn
converges to z. Given any δ ≥ 0 such that,
pn
| − z| ≤ δ
qn
Applying it the above equation and using x = pqnn we have,
Since
|
( pqnn − z)
2!
P 2 (z) + · · · +
( pqnn − z)d
d!
≤|
P d (z)| ≤
d
X
( pqn − z)i i
| n
||P (z)|
i!
i=1
pn
− z||P 0 (z)|
qn
Replacing this result in the inequality derived by using Triangle Inequility we have the
folowing,
pn
pn
pn
|P ( ) ≤ | − z||P 0 (z)| + | − z||P 0 (z)|
qn
qn
qn
pn
pn
|P ( )| ≤ 2| − z||P 0 (z)|
qn
qn
Using this we can write that,
1
pn
pn
≤
|P
(
)|
≤
2|
− z||P 0 (z)|
qnd
qn
qn
pn
1
1
≤ | − z|
d
0
qn 2|P (z)|
qn
Now from this Lemma the proof of the Louville’s theorem follows directly as some
changing of terms is just requires.From the above Lemma we have also estimated the
value of the constant c which we have proved that it can be found directly by computing
the derivatives of the irreducible polynomial at z.
We will now see some applications and use Louville’s theorem to construct some transcendental numbers. The essence and beauty of the Theorem lies in the fact that no
algebraic number can’t be approximated after a certain degree by a sequence of numbers
converging to it.
P
1
We consider the series ∞
n=1 10n!
P
1
Claim 3. The series ∞
n=1 10n! converges.
Proof. We take the partial sums of the first n terms and check for the convergence.
n
X
1
sn =
10k!
k=1
n
X
1
≤
10k
k=1
∞
X
1
≤
10k
k=1
=
1
10
1−
4
1
10
= 19
The partial sums are therefore
and they are montonic thus they converge to a
P∞ bounded
1
finite value and the series n=1 10n! is thus convergent.
Let
z=
∞
X
1
10n!
n=0
Claim 4. z is transcendental.
Proof. In the following expression,
n
X
1
sn =
10k!
k=0
Pn
sn =
So we can replace sn by
care what is pn .
pn
qn
k=0 10
10n!
n!
k!
where pn and qn are of the following form,we neednot need to
qn = 10n!
Assume that z is a algebraic number and then approach to prove it by contradiction
that it is transcendental. Let P (x) be irreducible polynomial which z satisfies where
P (x) ∈ Z[x]. So by the definiton we have that P (z) = 0. Further let d be the degree of
P (x).
∞
X
1
z − sn =
10k!
k≥n+1
By taking the modulus we get that,
|z − sn | ≤
1
10(n+1)!
1
1
1 − 10
Since ∀ k ≥ n + 1 we have
1
1
1
≤ (n+1)! ( n−k+1 )
k!
10
10
10
So our ploy is to use the theorem we have proved, i.e the Louville’s Theorem and then
we claim that for n big enough we have
1
1
10
≤ (n+1)! ( )(2|P 0 (z)|)
n!
d
(10 )
10
9
It is impossible as because as n → ∞
10n!d
→0
10(n+1)!
So it is proved that z is not algebraic.Thus we have that the number z is transcendental.
Remark : We can generalize the above proof for some particular numbers
by replacing 10 by any integer m 6= 1.
P
1
Claim 5. Any number of the form ∞
k=0 mk! where m 6= 1 is transcendental.
5
Proof. We will follow the same proof as the above just by replacing 10 by m as in the
previous proof there was no special application of the integer n thus obtain an abstract
genaralization of the above example for numbers of some certain form.
We take the partial sums of the first n terms and check for its convergence.
n
X
1
sn =
mk!
k=1
n
X
1
≤
mk
k=1
≤
=
∞
X
1
mk
k=1
1
m
1−
1
m
1
= m−1
Remark : The above shows why m was not chosen to be 1.
The partial sums are therefore bounded and they are montonic. So they converge to a
finite value. So the series is convergent.
Let
∞
X
1
l=
mk!
k=0
In the folowing expression
So we can replace sn by
pn
qn
n
X
1
sn =
mk!
k=0
Pn
n!
k!
k=0 m
sn =
mn!
where pn and qn are of the following form,
qn = mn!
Asssume that l is a algebraic number and then approach to prove it by the process of
contradiction that it is transcendental just by following the proof of the above example.
Let q(x) be an irreducible polynomial which l satisfies where q(x) ∈ Z[x]. As per the
definiton we have that q(l) = 0. Further let d be the degree of q(x).
l − sn =
∞
X
1
mk!
k≥n+1
By taking the modulus we get that,
|l − sn | ≤
1
m(n+1)!
1
1 − m1
Since ∀ k ≥ n + 1 we have
1
1
1
≤ (n+1)! ( n−k+1 )
k!
m
m
m
6
So our ploy is to use the theorem we have proved,i.e the Louville’s Theorem and then we
claim that for n big enough we have
1
1
m
)(2|P 0 (z)|)
≤ (n+1)! (
n!
d
(m )
m
m−1
It is impossible as because
mn!d
→0
m(n+1)!
as n → ∞. Thus we have shown that the number l is transcendental.
Remark : The above proof of the Louville’s Theorem is a proof ∀n ≥ N
where N depends on the value of the δ chosen.Thus by applying Taylor’s series
expansion we have found out the algebraic number to be approximated.
Transcendental Numbers and some open problems
(1) a if a is algebraic and nonzero (a consequence of the Lindemann-Weierstrass
theorem). In particular, e itself is transcendental.
(2) π is irrational.
(3) eπ Gelfond’s constant. (A consequence of the Gelfond-Schneider theorem.)
√
(4) 2 2 , the Gelfond-Schneider constant, or more generally ab where a 6= 0, 1 is
algebraic and b is algebraic but not rational (Gelfond-Schneider theorem and
Hilbert’s seventh problem).
(5) sin(a), cos(a) and tan(a) for any nonzero rational number a.
(6) ln(a) if a is positive, rational and 6= 1
(7) Γ(1/3), Γ(1/4), and Γ(1/6)
(8) Ω, Chaitin’s constant, and more generally: every non-computable number is transcendental (since all algebraic numbers are computable).
(9) Prouhet-Thue-Morse constant
(10)
P∞
k=0
k
10−bβ c ; where β > 1 and β 7→ bβc is the floor function.
(11) Sums, products, powers, etc. (except for Gelfond’s constant) of the number π
and the number e: π + e, π − e, πe, e/π, π/e, π π , ee , π e
(12) the Euler-Mascheroni constant γ (which has not even been proven to be irrational)
(13) Catalan’s constant, also not known to be irrational.
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All Liouville numbers are transcendental, however not all transcendental numbers are Liouville numbers. Any Liouville number must have unbounded terms
in its continued fraction expression, and so using a counting argument one can
show that there exist transcendental numbers which are not Liouville. Using the
explicit continued fraction expansion of e, one can show that e is not a Liouville
number. Kurt Mahler showed in 1953 that π is also not a Liouville number. It
is conjectured that all infinite continued fractions with bounded terms that are
not eventually periodic are transcendental (eventually periodic continued fractions
correspond to quadratic irrationals).
Biographical Sketch
I am a BSC-II year student in Chennai Mathematical Institute persuing hons in Math
and Computer Science.
Address for correspondence
Swarnava Mukhopadhyay
Email:[email protected]
Ph:09444889379
Chennai Mathematical Institute
Plot No. H1 SIPCOT IT Park,
PO. Padur,Siruseri-603103
Tamilnadu,India.
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Suggested Reading:
[1] G.H.Hardy and E.M.Wright, Introduction to the theory of numbers.
[2] G.H.Hardy, A course in Pure Mathematics.
[3] Allouche, J.P. and Cosnard, M. ”The Komornik-Loreti Constant Is Transcendental.”
Amer. Math. Monthly 107, 448-449, 2000
[4] Borwein, J. M.; Borwein, P. B.; and Bailey, D. H. ”Ramanujan, Modular Equations,
and Approximations to Pi or How to Compute One Billion Digits of Pi.” Amer. Math.
Monthly 96, 201-219, 1989
[5] Chudnovsky, G. V. Contributions to the Theory of Transcendental Numbers. Providence, RI: Amer. Math. Soc., 1984
[6] Pickover, C. A. ”The Fifteen Most Famous Transcendental Numbers.” J. Recr.
Math. 25, 12, 1993.
[7] David Hilbert, ”Uber die Transcendenz der Zahlen e und π”, Mathematische
Annalen 43:216-219 (1893).
[8] Nathan Jacobson, Basic Algebra (Vol-I)
[9] Serge Lang, Algebra
[10] A.O.Gelfond,Transcendental and algebraic numbers by A.O.Gelfond
[11] Alan Baker, Transcendental Number Theory
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