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In Mathematical, a transcendent number is a complex Real number or which is not root of any polynomial equation:

a_n~x^n + a_ {n-1} ~x^ {n-1} + \ ldots + a_1~x^1 + a_0 = 0 \,

where $n \ Ge 1 \,$ and the coefficients $a_i \,$ are integers (or, in manner equivalent, rational), of which at least one $a_n$ is nonnull. A real number or complex is thus transcendent if and only if it is not algebraic.

The existence of transcendent numbers is shown easily by an argument of cardinality (counting): there is a countable Infinité not of real numbers (or complexes), and only one countable infinity of algebraic numbers, therefore certain real numbers are not algebraic.

The transcendent numbers are never rational. Nevertheless, all the irrational numbers are not transcendent: the square Racine of 2 is irrational, but is a solution of the polynomial equation $x^2 - 2 = 0 \,$ .

The whole of all the transcendent numbers is indénombrable. The proof is simple: since the polynomials with whole coefficients are countable S, and since each one of these polynomials has a number finished of zero, the whole of the algebraic numbers is countable. But the Argument of the diagonal of Cantor establishes that the real numbers (and consequently complex numbers also) are indénombrables, therefore the whole of all the transcendent numbers must be indénombrable. In other words, there are much more transcendent numbers than of algebraic numbers. Nevertheless, only few classes of transcendent numbers are known and to prove that a given number is transcendent can be extremely difficult.

Results: Let us consider the unit has algebraic numbers. Then:

has is a subfield of $\ mathbb {R} \,$ . In particular, has is stable by addition and multiplication.
has is countable, which shows that has is different from the unit $\ mathbb {R} \,$ (the transcendent numbers exist well).

History

Leibniz was probably the first nobody to believe in the existence of the numbers which do not satisfy the polynomials with rational coefficients. The name " transcendants" from Leibniz in its publication of 1682 comes where it showed that sin ( X ) is not an algebraic function of X . The existence of the transcendent numbers was proven for the first time in 1844 by Joseph Liouville, which showed examples, including the constant of Liouville:

C = \ sum_ {j=1} ^ \ infty 10^ {- J!} = 0,110001000000000000000001000 \ ldots

in which N - ième figure after the comma is 1 if N is a Factorielle (i.e., 1,2,6,24,120,720,…., etc) and 0 if not. Liouville showed that this number is what we name now a Nombre of Liouville; this means primarily that it can be particularly well approximated by the rational numbers. Liouville showed that all the numbers of Liouville are transcendent.

Johann Heinrich Lambert, in its article proving the irrationality $\ pi \,$ conjectured that $e \,$ and $\ pi \,$ were transcendent numbers. The first transcendent number to be shown without to be built especially for that was E, by Charles Hermite in 1873. In 1874, Georg Cantor found the argument described above establishing the ubiquity of the transcendent numbers. In 1882, Ferdinand von Lindemann published a demonstration of the transcendence of $\ pi \,$ . It showed initially that $e$ with any nonnull algebraic power is transcendent, and since $e^ {I \ pi} = -1 \,$ is algebraic (see Identité of Euler), $i \ pi \,$ and consequently $\ pi \,$ must be transcendent. This approach was generalized by Karl Weierstrass with the Théorème of Lindemann-Weierstrass. The transcendence of $\ pi \,$ allowed the demonstration of the impossibility of several old geometrical constructions with the compass and the rule, including most famous of them, the Quadrature of the circle.

In 1900, David Hilbert raised an important question in connection with the transcendent numbers, known under the name of Seventh problem of Hilbert: “If has algebraic not no one and different from 1 and if B is an irrational algebraic number, then the number $a^b \,$ is it is a number necessarily transcendent? ” The affirmative response was given in 1934 by the Théorème of Gelfond-Schneider. One can easily obtain transcendent numbers thanks to him. For example $2^ {\ sqrt {2}} \,$ or $e^ {\ pi} \,$ .

This work was extended by Alan Baker in the Années 1960.

Known transcendent numbers and open problems

the number $\ pi \,$ (see the article pi).
the number E (Napierian Logarithme S bases)
$e^ {\ pi} \,$ Constante of Gelfond
$2^ {\ sqrt {2}} \,$ (Constante of Gelfond-Schneider) or more generally $a^b \,$ (see the Théorème of Gelfond-Schneider) where $a \ 0 \,$ and $a \ 1 \,$ is algebraic and B is algebraic but nonrational. The general case of the Seventh problem of Hilbert, i.e. to determine if $a^b \,$ is transcendent when $a \ 0 \,$ and $a \ 1 \,$ is algebraic and B is irrational, remains not-solved.
the value of the goniometrical Function $\ sin (1) \,$
$\ ln (a) \,$ if has is rational strictly positive and different from 1.
$\ Gamma \ left (\ frac {1} {3} \ right) \,$ and $\ Gamma \ left (\ frac {1} {4} \ right) \,$ (see Fonction Gamma of Euler).
the number of Champernowne 0,12345678910111213… obtained while writing after the integers in bases ten (theorem of Mahler, 1961)
$\ sum_ {k=0} ^ {+ \ infty} 10^ {- \ lfloor \ beta^ {K} \ rfloor}; \ qquad \ beta > 1 \; ,$

where

\ \ lfloor X \ rfloor

is the whole Partie

\ X \ in \ mathbb {R}

. For example, if

\ beta = 2 \,

, this number is 0,11010001000000010000000000000001000…

$\ Omega \,$ , Constant of Chaitin, and more generally: each calculable number not is transcendent (since all the algebraic numbers are calculable).

Constant of Prouhet-Thue-Morse

All Algebraic function not constant with a variable gives a transcendent value when a transcendent value is applied to him. Therefore, for example, to know that $\ pi \,$ is transcendent, we can immediately deduce that $5 \ pi \,$ , $\ frac {\ pi-3} {\ sqrt {2}} \,$ , $(\ sqrt {\ pi} - \ sqrt {3}) ^8 \,$ and $(\ pi^ {5} +7) ^ {\ frac {1} {7}} \,$ is also transcendent.

Nevertheless, an algebraic function with several variables can give an algebraic number when it is applied to the transcendent numbers if these numbers are not algebraically independent. For example, $\ pi \,$ and $1- \ pi \,$ are all the two transcendent ones, but $\ pi~+~ (1 \ pi) ~=~1 \,$ is obviously not it. One is unaware of if $\ pi~+~e \,$ , for example is transcendent, but at least one of $\ pi~+~e \,$ and $\ pi E \,$ must be transcendent. More generally, for two transcendent numbers has and B , at least one of has + B and has B must be transcendent. To see that, let us consider the polynomial $(x~-~a) (x~-~b) = x^2 - (a+b) X + has B \,$ . If ( has + B ) and has B are all the two algebraic ones, then this would be a polynomial with algebraic coefficients. Because the algebraic numbers form a Corps algebraically closed, this implies that the roots of the polynomial, has and B , must be algebraic. But this is a contradiction and thus, there must be the case where at least one of the two coefficients is transcendent.

The numbers which one is unaware of if it are transcendent or not include:

$(\ pi~+~e) \,$ , $(\ pi~-~e) \,$ , $(\ pi E) \,$ , $(\ frac {E} {\ pi}) \,$ , $(\ frac {\ pi} {E}) \,$ , $(\ pi^ {\ pi}) \,$ , $(e^e) \,$ , $(\ pi^e) \,$

the Constant of Euler-Mascheroni $\ gamma \,$ (which one is unaware of even if it is irrational)

the Constant of Catalan, which one is unaware of also if it is irrational

$\ zeta (3) \,$ , the Constant of Apéry

All the numbers of Liouville are transcendent, nevertheless the transcendent numbers are not all of the numbers of Liouville. Any number of Liouville must have terms not limited in their development in continued Fraction, and thus, by using an argument of enumeration, one can show that there exist transcendent numbers which are not numbers of Liouville. By using the explicit development in continued fraction of E , one can show that E is not a number of Liouville. Kurt Mahler showed in 1953 that $\ pi \,$ is not either a number of Liouville. It was conjectured that all the continued fractions with limited terms which are not possibly periodic are transcendent (the possibly periodic continued fractions correspond to irrational quadratic).

Outline demonstration of the transcendence of $e$

The first demonstration that

e

is transcendent goes back to 1873. We will now follow the strategy of David Hilbert (1862 - 1943) which gave a simplification of the original demonstration of Charles Hermite. The idea is the following one:

Let us suppose, with an aim of finding a contradiction, that $e$ is algebraic. Then, there exists a whole finished of whole coefficients $c_ {0}, c_ {1}, \ ldots, c_ {N} \,$ satisfying the equation:

$c_ {0} +c_ {1} e+c_ {2} e^ {2} + \ cdots+c_ {N} e^ {N} =0 \,$

and $c_0$ and $c_n$ are both different from zero.

Depending on the value of N , we specify a sufficiently large positive entirety K (for our need later) and multiply both with dimensions equation above by $\ int^ {\ infty} _ {0} \,$ , where the notation $\ int^ {B} _ {has} \,$ will be used in this demonstration as abbreviation of the integral:

$\ int^ {B} _ {has}: = \ int^ {B} _ {has} x^ {K} ^ {k+1} e^ {- X} \, dx \,$ .

We arrive at the equation:

$c_ {0} \ int^ {\ infty} _ {0} +c_ {1} E \ int^ {\ infty} _ {0} + \ cdots+c_ {N} e^ {N} \ int^ {\ infty} _ {0} = 0 \,$

who can now be written in the form

$P_ {1} +P_ {2} =0 \,$

where

$P_ {1} =c_ {0} \ int^ {\ infty} _ {0} +c_ {1} E \ int^ {\ infty} _ {1} +c_ {2} e^ {2} \ int^ {\ infty} _ {2} + \ cdots+c_ {N} e^ {N} \ int^ {\ infty} _ {N} \,$

P_ {2} =c_ {1} E \ int^ {1} _ {0} +c_ {2} e^ {2} \ int^ {2} _ {0} + \ cdots+c_ {N} e^{N} \ int^ {N} _ {0} \,

The plan of attack now is to show that for a K sufficiently large, the relations above are impossible to satisfy because

$\ frac {P_ {1}} {K!}\,$ is an entirety different from zero and $\ frac {P_ {2}} {K!}\,$ is not it.

The fact that $\ frac {P_ {1}} {K!}\,$ is an entirety different from zero results from the relation

$\ int^ {\ infty} _ {0} x^ {J} e^ {- X} \, dx=j! \,$

who is valid for entire positive J and can be proven by recurrence by means of a Intégration by parts.

To show that

$\ left|\ frac {P_ {2}} {K!}\ right|<1 \,$ for a K sufficiently large

we will note initially that $x^ {K} ^ {k+1} e^ {- X} \,$ is the product of the functions $^ {K} \,$ and $(x-1) (x-2) \ cdots (x-n) e^ {- X} \,$ . By using the upper limit for $|X (x-1) (x-2) \ cdots (x-n)|\,$ and $|(x-1) (x-2) \ cdots (x-n) e^ {- X}|\,$ on the interval and by employing the fact that

\ lim_ {K \ to \ infty} \ frac {G^k} {K!}=0 \,

for each real number G

is then sufficient to complete the demonstration.

A similar strategy, different from the original approach of Lindemann, can be used to show that the number $\ pi \,$ is transcendent. Moreover, the Function gamma, certain estimates for $e$ and of the facts in connection with the symmetrical polynomials play a vital part in the demonstration.

For detailed informations concerning the demonstrations of transcendences of $\ pi \,$ and $e \,$ , to see the external references and bonds.

See too

Theory of the transcendence, the study of the relative questions to the transcendent numbers

References

David Hilbert, “ Über die Transcendenz der Zahlen $e$ und $\ pi$ ”, Mathematische Annalen 43 : 216– 219 (1893).
Alan Baker, Transcendental Number Theory , Cambridge University Near, 1975.

External bonds

Démonstration that $e$ is transcendent
Démonstration that $e$ is transcendent
Démonstration that $\ pi$ is transcendent

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