Transcendental number

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In mathematics, a transcendental number is any complex number that is not algebraic, that is, not the solution of a non-zero polynomial equation with integer (or, equivalently, rational) coefficients. The most prominent examples of transcendental numbers are π and e.

Although transcendental numbers are never rational, some irrational numbers are not transcendental: the square root of 2 is irrational, but it is a solution of the polynomial x² − 2 = 0, so it is algebraic.

The transcendental numbers are uncountable. The proof is simple: Since the polynomials with integer coefficients are countable, and since each such polynomial has a finite number of zeroes, the algebraic numbers must also be countable. But Cantor's diagonal argument proves that the reals (and therefore also the complex numbers) are uncountable; so the set of all transcendental numbers must also be uncountable. In a very real sense, then, there are many more transcendental numbers than algebraic ones. However, only a few classes of transcendental numbers are known and proving that a given number is transcendental can be extremely difficult.

1 History
2 Known transcendental numbers and open problems
3 Proof sketch that e is transcendental
4 See also
5 References
6 External links

[edit] History

Leibniz was probably the first person to believe in the existence of numbers which do not satisfy polynomials with rational coefficients. The name "transcendentals" comes from Leibniz in his 1682 paper where he proved sin(x) is not an algebraic function of x. The existence of transcendental numbers was first proved in 1844 by Joseph Liouville, who exhibited examples, including the Liouville constant:

<math>\sum_{k=1}^\infty 10^{-k!} = 0.110001000000000000000001000\ldots</math> in which the nth digit after the decimal point is 1 if n is a factorial (i.e., 1, 2, 6, 24, 120, 720, ...., etc.) and 0 otherwise. Liouville showed that this number is what we now call a Liouville number; this essentially means that it can be particularly well approximated by rational numbers. Liouville showed that all Liouville numbers are transcendental.

Johann Heinrich Lambert conjectured that e and π were both transcendental numbers, in his paper proving the number π is irrational. The first number to be proven transcendental without having been specifically constructed for the purpose was e, by Charles Hermite in 1873. In 1874, Georg Cantor found the argument mentioned above establishing the ubiquity of transcendental numbers.

In 1882, Ferdinand von Lindemann published a proof that the number π is transcendental. He first showed that e to any algebraic power is transcendental, and since <math>e^{i\pi} = -1</math> is algebraic (see Euler's identity), <math>i\pi</math> and therefore <math>\pi</math> must be transcendental. This approach was generalized by Karl Weierstrass to the Lindemann–Weierstrass theorem. The transcendence of π allowed the proof of the impossibility of several ancient geometric constructions involving compass and straightedge, including the most famous one, squaring the circle.

In 1900, David Hilbert posed an influential question about transcendental numbers, Hilbert's seventh problem: If the number a is algebraic, but is neither 0 nor 1, and the number b is irrational and algebraic, is a^b necessarily transcendental? The affirmative answer was provided in 1934 by the Gelfond-Schneider theorem. This work was extended by Alan Baker in the 1960s.

[edit] Known transcendental numbers and open problems

Here is a list of some numbers known to be transcendental:

e^a if a is algebraic and nonzero (a consequence of the Lindemann-Weierstrass theorem). In particular, e itself is transcendental.
π.
e^π Gelfond's constant. (A consequence of the Gelfond-Schneider theorem.)
2^√2, the Gelfond-Schneider constant, or more generally a^b where a ≠ 0, 1 is algebraic and b is algebraic but not rational (Gelfond-Schneider theorem and Hilbert's seventh problem).
sin(a), cos(a) and tan(a) for any nonzero rational number a in radians.
ln(a) if a is positive, rational and ≠ 1
Γ(1/3), Γ(1/4), and Γ(1/6) (see gamma function).
the Champernowne constant is an example of a transcendental number that is normal in base 10.
Ω, Chaitin's constant, and more generally: every non-computable number is transcendental (since all algebraic numbers are computable).
Prouhet-Thue-Morse constant
<math>\sum_{k=0}^\infty 10^{-\lfloor \beta^{k} \rfloor};</math> where <math>\beta > 1</math> and <math>\beta\mapsto\lfloor \beta \rfloor</math> is the floor function.

Any non-constant algebraic function of a single variable yields a transcendental value when applied to a transcendental argument. So for example, from knowing that π is transcendental, we can immediately deduce that 5π, (π − 3)/√2, (√π − √3)⁸ and (π⁵ + 7)^1/7 are transcendental as well.

However, an algebraic function of several variables may yield an algebraic number when applied to transcendental numbers if these numbers are not algebraically independent. For example, π and 1 − π are both transcendental, but π+(1−π)=1 is obviously not. It is unknown whether π + e, for example, is transcendental, though at least one of π + e and π e must be transcendental. More generally, for any two transcendental numbers a and b, at least one of a + b and a b must be transcendental. To see this, consider the polynomial (x − a) (x − b) = x² − (a + b) x + a b. If (a + b) and ab were both algebraic, then this would be a polynomial with algebraic coefficients. Because algebraic numbers form an algebraically closed field, this would imply that the roots of the polynomial, a and b, must be algebraic. But this is a contradiction, and thus it must be the case that at least one of the coefficients is transcendental.

Numbers for which it is unknown whether they are transcendental or not include

Sums, products, powers, etc. (except for Gelfond's constant) of the number π and the number e: π + e, π − e, π·e, e/π, π/e, π^π, e^e, π^e
the Euler-Mascheroni constant γ (which has not even been proven to be irrational)
Catalan's constant, also not known to be irrational
Apéry's constant, ζ(3) (see Riemann zeta function)

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).

[edit] Proof sketch that e is transcendental

The first proof that the base of the natural logarithms, e, is transcendental dates from 1873. We will now follow the strategy of David Hilbert (1862–1943) who gave a simplification of the original proof of Charles Hermite. The idea is the following:

Assume, for purpose of finding a contradiction, that e is algebraic. Then there exists a finite set of integer coefficients <math>c_{0},c_{1},\ldots,c_{n},</math> satisfying the equation:

<math>c_{0}+c_{1}e+c_{2}e^{2}+\cdots+c_{n}e^{n}=0</math>

and such that <math>c_0</math> and <math>c_n</math> are both non-zero.

Depending on the value of n, we specify a sufficiently large positive integer k (to meet our needs later), and multiply both sides of the above equation by <math>\int^{\infty}_{0}</math>, where the notation <math>\int^{b}_{a}</math> will be used in this proof as shorthand for the integral:

<math>\int^{b}_{a}:=\int^{b}_{a}x^{k}[(x-1)(x-2)\cdots(x-n)]^{k+1}e^{-x}\,dx.</math>

We have arrived at the equation:

<math>c_{0}\int^{\infty}_{0}+c_{1}e\int^{\infty}_{0}+\cdots+c_{n}e^{n}\int^{\infty}_{0} = 0</math>

which can now be written in the form

where

<math>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}</math>

<math>P_{2}=c_{1}e\int^{1}_{0}+c_{2}e^{2}\int^{2}_{0}+\cdots+c_{n}e^{n}\int^{n}_{0}</math>

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

<math>\frac{P_{1}}{k!}</math> is a non-zero integer and <math>\frac{P_{2}}{k!}</math> is not.

The fact that <math>\frac{P_{1}}{k!}</math> is a nonzero integer results from the relation

<math>\int^{\infty}_{0}x^{j}e^{-x}\,dx=j!</math>

which is valid for any positive integer j and can be proved using integration by parts and mathematical induction.

To show that

<math>\left|\frac{P_{2}}{k!}\right|<1</math> for sufficiently large k

we first note that <math>x^{k}[(x-1)(x-2)\cdots(x-n)]^{k+1}e^{-x}</math> is the product of the functions <math>[x(x-1)(x-2)\cdots(x-n)]^{k}</math> and <math>(x-1)(x-2)\cdots(x-n)e^{-x}</math>. Using upper bounds for <math>|x(x-1)(x-2)\cdots(x-n)|</math> and <math>|(x-1)(x-2)\cdots(x-n)e^{-x}|</math> on the interval [0,n] and employing the fact

<math>\lim_{k\to\infty}\frac{G^k}{k!}=0</math> for every real number G

is then sufficient to finish the proof.

A similar strategy, different from Lindemann's original approach, can be used to show that the number π is transcendental. Besides the gamma-function and some estimates as in the proof for e, facts about symmetric polynomials play a vital role in the proof.

For detailed information concerning the proofs of the transcendence of π and e see the references and external links.

[edit] See also

Transcendence theory, the study of questions related to transcendental numbers

[edit] References

David Hilbert, "Über die Transcendenz der Zahlen <math>e</math> und <math>\pi</math>", Mathematische Annalen 43:216–219 (1893).
Alan Baker, Transcendental Number Theory, Cambridge University Press, 1975, ISBN 0-521-39791-X.