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Transcendental Numbers

Transcendental Numbers

The following section uses the notation from the chapter on Set Theory. For the moment, we shall consider the behaviour of inconcrete m, n ∈ ωℕ, and we define j, k ∈ ℕ.

Remark: From algebra, we know that the sum, difference, product, and quotient of two algebraic numbers of degree j and k are algebraic of degree at most jk, and that the 1/j-th power of an algebraic number of degree k is algebraic of degree at most jk.

Remark: Transcendental numbers can be viewed as the sum of an algebraic part and a transcendental remainder. When investigating whether a number is transcendental, if the remainder may be expressed as the limit value of a zero sequence (ak) (see Nonstandard Analysis), we cannot simply disregard the values of the sequence for large k: They are important. Transcendental numbers are the numbers that lie between algebraic numbers or on either side of them. By the counting theorem for algebraic numbers, if two distinct transcendental numbers (algebraic of degree j) are sufficiently close, there is no algebraic number (of degree < j) between them.

Bounding theorem for ω-transcendental numbers: Every non-zero complex number whose imaginary or real part has absolute value is ≤ 1/(ώ + 1) or ≥ ώ + 1, is automatically ω-transcendental.

Proof: In a polynomial equation, set am = 1 and ak = -ώ for k < m, then the claim in the real case follows from the geometric series formula after taking the reciprocal. We can find the exact limit value by replacing ώ + 1 by ω(m) = ώ + 1 - ώ/ω(m)m with ω := ώ + 1 - ώ/ωώ. In the complex case, substitutions of the form x = (1 + ib)(ώ + 1) with b ∈ ωℝ give the desired result.⃞

Coefficient theorem for ω-transcendental numbers: Every normalised irreducible polynomial and series such that |ak| ≥ ώ + 1 for at least one ak only has ω-transcendental zeros.

Proof: The zeros of normalised irreducible polynomials and series are pairwise distinct and uniquely determined. Since they are not ω-algebraic, they must be ω-transcendental.⃞

Approximation theorem for real algebraic numbers: Every real algebraic number of degree k > 1 may be approximated by a real ω-algebraic number of degree j < k with an average asymptotic error of π ζ(j + 1)/(2 ln j |ωℤ|j).

Proof: On the conventionally real axis, the number of ω-algebraic numbers approximately evenly distributed between fixed limits increases by a factor of approximately |ωℤ| per degree. The error corresponds to the distance between ω-algebraic numbers. The non-real ω-algebraic numbers are less dense.⃞

Conclusion: Two distinct real ω-algebraic numbers have an average distance of at least π/(ln(ώ + 1) |ωℤ|ώ). Determining this minimum distance exactly requires an infinite non-linear non-convex optimisation problem to be solved. Therefore, the c-algebraic numbers have an approximate order of O(c), disproving Roth's theorem, which essentially amounts to proving the (trivial) minimum distance between two rational numbers, and thus disproves the abc conjecture, but not Liouville's result.

Theorem: The maximum distance between two neighbouring real ω-algebraic numbers ≠ 0 amounts 1/ώ2 - O(1/ώ3).

Proof: The distance between two real ω-algebraic numbers ≠ 0 is largest around the points ±1. A ω-rational number r > 1 (0 < r < 1) may be better approximated by the real ω-algebraic x that satisfies the polynomial or series equation xm - rxm-1 = r/ώ (xm - ώx = -ώr). An analogous result holds for negative r. If we wish to approximate 1 by a larger real ω-algebraic x, it must necessarily satisfy a polynomial or series equation such that am = ώ - 1 and a0 = -ώ. By setting am = ώ - 1 and a0 = -ώ the claim follows, since the maximum distance can no longer be reduced.⃞

Remark: Together with the rational numbers, the infinite (complex-)rational numbers already numerically make up the entirety of ℝ (ℂ). Therefore, algebraic and transcendental numbers are (numerically) difficult to distinguish, and approximations are of little use when determining whether a number is algebraic (to a certain degree). The merit of distinguishing between them is therefore questionable. Real continued fractions that do not terminate as a rational number are ω-transcendental, since they are infinite rational. They can only be used as an approximation of ω-algebraic numbers.

Remark: Since all real numbers are (approximately) (infinite) rational numbers, we can compute them in real-time. In the case of ω-transcendental numbers we must satisfy ourselves with an arbitrarily precise infinitesimal number, unlike in the case of ω-algebraic numbers, where we can argue using the corresponding minimal polynomial or series. We can therefore represent them in the form of a rational quotient with infinite numerator and denominator.

Remark: However, when investigating the transcendence of a number using Liouville's approximation theorem, care should be taken to ensure that the approximating rational numbers are in fact conventionally rational and not infinitely rational, such as for example the number 10ώ+1. Similarly, confidence with infinite natural numbers is required when considering prime numbers ≥ ώ + 1 in proofs of transcendence. The following proofs can also be used to show ω-transcendence by replacing the set ωℕ by [ώ + 1, n]ℕ for some n that is not excessively large.

Theorem: The sum of all at least ω-natural powers of a complex-rational number x ≠ 0 that is not a root of unity (geometric series) is already ω-transcendental.

Proof: The modulus of either the numerator or denominator of xώ+1 is ≥ 2ώ/2. Subtracting 1 and dividing by 1 - x preserves ω-transcendence.⃞

Theorem: Euler's number e is ω-transcendental.

Proof: If we accept the exponential series as a representation for e, it follows that e = (kώ + 1)/ώ! for k ≥ ώ + 1. Therefore, the numerator and the denominator of this fraction must be ≥ ώ + 1, since neither ώ nor a prime divisor of k in the numerator simplifies with ώ!. However, if we accept the representation (1 + 1/ώ)ώ for e, the claim is trivial. Note that these two representations give different numbers.⃞

The greatest-prime criterion for ω-transcendental numbers: If a real number r may be represented as an irreducible fraction a/(bp) ± s/t where a, b, s and t are natural numbers, abst ≠ 0 and b + t > 2, and the (second-)greatest prime number p ∈ ωℙ, p ∤ a and p ∤ t, then r is ω-transcendental.

Proof: We have that r = (at ± bps)/(bpt) with denominator ≥ 2p ≥ 2ώ - O(ln ώ) > ώ + 1 by the prime number theorem.⃞

Theorem: Pi π is ω-transcendental.

Proof: This follows from its Wallis product representation, or its product representation using the gamma function with value -½, provided that we accept these representations. It should be noted that these two representations yield distinct numbers. Alternatively, we can apply the greatest-prime criterion to the Leibniz series, or the Taylor series of arcsin(x) at x = 1.⃞

Theorem: The constants of Artin (CArtin), Baxter (C2), Chaitin (ΩF), Champernowne (C10), Copeland-Erdős (CCE), Erdős-Borwein (E), Feller-Tornier (CFT), Flajolet and Richmond (Q), Glaisher-Kinkelin (A), Heath-Brown-Moroz (CHBM), Landau-Ramanujan (K), Liouville (£Li), Murata (CM), Pell (PPell), Prouhet-Thue-Morse (τ), Sarnak (Csa) and Stephen (CS) as well as the Euler resp. Landau totient constant (ET resp. LT), the twin prime constant (C2) and the carefree constants (K1, K2 and K3) are ω-transcendental, since an existing (large) power of a small or very large prime cannot be removed from numerator or denominator by simplifying.⃞

Remark: The claim for CCE clearly also holds for every base from ωℕ*.

Theorem: The constants of Catalan (G), Gieseking (π ln β), Smarandache (S1) and Taniguchi (CT) are ω-transcendental because of the greatest-prime criterion.⃞

Theorem: The trigonometric and hyperbolic functions and their inverse functions, the digamma function ψ, the Lambert-W function, the Ein function, the (hyperbolic) sine integral S(h)i, Euler's Beta function B, and, for positive natural numbers s and u and natural numbers t the generalised error function Et, the hypergeometric functions 0Ft, the Fresnel integrals C and S and the Bessel function It and the the Bessel function of the first kind Jt, the Legendre function χs, the polygamma function ψs, the generalised Mittag-Leffler function Es,t, the Dirichlet series Σu f(u)/us for u ∈ ωℕ* with maximally finite rational |f(u)|, the prime zeta function P(s), the polylogarithm Lis, and the Lerch zeta-function Φ(q, s, r) always yield ω-transcendental values for rational arguments and maximal finite rational |q| and |r| at points where their Taylor series converge.

Proof: The claim follows from the greatest-prime criterion, the Dirichlet prime number theorem, and the Wallis product. For the digamma function, the claim follows from the proof of ω-transcendence of Euler's constant below.⃞

Theorem: The gamma function Γ(z) := m!mz/(z(z + 1) (z + m)), where m = ώώ² and z ∈ ωℂ \ -ωℕ, is ω-transcendental for z ∈ ωℚ and for suitable supersets of ωℕ resp. ωℚ.

Proof: The values of Γ(z) are zeros of minimal polynomials or series with infinite integer coefficients.⃞

Theorem: For x ∈ ωℝ and k ∈ ωℕ*, let s(x) := Σk xk/k. If we define Euler's constant as γ = s(1) - ln ώ ∈ ]0, 1[ (since dx/⌊x⌋ - dx/x ≥ 0 and dx/⌊x + 1⌋ - dx/x ≤ 0 for x ∈ ω≥1 in the integral representation of γ) and accept m s(½) - s(j/2m) as a representation of ln ώ with ώ = 2m - j, m ∈ ωℕ and j ∈ [0, 2m-1[ with a precision of O(m/2ώ), then γ is infinite rational and therefore ω-transcendental.

Proof: We obtain -ln(1 - x) = s(x) + O(xώ/(1 - x)) + t(x)dx for x ∈ [-1, 1 - 1/c] and a real function t(x) such that |t(x)| < ώ + 1 by (exact) integration (see Nonstandard Analysis) of the geometric series. The claim follows from the greatest-prime criterion after applying Fermat's little theorem to the denominator of the k-th summand of s, for the largest or second-largest prime in ωℕ, whose product is greater than m 2m-1 + 2m - j by the prime number theorem.⃞

Remark: At all (higher) precisions, the numerators of the summands contain higher powers of two (after cancelling with the denominator where applicable), so the theorem remains valid. This in particular includes t(x)dx at almost any arbitrary precision, by successively reducing γ to a fraction. If we set dx to be the reciprocal of the maximal power of two in this configuration, with an infinite natural exponent n for infinite rational t(x), the claim even holds exactly. In any case, the statement is valid for infinitely many levels of infinity of n, and in particular holds from the conventional perspective.

Definition: When two numbers x, y ∈ ωℂ* or their reciprocals do not satisfy any polynomial or series equation p(x, y) = 0, so they are called ω-algebraically independent.

Theorem: The greatest-prime criterion, with e = (1 + 1/p)p for maximal p ∈ ωℙ and π as Wallis product, yields pairwise ω-algebraically independent representations of A, C2, γ, e, K and π.⃞

Theorem: The BBP series Σk p(k)/(q(k)bk) for b, k ∈ ωℕ and polynomials resp. series p and q ∈ ωℤ with q(k) ≠ 0 and deg(p) < deg(q) only yield ω-transcendental values.

Proof: We can reduce the sum to a smallest common denominator d ≥ bm > ω with d, m ∈ ℕ*.

Gelfond-Schneider theorem: For each α ∈ ωA\{0, 1} and β ∈ ωA\ωℚ, αβ is ω-transcendental.

Proof: In order for αβ to be zero of a minimal polynomial or series, since B := ωA is a field, we would have to perform the invalid substitution α := ξηq/β with ξ, η ∈ B* and q ∈ ωℚ*.⃞

The above theorem and the property that B is a field immediately imply the following result.

Conclusion: As before, given any α, there does not exist any β such that e = αβ, which may be seen by comparing the minimal polynomials or series of e (see above) and αβ. If there is precisely one γ ∈ B* for any given arbitrary ω-transcendental τ ∈ ωℂ such that τγ is ω-algebraic, then every τδ such that δ ∈ B\γωℚ is ω-transcendental.⃞

Definition: A rational number ≠ 0 is said to be power-free if it cannot be represented as the power of a rational number with integer exponent ≠ ±1.

Theorem: For any power-free q ∈ Q := ℚ>0, we have that qx ∈ Q, if and only if x ∈ ωℤ and |x| is not excessively large.

Proof: Since there is no contradiction when ±x ∈ ωℕ is not excessively large, we can similarly assume wlog that x ∈ Q\ωℕ. Since this implies qxωAR\Q with R := ℝ>0, we can similarly assume that x ∈ ωAR\Q. Then qx is ω-transcendental by the Gelfond-Schneider theorem. We can therefore assume that x := a/b ∈ ωℝ\ωAR where a, b ∈ ℤ are ω-transcendental. This implies that qa/b ∈ Q\ωℕ* and thus a = db with d ∈ ℤ, since q is power-free. We deduce the contradiction that qd = qx which proves the claim, since x must be real.⃞

Remark: The above theorem proves the Alaoglu and Erdős conjecture, which states that, given distinct p, q ∈ cℙ, px and qx are c-rational if and only if x is a c-integer.

Remark: These arguments can be extended to finite transcendental numbers by replacing cℕ with ωℕ and making the required adjustments. Inconcrete transcendence implies finite transcendence.

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