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Transcendental Numbers

Transcendental Numbers

In the following, the notations from the set theory are applied. The behaviour for inconcrete m, n ∈ ωℕ is considered and let be h, k ∈ ℕ.

Remark: Algebra teaches that the sum, difference, product and quotient of two algebraic numbers of degree h and k are algebraic of maximally degree hk, and the 1/h-th power of a algebraic number of degree h is algebraic of maximally degree hk.

Remark: Transcendental numbers consist of an algebraic main part and a transcendental remainder. If we examine the transcendence of a number and if the remainder is given by the limit value of a zero sequence (ak) (see nonstandardanalysis), we cannot thus simply omit the sequence values for great k: They are decisive. Transcendental numbers are all numbers that lie between the algebraic ones or beyond them. According to the number theorem of algebraic numbers, there is no algebraic number (of degree < h) between two different transcendental numbers (algebraic numbers of degree h), lying closely enough together.

Bound proposition for ω-transcendental numbers: Every complex number different from zero, whose absolute value of the real and/or imaginary part is ≤ 1/⌈ω⌉ or ≥ ⌈ω⌉, is already ω-transcendental.

Proof: If we set, in a polynomial equation, am = 1 and ak = -⌊ω⌋ for k < m, then the assertion follows in the real case from the geometric series, if we still build the reciprocal. We receive the exact limit values, if we replace ⌈ω⌉ in both cases by ω(m) = ⌈ω⌉ - ⌊ω⌋/ω(m)m with ω := ⌈ω⌉ - ⌊ω⌋/ω⌊ω⌋. In the complex case, statements of the form x = (1+bi)⌈ω⌉ with b ∈ ωℝ yield what we desire.⃞

Coefficient proposition for ω-transcendental numbers: All normalised irreducible polynomials where we have |ak| ≥ ⌈ω⌉ for at least one ak have only ω-transcendental zeros.

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

Product and sum proposition for ω-transcendental numbers: All products that cannot be further simplified of an infinite complex-rational number a and a complex algebraic or infinite algebraic number b (here also the sum) as well as sums and products of an infinite complex algebraic and a complex ω-algebraic (infinite complex algebraic) number with coprime polynomial degree are ω-transcendental.

Proof: Participating an infinite complex-rational number leads always to at least one coefficient modulus ≥ ⌈ω⌉ in the result of the corresponding minimal polynomial, demanding coprimeness does so always to a at least infinite algebraic number in the result and thus both to the assertion.⃞

Remark: The simplification is unique, if it also includes minimising the absolute value of the coefficients of the minimal polynomial of b and of each numerator and denominator of the infinite complex-rational number a. Together with the rational numbers, the infinite (complex-)rational numbers build numerically already complete ℝ (ℂ). Therefore, algebraic and transcendental numbers are (numerically) hardly to distinguish and approximations are little meaningful concerning the algebraicity (of a certain degree). This makes distinguishing them questionable. Real continued fractions that do not terminate as a rational number are ω-transcendental, since they are infinitely rational. They can only approximate ω-algebraic numbers.

Remark: For ω-transcendental numbers we are, unlike ω-algebraic numbers, where we should argue with the corresponding minimal polynomial, content with an arbitrarily precise infinitely small precision and can thus fix them as rational quotient with infinite numerator and denominator. Who wants to examine the transcendence of a number with Liouville's approximation theorem should ensure that the approximating rational numbers are actually still conventionally rational and not infinitely rational, such as the number 10⌈ω⌉. Also, who considers prime numbers ≥ ⌈ω⌉ in transcendence proofs should safely wander about the infinitely natural. The following proofs yield also ω-transcendental results, if the set ωℕ is replaced by [1, n]ℕ≥ ⌈ω⌉ with n ≪ |ℕ|.

Theorem: The sum of all at least conventionally natural powers of a complex-rational number x that is different from zero and no root of unity, (geometric series) is already ω-transcendental.

Proof: The absolute value of the numerator or denominator of x⌈ω⌉ is ≥ 2⌈ω⌉/2. The subtraction of 1 and the division by 1 - x do not change anything of the ω-transcendence.⃞

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

Proof: If we accept the exponential series as representation for e, it follows e = (k⌊ω⌋+1)/⌊ω⌋! with k ≥ ⌈ω⌉. Therewith, numerator and denominator of the stated fraction must be ≥ ⌈ω⌉, since in the numerator neither ⌊ω⌋ nor a prime divisor of k can be cancelled against ⌊ω⌋!. If we accept, on the other hand, (1+1/⌊ω⌋)⌊ω⌋ as representation for e, so the assertion is trivial. Notice that here two different numbers are present.⃞

Theorem: The twin prime constant C2, as product over all (1 - 1/(p - 1)2) for primes p ∈ ℙ>2, is ω-transcendental.

Proof: According to the prime number theorem, the greatest and second greatest prime number of ℕ - both = ⌈ω⌉ - |O(ln ⌈ω⌉)| - do not divide the denominator of C2.⃞

Theorem: The Landau-Ramanujan constant K, as product, divided by √2, over all (1 - 1/p2) with p ∈ ωℙ and p ≡ 3 mod 4, is ω-transcendental.

Proof: According to the (Dirichlet) prime number theorem, the greatest and second greatest prime number ≡ 3 mod 4 of ωℙ - both = ⌈ω⌉ - |O(ln ⌈ω⌉)| - do not divide the denominator of K.⃞

Theorem: The Glaisher-Kinkelin constant A = 11 22 33 ... ⌊ω⌋⌊ω⌋ (1+1/⌊ω⌋)⌊ω⌋³/4/⌊ω⌋(⌊ω⌋²/2+⌊ω⌋/2+1/12) is ω-transcendental.

Proof: After cancelling, the greatest prime number of ωℕ remains in the denominator with exponent > 2.⃞

Greatest-prime criterion for ω-transcendental numbers: If a real number r has, for cancelled fractions, the representation a/(bp) ± s/t with natural a, b, s and t, abst ≠ 0 and b + t > 2, as well as the (second) greatest natural prime number p ∈ ωℙ, p ∤ a and p ∤ t, so it is ω-transcendental.

Proof: It applies r = (at ± bps)/(bpt) with a denominator ≥ 2p ≥ 2⌈ω⌉ - |O(ln ⌈ω⌉)| > ⌈ω⌉ as consequence of the prime number theorem.⃞

Theorem: The circular constant π is ω-transcendental.

Proof: This follows from its representation as Wallis' product or the product representation of the gamma function for the value -½, provided we accept them. Notice that both representations specify different numbers. Alternatively, we apply the greatest-prime criterion to the Leibniz series or to the arcsin(x) Taylor series for x = 1.⃞

Theorem: The trigonometric and hyperbolic functions together with their inverse functions, the digamma function ψ, Lambert's W function, the function Ein, the (hyperbolic) sine integral S(h)i, the Euler beta function B and with natural positive s and u and natural t the error functions Et, the hypergeometric functions 0Ft, the Fresnel integrals C and S and the Bessel function It resp. the one of the first kind Jt, the Legendre function χs, the polygamma function ψs, the generalised Mittag-Leffler function Es,t, the Dirichlet series Σ f(u)/us with maximally finite rational |f(u)|, the prime zeta function P(s), the polylogarithm Lis as well as the Lerch zeta-function Φ(q, s, r) yield for rational arguments and maximally finite rational |q| and |r|, for which the corresponding Taylor series converges, only ω-transcendental values.

Proof: Applying the greatest-prime criterion yields with Dirichlet's theorem and the Wallis' product the assertion, which follows for the digamma function from the proof of the transcendence of Euler's constant further below.⃞

Theorem: The gamma function Γ(z) := m!mz/(z(z + 1) (z + m)) is, with m = ⌈ω⌉⌈ω⌉² and z ∈ ωℂ \ -ωℕ for z ∈ ωℚ ω-transcendental, and for in each case suitable supersets of ωℕ resp. ωℚ.

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

Theorem: Let be s(x) for x ∈ ωℝ the sum of xk/k over all k ∈ ωℕ*. If we define Euler's constant as γ = s(1) - ln ⌊ω⌋ ∈ ]0, 1[ (because of dx/⌊x⌋ - dx/x ≥ 0 as well as dx/⌊1+x⌋ - dx/x ≤ 0 for x ∈ ω≥1 in the integral representation of γ) and accept m s(½) - s(r/2m) as representation of ln ⌊ω⌋ with ⌊ω⌋ = 2m - r, m ∈ ωℕ and 2m-1 > r ∈ ℕ for a precision of O(m/2⌈ω⌉), so it is infinitely rational and therewith ω-transcendental.

Proof: We obtain -ln(1 - x) = s(x) + O(x⌈ω⌉/(1 - x)) + t(x)dx for x ∈ [-1, 1 - 1/κ] and a real function t(x) with |t(x)| < ⌈ω⌉ by (exact) integration (see nonstandardanalysis) from the geometric series. The assertion follows by applying Fermat's little theorem from the greatest-prime criterion, in each concerning denominator of the k-th addend of s, for the greatest or second greatest prime of ωℕ, whose product is greater than m 2m-1 + 2m - r, due to the prime number theorem.⃞

Remark: Since all (higher) precisions contain higher powers of two in the denominator of their addends, after contingently cancelling against its corresponding numerator, the theorem is also true here, thus especially for nearly arbitrary precision t(x)dx, if we successively summarise γ to a fraction. If we set dx as reciprocal of a power of two with infinitely natural exponent maximally possible in this constellation, for infinitely rational t(x), the assertion applies even exactly. What is sure is that the proposition is valid for infinitely many infinity levels of n, and especially according to conventional opinion.

Approximation proposition for real algebraic numbers: Every irrational algebraic number of degree k can be approximated by a real algebraic number of degree h < k with an average error that is asymptotically equal to π ζ(h+1)/(2 |ωℤ|h ln h).

Proof: On the conventionally real axis, the number of the ω-algebraic numbers increases, within unchanged limits almost evenly distributed, per degree more, circa by the factor |ωℤ|. The error corresponds to the distance of the ω-algebraic numbers among each other. The non-real ω-algebraic numbers lie less dense.⃞

Conclusion: Two different real ω-algebraic numbers have at least the average distance π/(|ωℤ|⌊ω⌋ ln ⌈ω⌉). The precise determination of the minimal distance requires solving an infinite non-linear non-convex optimisation problem. Therefore real κ-algebraic numbers have an approximation order of O(κ), which refutes Roth's theorem, where also is not more proved than the (trivial) minimal distance of two rational numbers, and therewith the abc conjecture, but does not affect the result by Liouville.

Theorem: The maximal distance of two real algebraic numbers ≠ 0 amounts 1/⌊ω⌋2 - |O(1/⌊ω⌋3)|.

Proof: The real ω-algebraic numbers ≠ 0 are farthest apart around ±1. A conventionally rational number r > 1 (0 < r < 1) can be better approximated by a real ω-algebraic x that satisfies the polynomial equation ⌊ω⌋/r xm - ⌊ω⌋xm-1 = 1 (xm - ⌊ω⌋x = -⌊ω⌋r). For negative r applies similar. If we want to approximate 1 by a greater real algebraic x, it must coercively satisfy a polynomial equation with am = ⌊ω⌋ - 1 and a0 = -⌊ω⌋. If we set still am = ⌊ω⌋ - 1 and a0 = -⌊ω⌋ follows the assertion, since the maximal distance cannot be further diminished.⃞

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

Proof: In order that αβ satisfies a minimal polynomial, it would be required, since B := ωA is a field, exactly the unfeasible substitution α := ξηq/β with ξ, η ∈ B* and q ∈ ωℚ*.⃞

From the preceding theorem and the field property of B, we get directly the

Conclusion: As before, there is no β for α with e = αβ, which results from comparing the minimal polynomials of e (see above) and αβ. Iff there is in each case one γ ∈ B* for every arbitrary ω-transcendental τ ∈ ωℂ with ω-algebraic τγ , all τδ with δ ∈ B\γωℚ are ω-transcendental.⃞

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

Theorem: For all power free q ∈ Q := ℚ>0, we have r := qx ∈ Q, iff x ∈ ωℤ and |x| is not too big.

Proof: Since we have for not too big ±x ∈ ωℕ no contradiction, we assume equally w. l. o. g. x ∈ Q\ωℕ*. Because we would have then qxωAR\Q with R := ℝ>0, we can equally assume x ∈ ωAR\Q. Then qx would be ω-transcendental according to the Gelfond-Schneider theorem. Thus we can equally assume x := a/b ∈ ωℝ\ωAR with a, b ∈ ℤ as ω-transcendental. Hence we would have qa/b ∈ Q\ωℕ* and thus a = cb with c ∈ ℤ, since q is power free. This leads to the contradiction qc = qx and hence to the assertion, since x can only be real. ⃞

Remark: The considerations above can be analogously transferred to concretely transcendental numbers, if κℕ instead of ωℕ resp. all further necessary adaptations from this are applied. Inconcrete transcendence implies finite one.

Remark: The preceding theorem proves the conjecture by Alaoglu and Erdős, according to which px and qx are κ-rational for different p, q ∈ κℙ, iff x is κ-integer.

Remark: The conventional differentiation and integration obliterates, by building the conventional limit, the precise distinction of transcendence and algebraicity. This is problematic, e.g. for the exact determination of zeros. Therefore, the nonstandardanalysis on this homepage goes another (more precise) way.

© 2009-2013 by Boris Haase


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