User talk:SHAWWPG19410425

λ== In the page on spherical trigonometry, crystallography is an important area of application. The Supplemental [Dual] Triangle enables the inverse solutions to be found in the cosine formula.SHAWWPG19410425 (talk) 18:00, 21 June 2011 (UTC) ==

Spherical trigonometry finds an important application in crystallography which is the study of the geometrical properties of the solid state of matter where the individual atoms and molecules are regularly distributed in a three dimensional Euclidean space.

Apparently, the topic of supplemental triangles has not been covered in the Spherical Trigonometry page. This is of utility in problems where the three angles are the information given and the three sides are to be computed.

The Supplemental Triangle [the Triangle of Supplements or Supplementary Triangle] Given the vertices [A, B, C] and the sides [a, b, c] of a spherical triangle, the straight lines from the spherical center through the vertices [A, B, C] may be taken as the normals to three planes that all intersect at the center. These planes form the supplemental triangle. It is readily shown using elementary geometry that the intersections of the three planes that constitute the supplemental triangle are the normals to the planes making the original spherical triangle. [The theorem applies that a straight line perpendicular to two other straight lines is therefore the normal to the plane containing the other two straight lines. Another theorem is useful, that the angle between the normals to two planes is the supplement of the angle between the planes.]

Calling the angles and the sides of the supplemental triangle respectively [A', B', C'] and [a', b', c'] then it is readily shown by consideration of a diagram or constructed model that:

A + a' = 180 degrees, B + b' = 180 degrees and C + c' = 180 degrees

and:    A'+ a  = 180 degrees, B' + b = 180 degrees and C' + c = 180 degrees

or using other consistent units - radians or grades for the angles and the sides, which are also angles.

The triangle and the corresponding supplemental triangle are reciprocally dual triangles. This means that if the triangle A'B'C' is by construction the supplemental triangle for the triangle ABC then consequently the triangle ABC is the supplemental triangle for triangle A'B'C'.

Using the supplemental/supplementary/dual triangle and the cosine formula, the class of problem cited above becomes solvable. SHAWWPG19410425 (talk) 18:00, 21 June 2011 (UTC)SHAWWPG19410425 (talk) 11:46, 22 June 2011 (UTC)SHAWWPG19410425 (talk) 23:15, 23 June 2011 (UTC)SHAWWPG19410425 (talk) 17:39, 24 June 2011 (UTC)SHAWWPG19410425 (talk) 19:17, 30 June 2011 (UTC)

The Riemann Zeta Function and Functional Equations.
The Riemann Zeta Function, the associated functional equation and ramifications have been much quoted along with the as yet not proved or disproved Riemann Hypothesis.

My first question is: Have the functional equations of the following functions been derived and their zeros computed and plotted?

[1] Sigma 1/[2^{k.s} = 1 + 1/2^s + 1/4^s + ... [with integer k running from zero to infinity by unit steps]

[2] Sigma 1/[p^{k.s} = 1 + 1/p^s + 1/p^{2.s} + ... [with integer k as above and p an uneven prime number taking

a value selected from the sequence: [3, 5, 7, 11, 13 ... ] in turn. The second function [2] really represents an infinitude of separate elementary functions, each involving just one uneven prime number.

My second question is: If the functional equations of the above functions [1] and [2] have been produced, how do these relate to the functional equations of the product[s] of the above functions ?

My third question is: How do the computed zeros of the above functions [1] and [2] relate to the zeros of the product[s] of the above functions ?

My fourth question, assuming that the previous questions have been answered affirmatively, is: How do the zeros of the product[s] of the above functions relate to the Riemann zeros, as the number of factors in the product function increases indefinitely [a] where the function [1] involving two is not included in the product and [b] where the function [1] involving two is included in the product ?

The prime two has not been merged with the class of uneven prime numbers.

My comment is: The answers to the questions above could provide valuable insights into the Riemann Hypothesis.SHAWWPG19410425 (talk) 12:53, 23 November 2011 (UTC)

DUAL MATRICES OF BINARY REMAINDERS UNIQUELY REPRESENTING THE UNEVEN INTEGERS Q[N] ARE READILY PROVED TO BE UNIVERSALLY NON-SINGULAR. [The letter Q denotes a quotient, that is an uneven integer, which is produced by division of an uneven integer, dividend, by two [2]. The remainder can only be either + 1 or - 1 and never zero with this kind of division.] These matrices were comprehensively derived in a WIKIVERSITY contribution, where it is found that the algebraic sum of the elements in a particular row [j] of the dual matrix D[N, n] of binary remainders is equal to the value of Q[N + 1 - j]. The integer [j] runs consecutively through the values: [1, 2, ... [N + 1]]. Matrices that uniquely represent uneven integers have associated dual matrices D of binary remainders, where the sign of two raised to some positive integral exponent is an appropriate binary remainder, r and goes to form an element of the matrix where either r = + 1 or r = - 1. A typical element of the matrix DN + 1], [N + 1 of binary remainders is: d[2, 4] = r[1].2^{N - 3} There are binary remainders: r[1], ... r[N - 1] multiplying various powers of two that represent Q[N], where Q[N] may be defined as Q[N] = 2.n - 1, so that the succession of uneven integers: 1, 3, 5 ... correspond to the ordinal numbers denoted by the integer n: 1, 2, 3 ... The successive rows of the matrices of binary remainders are the uneven integers Q[N], Q[N - 1], ..., Q[N - j], ... Q[1], Q[0} Clearly there are [N + 1] rows. There are also [N + 1] columns because the equation representing a general Q[j] in terms of binary remainders contains [N + 1] terms.

Generally, Q[N - [j]] = 1/2[Q[N - [j - 1]] - i^{1 + Q[N - [j - 1]}], where [i] satisfies i^{2} = - 1. Q[N] is a positive uneven integer in the interval: 2^{N} < Q[N] < 2^{N + 1}

The determinant of the dual matrix of remainders has [N + 1] rows and [N + 1] columns, by definition. To prove that the determinant is always non-zero, the following operations are carried out on the rows. [1]Row[1] is left unchanged. [2]Row[N] is multiplied by - 1 and added to row [N + 1] [3]Row[N - 1] is multiplied by - 1 and added to row[N] [j]Row[N - j] is multiplied by - 1 and added to row[N - j + 1] [N - 1]Row[1] is multiplied by - 1] and added to row[2] After the above has been done, the determinant entries [a] below the principal diagonal are all zero [b] on the principal diagonal are respectively, starting from the lower right corner: - 2^{1}, - 2^{2}, ... - 2^{N}, + 2^{N} and the entries above the principal diagonal are mixed, some being zero and others non-zero. The value of the determinant of the binary matrix of remainders is given by the product of the terms [all non-zero] on the principal diagonal. The matrix of binary remainders is therefore invertible or equivalently non-singular. The value of the determinant of the dual matrix of remainders is given by: |D| = [-1]^{N}.2^{[1/2].N.[N + 3]}

THE EIGENVALUE EQUATION. Placing the unknown λ into the primitive first found determinant of D and following the steps in the poof of non-singularity produces a much simpler form of the eigenvalue determinant. [a] above the principal diagonal there are various terms both zero and non-zero [b] on the principal diagonal there are terms in λ and constants [c] adjacent and immediately below the principal diagonal is a diagonal consisting entirely of λ's [d] the remaining lower left triangular area consists entirely of zeros.

In relatively elementary mathematics, as above, the assertions are verifiable by working through a typical example and generalising. SHAWWPG19410425 (talk) 20:00, 22 May 2012 (UTC)SHAWWPG19410425 (talk) 08:35, 24 May 2012 (UTC)SHAWWPG19410425 (talk) 11:37, 19 June 2012 (UTC)