Connections with Symmetry

We have already seen how the unique number sequences associated with the simple polynomial equation, 
(x – 1)ⁿ = 0, are intimately related in geometrical terms with perfectly symmetrical objects, based on each face representing an equilateral triangle.

So again, the unique numbers associated with (x – 1)³ = 0, are the triangular numbers 1, 3, 6, 10, 15, …

So for example, starting with one point (see Mathworld), when we then place a point at the vertices of a symmetrical i.e. equilateral triangle, we get 3 points in all.

Then when we place another point at the middle of each side (at an equal distance from the end points) we get 6 points in all. Then with additional points places at an equal distance from each other the total increases in accordance with the numbers in our unique number sequence for (x – 1)³ = 0.

And again the unique numbers associated with (x – 1)⁴ = 0 are the tetrahedral numbers 1, 4, 10, 20, 35, … (Note that the nth term of this sequence represents the sum of the first n terms of the previous sequence)!

So again starting with 1 point (see again Mathworld the successive terms in the series represent the equally placed points of the fully symmetrical 3-dimensional solid i.e. tetrahedron (as the number of equidistant points - initially with respect to each side of the object - increases.

Thus in the simplest case where we have one point at the vertex of each side of the tetrahedron (with 4 sides in all), we have 4 points.     

And we then extended this thinking more generally for n dimensions in space so that the unique number sequences associated with (x – 1)ⁿ = 0, represent the no. of equally spaced points for a “hyper” tetrahedron in (n – 1) space dimensions.

And such a simplex “hyper” tetrahedron (containing n sides) represents the least number of sides required to construct a fully symmetrical object in (n – 1) space dimensions.

Now, as we have seen, all these number sequences are intimately related to - what I refer to as - the Alt Zeta 2 function, which uses the reciprocals of these number sequences, which in turn complements the better known Zeta 1 (Riemann) function.

So again for example, the Zeta 1 (Riemann) function i.e. ζ(s), where s = 2 is

1 + 1/2² + 1/3² + 1/4² + ….     = 4/3 * 9/8 * 25/24 * …      = π²/6.

And the individual terms of the Zeta 1 can then be expressed in terms of infinite series based on representing the Alt Zeta 2 function.

For example the 1^st term in the product over primes expression above for ζ(2) is 4/3.

And when s = 2 with respect to the Zeta 1, n = 2² + 1 = 5 with respect to the Alt Zeta 2.

Therefore, the relevant Alt Zeta 2 expression represents the (infinite) sum of reciprocal terms based on the unique number sequences associated with (x – 1)⁵ = 0, i.e.

1 + 1/5 + 1/15 + 1/35 + 1/70 + …   = 4/3.

So then the next term in the product over primes expression for ζ(2)  i.e. 9/8 now corresponds to the infinite sum of reciprocals of the unique number sequences, where n = 3² + 1 = 10,  i.e. (x – 1)¹⁰ = 0,

i.e. 1 + 1/10 + 1/55 + 1/220 +  1/715 + …  = 9/8, and so on.

So again each (individual) term of the Zeta 1 (Riemann) function can be expressed in an ordered fashion, through the appropriate Alt Zeta 2 function. Though the Alt Zeta 2 function is more directly related to the product over primes expression of the Zeta 1 (Riemann) function (with s an integer > 1), it can equally be associated with the sum over natural numbers expression (through the subtraction of 1).

So for example the 2^nd term of ζ₁(2) in the sum over natural numbers expression = 1/4. This can in turn be related the Alt 2 expression (where n = s² + 2 = 6).

So the infinite series of reciprocals based on the unique number sequence associated with (x – 1)⁶ = 0, is

1 + 1/6 + 1/21 + 1/56 + 1/126 + …  = 5/4.

And when we subtract 1 we obtain 1/4, which is the required term in the Zeta 1 (Riemann) function.

However what is fascinating in this context is that the denominator numbers associated with the Zeta 1 (Riemann) function can themselves be directly related to fully symmetrical objects.

Thus in 2-dimensional space we have the square - as earlier the equilateral triangle - as another symmetrical object (with respect to angular rotation). (Again see Mathworld).

So again when s = 2, the denominator numbers associated with ζ₁(2) are 1, 4, 9, 16, 25, …

Starting with 1 point, we then move to the 2 dimensional symmetrical object of the square. So when we place a point at each vertex of the square we have 4 (i.e. 2²) points in all.

And then when we start by placing 3 equidistant points on each equidistant line of the square, the total number of points = 9 (i.e. 3²).

When we place 4 equidistant points on each equidistant line of the square, the total no. of points = 16 (i.e. 4²). 

And in general terms when we place s equidistant points on each equidistant line of the square, the total no. of points = s².

So what we have involved here in 2-dimensional space are the successive denominator terms of

ζ₁(2).

And then in 3-dimensional space for the symmetrical object of the cube, we use the successive denominator terms of ζ₁(3).

Thus starting with 1, the simplest case in 3 dimensions entails placing a single point at each vertex of the cube. And the total no. of points thereby involved = 8 (i.e. 2³)..

And then when we place 3 equidistant points on each side and line through the cube, the total no. of points = 27 (i.e. 3³).

And in general terms when we place s equidistant points on each equidistant line (outside and inside the cube) the total no. of points = s³.

Though impossible to properly visualise, we can then extend this into 4 space dimensions to obtain a hypercube (i.e. tesseract) where we now use the successive denominator terms of ζ₁(4). So when we now place a point at each vertex (i.e. 2 points at the end of each line) of the tesseract, we obtain 16 (i.e. 2⁴) in all.

And with 3 equidistant points on each line (outside and inside the tesseract) we obtain 81 (i.e. 3⁴) in all.

And in general terms for s equidistant points we obtain s⁴ in all.

And in even more general terms for s equidistant points with respect to each line of the n-dimensional hyper cube in space, we obtain sⁿ points in all. 

There is also an intimate connection here with the Zeta 2 function i.e. ζ₂(s).

So for example where 2 = 1/2, the denominator terms i.e. 1, 4, 8 represent the simplest case of just one point at each vertex, where starting with 1 point, the total no. of points involved as one moves from the 2-dimensional (4) to the 3-dimensional cube (8) to the 4-dimensional tesseract (16) and so on into progressively higher dimensions.