(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 + 1/2

^{2}+ 1/3^{2 }+ 1/4^{2}+ …. = 4/3 * 9/8 * 25/24 * … = π^{2}/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 = 2

^{2 }+ 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 = 3

^{2 }+ 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 ζ_{1}(2) in the sum over natural numbers expression = 1/4. This can in turn be related the Alt 2 expression (where n = s^{2 }+ 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 ζ

_{1}(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. 2

^{2}) 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. 3

^{2}).
When we place 4 equidistant points on each equidistant line
of the square, the total no. of points = 16 (i.e. 4

^{2}).
And in general terms when we place s equidistant points on
each equidistant line of the square, the total no. of points = s

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

ζ

_{1}(2).
And then in 3-dimensional space for the symmetrical object
of the cube, we use the successive denominator terms of ζ

_{1}(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. 2

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

^{3}).
And in general terms when we place s equidistant points on each
equidistant line (outside and inside the cube) the total no. of points = s

^{3}.
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 ζ

_{1}(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. 2^{4}) in all.
And with 3 equidistant points on each line (outside and
inside the tesseract) we obtain 81 (i.e. 3

^{4}) in all.
And in general terms for s equidistant points we obtain s

^{4 }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 s

^{n}points in all.
There is also an intimate connection here with the Zeta 2
function i.e. ζ

_{2}(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.

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