Further Interesting Connections

As we have seen the triangular nos. i.e. 0, 1, 3, 6, 10, 15, ... represent the unique digit sequence associated with the simple polynomial equation (x – 1)³ = 0.

And this number sequence is intimately related to the corresponding number sequence associated with the Zeta 1 (Riemann) function where s = – 2, i.e. ζ₁( – 2), i.e. 1, 4, 9, 16, 25, ...

Then when we look at the numbers in this later sequence, we can see that they are related to the  corresponding numbers of the former sequence in the following manner

1 = 0 + 1; 4 = 1 + 3; 9 = 3 + 6; 16 = 6 + 10; 25 = 10 + 15 and so on.

So in more general terms, t_{k – 1} + t_k (with respect to the former infinite sequence) = t_k (with respect to the latter).

With respect to the former sequence, 

1 – 0   = 1  and  1 + 0  = 1 (i.e. 1²);

3 – 1    = 2  and  3 + 1   = 4 (i.e. 2²);

6 – 3   = 3  and   6 + 3 =   9 (i.e. 3²);

10 – 6 = 4  and 10 + 6 = 16 (i.e. 4²);

So in general terms the t^th – (t – 1)^th term = t;   and t^th + (t – 1)^th term = t²

 
Again with respect to the former sequence, 

0 + 1   = 1    and 0² + 1²        = 1       (i.e. the 1^st term);  

1 + 3    = 4    and 1² + 3²        = 10     (i.e. the 4^th term);

3 + 6   = 9    and 3² + 6²       = 45     (i.e. the 9^th term);

6 + 10 = 16  and 6² + 10²   = 136    (i.e. the 16^th term);

Thus in general terms (t_{k – 1})² + (t_k ²) = the (t_{k – 1} + t_k)^th  term of the infinite sequence.

In like manner, with respect to the same sequence,

1 – 0    = 1 and  1² –  0 ²     = 1 (i.e. 1³)
 

3 – 1    = 2 and 3² –  1²     = 8 (i.e. 2³)

6 – 3    = 3 and 6² –  3²     = 27 (i.e. 3³)

10 – 6 = 4 and 10² –  4²    = 64 (i.e. 4³)

So once more in general terms,   

(t_k)² – (t_{k – 1})² = (t_k – t_{k – 1})³ for the infinite sequence.


With respect to the sequence  1, 3, 6, 10, 15, 21, 28, 36, 45, 55,  ..., the 1st term of each successive group of 3 leaves a remainder of 1 when divided by 9

So, 1 divided by 9  = 0 (+ remainder of 1);

     10 divided by 9 = 1(+ remainder of 1);

     28 divided by 9 = 3 (+ remainder of 1);

     55 divided by 9 = 6 (+ remainder  of 1); 


Thus continuing on, when we ignore the reminder of 1, we obtain the same series 0, 1, 3, 6,...

We have already noted the remarkable fact that with respect to the Zeta 1 (Riemann) function,

 _∞

∑{ζ₁(s) – 1} = .64493... + .20205... + .08232.. + .03692... + ...
s = 2

= 1/2 + 1/6 + 1/12 + 1/20 + …  = 1/2( 1 + 1/3 + 1/6 + 1/10 + …)  = 1.

And once again 1 + 1/3 + 1/6 + 1/10 + ... represents the sum of reciprocals of the unique number sequence associated with (x – 1)³ = 0.

Then 1/2 (1 + 1/3 + 1/6 + 1/10 + ...) +
         
         1/6 (1 + 1/4 + 1/10 + 1/20 + ...) +

        1/12 (1 + 1/5 + 1/15 + 1/35 + ... ) +

       1/20 (1 + 1/6 + 1/21 + 1/42 + ...) +
                          ...

  = 1/2(2/1) + 1/6(3/2) + 1/12(4/3) + 1/20(5/4) + ...                     ...

= 1 + 1/4 + 1/9 + 1/16 + ...

In other words by combining each successive term of 1/2 + 1/6 + 1/12 + 1/20 + ...
with the sum of reciprocals of the unique number sequences associated with (x – 1)ⁿ = 0, where n = 3, 4, 5, 6, ..., we obtain the Zeta 1(Riemann) function ζ₁(s), where s = – 2.

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.