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.

Symmetrical Geometrical Objects

I have been repeatedly referring to the (infinite) reciprocal sums of the unique number sequences associated with (x – k)ⁿ = 0 and especially (x – 1)ⁿ = 0.

And I refer to these reciprocal sums associated with (x – 1)ⁿ = 0 as the Alternative Zeta 2 function i.e. Alt ζ₂(n).

However it is possible now to provide a more general formula.

This is based on the fact that the nth term of these unique number sequences follow a definite pattern.

Then, for (x – 1)², the unique number sequence is 1, 2, 3, 4 with the kth = k and the

                            _{∞                                                                  ∞}

sum of  terms  = ∑ k/1! with (infinite) sum of reciprocals = ∑1!/k.    

                           ^{k=1                                                                n=1} 

For (x – 1)³, the unique number sequence is 1, 3, 6, 10, … with the kth term = k(k + 1)/2!

                                   _{∞                                                                       ∞}

And sum of terms  = ∑k(k + 1)/2! with sum of reciprocals = ∑ 2!/{k(k + 1)}.

                                  ^{k=1                                                                     k=1} 

Then, giving one more example to illustrate a consistent pattern, for (x – 1)⁴, the unique number sequence is 1, 4, 10, 20, …, with kth term = k(k + 1)(k + 2)/3!   

                                                             _∞

So the (infinite) sum of reciprocals = ∑ 3!/{k(k + 1)(k + 2)}.

                                                            ^k=1

Therefore for the general case (x – 1)ⁿ = 0, the (infinite) sum of reciprocals of the associated unique number sequences of this simple polynomial equation i.e. Alt ζ₂(n), is given as

_∞

∑ (k – 1)!/{k(k + 1)(k + 2) … (k + n – 2)}.

^k=1

Note, however that the Alt ζ₂(n) function remains undefined for n = 1!

A fascinating geometrical way of looking at the unique number sequences given by

(x – 1)ⁿ = 0, is through number patterns that are associated with the simplest perfectly symmetrical objects that can be constructed in the various dimensions.

For example the simplest symmetrical polygon figure (with least number of sides) that can be constructed in 2-dimensional space is a 3-sided equilateral triangle.

The diagram in Mathworld then illustrates how the triangular numbers - which correspond to the unique number sequences associated with (x – 1)³ = 0 - then arise.

So we start with 1 dot and with the construction of an equilateral triangle we then place a dot at each vertex = 3. Then by placing a dot midway along each line, the total no. of points = 6.

And then with 4 dots equally spaced out along each line (with another placed within the triangle), when the total no. = 10 (overall) we can preserve the perfect symmetry of equal distance between adjacent dots. 

Then in the final example we place 5 dots equally spaced on each side of the triangle with 3 more internally = 15 (overall). 

What is also interesting here is that we can join up the dots in each triangle to form a number of smaller triangles that represents the previous number in the series.

Thus in the simplest case in the 3 dot case we can join the points to form 1 triangle (which represents the previous number in the sequence). Then in the case of the 6 dot triangle we can join the dots to form 3 upright triangles (with same orientation to the 1 larger triangle contained). Then in the case of the 10 dot triangle we can form 6 upright triangles and in the 15 dot triangle 10 upright triangles of same orientation to larger triangle and so on.

Then the simplest 3-dimensional figure that is perfectly symmetrical with respect to its angular rotations is the tetrahedron which has the equilateral triangle as its base.

Though harder to represent (in 2-dimensional) terms, it is easy enough to see how the number of symmetrically arranged points (associated with this 4-sided figure follow the unique number sequence associated with (x – 1)⁴ = 0, i.e. 1, 4, 10, 20, …

So once again we start with 1 dot represented by the red dot at the top of the diagram (from Mathworld). Then to picture the simplest symmetrical tetrahedral figure we combine the 3 blue dots (representing the base of a tetrahedron with the red dot above) to obtain 4.

Then at the next level where each line of the base of the tetrahedron is divided in 2 we have now 6 green dots which combine the 3 blue and 1 red above to complete the new tetrahedron with 10 dots. Then with the next larger tetrahedron, we have 10 brown points at the base combined with the 6, 3 and 1 respectively above to complete the tetrahedron with 20 points.

So we can easily see here how this sequence of dots, representing the fully symmetrical nature of the 3-dimensional tetrahedron (with the least number of sides i.e. 4, possible in 3-dimensional space) is the unique digit sequence for (x – 1)⁴ = 0, i.e. 1, 4, 10, 20, 35, … 

However what is perhaps remarkable is that these insights can then be extended for the simplest fully symmetrical objects in n-dimensional space.

Thus - though impossible to properly visualise in 3 or less dimensions - we can equally envisage for example a 4-dimensional polytope, as a 5-sided equivalent in 4 space dimensions to the 4-sided tetrahedron in 3 space. This is known as a 5-cell or tetrahedral pyramid. (See Wikipedia entry).

Thus once again this contains the least number of sides i.e. 5, that a fully symmetrical object can possess in 4 space dimensions.  

So what we can now say therefore is that the unique number sequence associated with

(x – 1)⁵ = 0, i.e. 1, 5, 15, 35, 70, …  now describes the appropriate number of equally spaced points with respect to this object (as the number of points on each side progressively increases). 

Thus in general terms, the unique sequence of numbers associated with the polynomial equation
(x – 1)ⁿ = 0, encodes the manner, in which the equally spaced points of the simplex n-sided fully symmetrical geometrical objects, occurs in (n – 1) space dimensions.

Interesting Relationships Continued (2)

In the last entry, I considered the (infinite) reciprocal sum of the unique numbers associated with the polynomial equation xⁿ – kⁿ = 0, (where n = 2 and k prime) to find that the respective sums generate the successive terms of the product over primes expression for the (Riemann) Zeta 1 function i.e.
ζ₁(2), where s = 2, i.e.

4/3, 9/8, 25/24, …

And then when n = 3, the successive terms of ζ₁(3), where s = 3, are generated i.e.

8/7, 27/26, 125/124, …

However, when we now consider the respective reciprocal sums for xⁿ + kⁿ = 0 (where n = 2 and k prime) we generate the following terms,

4/5, 9/10, 25/26,…

And 4/3 * 9/8 * 25/24 * …   = π²/6, and

4/5 * 9/10 * 25/26 * …  = π²/15

Of course this means that multiplying the respective terms in each series,

16/15 * 81/80 * 625/624 * …  = π⁴/90 = ζ₁(4)

And in general, where k ranges over all the primes

Then (2ⁿ/2ⁿ – 1) * (3ⁿ/3ⁿ – 1) * (5ⁿ/5ⁿ – 1) * … = ζ₁(n),  when multiplied by

(2ⁿ/2ⁿ + 1) * (3ⁿ/3ⁿ + 1) * (5ⁿ/5ⁿ + 1) * … = ζ₁(2n).

However there is an easier of showing the true nature of the simple equation (x – k)ⁿ = 0 = 0

Let us initially - in relation to the this equation - consider the simple case where k = 1, 2, 3, … and n = 1

As we have already seen unique numbers associated with (x – 1)¹ = 0 are,

1, 1, 1, …

And of course the sum of reciprocals of these numbers diverges to infinity.

Then when (x – 2)¹ = 0, the unique numbers thereby associated are,

1, 2, 4, …

And the (infinite) sum of reciprocals of these numbers

= 1 + 1/2 + 1/4 + …    = 2. (i.e. 2/1)

And when (x – 3)¹ = 0,  the unique numbers associated are

1, 3, 9, … and the corresponding (infinite) sum of reciprocals =

1 + 1/3 + 1/9 + … = 3/2.

Thus in general terms, the (infinite) sum of reciprocals of the unique numbers associated with (x – k)¹ = 0, = k/k – 1).

Let us now again in relation to the alternative equation (x – k)ⁿ = 0, consider the reverse complementary situation where k = 1 and n = 1, 2, 3, …

Thus again we start with (x – 1)¹ = 0.

And as we have seen the infinite reciprocal sum of unique numbers associated with this equation = 1 + 1 + 1 + … (which diverges to infinity).

Then when k = 1 and n = 2 we obtain (x – 1)² = 0.

And the unique number sequence - as seen in previous entries - of this equation is the set of natural numbers 1, 2, 3, …

And the (infinite) sum of reciprocals of these numbers (i.e. the harmonic series = 1 + 1/2 + 1/3 + … (which diverges to infinity).

Then when k = 1 and n = 3 we obtain (x – 1)³ = 0.

And again as we have seen - in previous entries - the unique numbers associated with this equation are 1, 3, 6, …

And the (infinite) sum of reciprocals of these numbers = 1 + 1/3 + 1/6 + … = 2 (i.e. 2/1).

And taking one more example, when k = 1 and n = 4, we obtain (x – 1)⁴ = 0 with the unique numbers associated 1, 4, 10, …

And the (infinite) sum of reciprocals of these numbers = 1 + 1/4 + 1/10 + … = 3/2.

So in general, the (infinite) sum of reciprocals = (n – 1)/(n – 2).

Now I have referred before to the first formulation i.e. the (infinite) sum of reciprocals of unique number sequences associated with (x – k)¹ = 0 i.e. k/(k – 1) as the Zeta 2 function i.e. ζ₂(1/k).

Thus when k = 2, 1/k = 1/2;

So ζ₂(1/k) = 1 + (1/k)¹ + (1/k)² + (1/k)³ + …  = 1 + (1/2)¹ + (1/2)² + (1/2)³ + …  = 2.                

And the alternative formulation, i.e. the (infinite) sum of reciprocals of unique numbers associated with (x – 1)ⁿ is expressed as Alt ζ₂(n).

However as both k and (x – 1)ⁿ range over the same natural numbers 1, 2, 3, …

then (x – 1)ⁿ = (x – 1)^k and Alt ζ₂(n) = Alt ζ₂(k)

So ζ₂(1/k) = Alt ζ₂(k + 1).

Thus, just as there are two formulations of the Zeta 1 (Riemann) function as the sum over natural numbers and product over primes expressions, equally there are two formulations of the Zeta 2 function.

And the key to understanding the two Zeta 2 formulations is that they represent complementary notions of number (both in terms of base and dimensional formulations) which are analytic (quantitative) and holistic (qualitative) with respect to each other. In other words, number keeps switching as between both its particle (independent) and wave (interdependent) expressions.

And these two expressions can only be properly understood in a dynamic interactive manner.

So it is equally true that with respect to the macro formulation of the number system, that the sum over natural numbers and product over primes formulations represent complementary expressions that are analytic (quantitative) and holistic (qualitative) with respect to each other.

So for example in macro terms when s = 2, in the sum over natural numbers expression,

ζ₁(2) = 1 + 1/4 + 1/9 + …  = π²/6

However in the product over primes expression,

ζ₁(2) = 4/3 * 8/9 * 24/25 + …  = π²/6.  

And properly understood i.e. in dynamic interactive terms, these two expressions are analytic (quantitative) and holistic (qualitative) with respect to each other.

However equally in micro terms when k = 1/2²,

ζ₂(1/2²) = 1 + 1/2² + 1/2⁴ + … = 4/3

So we see how the 1^st term of the product over primes expression of the Zeta 1 represents a corresponding Zeta 2 expression.

And equally this is true of every other term of the Zeta 1, which can be expressed in Zeta 2 terms.  

And also each individual term of the Zeta 1 can be expressed in Alt Zeta 2 terms

So when k = 2² + 1, Alt ζ₂(5) = 1 + 1/5 + 1/15 + 1/35 + … = 4/3.

And equally, this is true of every other term of the Zeta 1, which can be expressed in Alt Zeta 2 terms.

This is also true - though slightly more convoluted - in relation to the sum over natural numbers expression of the Zeta 1.

For example the 2^nd term here = 1/4 = 5/4 – 1

So ζ₂{(1/(2² + 1)} = 1 + 1/5 + 1/5² + … = 5/4

Thus 1/4 as 2^nd term of ζ₁(2), = ζ₂{(1/(2² + 1)} – 1

And similar Zeta 2 type expressions are available for ever other term of Zeta 1 and by extension ζ₁(s) where s > 2.

Finally each term of Zeta 1 (sum over natural numbers expression) can be expressed in appropriate Alt Zeta 2 terms.

So Alt ζ₂ (2² + 2) = Alt ζ₂ (6)  = 1 + 1/6 + 1/21 + 1/56 + …  = 5/4.

Thus 1/4 = Alt ζ₂ (2² + 2) – 1.

And every other individual term of ζ₁(2) in a sum over natural numbers expression, and by extension ζ₁(s) where s > 2, can be given an Alt Zeta 2 formulation.

So a full understanding of Riemann’s Zeta function requires not only that the Zeta 1 function can be expressed in two different ways (which in dynamic terms are complementary), but that each individual term of the Zeta 1 (for both expressions) can equally be expressed in two different ways i.e. through the Zeta 2 and Alt Zeta 2 functions respectively.

Interesting Relationships Continued

Initially we started with the utterly simple equation x = 1, to quickly show how it possesses both linear (particle) and circular (wave) expressions, that correspond in fact to two distinct notions of mathematical dimensions.

Thus if x = 1, then x – 1 = 0 and therefore (x – 1)ⁿ = 0. (1)

Equally if x = 1, then xⁿ = 1 so that xⁿ – 1 = 0. (2)

Both relationships conform to the use of conventional accepted mathematical procedures.

However, (x – 1)ⁿ = 0 and xⁿ – 1 = 0 are no longer equivalent expressions from this perspective!

So, in fact the attempted reconciliation of (1) and (2) requires a much more refined mathematical interpretation that entails the combined interplay of both quantitative (analytic) and qualitative (holistic) aspects that are dynamically complementary with each other.

Then in yesterday’s blog entry, I extended (1) to consideration of a particular class of the more general case, where (x – k) ⁿ = 0, with k an integer > 1 (and n = 2).   

Finally, I also considered the related case of (x + k)ⁿ with k an integer ≥ 1 (and again n = 2).

So now in this entry, we will switch to consideration of (2).

So now we start with x = k. Therefore xⁿ = kⁿ so that here xⁿ – kⁿ = 0,

And initially we will confine ourselves to one particular class of this general expression (where n = 2).

Thus x² – k² = 0.

And when k = 2, this implies that x² – 4 = 0.

Then the unique number sequence, associated with this polynomial equation is,

1, 0, 4, 0, 16, 0, 64, …

Then ignoring the terms with 0 we obtain the (infinite) sum of reciprocals of these numbers to obtain 1 + 1/4 + 1/16 + 1/64 + …  = 4/3.  (i.e.  r = 1/4)

Then when k = 3, this implies that x² – 9 = 0.

Then the unique number sequence, associated with this polynomial equation are,

1, 0, 9, 0, 81, 0, 729, …

Then again ignoring terms in 0, the (infinite) sum of reciprocals of these numbers are

1 + 1/9 + 1/81 + 1/729 + … = 9/8 (i.e. r = 1/9)

And using just one more case to illustrate, when k = 5 (the next prime), this implies that x² – 25 = 0.

The unique number sequence then associated with this equation is

1, 0, 25, 0, 625, 0,  …

And again ignoring terms in 0, the corresponding (infinite) sum of reciprocals of these terms is

1 + 1/25 + 1/625 + …    = 25/24 (i.e. r = 1/25).

Now if we examine the product over primes expression for the Zeta 1 function, where s = 2, then

ζ₁(2) = 4/3 * 9/8 * 25/24 * …    = π²/6

In other words, each individual term of the Zeta 1 (Riemann) function i.e. ζ₁(2), corresponds to a Zeta 2 function, that is directly related to the simple polynomial expression xⁿ – kⁿ  = 0 (where n = 2) .

And this correspondence can be fully generalised for all values of n > 1.

So for example,

ζ₁(3) = 8/7 * 27/26 * 125/124 * …    = 1.20205693…

And when n = 3, x³ – k³  = 0.

Thus where k = 2, x³ – 8  = 0.

And the unique numbers associated with this polynomial equation are

1, 0, 0, 8, 0, 0, 64, …

Therefore, again ignoring the terms in 0, the (infinite) sum of reciprocals of this number sequence,

= 1 + 1/8 + 1/64 + …  = 8/7 (i.e. r = 1/8)

And as we can see this is the first term in the product sequence for ζ₁(3)!

All the other terms of ζ₁(3) will correspond to the (infinite) reciprocal sum of the unique numbers (excepting 0’s) related to the polynomial expression xⁿ – kⁿ = 0 (n = 3; k prime). 

And this can be readily extended to all positive integer values of ζ₁(s), where s > 3, with corresponding Zeta 2 expressions  for each individual terms related to xⁿ – kⁿ = 0 (where n > 3).

However there is an important additional point to be made here, which is crucial for proper understanding the true nature of the product over primes expressions, associated with the Zeta 1 function (where s is an integer > 1).

In establishing the equivalence of each individual term of the Zeta 1 with a corresponding Zeta 2 expression, we edited out the role of the 0’s (where meaningful reciprocals cannot be given).

However, the key point about the 0’s is that they point directly to the holistic - rather than analytic - interpretation of multiplication.

So for example, in conventional mathematical terms, when 2 (or more) numbers are multiplied together, interpretation is given solely in a reduced analytic i.e. quantitative manner.

So 2 * 3 = 6, with each number interpreted in a 1-dimensional manner (as a point on the real number line).

However, properly understood, a qualitative transformation is also necessarily involved. Thus in geometrical terms, we can easily appreciate how 2 * 3 leads to a rectangular figure (that is correspondingly measured in square i.e. 2-dimensional units).

Thus, though the result = 6 (from a quantitative perspective), a qualitative transformation has also taken place (from 1-dimensional to 2-dimensional) in the nature of the units involved.

However this in itself represents but an analytic interpretation of such dimensional transformation.

A much subtler holistic transformation is also required, which I will now again briefly illustrate.

Imagine we have two rows of coins (with 3 in each row). 

In conventional mathematical terms the independent nature of each unit is solely recognised!

So 3 = 1 + 1 + 1 (with the units independently interpreted in a homogeneous manner that lacks any qualitative distinction).

Now we could independently add up the units (in each row) to achieve a total of 6.

However the very basis of multiplication is to recognise a common similarity as between the rows (or equally the columns).

Thus when we recognise for example that the two rows are similar (with 3 units in each row), we can use the operator 2 with 3 to more quickly attain our results.

So 2 * 3 = 6.

However the crucial point here - which is completely overlooked in conventional mathematical interpretation - is that we have thereby moved from the quantitative (analytic) notion of number independence to the corresponding qualitative (holistic) notion of number interdependence.

Thus without both the analytic notion of number units as independent of each other, and the holistic notion of numbers as interdependent with each other (i.e. in sharing a similar identity) the very notion of multiplication cannot be meaningfully interpreted.

So the very number sequences that we generate in the above examples (all entailing 0’s), when properly interpreted, imply the holistic aspect related to multiplication. And the higher the
dimension, the more holistic is the corresponding interpretation involved (indirectly signified by the increased repetition of 0’s in the unique number sequences generated).

Properly interpreted therefore the key significance of the dual relationship as between the sum over natural numbers and product over primes expressions of the Zeta 1 (Riemann) function - and indeed the entire plethora of associated L functions - relates to the fact that addition and multiplication are, relatively, analytic (quantitative) and holistic (qualitative) with respect to each other.

And this can only be properly appreciated in a dynamic interactive manner, where both aspects - of equal importance - are understood as complementary with each other

If you can even vaguely appreciate the significance of what is stated above, then you will have no option but to conclude that the present accepted understanding of number (and all associated number relationships) is simply not fit for purpose.

Interesting Log Relationships (1)

We have already considered the (infinite) reciprocal sequences of the unique numbers associated with (x – 1)ⁿ  = 0.

So once again, for example, where n = 3, the unique numbers associated with the polynomial equation (x – 1)³  = 0, i.e. x³ – 3x²  + 3x – 1 = 0 are,

1, 3, 6, 10, 15, 21, …

And the sum of the (infinite) reciprocal sequence of these numbers is

1 + 1/3 + 1/6 + 1/10 + 1/15 + 1/21 + …  = 2. 

We now consider (x – k)ⁿ = 0, with k an integer > 1, the sum of the (infinite) reciprocal sequences of the corresponding unique numbers associated with (x – k)ⁿ = 0, where n = 2.

For example when k = 2, we have (x – 2)² = 0, i.e. x²  – 4x + 4 = 0, and the unique numbers associated with this polynomial equation are, 

1, 4, 12, 32, 80, …

So the sum of the (infinite) reciprocal sequence of these numbers =

1 + 1/4 + 1/12 + 1/32 + 1/80 + …  = 1.3862…  = log 4 = 2(log 2/1) = 2(log 2).

Then, when k = 3, we have (x – 3)² = 0, i.e. x²  – 6x + 9 = 0, and the unique numbers associated with this polynomial equation are, 

1, 6, 27, 108, 405, 1458, …

Thus the sum of the (infinite) reciprocal sequence of these numbers,

= 1 + 1/6 + 1/27 + 1/108 + 1/405 + 1/1458 + … = 1.2163… = log 27/8 = 3(log 3/2).

In fact there is a general pattern at work here!

So log 4 and log 27/8 = log{k^k/(k – 1)^k} = k{log k/(k – 1)}, where k = 2 and k = 3 respectively.

This implies that the sum of the (infinite) reciprocal sequence of the unique numbers associated for example with (x – 4)² = 0,  = log {4⁴/3⁴} = log 256/81 = 4(log 4/3).

And (x – 4)² = 0 implies x²  – 8x + 16 = 0, and the unique numbers associated with this polynomial equation are 1, 8, 48, 256, 1280, 6144, …

Thus the sum of the (infinite) reciprocal sequence, 

= 1 + 1/8 + 1/48 + 1/256 + 1/1280 + 1/6144 + … = 1.1507… = 4(log 4/3).

Incidentally this pattern indicates clearly why the sum of the reciprocals of the natural numbers i.e. the harmonic series diverges to infinity.

The sum of this sequence is associated with (x – 1)² = 0, and the answer thereby according to the formula is log (1/0) i.e. log ∞ .  

There are also fascinating patterns associated with (x + k)² = 0, where k is now an integer ≥ 1.

These simple polynomial equations lead to the same unique number sequences associated with
(x – k)² = 0, except that the numbers keep alternating as between positive and negative signs.

Therefore in the case where k = 1, the unique number sequence associated with (x + 1)² = 0, 

i.e.  x²  + 2x + 1 = 0, are 1, – 2, 3, – 4, 5, – 6 …and the (infinite) sum of its reciprocals,

= 1 – 1/2 + 1/3 – 1/4 + 1/5 – 1/6 + …   = log 2.

Then when k = 2, the unique number sequence associated with (x + 2)² = 0, i.e.   

x²  + 4x + 4 = 0, is 1, – 4, 12, – 32, 80, – 192, …

Thus the corresponding (infinite) sum of its reciprocals

= 1 – 1/4 + 1/12 – 1/32 + 1/80 – 1/192 + …  = log 9/4 = 2(log 3/2)

Again a fascinating general pattern is at work here.

For example log 9/4 = log{(k + 1)^k/k^k} = k{log (k + 1)/k}, where k = 2  

This implies that when k = 3, the (infinite) sum of reciprocals associated with the unique number sequence corresponding to (x + 3)² = 0,  i.e. x²  + 6x + 9 = 0, = log{4³/3³}

= 3 log 4/3.

And the unique number sequence is 1, – 6, 27, – 108, 405, – 1458, …

Therefore the corresponding (infinite) sum of reciprocals

= 1 – 1/6 + 1/27 – 1/108 + 1/405 – 1/1458 + … = .8630 … = 3(log 4/3).

So the famous case of 1 – 1/2 + 1/3 – 1/4 + 1/5 – 1/6 + …   = log 2, represents just a special example corresponding to the (infinite) sum of reciprocals of the unique number sequence associated with the simple polynomial expression (x + k)² = 0, where k = 1.

And again the general solution for all integer values of k ≥ 1, is log{(k + 1)^k/k^k}

 = k{log (k + 1)/k}.