Thursday, November 9, 2017

More 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 representing the Alt Zeta 2 function.

For example the 1st 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 2nd 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.

Thursday, October 26, 2017

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 tehrahedral 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.

Tuesday, October 24, 2017

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 1st 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 2nd term here = 1/4 = 5/4 – 1

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

Thus 1/4 as 2nd 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.

Friday, October 20, 2017

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.

Thursday, October 19, 2017

Interesting Log Relationships

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 (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 (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 (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 (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 {kk/(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 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/kk} = 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 case 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/kk}

 = k{log (k + 1)/k}.

Friday, September 29, 2017

Some Number Magic (2)

In yesterday’s blog entry, I sought to show how the sums of reciprocals of the unique number sequences associated with (x – 1)n = 0 (where n is an integer > 1) - what I refer to more simply as the alternative Zeta 2 function - can be shown to have fascinating connections with the corresponding Zeta 1 (Riemann) function.

In particular, I showed how to generate the respective terms of the Zeta 1 function i.e. ζ1(s), for s = 2 from the alternative Zeta 2 function.

However, it is time now to attempt to generalise this procedure for to all values of ζ1(s) where s is an integer ≥ 1.

Indeed the procedure appears trivial where s = 1.

So ζ1(1) = 1 + 1/2 + 1/3 + 1/4 + ….

And this equates directly with the sum of reciprocals of the unique digit sequence for (x – 1)2 = 1 + 1/2 + 1/3 + 1/4 + ….

Thus for simplicity we will refer to this latter expression of the same harmonic series as Alt ζ2(2) i.e. where n in the polynomial expression (x – 1)n, from which the reciprocals of the unique number sequence involved is derived = 2.

However there is an alternative way of relating the two zeta expressions, which will be very useful for deriving the terms of ζ1(s), where s > 1.

ζ1(1), as the sum of  the terms of the harmonic series, diverges to infinity.

However we can express each term of this expression as half the sum of all Alt ζ2(s), where s > 3.

So Alt ζ2(3) = 2/1;  Alt ζ2(4) = 3/2; Alt ζ2(5) = 4/3; ; Alt ζ2(6) = 5/4 and so on.

Therefore ζ1(1) = 1/2{∑Alt ζ2(s)} for s ≥ 3.    

So Alt ζ2(2) = 1/2{∑Alt ζ2(s)} for s ≥ 3.

Then as we saw yesterday to obtain the appropriate “conversion” for the terms of ζ1(2),
we then use the respective terms of Alt ζ2(3) multiplied by 1/2. Now 1/2 in this context can be more precisely expressed as 1/2!

So 1/2!{Alt ζ2(3)} = 1/2(1 + 1/3 + 1/6 + 1/10 + 1/15 + …)

= 1/2 + 1/6 + 1/12 + 1/20 + 1/30 + …

We then multiply each successive term by the corresponding sum of infinite series values for Alt ζ2(3), Alt ζ2(4), Alt ζ2(5), Alt ζ2(6), …
So   1/2{Alt ζ2(3)} =   1/2 * 2/1 = 1       i.e. (1/12)
       1/6{Alt ζ2(4)} =   1/6 * 3/2 = 1/4    i.e. (1/22)
     1/12{Alt ζ2(5)} = 1/12 * 4/3 = 1/9    i.e. (1/32)
     1/20{Alt ζ2(6}   = 1/20 * 5/4 = 1/16  i.e. (1/42)
                                                     ….     

And we can continue on in this general manner to generate the further terms in the infinite series for ζ1(2).


Then to generate the respective terms of ζ1(3), we move on to consideration of the respective terms of Alt ζ2(4) i.e. 1 + 1/4 + 1/10 + 1/20 + 1/35 + …

However in this case each individual term is now multiplied by 1/3! = 1/6.

So we now have 1/6 + 1/24 + 1/60 + 1/120 + 1/210 + …

Each of these individual terms is now successively multiplied by Alt ζ2(4), Alt ζ2(5),
Alt ζ2(6), Alt ζ2(7) … (Note that Alt ζ2(3) is now omitted).

Now the value of Alt ζ2(4)= 3/2. So we multiply the 1st term by this value; however we also multiply by (3 – 2)/2 = 1/2.

So in general terms we multiply each successive term by,

 (n – 1)/(n – 2)} * (n – 3)/(n – 2)

Therefore using the 1st term,        1/6 * 3/2 * 1/2     = 1/8 (i.e. 1/23).
Then with the 2nd term we have    1/24 * 4/3 * 2/3 =  1/27 (i.e. 1/33).
With the 3rd term, we have          1/60 * 5/4 * 3/4   = 1/64  (i.e. 1/43).
Then with the 4th, we have        1/120 * 6/5 * 4/5 =  1/125 (i.e. 1/53).
                                                                                        

Though the derivation might appear arbitrary, once the correct procedure is used, it works universally to generate all further terms.

Also this procedure preserves unique hidden number patterns.

For instance the 1st case 1/6 * 3/2 * 1/2 = 1/23 .

Now if we replace 3/2 with 3 * 2 we obtain 1/6 * (3 * 2) * 1/2 = 1/2.

Alternatively if we replace 1/2 by 1 * 2, we get 1/6 * 3/2 * 2 = 1/2 

And this pattern will universally hold.

For example in the last case we have 1/120 * 6/5 * 4/5 =  1/53.
If we replace 6/5 by 6 * 5 we get 1/120 * (6 * 5) * 4/5 = 1/5

Then if we replace 4/5 by 4 * 5 we get 1/120 * 6/5 * 4 * 5 = 1/5.

So in this way we generate all the terms of the original harmonic series (except 1).


Admittedly it does get more cumbersome attempting to generate terms for ζ1(s), when s > 2.

However a structured pattern for doing this exists in every case.

For example to create the individual terms of ζ1(4) - less the 1st two terms - we move to consideration of the corresponding terms of Alt ζ2(5) i.e.

1 + 1/5 + 1/15 + 1/35 + 1/70 + …

We now multiply each of these terms by 1/4! = 1/24 to obtain

1/24 + 1/120 + 1/360 + 1/840  + 1/1680 + …

Alt ζ2(5) = 4/3.

We now use a more complex conversion factor based on this value which in general terms is given as,

 (n – 1)/(n – 2) * (n – 3)/(n – 2) * (n – 4)/(n – 2)

So for Alt ζ2(5), this gives 4/3 * 2/3 * 1/3

So we using the 1st term, 1/24 we obtain 1/24 * 4/3 * 2/3 * 1/3 = 1/81 (i.e. 1/34)

Once again there are lovely hidden number patterns preserved in this formulation.

So replacing 4/3 by 4 * 3 we get 1/24 * (4 * 3) * 2/3 * 1/3 = 1/9 (i.e. 1/32).

Then replacing 2/3 by 2 * 3 we get 1/24 * 4/3 * (2 * 3) * 1/3 = 1/9 (i.e. 1/32).

Finally replacing 1/3 by 1 * 3 we get 1/24 * 4/3 * 2/3 * (1 * 3) = 1/9 (i.e. 1/32).


And a similar pattern presents itself with all subsequent “conversions” so that we have the emergence of ζ1(2), less the 1st two terms.

Just to illustrate further, using the 2nd term 1/120,

we obtain 1/120 * 5/4 * 3/4 * 2/4 = 1/256 (i.e. 1/44).

Using the 3rd term 1/360, we obtain,

1/360 * 6/5 * 4/5 * 3/5 = 1/625 (i.e. 1/54).

Then using the 4th term 1/840, we obtain

1/840 * 7/6 * 5/6 * 4/6 = 1/1296 (1/64).

And again we can continue in this manner to generate the further terms for
ζ1(4).


I will just finish by setting up the “conversion” to generate the computable term
i.e. the 4th term of ζ1(5).

Here we use the next series for Alt ζ2(6) i.e.

1 + 1/6 + 1/21 + 1/56 + 1/126 + …

Each of these individual terms is now multiplied by 1/5! = 1/120 to obtain

1/120 + 1/720 + 1/2520 + 1/6720 + 1/15120 + …

The general conversion factor now contains 4 product terms

i.e. (n – 1)/(n – 2) * (n – 3)/(n – 2) * (n – 4)/(n – 2) * (n – 5)/(n – 2)

So when n = 6, this conversion factor = 5/4 * 3/4 * 2/4 * 1/4.

Therefore using the 1st term, we get,

1/120 * 5/4 * 3/4 * 2/4 * 1/4 = 1/1024 = (i.e. 1/45).

Again we can illustrate even more impressively this time, the number magic preserved within this formulation.

So replacing 5/4 with 5 * 4, we obtain,
1/120 * (5 * 4) * 3/4 * 2/4 * 1 /4 = 1/64 (i.e. 1/43).

Then replacing 3/4 with 3 * 4 we obtain

1/120 * 5/4 * (3 * 4) * 2/4 * 1/4  = 1/64 (i.e. 1/43).

Replacing 2/4 with 2 * 4 we obtain

1/120 * 5/4 * 3/4 * (2 * 4) * 1/4  = 1/64 (i.e. 1/43).

Finally replacing 1/4 with 1 * 4 we obtain,

1/120 * 5/4 * 3/4 * 2/4 * (1 * 4) = 1/64 (i.e. 1/43).

Thursday, September 28, 2017

Some Number Magic

It struck me when I had finished the last entry that a ready means thereby existed for solving another interesting feature regarding the sum of the Riemann Zeta functions (less 1) for both odd and even integer values of s (> 1).

Now empirical information had long suggested to me, that where s is odd that the sum = .25 and where s is even, the sum = .75.

For example where s = 3, 5, 7 and 9, ζ1(s) – 1 = .20205…, .03692…, .00834… and .002008… respectively.

So the sum of these values = .2493…, which is already very close to .25. And as ∑{ζ1(s) – 1} where s ≥ 2 = 1, this would imply that the corresponding sum for even values of s = .75.

However, we now have a ready means using the approach of yesterday to conveniently prove this result.

So once again we have,

ζ1(2) – 1       = 1/22 + 1/32 + 1/42 +  …       =  .64493…
ζ1(3) – 1       = 1/23 + 1/33 + 1/43 +  …       =  .20203…
ζ1(4) – 1       = 1/24 + 1/34 + 1/44  + …        =  .08232…
ζ1(5) – 1       = 1/25 + 1/35 + 1/45  + …        =  .03692…
    …                           …                                     …
    …                           …                                     …

Now if we concentrate on vertical columns associated with odd numbered values of s (> 2), again we can see that can be defined in Zeta 2 terms in the form of simple geometric series.

So the 1st series = 1/23 + 1/25 + 1/27 + …  = (1/8)/1 – 1/4) = 1/8 * 4/3            = 1/6
The 2nd series = 1/33 + 1/35 + 1/37 + …  =  (1/27)(1– 1/9)  = 1/27 * 9/8           = 1/24
The 3rd series  = 1/43 + 1/45 + 1/47 + …  = (1/64)(1– 1/16)  = 1/64 * 16/15     = 1/60
The 4th series = 1/53 + 1/55 + 1/57 + …  =  (1/125)(1– 1/25)  = 1/125 * 25/24 = 1/120

So the sum of all the vertical columns = corresponding sum of all horizontal rows

= 1/6 + 1/24 + 1/60 + 1/120 + …

= 1/6(1 + 1/4 + 1/10 + 1/20 + …)

And the numbers inside the brackets correspond to the reciprocals of the tetrahedral numbers i.e. the unique number sequence associated with (x – 1)4 = 0

Now 1 + 1/4 + 1/10 + 1/20 + … = 3/2

Therefore 1/6(1 + 1/4 + 1/10 + 1/20 + …) = 1/6 * 3/2 = 1/4.

So ∑{ζ1(s) – 1} where s is odd (>1) = 1/4 (.25).

Thus ∑{ζ1(s) – 1}where s is even (> 0) = 3/4 (.75).


There are many fascinating number patterns in evidence.

The series used to prove that ∑{ζ1(s) – 1} = 0 for all s (≥ 1) is

1/2 + 1/6 + 1/12 + 1/20 + 1/30 + …

= 1/2(1 + 1/3 + 1/6 + 1/10 + 1/15 + … )

And the 1/2 (outside the brackets) corresponds with the 1st term of the series inside the brackets). 

So the numbers inside the brackets correspond to the reciprocals of unique number sequence associated with (x – 1)n = 0 (where s = 3),

Therefore  = 1/2 * 2/1 = 1 (i.e. 1/12).

Then the series used to prove that ∑{ζ1(s) – 1} where s is odd (>1) = 1/4, where s is odd (>1) is,

1/6(1 + 1/4 + 1/10 + 1/20 + 1/35 +…).

So the number outside the brackets corresponds to the 2nd term in the original series i.e.
1/2 + 1/6 + 1/12 + 1/20 + 1/30 + …, while the numbers inside, correspond to the reciprocals of the unique digit sequence associated with (x – 1)n = 0 (where s = 4)
= 1/6 * 3/2 = 1/4 (i.e. 1/22).

If we continue on this manner the next number to be placed outside the bracket = 1/12 (the 3rd term of the original series) and the numbers inside the bracket will then represent the reciprocals of the unique number sequence associated with (x – 1)n = 0 (where s = 5) = 1/12 (1 + 1/5 + 1/15 + 1/35 + 1/70 + …) = 1/12 * 4/3 = 1/9 (i.e. 1/32).

Using just one more example to illustrate the next number to be placed outside the brackets = 1/20 (the 4th term of the original series) and the numbers inside corresponds to the reciprocals of the unique number sequence associated with (x – 1)n = 0 (where s = 6)

= 1/20(1 + 1/6 + 1/21 + 1/56 + 1/126 + …) = 1/20 * 5/4 = 1/16 (i.e. 1/42).

Note that there is a unique feature to the nature of these products.

Take the last one for example where we have 1/20 * 5 * 1/4. Now if we replace the last fraction (1/4) by its reciprocal we obtain 1/20 * 5 * 4 = 1. So instead of dividing 5 by 4 (the sum of the series inside the brackets) we multiply 5 * 4 the resulting product = 1
And this will always be the case!So in the next example, 1/30 * 6/5 = 1/25 (i.e. 1/52). and 1/30 * 6 * 5 = 1!

There is also another interesting feature. For example in the first case,

1/6 * 3/2 = 1/4 (i.e. 1/22). However if alternatively, we divide 1/6 by 3/2,

(1/6)/(3/2) = 1/6 * 2/3 = 1/9 (i.e. 1/32) . And again this feature is universal, where in the former case 1/n2 (n = 1, 2, 3,...) results and in the latter case 1/(1 + n)2.

Thus as we have seen, 1/20 * 5/4 = 1/16 (i.e. 1/42). And, 1/20 * 4/5 = 1/25 (i.e. 1/52). 


By progressing in the manner above we generate the series,

1/12 + 1/22 + 1/32 + 1/42 + … i.e. ζ1(2) i.e. the Riemann Zeta function for s = 2.

So we started by demonstrating the close links as between the Zeta 1 (Riemann) as terms in horizontal rows and the Zeta 2 function as the corresponding terms in the vertical columns.

And now we have been able to demonstrate very close links as between the alternative Zeta 2 function - as the sums of reciprocals corresponding to the unique number sequences associated with the general polynomial expression (x – 1)n = 0 - and the Zeta 1 (Riemann) function.