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.

Wednesday, September 27, 2017

Some Interesting Connections

In an earlier entry, I showed a simple way to prove that with respect to the Zeta 1 (Riemann) function,
 
∑{ζ1(s) – 1} = 1.
 2
So we will now illustrate this for the first few positive integer values of s (i.e. where s ≥ 2).

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…
                                                                   
                                                                   

Thus the Zeta 1 expressions here result from reading the terms here across the respective horizontal rows.
However the corresponding Zeta 2 expressions result from reading terms down the corresponding vertical columns (representing geometric series) .

So the 1st column here =  1/4 +  1/8 +   1/16 +   1/32  +      = 1/2
The 2nd column then    =  1/9 +   1/27 + 1/81 +   1/243 + …  = 1/6
The 3rd column             =  1/16 + 1/64 + 1/256 + 1/024 + …  = 1/12

So the sums of these vertical columns i.e. 1/2 + 1/6 + 1/12 + … by definition equals the corresponding sums of the horizontal rows i.e. .64493… + .20205… + .08232… +…

And 1/2 + 1/6 + 1/12 + … represents half the values of the corresponding reciprocals of the triangular numbers i.e. 1 + 1/3 + 1/6 + …  = 2.

Therefore the sum of  1/2 + 1/6 + 1/12 + … = 1.

Likewise therefore the sum of .64493… + .20205… + .08232… + …  = 1.

So as well as illustrating the close complementary links as between the Zeta 1 and Zeta 2 functions (which are horizontal and vertical with respect to each other), this also helps to prove a very interesting feature with respect to the sums of the Zeta 1 (Riemann) series (for real integer values of s ≥ 2)


Now as we have seen the sum of reciprocals of the triangular numbers is directly associated with the unique number sequence for (x – 1)n = 0 (where n = 3).

And as we have seen in general terms that the sums of reciprocals associated with the unique number sequences of (x – 1)n = 0, = (n – 1)/(n – 2) i.e. 2/1, 3/2, 4/3, 5/4 and so on.

And then when we subtract 1 from each of these values we obtain 1, 1/2, 1/3, 1/4 …

In other words we obtain the harmonic series, that from one perspective represents the sum of reciprocals of the unique number sequence associated with (x – 1)n = 0 (where n = 2).

Equally from another perspective it represents the Zeta 1 (Riemann) function i.e. ζ1(s), where s = 1.

Therefore we can now perhaps better appreciate the intimate connections as both between the both The Zeta 1 and Zeta 2 functions and this new alternative function based on the the sums of reciprocals of the unique number sequences associated with (x – 1)n = 0.    

Thursday, September 21, 2017

New Perspective on Zeta Function

In yesterday's entry, we illustrated how each individual term both in the sum over natural numbers and product over primes expressions respectively for the Zeta 1 (Riemann) function, can equally be expressed as an infinite sum of product terms using the complementary Zeta 2 function. 

However, this then raises the important question as to whether we can equally express each of the individual terms of the Zeta 1 (Riemann) function - both in the sum over natural numbers and product over primes expressions - in corresponding Zeta 2 terms, as an infinite sum of additive terms! 


And remarkably, through our recent investigation of the unique number sequences associated with the general polynomial expression i.e. (x – 1) n = 0, this can now be made possible.

To make this a little easier to illustrate, I will start with the product over primes expression for ζ1(s), where s = 1.

Thus ζ1(1)  = 2/1 * 3/2 * 5/4 * 7/6 * …

Of course this expression which equates to the harmonic series (in the sum over natural numbers expression) does not converge to a finite answer!

However we can still equate each individual term with an infinite series based of the sum of the reciprocals of the unique digit sequences associated with (x – 1)n = 0.

So when n = 3, the unique digit sequence associated with (x – 1)3 = 0 is

1, 3, 6, 10, 15, 21, …

Thus the infinite series of the sum of the reciprocals of these numbers

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

And this in turn represents the 1st term of the Zeta 1 (Riemann) product expression over the primes for s = 1.

Now it is easier to demonstrate that our new reciprocal expression does in fact represent a sum over all the natural numbers in the following manner.

So 1 + 1/3 + 1/6 + 1/10 + 1/15 + 1/21 + … 

= 1/1 + 1/(1 +2) + 1/(1 + 2 + 3) + 1/(1+ 2 + 3 + 4) + 1/(1 + 2 + 3 + 4 + 5) + …

Thus in general terms the denominator of the nth term represents the sum of the first n natural number terms!

And further reciprocal expressions with respect to the unique digit expressions of
(x – 1)n = 0 for (n > 3) involve in their denominators, compound combinations involving all the natural numbers (up to n).

For example, the 2nd term in the product expression for ζ1(1) = 3/2.

Now this in turn equates with infinite sum of the reciprocals associated with the unique number sequence for (x – 1)4 = 0, i.e.

1 + 1/4 + 1/10 + 1/20 + 1/35 + …  = 3/2.

And the denominators of any term t, represents compound expressions entailing all the natural numbers to t.

For example the 3rd term = 1/10.

And the denominator 10 = 1 + (1 + 2) + (1 + 2 + 3)!

Finally, to fully illustrate this point, the 3rd term in the product expression for ζ1(1) = 5/4    

This in turn equates with infinite sum of the reciprocals associated with the unique number sequence associated with (x – 1)6 = 0, i.e.

1 + 1/6 + 1/21 + 1/56 + 1/121 + …  = 5/4.

The denominator of the 2nd term - which in this case is easiest to illustrate - then represents a compound expression entailing the first two natural numbers i.e.
1 + {1 + [1 + (1 + 2)]}.


Though we have illustrated here with respect to the Zeta 1 (Riemann) product expression over all the primes for s = 1, corresponding further reciprocal expressions, based on the unique number sequences associated with (x – 1)n = 0 can be found for all Zeta 1 product expressions where s is an integer > 1.

We can likewise associate each of the individual terms in the sum over natural numbers Zeta 1 (Riemann) expressions with infinite series based on the reciprocals of the unique digit sequences associated with (x – 1)n = 0.

Again for example in the simplest case where s = 1.

ζ1(1) = 1 + 1/2 + 1/3 + 1/4 + …   (the harmonic series)

Now the 1st term here can be expressed through the infinite reciprocal sequence - already considered - associated with (x – 1)3 = 0.

So 1 = (1 + 1/3 + 1/6 + 1/10 + …) – 1.

Then the 2nd term 1/2 can be expressed through the infinite reciprocal sequence associated with 
(x – 1)4 = 0 i.e.

1/2 = (1 + 1/4 + 1/10  + 1/20 + …) – 1.

Then the 3rd term 1/3 can be expressed through the infinite reciprocal sequence associated with 
(x – 1)5 = 0 i.e.

1/3 = (1+ 1/5 + 1/15 + 1/35 + …) – 1.

And we can continue on indefinitely in this manner with all further terms for real integer values of
ζ1(s), where s > 1.

Wednesday, September 20, 2017

Surprising Connections to the Zeta Function

In yesterday’s entry we saw that the infinite sum of reciprocals of the unique digit sequences associated with the general polynomial equation (x – 1)n = 0 is related solely to the value of n and given by the simple expression (n – 1)(n – 2).

So once again to illustrate the unique digit sequence associated with (x – 1)5 = 0 is

1, 5, 15, 35, 70, 126, 210, … and the corresponding infinite sum of reciprocals i.e.

1 + 1/5 + 1/35 + 1/70 + 1/126 + 1/210 + … = (5 – 1)/(5 – 2) = 4/3.

Now there is a surprisingly important significance to these results which may not be immediately apparent.

In discussions on the Riemann Hypothesis, I have repeatedly drawn attention to the importance of the Zeta 2 (which fully complements the well-known Zeta 1 i.e. Riemann) zeta function.

I was at pains in those discussions how each individual term both in the sum over (all) natural numbers and product over primes infinite expressions can in turn be expressed by a Zeta 2 infinite series.

So for example in the Zeta 1 (Riemann) function where s = 2, the sum over natural numbers is given as

1 + 1/22 + 1/32 + 1/42 + … = 1 + 1/4 + 1/9 + 1/16 and the product over primes,

= 4/3 * 9/8 * 25/24 * 49/48 * … = π2/6.

However each of these individual terms, both for the sum over natural numbers and product over primes expressions respectively, can be expressed by an infinite Zeta 2 function.

Therefore the Zeta 1 (Riemann) function can equally be expressed as 1) the overall sum of an infinite sequence of Zeta 2 functions and 2) the overall product of an infinite sequence of Zeta 2 functions.

So in general terms, the (infinite) Zeta 2 function is given as

ζ2(s) = 1 + s + s2 + s3 + …   = 1/(1 – s)

So, to illustrate the 1st term i.e. 1, in the sum over natural numbers can be given as,

ζ2(s) – 1, where s = 1/2 (i.e. 1/(12 + 1).

Then the next term i.e. 1/4 can be given as ζ2(s) – 1, where s = 1/5 (i.e. 1/(22 + 1).

In this manner, the denominator can always be expressed as the square of a prime + 1.


In like manner, all the remaining terms can be expressed in the form of ζ2(s) – 1, with s given an appropriate value, based on the square of consecutive primes.

Then the 1st term, i.e. 4/3, in the product over natural numbers expression can be given as
ζ2(s), where s = 1/4 (i.e. 1/22).

Then the 2nd term, i.e. 9/8 can be given as ζ2(s), where s = 1/9 (i.e. 1/32).

And again in like manner, all the remaining terms can be expressed in the form of ζ2(s), with s given an appropriate value.

However just as the Zeta1 (Riemann) function can be given two expressions as a sum over natural numbers and product over primes respectively, likewise the Zeta 2 can be given two similar complementary expressions.

So far however, we have only considered the product version of this (both as a product over primes and product over natural numbers respectively).

Therefore when again to illustrate, we take for the Zeta 1 (Riemann) function the product over primes expression (for s = 2), i.e. 4/3 * 9/8 * 25/24 * 49/48 * …, each of these individual terms can equally be expressed as an infinite sum of product over primes terms, entailing the Zeta 2 function.

Thus 4/3 = 1 + 1/22 + 1/24 + 1/28  + …

i.e. 1 + 1/(2 * 2) + 1/(2 * 2 * 2 * 2) + 1/(2 * 2 * 2 * 2 * 2 * 2 * 2 * 2) + …

So for the 1st term here (4/3), the prime number 2 is involved; then for the next term  (9/8) the prime number 3 is involved in the denominator and so on indefinitely over all the primes in this manner.


Then again with respect to the Zeta 1 (Riemann) function, we take the sum over natural numbers expression (for s = 2) i.e. 1 + 1/4 + 1/9 + 1/16 + …, each of these individual terms can equally be expressed as an infinite sum of products over a natural number entailing the Zeta 1 function.

Thus the 1st term, i.e. 1 =  {1 + 1/2 + 1/22 + 1/23 + …} – 1

= {1 + 1/(12 + 1)  + 1/(12 + 1)2 + 1/(12 + 1)3 + …} – 1.

So the 1st natural number 1 is used here in the denominator.

The 2nd term i.e. 1/4 = {1 + 1/5 + 1/52 + 1/53 + …} – 1

= {1 + 1/(22 + 1) + 1/(22 + 1)2 + 1/(22 + 1)3 + …} – 1.

So the 2nd natural number 2 is used here in the denominator.

The 3rd term i.e. 1/9 = {1 + 1/10 + 1/102 + 1/103 + …} – 1

= {1 + 1/(32 + 1) + 1/(32 + 1)2 + 1/(32 + 1)3 + …} – 1.

So the 3rd natural number 3 is used here in the denominator. And we can continue on indefinitely in this manner with all the natural numbers appearing in the denominator terms.

However in both cases here, we are obtaining the infinite sum of product terms, entailing with respect to the denominator all the primes and natural numbers respectively.

Monday, September 18, 2017

Ratios and Sums of Reciprocals

In previous entries, I have shown the first 9 terms (as an illustration) in the respective unique number sequences for the 9 equations from (x – 1)1 = 0 to (x – 1)9 = 0 .
    1
   1
   1
    1
     1
    1
    1
     1
    1
    1
   2
   3
    4
     5
    6
    7
     8
    9
    1
   3
   6
   10
    15
   21
   28
    36
   45
    1
   4
  10
   20
    35
   56
   84
   120
  165
    1
   5
  15
   35
    70
  126
  210
   330
  495
    1
   6
  21
   56
   126
  252
  462
   792
 1287
    1
   7
  28
   84
   210
  462
  924
  1716
 3003
    1
   8
  36
  120
   330
  792
 1716
  3432
 6435
    1
   9
  45
  165
   495
 1287
 3003
  6435
12870


Now we can always derive the unique number sequences for (x – 1)n + 1 from the corresponding sequences for (x –1)n through application of the fact that the kth term in the former = the sum of the first k terms in the latter sequence.

Therefore as we can see in the unique number sequence for (x – 1)3 i.e. 1, 3, 6, 10, …, the first term 1, represents the sum of the first single term in the corresponding sequence for sequence for (x – 1)2 i.e. 1, 2, 3, 4, …; the second term, 3 then represents the sum of the first 2 terms in the corresponding sequence i.e. 1 + 2; the 3rd term, 6 represents the sum of the first 3 terms in the previous sequence i.e. 1 + 2 + 3; the fourth term, 10 then represents the sum of the first 4 terms in the previous sequence i.e. 1 + 2 + 3 + 4 and so on indefinitely.

However the limitation of this procedure is that we must already know the unique number sequence corresponding to (x – 1)n to be able to calculate the corresponding sequence for (x – 1)n + 1.    


However there is a simple way to calculate independently the unique number sequence corresponding to (x – 1) n for any given n.

The bais for this calculation is that in general terms the ratio of the (k + 1)th to the kth term
i.e. (k + 1)th/ kth = (k + n – 1)/k

So for example in the sequences above, when n = 4, the unique digit sequence for (x – 1)4
is given by 1, 4, 10, 20, 35, 56, 84, 120, 165,…

So if for example we take k = 6 then the  ratio of the 7th to the 6th term i.e. 7th/6th = (6 + 4 – 1)/6 = 9/6 (i.e. 3/2).

And as we can see, this is indeed true for the 7th term = 84 and the 6th term = 56 and 84/56 = 3/2.

So aided with this simple general fact, regarding the ratio of successive terms, to illustrate, I will now calculate the unique number sequence corresponding to (x – 1)12.
Now the first term is - as always - 1.
Therefore the  2nd term  =1 * (1 + 12 – 1)/ 1 = 12.
The 3rd term then = 12 * (2 + 12 – 1)/2 = 12 * 13/2 = 78.
The 4th term = 78 * (3 + 12 – 1)/3 = 78 * 14/3 = 364.
The 5th term = 364 * (4 + 12 – 1)/4  = 364 * 15/4 = 1365.
The 6th term = 1365 * (5 + 12 – 1)/5 = 1365 * 16/5 = 4368.
The 7th term = 4368 * (6 + 12 – 1)/6 = 4368 * 17/6 = 12376.

So the unique digit sequence corresponding to (x – 1)12 is

1, 12, 78, 364, 1365, 4368, 12376, …

Now we already know that the 12 roots of this equation are 1.

However if we attempt to approximate these roots through the ratio of (k + 1)th/kth terms, we must include a great number of terms so as to get a valid approximation.

So ultimately (k + 1)th/kth  term  ~ 1 (when k is sufficiently large).    


I then looked at the sum of reciprocals for these unique number sequences to find that an interesting general pattern was at work.

Clearly from a conventional perspective, the sums of reciprocals of the numbers associated with the first two sequences for (x – 1)1 and (x – 1)2 respectively, diverge.

However the sum of reciprocals of the sequence, corresponding to (x – 1)3
i.e. 1 + 1/3 + 1/6 + 1/10 + 1/15 + …  converges to 2.

In fact a general result can be given for all such convergent sequences with respect to the sums of reciprocals of all the unique number sequences associated with (x – 1)3 where n ≥ 3.

This result in fact depends solely on the value of n (as the dimensional power or index) and is given simply as (n – 1)/(n – 2).


So for example, the sum of reciprocals of the next number sequence, corresponding to
(x – 1)4, i.e.

1 + 1/4 + 1/10 + 1/20 + 1/35 + …, converges to (4 – 1)/(4 – 2) = 3/2.

And the sum of reciprocals corresponding to the number sequence uniquely associated with
(x – 1)12, that we earlier calculated i.e.

1 + 1/12 + 1/78 + 1/364 + 1/1365 + 1/4368 + 1/12376 + … thereby converges to (12 – 1)/(12 – 2)  = 11/10 = 1.1.

In fact the sum of the first 7 terms above = 1.09994…, which is already very close to the postulated answer.

Monday, September 4, 2017

Imaginary Numbers

Though imaginary numbers are now widely employed in Mathematics, as yet there appears to be very little appreciation as to their holistic significance (which is crucially important).

Expressed in an equivalent manner, through the importance of imaginary numbers is now well established in quantitative terms, their corresponding qualitative significance is not yet formally recognised.

However we will now indirectly probe the true nature of imaginary numbers through the use of the unique digit sequence associated with the polynomial equation x2 = – 1, i.e

x2 + 1 = 0.

Now the unique digit sequence associated with this equation is 0, 1, 0, – 1, 0, 1, 0, – 1, …

Now when we take the ratio of nth/(n – 2)th terms, from one perspective we obtain – 1/1 or alternatively 1/– 1 = – 1.

This therefore suggests indeed that x2  = – 1 (which was our starting point).

However from an alternative perspective, the ratio of the nth/(n – 2)th terms = 0/0.

Again, as we saw in the last entry, this result is somewhat meaningless from a Type 1 linear perspective (where number is viewed in an independent manner).

So the result of 0/0 suggests a relative relationship involving interdependent - rather than independent - units.

In this respect, x2 + 1 = 0 with the two imaginary roots of x, is similar to the the two real roots of x2 – 1 = 0 (which we dealt with in the last entry) where interpretation properly entails both linear notions of independence and circular notions of interdependence respectively.

The difference however with respect to the imaginary roots is that the sign of 1 keeps switching alternatively as between positive and negative values, and herein lies the true clue as to the nature of imaginary numbers.


Once again in conventional mathematical terms, based on linear rational notions of quantitative independence, positive (+) and negative (–) signs are considered as absolutely independent. So as we have seen from this perspective, a left turn at a crossroads (+ 1) is thereby clearly separated from a right turn (+ 1).

However there is equally - a largely unrecognised - qualitative interdependent interpretation of number, of a purely relative nature, where each ordinal position is fully interchangeable with each other ordinal position.

So in this case of two items, (+) and negative (–) signs become fully interchangeable.

Thus as we have seen in our crossroads example, when one envisages the approach to the crossroads from two opposite directions simultaneously, a left turn (+ 1) cannot be distinguished from a right turn (– 1) and vice versa. 
Therefore left and right turns are now purely relative.

So the former understanding of absolute independence corresponds to linear rational type appreciation; however the latter understanding of relative interdependence corresponds to circular intuitive type appreciation.
And we can refer to the former as analytic and the latter as holistic type understanding respectively.

Now, both of these two types of understanding (analytic and holistic) are necessarily involved in the interpretation of left and right turns at a crossroads!

However in conventional terms - consistent with the absolute quantitative interpretation of number as points on a real number line - the analytic is solely recognised in formal terms.

However, the latter holistic aspect is then recognised in conventional mathematical terms, through “conversion” into analytic meaning of an “imaginary” rather than “real” nature.

Therefore though the imaginary points directly to the holistic meaning of interdependence of a qualitative nature, through being then indirectly converted in an analytic independent manner, this enables the new imaginary aspect of number to be treated in a quantitative manner.

So in analytic terms i (as the imaginary unit) = √– 1 i.e. (– 1)1/2.

This therefore provides the indirect quantitative means of expressing a 2-dimensional (circular) notion as a point on the unit circle (in the complex plane) in a 1-dimensional i.e. linear manner. 

Now if we look at 2 holistically i.e. through the Type 2 aspect of the number system, it is represented as 12. So 2 here represents the 1st and 2nd dimensions respectively that are  + 1 and – 1 respectively.

Now + 1 as the 1st dimension is independent (like the unambiguous identification of a left or right turn at a crossroads). However the 2nd dimension then properly relates to the interdependence of two units (as in the case of the crossroads the interdependence of both left and right turns).

Properly understood when we use – 1, in this holistic interdependent context, it becomes dynamically related to + 1. And the interaction of both + 1 and – 1, just like  matter and anti-matter particles in physics, leads to a fusion in energy.
And we refer to this psycho-spiritual fusion in understanding (where opposite polarities are recognised as complementary) as intuition.

From an equivalent perspective, in holistic terms + 1 (where literally a phenomenon is posited in understanding) represents the conscious direction of understanding.

– 1, then as the corresponding (dynamic) negation of the posited unit, represents the unconscious direction of understanding.

So both conscious and unconscious aspects interact in mathematical understanding through the corresponding interaction of rational and intuitive understanding.

However, though the importance of intuition - especially for creative work - is recognised in conventional terms, invariably it is reduced in a merely conscious rational manner. And strictly this greatly distorts the true nature of mathematical understanding!


So the key function of the imaginary notion in Mathematics is to enable the unconscious aspect of understanding with respect to number (which is of a holistic intuitive nature) be indirectly assimilated in a conscious rational manner.

Put yet again in an equivalent manner, the imaginary notion serves to convert the qualitative aspect of number indirectly in a quantitative manner.

So for example 1 (one) represents the real conscious understanding of the quantitative notion of a unit in a rational manner.

Then 1 (oneness) represents the corresponding unconscious appreciation of the qualitative notion of a unit in an intuitive manner.

And in the dynamics of understanding quantitative and qualitative notions are positive (+) and negative (–) with respect to each other. 

Then √– 1 (=  + i or – i) indirectly represents in a quantitative rational manner the corresponding qualitative aspect of the unit (which is inherently of a holistic intuitive nature). And this indirect understanding represents the imaginary - as opposed to the real - aspect of number.

However just as all analytic quantitative notions have holistic qualitative counterparts in Mathematics, likewise this also implies to the imaginary notion.


Thus when appropriately understood i.e. in a dynamic interactive manner, the Riemann Hypothesis can be given a simple - yet compelling - interpretation.

As is well known this postulates that all the non-trivial zeros of the Riemann zeta function lie on an imaginary line (through ½).

Now in conscious rational terms it is assumed that all the real numbers lie on a straight line.
However this begs the question as to whether the assumption is also valid from an unconscious holistic perspective.

Again in conventional mathematical terms - because of conscious reductionism - it is just blindly assumed that the qualitative aspect corresponds with the quantitative.

So no distinction is made as between numbers as independent quantitative entities and the interdependent relationship as between these numbers (which is strictly of a qualitative nature).

However by resorting to the imaginary notion, it is thereby possible to isolate both quantitative and qualitative aspects (by indirectly representing the qualitative aspect in an imaginary quantitative manner).

Thus the requirement that all the (non-trivial) zeros lie on an imaginary line, in effect amounts to the requirement that the unconscious qualitative aspect of mathematical understanding - which is inherent in all interdependent relationships as between numbers - corresponds in a consistent manner with the conscious quantitative aspect (where numbers lie on a real straight line).

Therefore if all the zeros do not lie on an imaginary line, we can no longer have faith in the very consistency of the number system, with the entire mathematical edifice thereby built on faulty foundations.

However, clearly this most fundamental question of all cannot be proven within the accepted mathematical axioms (as they already blindly assume the truth of the Riemann hypothesis).