## 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.