## Monday, October 2, 2017

### Product Formulation of Alternative Zeta 2 Function

In a previous entry I drew attention to a complementary (alternative) version of the Zeta 2 function which I now refer to as Alt Zeta 2.

So the Alt Zeta 2 function is associated with the (infinite) sum of reciprocals of the unique number sequences associated with the general polynomial equation (x – 1)n = 0.

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

1, 3, 6, 10, 15, …

And the corresponding (infinite) sum of reciprocals is

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

So I refer to this sequence as Alt ζ2(3).

Now in general terms, I represent the Zeta 2 function as  ζ2(s2) = 1 + s21 + s22 + s23 + …

And where n is an integer > 2,  s2 = 1/(n – 1) and ζ2(s2) = Alt ζ2(n).

Therefore, when n = 3, s2 = 1/2;

So ζ2(1/2) = 1 + (1/2)1 + (1/2)2 + (1/2)3 + …  = 2;

Thus ζ2(1/2) = Alt ζ2(3), and in general terms ζ2{1/(n – 1)} = Alt ζ2(n).

Now I have commented before on the fact that Alt ζ2(n) represents numbers (as denominators) that result from an ordered summation of natural numbers.

Thus for example with respect to Alt ζ2(3) = 1 + 1/3 + 1/6 + 1/10 + 1/15 + …

1                               = 1
3   = 1 + 2                = 1 + (1 + 1)
6   = 1 + 2 + 3          = 1 + (1 + 1) + (1 + 1 + 1)
10 = 1 + 2 + 3 + 4    = 1 + (1 + 1) + (1 + 1 + 1) + (1 + 1 + 1+ 1) .

So we can see a definite order to these numbers where the denominator of the kth term represents the sum of the first k natural numbers.

And when n > 3 the denominator of the kth term still represents a compound ordered sum involving the first k natural numbers.

For example when n = 4, the (infinite) sum of unique associated reciprocals is,

1 + 1/4 + 1/10  + 1/20 + 1/35 + …

Now if to illustrate, we take the denominator of the 3rd term it can be shown to represent a compounded ordered sum entailing the first 3 natural numbers.

So 10 = 1 + 3 + 6 (the sum of the first 3 denominators terms of the previous sequence for n = 3.

Thus 10 = 1 + (1 + 2) + (1 + 2 + 3).

However there is another remarkable feature associated with these denominators entailing the ordered product of natural numbers.

So for example 3 = 3/1;

6   =  (4 * 3)/(1 * 2);

10 = (5 * 4 * 3)/(1 * 2 * 3);

15 = (6 * 5 * 4 * 3)/(1 * 2 * 3 * 4).

So starting with the initial value of denominator (= n/1), subsequent values are given as {(n + 1) * n}/(1 * 2), {(n + 2) * (n + 1) * n}/(1 *2 * 3), {(n + 3) * (n + 2) * (n + 1) * n}/(1 * 2 * 3 * 4), ...

In fact we can express these in an alternative fashion which shows that each number (as denominator) represents a unique combination.

So in general terms nCr  = the number of r possible combinations taken from a group of n items.

Now each number as denominator likewise represents a unique combination with respect to a certain number of items (defined by the value of n1) = n1!/{r!(n1 – r)!}

So for example 6 = (4 * 3 * 2 * 1)/(2 * 1)(1 * 2)  = 4!/{2!(4 – 2)!}

Thus in general terms, n1 = (n + k – 2) where k = cardinal number of kth term and n represents the denominator of the 2nd term in the infinite series associated with (x – 1)n = 0, r = k – 1 and n1 – r = n – 1 respectively.

Thus in terms of our definitions, n1!/{r!(n1 – r)!} = (n + k – 2)!/{(k – 1)! * (n – 1)!}

Thus when n = 3, the denominator of the 4th term (i.e. where k = 4)
= (3 + 4 – 2)!/(3! * 2!)

= 5!/(3!* 2!) = (5 * 4)/(1 * 2) = 20/2 = 10.

So the denominator, i.e. 10  represents the total number of possible combinations of 3 taken from 5 items.

And by definition the denominator of every term, in the infinite series of reciprocals of the unique numbers associated with (x – 1)n = 0, represents a unique combination of numbers.

This then provides a ready means for calculating any term in the infinite sequence for a given value of n.

Thus for example when n = 6, the denominator of the 5th term,

= (6 + 5 – 2)!/(5 – 1)!(6 – 1)!   =  9!/(4! * 5!)

= (9 * 8 * 7 * 6)/(1 * 2 * 3 * 4) = 3024/24  = 126.

Thus the 5th term of the (infinite) sum of reciprocals based on the unique numbers associated with
(x – 1)6 = 0 = 1/126.

Thus to sum up, the Alt ζ2(n) series can be defined both in terms of an ordered sum of the 1st k natural numbers (to a given term k) relating to the denominators of the terms of its associated series and an ordered product of the natural numbers representing the factorials of numbers involved in the respective unique combinations associated with (n + k – 2)!/(k – 1)!(n – 1)!

And the importance of these formulations in turn relates to the fact that every individual term of the Zeta 1 (Riemann) function i.e. ζ2(s1) with s1 > 1 - both in its sum over natural numbers and product over primes formats - can be expressed in terms of the Alt ζ2(n) series.