## Sunday, May 6, 2018

### Pentagonal Number Theorem (3)

Let us return briefly to the formula given in the previous entry i.e.

(1 – x)(1 – x2)(1 – x3)(1 – x4) … = 1 – 1/{(1/x – 1)} + 1/{(1/x – 1)(1/x2 – 1)} –
1/{(1/x – 1)(1/x2 – 1)(1/x3 – 1)} + 1/{(1/x – 1)(1/x2 – 1)(1/x3 – 1)(1/x4 – 1)} – …

We have seen that when x = 1/2, this results in the relationship,

1/2 * 3/4 * 7/8 * 15/16 * …  = 1 – 1/(1) + 1/(1 * 3) – 1/(1 * 3 * 7) + 1/(1 * 3 * 7 * 15) – …

Then when x = 1/3, the following relationship results,

2/3 * 8/9 * 26/27 * 80/81 … = 1 – 1/(2) + 1/(2 * 8) – 1/(2 * 8 * 26) + 1/(2 * 8 * 26 * 80) – …

When x = 1/5, we get,

4/5 * 24/25 * 124/125 * 624/625 … = 1 – 1/(4) + 1/(4 * 24) – 1/(4 * 24 * 124) +
1/(4 * 24 * 124 * 624) – …

And when x = 1/7, we get,

6/7 * 48/49 * 342/343 * 2400/2401 … = 1 – 1/(6) + 1/(6 * 48) – 1/(4 * 48 * 342) +
1/(4 * 48 * 342 * 2400) – …

If we now look at the product over primes version of the Riemann Zeta Function for ζ(s)
s=1
we get the following

ζ(1) = 2/1 * 3/2 * 5/4 * 7/6 * …

ζ(2) = 4/3 * 9/8 * 25/24 * 49/48 * …

ζ(3) = 8/7 * 27/26 * 125/124 * 343/342 * …

ζ(4) = 16/15 * 81/80 * 625/624 * 2401/2400 * …

What is remarkable here is that what represent rows of numbers with respect to the Riemann Zeta function, represent similar columns with respect to the reciprocals of the products generated through our formula i.e.

(1 – x)(1 – x2)(1 – x3)(1 – x4) … = 1 – 1/{(1/x – 1)} + 1/{(1/x – 1)(1/x2 – 1)} –
1/{(1/x – 1)(1/x2 – 1)(1/x3 – 1)} + 1/{(1/x – 1)(1/x2 – 1)(1/x3 – 1)(1/x4 – 1)} – …
So for example the 1st row with respect to the Riemann Zeta function is,

2/1 * 3/2 * 5/4 * 7/6 * …

And the corresponding 1st column with respect to the products arising from the formula is

1/2 * 2/3 * 4/5 * 6/7 * …

This suggests therefore that there is an intimate relationship here as between the stated formula (that generates the product results) and the corresponding formula that generates the product results for the Riemann zeta function.

Thus is we were to generate a n * n grid (i.e. with n rows and n columns) with respect to the product entries for both our formula and the Riemann zeta function, the combined products of products (whereby the values of all rows and then values of all columns are multiplied together) in either case would represent the reciprocal of the alternative result. However we would need here to exclude the 1st row and 1st column to ensure a finite result!

And it has to be remembered in this context that likewise that each product over primes expression can be given an alternative sum over the natural numbers expression.

So in the case of the Riemann zeta function, the corresponding sum over the natural numbers expression for the corresponding product over primes expression is the harmonic series i.e.

1/2 * 3/2 * 5/4 * 7/6    =   1/11 + 1/21 + 1/31 + 1/41 + … (which in this one particular case is divergent).

Thus strictly speaking we are obtaining here the sum over the natural numbers for the base aspect of number where by contrast the dimensional aspect is fixed in each case to just one of the natural numbers (in this case 1).

Then with respect to our formula the sum over natural numbers expression again is given as
1 – 1/{(1/x – 1)} + 1/{(1/x – 1)(1/x2 – 1)} –
1/{(1/x – 1)(1/x2 – 1)(1/x3 – 1)} + 1/{(1/x – 1)(1/x2 – 1)(1/x3 – 1)(1/x4 – 1)} – …

So here in complementary fashion it is the dimensional aspect that varies over the natural numbers with the base aspect in each given case fixed.

Thus when x = 1/2 with 1/x = 2, we have

1/2 * 3/4 * 7/ 8 * 15/16 * …  =  1 – 1 + 1/3 – 1/21 + 1/315 – …

We have seen already how the Zeta 2 function can be used to provide a similar vertical (column) readout of the entries for the Riemann Zeta function where terms are added (rather than multiplied)

Now excluding values for s = 1, the sum of 1st column will be 1/4 + 1/8 + 1/16 + 1/32 +
= (1 + 1/4 + 1/8 + 1/16 + 1/32 + …)  – 1.

And this corresponds to the Zeta 2 function (1 + x + x2 + x3 + …) – 1, with x = (1/p2 – 1) (where p = 2) = 1/2.

And we used this earlier relationship to prove that ∑{ζ(s) – 1)} = 1.
s=2