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/{(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
No comments:
Post a Comment