## Monday, March 13, 2017

### Extending to Negative Values

As we have seen, each of the terms in the product over primes formula for the Zeta 1 (Riemann) function (where s is a positive integer) can be expressed through the complementary Zeta 2 function.

In fact this is equally true with respect to each term of the Zeta 1 sum over natural numbers expression.

In general terms,

ζ(s1) = 1/1s1 + 1/2s1+ 1/3s1 + 1/4s1 + …

For example when s1 = 2,

ζ(2) = 1/12 + 1/22+ 1/32 + 1/42 + … = 1 + 1/4 + 1/9 + 1/16 + …

Now each individual term can be expressed in a Zeta 2 manner.

In general terms,

ζ(s2) =  1 + s21 + s22  + s23 + ...

Thus ζ(s2 – 1 = s21 + s22  + s23 + ...

To convert each term in the Zeta 1 expression to the corresponding Zeta 2 format,

let s2 = 1/(1 + ns1), where n = 1, 2, 3, ...

So, as in our example, when n = 2 and s1 = 2, ,

s2 = 1/(1 + 22), = 1/(1 + 4) = 1/5

Then in this case, ζ(s2 – 1 =  1/5 + 1/52 + 1/53 +... = 1/4.

For the next term in our example, s1 = 2, and n = 3.

So s2 = 1/(1 + 32)  = 1/10, with ζ(s2 – 1 = 1/10 + 1/102 + 1/103 +... = 1/9.

Therefore we can universally express all the individual variable terms (i.e. other than 1) in the Zeta 1 (Riemann) function, for all positive integers of s, through corresponding Zeta 2 expressions, (i.e. strictly ζ(s2 – 1, expressions).

In principle, it is possible to extend these notions for negative integers of s in the Zeta 1 (Riemann) function.

So when again for example, s1 = 2,
ζ( 2) = 1/1 2 + 1/2 2+ 1/32 + 1/42 + … = 1 + 4 + 9 + 16 + …

Then, when s2 = 1/(1 + ns1) with n = 2 and  s1 = 2,

s2 = 1/(1 + 2 2) = 1/(1 + 1/4) = 5/4

So, ζ(s2  = – {5/4 + (5/4)2 (5/4)3 + ...} = (1 5/4) = 1/4 = 4, where ζ(s2) for negative values = 1 {ζ(s2) – 1}.

Thus we have now replicated the corresponding term in the Zeta 1 (Riemann) function.

We can also express the product over primes expression for each individual term in the Zeta 1 (Riemann) function through a corresponding Zeta 2 expression.

So the general formula for the product over primes expression of the Zeta 1 function is given as

∏(1/(1 – 1/ps1)

So with s1 = 2, we obtain,

1/(1 1/22) * 1/(1 1/32) * 1/(1 1/52) * ...

= (– 1/3) * (– 1/8) * (– 1/25) * ....

So with p = 2 and  s1 = – 2, the 1st term in the product over primes expression is given as 1/3.

Then to express each term as a Zeta 2 infinite series, we let s2 = 1/ps1.

Thus again with p = 2 and s1 = – 2,

ζ(s2) =  1 + 4 + 42  + 43 + ...

= 1/(1 – 4) = – 1/3