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+ 1/3–2 + 1/4–2 + … = 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
So with s1 = – 2, we obtain,
1/(1 – 1/2–2) * 1/(1 – 1/3–2) * 1/(1 – 1/5–2) * ...
= (– 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
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+ 1/3–2 + 1/4–2 + … = 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 – 1/2–2) * 1/(1 – 1/3–2) * 1/(1 – 1/5–2) * ...
= (– 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
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