- #1
mustang19
- 75
- 4
Consider
Z(s)=Sum(1/N^s)
For n=1 to infinity.
Let s=(xi+1/2).
The divisor is then:
N^(xi+1/2)
This is equivalent to
(N^xi)(N^(1/2))
As n increases, the n^1/2 term will more greatly slow the increase of the divisor and accelerate z(s) away from zero. This means that zeroes will occur less often for larger n. The 1/2 term is necessary so that the zeta function z(s) will encounter fewer primes as it iterates.
Is this a valid understanding of z(s)' behavior?
Z(s)=Sum(1/N^s)
For n=1 to infinity.
Let s=(xi+1/2).
The divisor is then:
N^(xi+1/2)
This is equivalent to
(N^xi)(N^(1/2))
As n increases, the n^1/2 term will more greatly slow the increase of the divisor and accelerate z(s) away from zero. This means that zeroes will occur less often for larger n. The 1/2 term is necessary so that the zeta function z(s) will encounter fewer primes as it iterates.
Is this a valid understanding of z(s)' behavior?