Question about Riemann Zeta Function

willr12
Messages
17
Reaction score
2
I understand how to calculate values of positive values ζ(s), it's pretty straightforward convergence. But when you expand s into the complex plane, like ζ(δ+bi), how do you assign a value with i as an exponent? Take for example
ζ(1/2 + i)
This is the sequence
1/1^(1/2+i) + 1/2^(1/2+i) + 1/3^(1/2+i) ...
How do you assign a value to this? do you have to use euler's identity to calculate it? Or am I looking at it all wrong?
P. S. Try to dumb it down for me. I'm in algebra 2 right now...
 
Mathematics news on Phys.org
First, Euler's Identity has nothing to do with the Reimann-Zeta function. Euler's Identity is just
##e^{i \pi} + 1 = 0##
Euler's Formula is probably what you're thinking of (which is where the above identity comes from):
##e^{ix} = \cos x + i \sin x##
But there's no exponential function (it works that way because the exponential function is re-written with a power series to get the above formula), and ##i## would have to be factored out of the exponent entirely to rewrite it that way (it's only in one of the terms).
But, in the case you're talkng about where ##s## is defined as ##\sigma + bi## for ##0 < \sigma < 1##, then it can be left as is, but the imaginary zeroes are plotted along the "critical line", ##\Re(s) = \frac{1}{2}##

http://mathworld.wolfram.com/RiemannZetaFunction.html
 
Last edited:
I'd say, calculate the first m terms, and the first m+1, m+2... and see what it converges to.
 
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...
I'm interested to know whether the equation $$1 = 2 - \frac{1}{2 - \frac{1}{2 - \cdots}}$$ is true or not. It can be shown easily that if the continued fraction converges, it cannot converge to anything else than 1. It seems that if the continued fraction converges, the convergence is very slow. The apparent slowness of the convergence makes it difficult to estimate the presence of true convergence numerically. At the moment I don't know whether this converges or not.
Back
Top