Question about Riemann Zeta Function

Click For Summary
SUMMARY

The discussion centers on the Riemann Zeta Function, specifically the evaluation of ζ(s) for complex values such as ζ(1/2 + i). Participants clarify that Euler's Identity is not applicable, while Euler's Formula, e^(ix) = cos(x) + i sin(x), is relevant for understanding complex exponentiation. The critical line for the imaginary zeros of the function is defined as Re(s) = 1/2. To compute ζ(1/2 + i), it is suggested to calculate the first m terms of the series and observe convergence.

PREREQUISITES
  • Understanding of complex numbers and their properties
  • Familiarity with the Riemann Zeta Function
  • Knowledge of Euler's Formula and its implications
  • Basic calculus concepts related to series convergence
NEXT STEPS
  • Study the properties of the Riemann Zeta Function in the complex plane
  • Learn about series convergence and techniques for evaluating infinite series
  • Explore Euler's Formula in depth and its applications in complex analysis
  • Investigate the significance of the critical line in the context of the Riemann Hypothesis
USEFUL FOR

Mathematicians, students studying complex analysis, and anyone interested in the properties and applications of the 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...
 
Physics 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.
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 17 ·
Replies
17
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 0 ·
Replies
0
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 10 ·
Replies
10
Views
4K