My Math Forum  

Go Back   My Math Forum > Math Forums > Math

Math General Math Forum - For general math related discussion and news

LinkBack Thread Tools Display Modes
September 10th, 2017, 06:40 PM   #1
Joined: Jul 2014
From: Taiwan

Posts: 8
Thanks: 0

Riemann Hypothesis

P: ζ(sᵢ)=0
Q: sᵢ=σ+it
R: t≠0
S: σ=1/2

A: P≡[Q≡(R≡S)]

It is that ζ(1/2)=0 if the Proposition A doesn't hold, but actually, ζ(1/2)≈-1.4603545; therefore RH does hold.
777HENRY is offline  
September 11th, 2017, 06:38 PM   #2
Global Moderator
Joined: May 2007

Posts: 6,379
Thanks: 542

I am not sure what you are trying to say. However (using your notation) the Riemann hypothesis states that all zeroes for 0<σ<1 are on the line σ=1/2+it, not necessarily for t=0.
mathman is offline  
October 20th, 2017, 11:44 AM   #3
Math Team
Joined: Jan 2015
From: Alabama

Posts: 2,824
Thanks: 752

I presume you mean "for 0< |σ|<1".
Country Boy is online now  
October 20th, 2017, 02:13 PM   #4
Global Moderator
Joined: May 2007

Posts: 6,379
Thanks: 542

Originally Posted by Country Boy View Post
I presume you mean "for 0< |σ|<1".
No, sorry. I meant all σ = x + iy where 0 < x < 1.
mathman is offline  

  My Math Forum > Math Forums > Math

hypothesis, riemann

Thread Tools
Display Modes

Similar Threads
Thread Thread Starter Forum Replies Last Post
Riemann hypothesis Tkipp Number Theory 1 July 8th, 2017 06:08 PM
Riemann Hypothesis. mathbalarka Number Theory 0 October 31st, 2013 01:54 AM
Do you believe the Riemann hypothesis? Eureka Number Theory 1 September 12th, 2011 02:07 PM
The Riemann Hypothesis and what it say MyNameIsVu Complex Analysis 2 April 16th, 2009 08:51 AM
Riemann Hypothesis Whoever Number Theory 72 February 11th, 2009 07:03 AM

Copyright © 2017 My Math Forum. All rights reserved.