Oddball Zeta functional equation: uh?

  • Context: Graduate 
  • Thread starter Thread starter benorin
  • Start date Start date
  • Tags Tags
    Functional
Click For Summary
SUMMARY

The discussion centers on a specific functional equation related to the Riemann zeta function, as found in "Table of Higher Functions, 6th ed." by Jahnke, Emde, & Losch. The equation is expressed as z(z+1)\frac{\zeta (z+2)\zeta (1-z)}{\zeta (z)\zeta (-1-z)}=-4\pi ^2. Participants highlight the analytic continuation of the zeta function, noting its validity across the complex plane except at z=1. The relationship between the zeta function and the chi function, defined as \chi(s)=\pi^{s-1/2}\frac{\Gamma(\frac{1-s}{2})}{\Gamma(\frac{s}{2})}, is also discussed, emphasizing the application of the functional equation for different values of s.

PREREQUISITES
  • Understanding of the Riemann zeta function and its properties
  • Familiarity with analytic continuation in complex analysis
  • Knowledge of the Gamma function and its functional equations
  • Basic grasp of functional equations in mathematical analysis
NEXT STEPS
  • Study the properties of the Riemann zeta function and its analytic continuation
  • Explore the Gamma function and its role in functional equations
  • Investigate the implications of the functional equation \chi(1-s)=\frac{\zeta(1-s)}{\zeta(s)} in detail
  • Review the derivation of functional equations for various values of s
USEFUL FOR

Mathematicians, researchers in number theory, and students studying complex analysis who are interested in the properties and applications of the Riemann zeta function.

benorin
Science Advisor
Insights Author
Messages
1,442
Reaction score
191
I found this functional equation for the Riemann zeta function in Table of Higher Functions, 6th ed. by Jahnke, Emde, & Losch on pg. 40:

[tex]z(z+1)\frac{\zeta (z+2)\zeta (1-z)}{\zeta (z)\zeta (-1-z)}=-4\pi ^2[/tex]​

any suggestions as to how one might consider such an equation, much less derive it? There is, for example, an analytic continuation of the zeta function to all points in the complex plane except z=1, see this thread for the particular continuation of which I speak.
 
Physics news on Phys.org
It's just the product of one of the more usual functional equations for different values of s to cancel out the gamma factors.

[tex]\chi(1-s)=\frac{\zeta(1-s)}{\zeta(s)}[/tex]

where

[tex]\chi(s)=\pi^{s-1/2}\frac{\Gamma(\frac{1-s}{2})}{\Gamma(\frac{s}{2})}[/tex]

so

[tex]\frac{\zeta(z+2)\zeta(1-z)}{\zeta(-1-z)\zeta(z)}=\chi(z+2)\chi(1-z)[/tex]

Where we've used the usual functional equation once with s=-1-z and once with s=z. The rest is straightforward with the usual functional equation of Gamma.
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 5 ·
Replies
5
Views
4K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
Replies
2
Views
6K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K