How Do You Calculate and Apply the Riemann Zeta Function?

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SUMMARY

The Riemann Zeta Function, specifically for n = 0, yields a value of -1/2, which involves Bernoulli numbers. The formula for even integers is given by \(\zeta(2n) = \frac{(-1)^{n+1}B_{2n}(2\pi)^{2n}}{(2)(2n)!}\). The proof for the sum \(\frac{\pi^2}{6} = \sum_{n=1}^\infty \frac{1}{n^2}\) utilizes complex analysis and Cauchy's Generalized Integral Theorem, as detailed in Riemann's original paper. Euler's method for this sum involves expanding the sine product and equating coefficients with the Taylor series.

PREREQUISITES
  • Understanding of Riemann Zeta Function
  • Familiarity with Bernoulli numbers
  • Knowledge of complex analysis
  • Experience with Taylor series expansions
NEXT STEPS
  • Study the derivation of Bernoulli numbers and their applications in number theory
  • Learn about Cauchy's Generalized Integral Theorem
  • Explore the relationship between the Riemann Zeta Function and quantum mechanics
  • Investigate Euler's methods for series summation and sine function expansions
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Mathematicians, physicists, and students interested in advanced calculus, number theory, and the applications of the Riemann Zeta Function in various fields.

der.physika
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I was wondering how do you calculate the Riemann value, of a Riemann Zeta Function, for example the riemann zeta function for n = 0, is -1/2, which envolves a bernoulli number (what is a bernoulli number and what roll does it play in the Riemann Zeta Function), can anyone explain that to me? Also what do you apply the riemann zeta function to, besides Quantum Mechanics.

And also how do you go about finding the sum of a series (non geometrical)?

\zeta(2n)=\frac{(-1)^{n+1}B_{2n}(2\pi)^2^n}{(2)(2n)!}

<br /> \frac{\pi^2}{6}=\sum_{n=1}^\infty \frac{1}{n^2}<br />

Specifically, this one I want to know the proof for this sum, can anyone nice enough out there, please show me!
Step by step process
 
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The proof that I have seen uses a bit of complex analysis on the contour integral found in Riemann's original paper. http://www.maths.tcd.ie/pub/HistMath/People/Riemann/Zeta/EZeta.pdf is a good translation (the integral is found in the middle of the third page).
The integrand there is a power of x multiplied by the function whose derivatives give the Bernoulli numbers, so Cauchy's Generalized Integral Theorem can then find values for negative integers. Then the functional equation is used to find the values for positive even numbers.
I hope that other people can help you find a more elementary proof, but this is all I've seen.

The specific sum you ask about was found by Euler by expanding the product for sine (http://en.wikipedia.org/wiki/Wallis_product#Derivation_of_the_Euler_product_for_the_sine also has this proof on the same page) in terms of powers of x then equating coeficients with the Taylor series for sine.
 

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