Converging to the Basel Problem: Solving for Poles on the Real Axis

  • Context: MHB 
  • Thread starter Thread starter alyafey22
  • Start date Start date
Click For Summary
SUMMARY

The discussion centers on the Basel Problem, specifically the convergence of the series $$\sum_{k\geq 1} \frac{1}{k^2}$$ which equals $$\frac{\pi^2}{6}$$. Various methods to solve this problem are explored, including Euler's use of Bernoulli numbers and a modified theorem for poles on the real axis. The adjusted theorem states that $$\sum_{k\leq -1} \frac{1}{k^2} + \sum_{k \geq 1} \frac{1}{k^2} = - \text{Res}\, \left(\frac{\pi \cot(\pi z)}{z^2};0 \right)$$, leading to the conclusion that $$\sum_{k\geq 1} \frac{1}{k^2} = \frac{\pi^2}{6}$$.

PREREQUISITES
  • Understanding of analytic functions and their properties
  • Familiarity with residue theory in complex analysis
  • Knowledge of Bernoulli numbers and their applications
  • Basic grasp of series convergence and divergence
NEXT STEPS
  • Study the application of residue theory in complex analysis
  • Explore the properties and applications of Bernoulli numbers
  • Learn about analytic continuation and its implications for series
  • Investigate alternative proofs of the Basel Problem
USEFUL FOR

Mathematicians, students of complex analysis, and anyone interested in advanced series convergence techniques and the Basel Problem.

alyafey22
Gold Member
MHB
Messages
1,556
Reaction score
2
$$\sum_{k\geq 1} \frac{1}{k^2} = \frac{\pi^2}{6}$$

Let us see how many different methods can we get (Cool)
 
Physics news on Phys.org
Euler used the Bernoulli numbers to solve the problem

But there are other ways

Fourier series , Complex analysis
 
$$\sum_{k = -\infty}^{\infty} f(k) = - \sum_{i\geq 0} \text{Res}\, \left( \pi \cot(\pi z) f(z);z_i \right) $$

This requires the function to be analytic on the real axis but $$\frac{1}{k^2}$$ has a pole of order $$2$$ at the origin .

So we can adjust the theorem to solve for poles on the real axis

$$\sum_{k\leq -1} \frac{1}{k^2} +\sum_{k \geq 1} \frac{1}{k^2}= - \text{Res}\, \left(\frac{\pi \cot(\pi z)}{z^2};0 \right)$$

$$\sum_{k\geq 1} \frac{1}{k^2} +\sum_{k\geq 1} \frac{1}{k^2}= \frac{\pi^2}{3}$$

$$\sum_{k\geq 1} \frac{1}{k^2} = \frac{\pi^2}{6}$$

I deleted the proof of the modification of the theorem above . If someone is interested I will try to post it.
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 21 ·
Replies
21
Views
4K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 9 ·
Replies
9
Views
2K