Start at Basel Problem

In 18th century, Euler solved a well known problem, even Leibniz cannot solve it, either. This event made him become famous and confident. Basel problem shows \(\displaystyle \lim_{n \to +\infty} \left(\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\cdots+\dfrac{1}{n^2}\right)= \sum_{m=1}^{\infty} \dfrac{1}{m^2}=\dfrac{\pi^2}{6}\). It is a wonderful theory, but I’m not satisfied with this. We can use it to find more interesting conclusions. \[\begin{align*} \because \dfrac{\pi^2}{6} &=\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\cdots \\ \therefore \dfrac{1}{2^2} \times \dfrac{\pi^2}{6} &=\dfrac{1}{2^2} \times \dfrac{1}{1^2}+\dfrac{1}{2^2} \times \dfrac{1}{2^2}+\dfrac{1}{2^2} \times \dfrac{1}{3^2}+\cdots \\ \therefore \dfrac{1}{2^2} \times \dfrac{\pi^2}{6} &=\dfrac{1}{2^2}+\dfrac{1}{4^2}+\dfrac{1}{6^2}+\cdots \\ \therefore \left(1-\dfrac{1}{2^2}\right) \dfrac{\pi^2}{6} &=\dfrac{1}{1^2}+\dfrac{1}{3^2}+\dfrac{1}{5^2}+\cdots \\ \therefore \dfrac{\pi^2}{8} &=\dfrac{1}{1^2}+\dfrac{1}{3^2}+\dfrac{1}{5^2}+\cdots \\ \dfrac{\pi^2}{24} &=\dfrac{1}{2^2}+\dfrac{1}{4^2}+\dfrac{1}{6^2}+\cdots \\ \therefore \left(\dfrac{1}{8}-\dfrac{1}{24}\right)\pi^2 &=\dfrac{1}{1^2}-\dfrac{1}{2^2}+\dfrac{1}{3^2}-\cdots \\ \therefore \dfrac{\pi^2}{12} &=\dfrac{1}{1^2}-\dfrac{1}{2^2}+\dfrac{1}{3^2}-\cdots \\ \end{align*}\] It’s true that Basel problem has an important relationship with Riemann \(\zeta\) function, because Basel problem is in order to find \(\zeta(2)\). About 100 years later, Riemann used one equation to build a bridge between functions and prime numbers. The equation is \(\displaystyle \zeta(s) =\sum_{n=1} ^{\infty} \dfrac{1}{n^s}= \prod_{p \text{ prime}} ^{\infty} (1-p^{-s})^{-1}\). Here is Euler's proof. \[\begin{align*} \because \zeta(s) &=\dfrac{1}{1^s}+\dfrac{1}{2^s}+\dfrac{1}{3^s}+\cdots \\ \therefore \dfrac{1}{2^s} \zeta(s) &=\dfrac{1}{2^s}+\dfrac{1}{4^s}+\dfrac{1}{6^s}+\cdots \\ \therefore \left(1-\dfrac{1}{2^s}\right)\zeta(s) &=\dfrac{1}{1^s}+\dfrac{1}{3^s}+\dfrac{1}{5^s}+\cdots \\ \because \dfrac{1}{3^s} \times \left(1-\dfrac{1}{2^s}\right)\zeta(s) &=\dfrac{1}{3^s} \times \dfrac{1}{1^s}+\dfrac{1}{3^s} \times \dfrac{1}{3^s}+\dfrac{1}{3^s} \times \dfrac{1}{5^s}+\cdots \\ \therefore \left(1- \dfrac{1}{3^s}\right) \left(1-\dfrac{1}{2^s}\right)\zeta(s) &=\dfrac{1}{1^s}+\dfrac{1}{5^s}+\dfrac{1}{7^s}+\cdots \\ \therefore \left(1-\dfrac{1}{5^s}\right)\left(1- \dfrac{1}{3^s}\right)\left(1-\dfrac{1}{2^s}\right)\zeta(s) &=\dfrac{1}{1^s}+\dfrac{1}{7^s}+\dfrac{1}{11^s}+\cdots \\ \therefore 1 &=\zeta(s)\left(1-\dfrac{1}{2^s}\right)\left(1- \dfrac{1}{3^s}\right)\left(1-\dfrac{1}{5^s}\right)\cdots \\ \therefore \zeta(s) &=\dfrac{1}{\left(1-\dfrac{1}{2^s}\right)\left(1- \dfrac{1}{3^s}\right)\left(1-\dfrac{1}{5^s}\right)\left(1-\dfrac{1}{7^s}\right)\cdots} \\ \therefore \sum_{n=1} ^{\infty} \dfrac{1}{n^s} &= \prod_{p \text{ prime}} ^{\infty} (1-p^{-s})^{-1} \end{align*}\]

Note by Edward Christian
1 year, 5 months ago

No vote yet
1 vote

  Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

  • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
  • Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
  • Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
  • Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link]( link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 i=13 \sum_{i=1}^3
\sin \theta sinθ \sin \theta
\boxed{123} 123 \boxed{123}


Sort by:

Top Newest

Log in to reply

Are you actually 13?

Joshua Olayanju - 1 year, 1 month ago

Log in to reply

My friend, you’re just 12, too.

Edward Christian - 10 months, 3 weeks ago

Log in to reply


Problem Loading...

Note Loading...

Set Loading...