The Basel Problem: A double integral solution

Рет қаралды 22,525

Күн бұрын

The Basel problem is an all-time classic math problem. Among all of its solutions, my favorite is based around the computation of a double integral.
Here is a link to the source of this solution: www.researchgate.net/publicat.... In fact, using the same technique from the video, one can evaluate the infinite sum of 1/n^(2k) for any (positive integer) k!
A slightly different double integral solution can be found here: analysisofmath.wordpress.com/....
Obviously, more reading about the problem can be found here: en.wikipedia.org/wiki/Basel_p...

Пікірлер: 54

@uhbayhue 3 жыл бұрын

My mind is blown, this is incredibly elegant

@hosz5499 25 күн бұрын

GOAT change of variable is amazing! Its logic would be simpler if defining (x=sinu/sinv, y=cosv/cosu) in a triangle 0

@awindwaker4130 2 жыл бұрын

I wouldn't have thought of that... but wow, my new favorite way to solve the Basel problem.

@mohitsinha2732 Жыл бұрын

Absolutely Brilliant! The substitution is simply Mind-blowing!

@Someone-cr8cj 3 жыл бұрын

you are criminally undersubscribed to. this is amazing.

@hymanimy 2 жыл бұрын

This is such a sick way to solve the Basel Problem. I'm always looking for new ways to solve it other than the way Euler did it with the function sin(x)/x. I've got a video on my channel solving the problem by exploiting complex numbers but this solution is so much more elegant. Plus I love the calm way you lead us through the proof - it makes all the steps feel incredibly natural yet clever. Awesome video!

@strikerstone 5 ай бұрын

i have never seen such good solution with so nice quality for basel problem , thank you and keep making math videos

@izaret 2 жыл бұрын

Unity between substance and form which underlies great art is remarkable. The elegance of the solution is mirrored by the style of the video, calmly going through the proof with chill musical background. The explanation both takes the time to decompose the steps yet does not over-explain. Well-pitched. Like Grant of 3b1b, Mathologer or Math Parker, you have a unique style which is likely to help people discover math ideas under a new light even if the subject has been treated somewhere else. Keep going, you have something.

@dragonheart2696 3 жыл бұрын

Thank you for your well-done job

@ayushdeep7900 3 жыл бұрын

You will do great man, keep doing this

@rkmoitra123 Жыл бұрын

Marvellous solution... just the best solution to the Basel problem..

@emmanuelldx7788 Жыл бұрын

Excellent video, very well explained.

@nishchayy 2 жыл бұрын

So cool. Thank you for sharing this. Amazing

@HaliPuppeh 10 ай бұрын

What the holy actual F??? You made this insanely easy to understand and this trick is awesome.

@helmutriedel2647 2 жыл бұрын

Dear Joe Breen, wow, what a brilliant solution of the Basel problem. Did you ever think about generalizing this technique to larger n? While there is a formula for even n, there are no closed expressions for odd n to my knowledge. If I rewrite the so-called Apery constant zeta(3) using your technique, I get the triple integral zeta(3) = 8/7 int_0^1 int_0^1 int_0^1 f(x,y,z) dx dy dz with f(x,y,z) = 1/(1-x^2 y^2 z^2) Now, a variable substitution is needed, such that the determinant of the Jacobian is just 1/f(x,y,z) leading to the integrad of 1. Then, zeta(3) would just be the volume defined by the transformed boundaries. In analogy to your solution for n=2, I tried x = sin u / cos v, y = sin v / cos w, z = sin w / cos u. The transformed integrand is g(u,v,w) = 1 / (1 - tan^2 u tan^2 v tan^2 w), but the determinant of the Jacobian is det(J) = 1 + tan^2 u tan^2 v tan^2 w. I could not resolve the issue with the different signs so far. What do you think about this approach?

@b.afreeshooters146 3 ай бұрын

try inverting the sign one of the variable substitutions, just a desperate guess.

@ilafya 2 жыл бұрын

You are the man Man

@nasim09021975 Жыл бұрын

Very nice video 🤟😄 Very clear explanation 😊

@gavintillman1884 Жыл бұрын

Amazing. To think discoveries like this being made after I did my maths degree.

@didierchaumet 2 жыл бұрын

Brilliant!

@jozsefgurzo8777 Жыл бұрын

This change of variables is just soo sick, blew my mind. However there's something I'm not entirely sure about. When you solved the inequalities, you didn't cover the case when the cosines are negative and thus the sinses are negative as well. I guess it would lead to a contradiction, cause the result is correct, but it's such a bittersweet feeling. After all of these incredible ideas, it just doesn't feel right to neglect this case. Thanks for the effort tho! I really enjoyed the video!

@NateROCKS112 11 ай бұрын

Since we don't want any duplicates when performing our map, we need to define a suitable inverse function. The standard codomain of arcsin is [-pi/2, pi/2], and it's [0, pi] for arccos. You could restrict them differently, but they need to actually function as left and right inverses. While I don't believe it'd lead to a contradiction if you defined arcsin and arccos differently, you'd have to carefully the relationships between u and v when computing the area. It's a lot easier to just use the principal branch.

@jlsown 4 жыл бұрын

This is some cool stuff

@markferrufino5038 4 жыл бұрын

We love a cool math video 🤝

@joebreenmath7442 4 жыл бұрын

Gotta provide for the world in quarantine

@ivywoodxrecords 3 жыл бұрын

The 4/3 here is indicative of a basis of 3D volumes. Consider the generation of sequential growth resulting in volumetric definition of geometries in 3D yet defined in the 2D reals. Holographic representation of an additional dimension, via the definition of circular geometries within the summation of zeta(2)

@revanthk61 Жыл бұрын

Impressive bro

@kailasnathastro 8 ай бұрын

Amazing

@costelnica3988 Жыл бұрын

Wow! Super!

@hosz5499 24 күн бұрын

how do you generalise to evaluate zeta(4) with integrate_0^1 1/(1-xyzw) dx dy dz dw --> Pi^4/90?

@redroach401 2 ай бұрын

can someone explain the triangle thing to me please and what would the new bounds be?

@ms070965 Жыл бұрын

Vert nice

@jasonlin5884 Ай бұрын

But the mapping of domain to codomain ( or vice versa) seems very strange and hard to understand.(for example What would be the value of (u,v) when x=0,y=0). It is not one to one mapping. Thus the equality of integration (with Jacobian factor ) can still hold true?