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!
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.
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?
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!
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.
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)
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?
My mind is blown, this is incredibly elegant
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!
GOAT change of variable is amazing! Its logic would be simpler if defining (x=sinu/sinv, y=cosv/cosu) in a triangle 0
i have never seen such good solution with so nice quality for basel problem , thank you and keep making math videos
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.
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?
try inverting the sign one of the variable substitutions, just a desperate guess.
What the holy actual F??? You made this insanely easy to understand and this trick is awesome.
you are criminally undersubscribed to. this is amazing.
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!
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.
Absolutely Brilliant! The substitution is simply Mind-blowing!
Marvellous solution... just the best solution to the Basel problem..
Amazing. To think discoveries like this being made after I did my maths degree.
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)
Thank you for your well-done job
We love a cool math video 🤝
Gotta provide for the world in quarantine
Excellent video, very well explained.
You will do great man, keep doing this
You are the man
Man
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?
Very nice video 🤟😄
Very clear explanation 😊
So cool. Thank you for sharing this. Amazing
how do you generalise to evaluate zeta(4) with integrate_0^1 1/(1-xyzw) dx dy dz dw --> Pi^4/90?
can someone explain the triangle thing to me please and what would the new bounds be?
This is some cool stuff
Amazing
Brilliant!
I tried the double integral with partial fraction and it is a nonelementary integral.
Impressive bro
Thats you
Wow! Super!
Vert nice
Yes, very nice. But too clever. Could the integral be solved with simpler less clever method?
Of course! There are countless methods to solve double integrals, and some are much more elementary, albeit slower.
Apostol's solution to the Basel problem??!!
Chuks from Nigeria.
Cool
X=2531
=022.0.50
😮
But how large is pi?
-4
log(2cos(x))dx much better)
This is more elementary. No complex analysis.
=
=\ the three you give us negative one on top on you is far
Very nice - except for this horrible background music…
Man, loose the background music!
Musiclove|
-1000000000000000000000db