One important fact my students tend to forget is sqrt(x^2)=|x|, so I got into the habit of saying sqrt(cos^2(x))=|cos(x)|, and then explaining why the absolute operation can be removed. Just removing the absolute value operation leads to errors when, say, the interval of integration contains regions in which a negative sign is needed to remove the absolute operation. Nice proof, though.
A simpler proof is the infinite sum is equal to.the integral of 1/n squared dn so why nkt just do thst and the inegral.of that is 1/n..why not just do that instead of introducing x and y?
@@leif1075 because that isn’t correct. You can write a sum as an integral, but the differential would be d[n] and you have to handle the [n] appropriately. Your explanation skips over this entirely and wouldn’t give the proper value for the series.
@@briemann4124 I forgot but you have mhlriply.by dn and take limit as dn goes to zero then it is correct..and an integral is basically just an infinite sum/prodict anyway though..
@@leif1075 it doesn’t change that your approach is incorrect. Yes, the series can be written as an integral using Riemann-Stieltjes. But you can’t evaluate it nearly as simply as you imply.
@@briemann4124 it may not be simple but that doesnt meanmy approach is incorrect. You can write it as that integral.and then maybe worm.from tjere..why would say that is incorrect at all??
@@NC-hu6xd why are you so toxic? i appreciate prof. Michael Penn for his videos however, sometimes people might have no clue what’s going on, so my intention was only to emphasize the actual value of determinant, nothing more anyway I didn’t want to offense anyone, so never mind if i expressed it wrongly
this was a question on an analysis exam at my university once. of course the right integral transformation was stated as a hint - still i found it very cool to be proving something as famous as this on an exam. caught us a bit by suprise haha
The most sus aspect is perhaps changing the order of integration and summation, because you need to use lebesgue integrals and the like to prove it is a valid step, and that requires some work. But apart from that, all the steps are things that the average calc student could do by themselves, which is honestly amazing
I discovered a solution to the Basel Problem involving NO calculus at all. It only uses Year 10 mathematics. It only uses sinx/x->1, (which can be proven with geometry), the zeroes of the sin and cos functions and Pythagorus Theorem. That's it. It gives π²/6.
and if you calculate out the O(x⁴), O(x⁶) and higher terms, you can find all of the higher order Basel sums. I believe my proof is the simplest proof of Basel, and all higher orders, and can be followed by a child and you see exactly where the π², sum and 6 come from.
This is a quite tricky way to solve the Basel-problem . Another way is to use Fourier series: if you expand the function f(x) =x *(Pi -x) in the interval (0 ,Pi) in cosine -functions you get f(x) = Pi^2/6 - (Cos2x/1 +Cos4x/4 +Cos6x/9 + ....). For x = 0 you have the solution of the Basel problem.
Another great video, Michael! Thanks :) I'm going to go read more about Jacobian thingys in multi-variable integral substitution stuff, because I'm not fully aware of them, but everytime I see you and Flammable-Maths do them, things seem to evaluate really nicely.
I still have my used copy of Apostol's Calculus volume 1. It's a somewhat unorthodox text. It introduces integral calculus before differential calculus and then closes it off with linear algebra topics
Another excellent solution to the Basel problem! Thanks Michael ! Another close way, is to use the substitution x=sin(u) and y=cos(v) instead the one proposed in the video as the integrand becomes simply 1 and the integral is just the Area of the integration domain, easy to compute as it is a triangle.
I also have them from Dr Philip Curtis' UCLA class in 1960. I still remember that I was astounded that you had to prove the intermediate value theorem.
@@silvio4386 Would be great if he issued a notice of correction. He skips a lot of steps, which is fine, I do that too - working mentally a lot, but you have to know you're right. If wrong, please remove confusion by issuing a correction.
also it's marvelous way to transform bound of integration (x,y) -> (u,v) 6:10, but i prefer other one in the video about average distance between two points P, Q in [0,1] * [0,1] square
@@miloszforman6270 when you need the integral of 1/(1-x^2 y^2) on unit square, such substitution's Jacobian coincide with 1-x^2 y^2 that will cancel out, then you only need to find is the area after substitution, which mapping to a isosceles right triangle with side length π/2.
@@jitzukinanaya4626 Unfortunately, that's not too much of additional information which you are offering. And I have some doubts that this will work as advertised. An isosceles right triangle with side length π/2 has an area of (π/2)²/2 = π²/8, which is not _exactly_ the same as π²/6. It is true that the Jakobian of the mapping in your first posting does cancel out with 1/(1-x²y²). However, we need the integral of 1/(1-xy), not 1/(1-x²y²), so there is still an integrand (1 - x²y²)/(1 - xy) = 1+xy left. This "1" yields our π²/8, but xy = sin(u)sin(v)/cos(u)/cos(v) = tan(u)tan(v), which has to be integrated over the said triangle. How do you do that? Ok, I just noticed that this integral 1/(1-x²y²) gives us the sum of the inverse squares of the _uneven_ numbers, and this sum is indeed π²/8. That's the tricky bit. From there we can easily get the sum of the inverse squares of all natural numbers, which had been shown in many other proofs of the Basel problem. Still it has to be proved that the above mapping is indeed a bijection from {(u, v) | 0≤u≤π/2, 0≤v≤π/2, u+v≤π/2} to [0, 1]².
That’s a quite interesting problem close to the one that was presented in this video: Find the sum of 1/( (2n+1)(3n+1) ) Where n goes from 0 to infinity
From 5:48 onwards I suggest directly setting u = sin(theta) v=cos(theta)sin(alpha) whose jacobian cancels out the integrand, and sends D to a kite-shape of area pi^2/12 in the theta-alpha plane. It's the same idea, expressed differently.
A great proof! Some minor details: The determinant is +2, not -2 but it does not matter. Also, the line of symmetry that is used is horizontal one, but this is not too confusing either, probably most viewers haven't even noticed.
Hi, Amazing result, it means that there is no need to invoke the formula of Euler, so disputed: sin pi x /(pi x) = (1 - x^2) (1 - x^2/4) (1 - x^2/9) ...(1 - (x/n)^2)...
I wouldn't say that the formula itself is disputed - it comes out quite naturally from Infinite product theory of complex analysis. It's more the usage without the proper rigorous justification (which didn't exactly exist at the time!)
@@angelmendez-rivera351 you're right, I just wanted to say that Euler didn't worry much about convergence. But I do not question the genius of the find.
The double integral of 1/(1-xy) over the unit square can be expressed as the integral from 0 to 1 of (-1/x)*log(1-x). That integral must be π^2/6 but I have no idea how to prove it.
They dont, he misspoke a bit there. All he is doing is using Fubini Theorem, so it is enough that x^n.y^n be continuous in R^2, which it is. Maybe he was thinking about justifying the summation/integration change but mixed things up.
I see this is all true, but what motivated putting the original problem into a form with x and y in the first place (after the 2nd equals)? Yes, it looks like the integral of something, but why would anyone try that?
Bisect an equilateral triangle with unit sides into two congruent right triangles and consider the trig ratios of each of the angles of one of the right triangles.
Most torturous topic for any beginning graduate student in pure math is general topology. words like box topology, lindeloff , limit point compact makes me feel so unhappy. We need enormous patience in case of general topology. It is all set theory and requires patience.
Somebody discovered it, probably when playing around with the middle or later parts of the proof, not necessarily thinking of the Basel problem. E. g. when trying to calculate this integral ∫∫ [x, y ∈ [0,1]² 1/(1-xy) dx dy Sometimes people discover things when they are trying to invent exercises for their students. The above integral could well have been such an example. Turning the range of integration by 45 degrees (and stretching it) is an easy exercise on how to use the Jakobian.
I would like to suggest a basic algebra problem. Show that 1/9801 = 0.00 01 02 03 … 96 97 99 00 01 … . (I didn’t forget the group 98). Then maybe find another example of this phenomenon.
Is the skipping of 98 because it's actually 98 99 100 but the 1 and 9 overlap so you get 98 100 00 but the 8 and 1 overlap so you get 99? Or is it another reason?
For anyone who is confused why 1/(n+1)² is rewritten as [x^(n+1)/(n+1)] etc: To be clear, any nonzero exponent of x (resp. y) could have been chosen, as long as they are independent of x (resp. y). x²⁰²² could have validly been chosen, but n+1 was chosen for convenience.
5:27 Finicky thing: the determinant is already 2 not -2 because (1)(1) - (-1)(1) = 1 - (-1) = 2.
But it doesn't matter because the absolute value.
I dont understand the part where he starts the iterated integral to 1/2
I have a pretty good idea what Michael Penn will say while taking his final breath…
Should be on his gravestone!
"Okay, nice. And that's a good place to stop."
The first tombstone to have a tombstone written on it.
🥺🥺🥺🥺😭
We need to make sure he has a chalkboard nearby so he can do one last filled square: ⬛️
One important fact my students tend to forget is sqrt(x^2)=|x|, so I got into the habit of saying sqrt(cos^2(x))=|cos(x)|, and then explaining why the absolute operation can be removed. Just removing the absolute value operation leads to errors when, say, the interval of integration contains regions in which a negative sign is needed to remove the absolute operation.
Nice proof, though.
A simpler proof is the infinite sum is equal to.the integral of 1/n squared dn so why nkt just do thst and the inegral.of that is 1/n..why not just do that instead of introducing x and y?
@@leif1075 because that isn’t correct. You can write a sum as an integral, but the differential would be d[n] and you have to handle the [n] appropriately. Your explanation skips over this entirely and wouldn’t give the proper value for the series.
@@briemann4124 I forgot but you have mhlriply.by dn and take limit as dn goes to zero then it is correct..and an integral is basically just an infinite sum/prodict anyway though..
@@leif1075 it doesn’t change that your approach is incorrect. Yes, the series can be written as an integral using Riemann-Stieltjes. But you can’t evaluate it nearly as simply as you imply.
@@briemann4124 it may not be simple but that doesnt meanmy approach is incorrect. You can write it as that integral.and then maybe worm.from tjere..why would say that is incorrect at all??
Jacobian Determinant is 2 even before absolute value!
yeah
| a b |
| c d | = ad - bc = 1 * 1 - (-1) * 1 = 2 (our case)
Wow REALLY ???? Peak midwitt comment
@@NC-hu6xd why are you so toxic? i appreciate prof. Michael Penn for his videos
however, sometimes people might have no clue what’s going on, so my intention was only to emphasize the actual value of determinant, nothing more
anyway I didn’t want to offense anyone, so never mind if i expressed it wrongly
this was a question on an analysis exam at my university once. of course the right integral transformation was stated as a hint - still i found it very cool to be proving something as famous as this on an exam. caught us a bit by suprise haha
wow this might be my favorite solution to basel problem so far, it doesn't involve any super crazy tricks
The most sus aspect is perhaps changing the order of integration and summation, because you need to use lebesgue integrals and the like to prove it is a valid step, and that requires some work. But apart from that, all the steps are things that the average calc student could do by themselves, which is honestly amazing
I discovered a solution to the Basel Problem involving NO calculus at all. It only uses Year 10 mathematics. It only uses sinx/x->1, (which can be proven with geometry), the zeroes of the sin and cos functions and Pythagorus Theorem. That's it. It gives π²/6.
@@idjles well let's see it !
@@idjles That sounds interesting. Where can I find such a solution?
and if you calculate out the O(x⁴), O(x⁶) and higher terms, you can find all of the higher order Basel sums. I believe my proof is the simplest proof of Basel, and all higher orders, and can be followed by a child and you see exactly where the π², sum and 6 come from.
This is a quite tricky way to solve the Basel-problem . Another way is to use Fourier series:
if you expand the function f(x) =x *(Pi -x) in the interval (0 ,Pi) in cosine -functions you get f(x) = Pi^2/6 - (Cos2x/1 +Cos4x/4 +Cos6x/9 + ....).
For x = 0 you have the solution of the Basel problem.
Another great video, Michael! Thanks :) I'm going to go read more about Jacobian thingys in multi-variable integral substitution stuff, because I'm not fully aware of them, but everytime I see you and Flammable-Maths do them, things seem to evaluate really nicely.
I still have my used copy of Apostol's Calculus volume 1. It's a somewhat unorthodox text. It introduces integral calculus before differential calculus and then closes it off with linear algebra topics
Another excellent solution to the Basel problem! Thanks Michael ! Another close way, is to use the substitution x=sin(u) and y=cos(v) instead the one proposed in the video as the integrand becomes simply 1 and the integral is just the Area of the integration domain, easy to compute as it is a triangle.
Doesn't seem to work. Could you please demonstrate that?
I still have Tom Apostol’s Calculus I & II in my bookshelf. We called them “Tommy One” and “Tommy Two” when I was an undergrad.
I also have them from Dr Philip Curtis' UCLA class in 1960. I still remember that I was astounded that you had to prove the intermediate value theorem.
Same. Tom Apostol was one of my best instructors at Caltech.
How are you getting the determinant to equal negative 2? I just get 2. Even before the absolute value operation. (1)(1) - (-1)(1) = 2.
@Light Yagami Me too! He made a slight miscalculation no doubt. 🧠
Yeah, it's 2.
@@silvio4386 Would be great if he issued a notice of correction. He skips a lot of steps, which is fine, I do that too - working mentally a lot, but you have to know you're right. If wrong, please remove confusion by issuing a correction.
also it's marvelous way to transform bound of integration (x,y) -> (u,v) 6:10, but i prefer other one in the video about average distance between two points P, Q in [0,1] * [0,1] square
another very elegant substitution can be utilized in the problem is let x=sin u/cos v, and y=sin v/cos u, and this integral becomes very simple.
Could you please demonstrate this?
@@miloszforman6270 when you need the integral of 1/(1-x^2 y^2) on unit square, such substitution's Jacobian coincide with 1-x^2 y^2 that will cancel out, then you only need to find is the area after substitution, which mapping to a isosceles right triangle with side length π/2.
@@jitzukinanaya4626
Unfortunately, that's not too much of additional information which you are offering. And I have some doubts that this will work as advertised.
An isosceles right triangle with side length π/2 has an area of (π/2)²/2 = π²/8, which is not _exactly_ the same as π²/6.
It is true that the Jakobian of the mapping in your first posting does cancel out with 1/(1-x²y²). However, we need the integral of 1/(1-xy), not 1/(1-x²y²), so there is still an integrand (1 - x²y²)/(1 - xy) = 1+xy left. This "1" yields our π²/8, but xy = sin(u)sin(v)/cos(u)/cos(v) = tan(u)tan(v), which has to be integrated over the said triangle. How do you do that?
Ok, I just noticed that this integral 1/(1-x²y²) gives us the sum of the inverse squares of the _uneven_ numbers, and this sum is indeed π²/8. That's the tricky bit. From there we can easily get the sum of the inverse squares of all natural numbers, which had been shown in many other proofs of the Basel problem.
Still it has to be proved that the above mapping is indeed a bijection from
{(u, v) | 0≤u≤π/2, 0≤v≤π/2, u+v≤π/2} to [0, 1]².
That’s a quite interesting problem close to the one that was presented in this video:
Find the sum of 1/( (2n+1)(3n+1) )
Where n goes from 0 to infinity
Beautiful demonstration. You are a very good speaker! You explain nicely, cleanly, clearly, w/ patience and logic.
Thanks a lot, Michael! I've learned one more way how to to determine that sum, owing to your brilliant explanation. That's just great!
From 5:48 onwards I suggest directly setting u = sin(theta) v=cos(theta)sin(alpha) whose jacobian cancels out the integrand, and sends D to a kite-shape of area pi^2/12 in the theta-alpha plane.
It's the same idea, expressed differently.
Could you please demonstrate this?
A great proof!
Some minor details:
The determinant is +2, not -2 but it does not matter.
Also, the line of symmetry that is used is horizontal one, but this is not too confusing either, probably most viewers haven't even noticed.
Apostol's analytic number theory book is brilliant.
Beautiful video, the animations give an abstract but enlightening aura.
13:24
just asking for a friend are you a mixhael alt account or a follower
@@birdbeakbeardneck3617 Not related at all. I’ve started this account for a joke, and I can’t stop doing it 😂
@@goodplacetostop2973 and now you're an integral part of the community :D
@@goodplacetostop2973 no asked cz u fast hh
@@vinvic1578 Good Place To Stop make a big differential :)
Much like this channel, Apostol's proofs in his books often made things way more computationally complicated and obtuse than they needed to be.
Now i got the idea why euler is a genius.
He solved in a most creative but easy way😂😅
Hi,
Amazing result, it means that there is no need to invoke the formula of Euler, so disputed:
sin pi x /(pi x) = (1 - x^2) (1 - x^2/4) (1 - x^2/9) ...(1 - (x/n)^2)...
I wouldn't say that the formula itself is disputed - it comes out quite naturally from Infinite product theory of complex analysis. It's more the usage without the proper rigorous justification (which didn't exactly exist at the time!)
@@angelmendez-rivera351 you're right, I just wanted to say that Euler didn't worry much about convergence.
But I do not question the genius of the find.
@@peterclark5244 that's it.
I had calculus from Apostol (with multivariable from Spivak's Calculus on Manifolds. 3 in the class got Math PhD's
The double integral of 1/(1-xy) over the unit square can be expressed as the integral from 0 to 1 of (-1/x)*log(1-x). That integral must be π^2/6 but I have no idea how to prove it.
Very cool 👍🏼
Elementary solutions may be simple but there’s no thrill in using it
I would completely disagree
Btw the determinant is two. But it doesn't change anything.🙂
Uniform convergence gone boom
Never this method awesome
🔥🔥🔥🔥🔥🔥
Great !!
3:00 Why do they need to be over the same interval? For example why can't y span from -1 to 1 instead?
They dont, he misspoke a bit there. All he is doing is using Fubini Theorem, so it is enough that x^n.y^n be continuous in R^2, which it is.
Maybe he was thinking about justifying the summation/integration change but mixed things up.
@@juyifan7933 the interchange can be justified pretty easily by monotone convergence
bravo
I’ll be sure to show this video to someone if they ever wonder what I mean after I tell them I like math.
At min. 3:40 you really don’t explain why you can change order of integration and summation
By the dominated convergence theorem, since (xy)^n is bounded on [0,1]^2
@@TheEternalVortex42 He should just state that. I was taught to justify that inversion pretty intensively back in the days.
He never does.
Yes I know that, but just state it since it’s not a general thing one can do
I want to work out why this method fails for zeta(3), and I'm assuming it must or someone would have succeeded in it by now
I think there's no nice change of variables that causes the integrals to simplify.
I see this is all true, but what motivated putting the original problem into a form with x and y in the first place (after the 2nd equals)? Yes, it looks like the integral of something, but why would anyone try that?
Can we have negative differential form 🤔
Why is he allowed to switch the integration and summation?
very nice
Such a nice proof. So many ways to solve the Basel problem. This seems one of the simplest. 🍉
Great video, as usual, but please don't pronounce Basel (the city) as if it was basil (the herb). The first syllable of Basel is pronounced "bah".
How do we now that sin(pi/6) = 1/2. Is there any simple proof?
know
here is a proof th-cam.com/video/y2JTtcgQ5Wo/w-d-xo.html
One must take in consideration the definition of pi/2 as the first root of the cosine function
Are you familiar with the 30-60-90 right triangle and its properties?
Bisect an equilateral triangle with unit sides into two congruent right triangles and consider the trig ratios of each of the angles of one of the right triangles.
Most torturous topic for any beginning graduate student in pure math is general topology. words like box topology, lindeloff , limit point compact makes me feel so unhappy. We need enormous patience in case of general topology. It is all set theory and requires patience.
Nothing to do with the Basel problem but pls pls try this integral from 0 to inf (1/((x^2)-tanh(ln(x))) dx result is pretty nice
(u+v)(u-v) = u²-v², you got u²+v² somehow
Where did you see this?
i kinda dislike this solution because idk how one would ever come up with it
Somebody discovered it, probably when playing around with the middle or later parts of the proof, not necessarily thinking of the Basel problem. E. g. when trying to calculate this integral
∫∫ [x, y ∈ [0,1]² 1/(1-xy) dx dy
Sometimes people discover things when they are trying to invent exercises for their students. The above integral could well have been such an example. Turning the range of integration by 45 degrees (and stretching it) is an easy exercise on how to use the Jakobian.
Woww
Enbek aztán sok értelme volt phuuuuuu
I would like to suggest a basic algebra problem. Show that 1/9801 = 0.00 01 02 03 … 96 97 99 00 01 … . (I didn’t forget the group 98). Then maybe find another example of this phenomenon.
Is the skipping of 98 because it's actually 98 99 100 but the 1 and 9 overlap so you get 98 100 00 but the 8 and 1 overlap so you get 99? Or is it another reason?
For anyone who is confused why 1/(n+1)² is rewritten as [x^(n+1)/(n+1)] etc:
To be clear, any nonzero exponent of x (resp. y) could have been chosen, as long as they are independent of x (resp. y). x²⁰²² could have validly been chosen, but n+1 was chosen for convenience.
What? I can't see any relevant sense in this comment. Some hidden sense?
The shakeup is evolution basically
i already liked commented and subscribed with no respones. sorry. something clearly doesnt work properly with algarithms
asnwer=1 os isiti 🤣🤣🤣🤣
great video, but what i know is that this is definitely not elementary...
Nice, but you lost me here. 😀
Tom M. Apostol ; en.wikipedia.org/wiki/Tom_M._Apostol