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.
@2kchallengewith4video Жыл бұрын
I dont understand the part where he starts the iterated integral to 1/2
@tcmxiyw2 жыл бұрын
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.
@leif10752 жыл бұрын
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?
@briemann41242 жыл бұрын
@@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.
@leif10752 жыл бұрын
@@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..
@briemann41242 жыл бұрын
@@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.
@leif10752 жыл бұрын
@@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??
@gilamesh34172 жыл бұрын
Jacobian Determinant is 2 even before absolute value!
@viktoryehorov43142 жыл бұрын
yeah | a b | | c d | = ad - bc = 1 * 1 - (-1) * 1 = 2 (our case)
@NC-hu6xd2 жыл бұрын
Wow REALLY ???? Peak midwitt comment
@viktoryehorov43142 жыл бұрын
@@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
@benheideveld46172 жыл бұрын
I have a pretty good idea what Michael Penn will say while taking his final breath…
@warrickdawes79002 жыл бұрын
Should be on his gravestone!
@damascus212 жыл бұрын
"Okay, nice. And that's a good place to stop."
@dethmaiden19912 жыл бұрын
The first tombstone to have a tombstone written on it.
@dalitlegreenfuzzyman2 жыл бұрын
🥺🥺🥺🥺😭
@marcushendriksen84152 жыл бұрын
We need to make sure he has a chalkboard nearby so he can do one last filled square: ⬛️
@WhiterockFTP2 жыл бұрын
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
@lightyagami17522 жыл бұрын
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.
@Jack_Callcott_AU2 жыл бұрын
@Light Yagami Me too! He made a slight miscalculation no doubt. 🧠
@silvio43862 жыл бұрын
Yeah, it's 2.
@lightyagami17522 жыл бұрын
@@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.
@renesperb2 жыл бұрын
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.
@backyard2822 жыл бұрын
wow this might be my favorite solution to basel problem so far, it doesn't involve any super crazy tricks
@ClaraDeLemon2 жыл бұрын
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
@idjles2 жыл бұрын
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.
@moutonso2 жыл бұрын
@@idjles well let's see it !
@mustafaahmed93802 жыл бұрын
@@idjles That sounds interesting. Where can I find such a solution?
@idjles2 жыл бұрын
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.
@sniperwolf502 жыл бұрын
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
@TimMaddux2 жыл бұрын
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.
@cutcrew27432 жыл бұрын
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.
@roblangston3192 жыл бұрын
Same. Tom Apostol was one of my best instructors at Caltech.
@alainleclerc2332 жыл бұрын
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.
@miloszforman62703 ай бұрын
Doesn't seem to work. Could you please demonstrate that?
@sachatostevin64352 жыл бұрын
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.
@il_caos_deterministico2 жыл бұрын
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
@jitzukinanaya46262 жыл бұрын
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.
@miloszforman62703 ай бұрын
Could you please demonstrate this?
@jitzukinanaya46263 ай бұрын
@@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.
@miloszforman62703 ай бұрын
@@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.
@viktoryehorov43142 жыл бұрын
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
@qbqbqdbq2 жыл бұрын
Apostol's analytic number theory book is brilliant.
@koenth23592 жыл бұрын
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.
@sergeipetrov55722 жыл бұрын
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!
@3x3-x3x-oXo2 жыл бұрын
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.
@miloszforman62703 ай бұрын
Could you please demonstrate this?
@Abhi-kr6df2 жыл бұрын
Now i got the idea why euler is a genius. He solved in a most creative but easy way😂😅
@rafabenjumea46832 жыл бұрын
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.
@mihaipuiu6231 Жыл бұрын
Beautiful demonstration. You are a very good speaker! You explain nicely, cleanly, clearly, w/ patience and logic.
@goodplacetostop29732 жыл бұрын
13:24
@birdbeakbeardneck36172 жыл бұрын
just asking for a friend are you a mixhael alt account or a follower
@goodplacetostop29732 жыл бұрын
@@birdbeakbeardneck3617 Not related at all. I’ve started this account for a joke, and I can’t stop doing it 😂
@vinvic15782 жыл бұрын
@@goodplacetostop2973 and now you're an integral part of the community :D
@birdbeakbeardneck36172 жыл бұрын
@@goodplacetostop2973 no asked cz u fast hh
@PeperazziTube2 жыл бұрын
@@vinvic1578 Good Place To Stop make a big differential :)
@CM63_France2 жыл бұрын
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)...
@peterclark52442 жыл бұрын
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!)
@CM63_France2 жыл бұрын
@@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.
@CM63_France2 жыл бұрын
@@peterclark5244 that's it.
@TomJones-tx7pb2 жыл бұрын
Much like this channel, Apostol's proofs in his books often made things way more computationally complicated and obtuse than they needed to be.
@timothybrown83952 жыл бұрын
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
@TheEternalVortex422 жыл бұрын
I think there's no nice change of variables that causes the integrals to simplify.
@Calcprof4 ай бұрын
I had calculus from Apostol (with multivariable from Spivak's Calculus on Manifolds. 3 in the class got Math PhD's
@Denosophem2 жыл бұрын
Elementary solutions may be simple but there’s no thrill in using it
@replicaacliper2 жыл бұрын
I would completely disagree
@Andbovi2 жыл бұрын
Btw the determinant is two. But it doesn't change anything.🙂
@ecoidea1002 жыл бұрын
Beautiful video, the animations give an abstract but enlightening aura.
@scottmiller25912 жыл бұрын
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?
@loopingdope2 жыл бұрын
Why is he allowed to switch the integration and summation?
@il_caos_deterministico2 жыл бұрын
At min. 3:40 you really don’t explain why you can change order of integration and summation
@TheEternalVortex422 жыл бұрын
By the dominated convergence theorem, since (xy)^n is bounded on [0,1]^2
@ladyloose2 жыл бұрын
@@TheEternalVortex42 He should just state that. I was taught to justify that inversion pretty intensively back in the days.
@qdrtytre2 жыл бұрын
He never does.
@il_caos_deterministico2 жыл бұрын
Yes I know that, but just state it since it’s not a general thing one can do
@chrkts41912 жыл бұрын
Uniform convergence gone boom
@alanwhite31542 жыл бұрын
Very cool 👍🏼
@Coldgpu2 жыл бұрын
Never this method awesome 🔥🔥🔥🔥🔥🔥
@jellyfrancis2 жыл бұрын
Can we have negative differential form 🤔
@abrahammekonnen2 жыл бұрын
3:00 Why do they need to be over the same interval? For example why can't y span from -1 to 1 instead?
@juyifan79332 жыл бұрын
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.
@peterclark52442 жыл бұрын
@@juyifan7933 the interchange can be justified pretty easily by monotone convergence
@robert-skibelo2 жыл бұрын
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".
@kalamardesk3666 Жыл бұрын
bravo
@tubamazouz2 жыл бұрын
Great !!
@mayonakao24882 жыл бұрын
I’ll be sure to show this video to someone if they ever wonder what I mean after I tell them I like math.
@firemaniac1002 жыл бұрын
How do we now that sin(pi/6) = 1/2. Is there any simple proof?
@firemaniac1002 жыл бұрын
know
@firemaniac1002 жыл бұрын
here is a proof kzbin.info/www/bejne/r2OthaeZnLZojdE
@firemaniac1002 жыл бұрын
One must take in consideration the definition of pi/2 as the first root of the cosine function
@timothybrown83952 жыл бұрын
Are you familiar with the 30-60-90 right triangle and its properties?
@lightyagami17522 жыл бұрын
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.
@Jack_Callcott_AU2 жыл бұрын
Such a nice proof. So many ways to solve the Basel problem. This seems one of the simplest. 🍉
@srikanthtupurani63162 жыл бұрын
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.
@peterwaksman91792 жыл бұрын
very nice
@funktorial4 ай бұрын
i kinda dislike this solution because idk how one would ever come up with it
@miloszforman62703 ай бұрын
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.
@otiswebb57832 жыл бұрын
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
@RamsLiff2 жыл бұрын
(u+v)(u-v) = u²-v², you got u²+v² somehow
@miloszforman62703 ай бұрын
Where did you see this?
@Denosophem2 жыл бұрын
i already liked commented and subscribed with no respones. sorry. something clearly doesnt work properly with algarithms
@MGSchmahl2 жыл бұрын
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.
@miloszforman62703 ай бұрын
What? I can't see any relevant sense in this comment. Some hidden sense?
@charlienyikos15642 жыл бұрын
Enbek aztán sok értelme volt phuuuuuu
@andreamartelli8232 жыл бұрын
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.
@QuantSpazar2 жыл бұрын
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?
@bilalabbad79542 жыл бұрын
Woww
@Denosophem2 жыл бұрын
The shakeup is evolution basically
@emmepombar33282 жыл бұрын
Nice, but you lost me here. 😀
@와우-m1y2 жыл бұрын
asnwer=1 os isiti 🤣🤣🤣🤣
@xaxuser50332 жыл бұрын
great video, but what i know is that this is definitely not elementary...
@charleyhoward45942 жыл бұрын
Tom M. Apostol ; en.wikipedia.org/wiki/Tom_M._Apostol