An interesting approach to the Basel problem!
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- Опубліковано 16 жов 2024
- We present an interesting and (I think) fairly unique approach to the famous Basel problem. That is, finding the sum of the reciprocal of the squares of all natural numbers.
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Not gonna lie, the magical cancellations were really satisfying!!
What is even more satisfying is calling it a "carnage" :-) (at 15:30)
Rewriting differences as an integral always blows my mind. Just like how would someone ever see that while thinking of how to solve the problem??
After learning Fubini's and the dominated convergence theorem you become a wizard that simplifies problems by creating integrals and summations.
my thought exactly :D this trick always looks like magic or genius or both ^^
Doing lots of problems that's how!
people who are able to come up with those tricks are on a whole other level.
You can't intuitively discover that you should use that exact limit. Rather, you would come up with an idea (replace the natural log with integral of something) and then try to force that to work. I guarantee that the natural log limit substitution was caused by trying a more general substitution until the exponent on y was forced to cause these cancellations.
How did someone ever think of this??? I understand all the steps, but can't imagine discovering this method myself.
I've seen Michael Penn use this method a lot in his vids if he has to rewrite an integral/sum. I guess it involves finding an integral that resolves to the sum formula, and then using the dominated convergence theorem to swap the sum & integral. Then you can simplify it using a taylor series substitution or solve the integral. I guess you do it enough that you think of it naturally?
You may start from another side, and think to compute the integral of 1/(1+x)(1+xy^2) on [0,oo)X[0,1] . Computing it as iterated integral by Tonelli in the two ways, you end up naturally with the final equality (the integration by series is also quite natural for the function log(x)/(1-x^2), since it is the easiest thing to do.
Of course, this does not answer the question: How did someone think of this, starting from the Basel problem. But, as a matter of fact, many problems are solved just because we took notice of some phenomenon before. Of course this is just bricolage, not technology, but it is cheap. For instance, many antiderivatives are known just because we did many derivatives before and took note of them :)
Informally, integrals are essentially sums/series with infinite "grain". So there's a direct relationship between them that would hint to using them together.
But yeah, I agree it's still pretty nuts :P
@@garrycotton7094 yeah but in this particular case, the sum that was originally used for 1/n² turns into the sum of x^(2n). This has nothing to do with any of the integrals that are involved; in this problem there's no transformation of a sum into an integral directly or the converse.
@djttv Ditto. Still beats me. I understand all the steps, but to think of his methodology is just beyond me. 😔
I own page 235 of Euler´s slip of paper
(and more from him). It´s from year 4 (1729) and worth around 450€ today. It took him 9 years to solve it and he left behind around 900 pages of this.
Wow!
Do you have the copy, or the original? It would cost a fortune if it is original manuscript.
10:35 - you had the 1/2 multiplying the ln x^2 only but then you took it out to apply to ln 1. it still works since ln(1)=0 but it was jarring 😁. Great video as always!
There's always that moment when you suddenly realize what's gonna happen next, amazing video
When that arctan appears it all comes together
I really admire how dedicated you are to do these problems with such patience.
amazing Sir
Really cool approach to solve this in a new way. Still remember how awesome it felt to solve it for the first time using Fourier series.
I am breathless. How could anyone think of this? This is so undeniably insane and magic at the same time.
The the final integral y/(1+x²y²)(1+y²) dxdy over {0
Love the u substitution in the video! Great job!
Hey Michael. I watch your Channel for a few months and I love it. Watch every video. I am in 11ths class in Germany and really look forward to studying Maths. I love your real analysis course because there I can feel like already studying. Keep going. I had an interesting problem in a German Maths contest. I would appreciate you explaining it.
A sequence is recursively defined as a1=0, a2=2, a3=3, an=max(0
CraftexX Hi there! here is my thought.
This problem is similar to the following statement “for x, y are integers. Find a_n = max{2^x*3^y | x*2 + y*3 = n}”
You may notice that if we want to find a_n then we should make y as huge as possible since 2^3 < 3^2.
so I think the answer you are looking for is 3^(19702020/3)
It's awesome to see a proof of the Basel problem that just uses some basic calculus (and black magic cancellation)!
The first proof was so simple yet so elegant.
Very good calculus. You could probably do this as a fun Calc 2 problem for your class. Maybe go over dominated convergence, geometric series, and p-series, and they should be ready. Partial derivatives might be a bit scary for them, but it’s not too bad
You mean "gloss over" dominated covergence, since measure theory is a bit advanced for a Calc 2 class.
@@renerpho Yeah good point
Beautiful 👏🏻 👏🏻 I never used that Ln limit in Calculus though. Nice! Thanks!!
Whoa! Really nice derivation! The venturing into calculus seemed to complicate things initially, but then suddenly everything falls in place, and π^2 finally emerges towards the very end! Big kudos!!
Just...wow. This is really good. I love your content. This inspired me to study a long neglected book in my library (Real Infinite Series).
Did you come up with it with yourself? So brilliant! Keep on going
I loved the method and the way you show it . Thanks .
In a sense, the second lemma *does* apply in the limit as m→-1 from the right: The left side approaches +∞, while an antiderivative for the right side turns out to be ½(ln x)^2, from which the improper integral is +∞ (the integral also does not converge for m
7:28 i think that you can bring the sum into the integral after you check about the (uniform/point-wise) convergence of the series of function. i think it's important to show more
I feel that there are some equalities in the derivation where you need to consider improper integrals instead of the usual integral. For example, when you apply the closed form 1/(1 - x^2) of the geometric series Sigma(x^(2n), n from 0 to infinity), an implicit assumption is 0 < x < 1; the closed form doesn’t apply to x = 0 or x = 1. This makes the integral from x = 0 to x = 1 indeed an improper integral from x = 0+ to x = 1-.
Anyway, that’s a nice video with an excellent explanation. Thanks for sharing! 👍
Ammazing way of proof . Thanks for this
As it goes, its soo satisfying to watch the expected answer revealing itself. Great video as always.
Very impressive. Thanks for the awesome video!
Amazing demonstration !!
Euler would be pleased
19:21
As usual!
Always so helpful!
Great video as always, thanks for your work! You really inspire me to keep on improving my math skills!
I like this problem , I like you make good solution in this video also.
Best math teacher 😍😍😍
Nice, but my question is, when you drag in logarithms, arctan, integrations, and a bunch of related theorems, how do you make sure that the desired result (or, more difficult to see, some equivalent statement) was not already used to prove one of the premises? Clearly reasoning via limits of sums was probably the base for most of these.
I don't think that in this proof there is any claim or assumption that is related to the conclusion itself; the conclusion is only the exact value of the limit of partial sum of the squared reciprocals of natural number; the existence of this limited is granted by a general criterion of series convergence; then there is a correct application of the theorem on integration of absolute convergent series; then, the evaluation of the limit of a geometric series is surely not related to the present problem; analogously, the clever solution of the final integral is certainly based on the application of general theorems (e.g. Fubini) and some closed form primitive evaluations which are surely not theoretically consequent to the computation of our series! Consider that the modern rigorous theory about goniometric functions starts from complex convergent series; the exp(z) function is defined as a series, and it is verified that its restriction in R is coincident with the real exponent function; then goniometric functions are axiomatically defined by combination of complex exponentials in order to rigorously verify all the classic "intuitive" properties
Typically you would structure the proof carefully, state all the assumptions you're making and the notation you're using explicitly and completely. That being said in a short video format like this it would be impossible to do that and some knowledge is assumed on the part of the viewer. In a paper or textbook it's easier to explain in detail since you don't have a time limit.
Would it be possible to do a video on dominated convergence theorem and fubini's theorem?
Awesome. I didn't enjoyed like this since a long time. Congrats.
What a beautiful way of approaching the Basel problem! If it doesn't bother you: could you please do an introduction about the Bernoulli's numbers? Thank for your content
Outstanding. Bravo.
Wow I’m just looking at this now but who comes up with these crazy work around solutions
Maths really is divine man when that arctan substitution happened I lost it
Great job bringing it to a definite integral. I wonder if there’s another way to bring it home
Very motivating ty !
Hi,
Fanstastic! You have done it "normally", without any "trick" like the one of Euler (infinite product of sin pi x / pi x), I thought that was not possible!
For a moment I have been wondering how pi would apears from the hat.
Man this is the Oxford interview question ! Thx for explaining it
if you got this in an oxford interview i feel sorry for you mate
Well done. For not so bright minded folks there is still more peasant minded way to prove this via Fourier series method (supplemented with Parseval identity, depending on which model of periodic function one starts with).
This is a very complicated proof! I much prefer the derivation involving Bernoulli numbers of the formula for ζ(2n), where n is an arbitrary positive integer. Good explanation though!
Much amazing!
This was a really appealing approach.
I'm just wondering how many different ways of Basel problem are known nowadays. Now, I'm aware of three of them. Upper one is a masterpiece. Definitely it's one of my favourite. Best regards Michael! Ok, great :)
At about 7:45, when we're pushing the sum through the integral, would it be correct to say that we could also justify this with the uniform convergence of the sum?
Possible but not necessary
Wonderful!
Oh, man. That was awersome.
Excellent!
Awesome… and so difficult to imagine by myself… 👌
Nice one !
Absolutely phenomenal video. Never seen this approach to the Basel problem. Can the sum of the reciprocal fourth powers be evaluated via the same approach, since the inverse fourth power has a similar expression in terms of the integral of x^m ln^3(x)?
Is it possible to come up with an approach where you have a sum of integrals where each integral has limits of n and n+1?
I wouldn't have thought of the replacement at 9:30 in a zillion years.
Are you going to make videos about different modes of convergence in the real analysis playlist?
Thanks for this vdo it was so easy and beautiful proof.👍👍👍👍👍
As you said...Fantastic !,...I say the same.
Gorgeous proof! And using just basic calculus (and a massive amount of genius, I guess:))
an interesting method, and that's a good place to push the like button
That was a fun ride!
I feel smarter just watching this thanks for sharing your genius with us 😁👍
clever.. no doubt
thanks for sharing
I really enjoy watching those videos, how do the authors of these proof come with the idea ?
A very nice and elegant solution! One question though: In 7:50, how could you know abs(x^2)
the integral is from 1 to 0, so x is between these two values. thats why x^2 < 1
@@timohiti8386 I thought so at first but can't x be smaller or equal to one and then there's the case when x=1?
@@itamargolomb8530 since the integral does not change when you leave out the borders, you can exclude the case x=1 in the inside of the integral
@@timohiti8386 According to what rule can I do it? (I never took a calculus class but I watched enough videos to have some knowledge)
@@itamargolomb8530 the definition of integrals: int from x=1 to x=1 is 0 independent of the term in the integral. after integration you would subtract the same value from itself since the upper and lower border of the integral are the same. The rule is "integration of a point" but I dont think that this has a special name
Superb!! Love it.
6:52 , why it change to 2n + 1 ? kindly why ? thank you sir michael pen
Hats off sir👍👍👍👍
Beautiful video- one of my favourite proofs. Your videos keep getting better! Sorry for all the nit picky comments too, but it's good to at least mention the dominated convergence theorem or whatever you need to use (which you did!).
Magical
7:26 -- Dominated convergence theorem
en.wikipedia.org/wiki/Dominated_convergence_theorem
15:48 -- Fubini's theorem
en.wikipedia.org/wiki/Fubini%27s_theorem
Can you use the first bullet to show the value of the infinite sum? Aren't you assuming that it converges when you do algebra on the value of the sum?
Hence why he says in the beginning that it is fairly easy to show that it converges. You can use the integral test for convergence.
Great video
amazing!
At 16:40, it is more proper to write "x=0 to inf" instead of just "0 to inf" since there are x and y in play.
the closed-form expression for that infinite sum converges only for (-1,1). How can you substitute that in if the bounds of the integral are from [0,1]? Do you need a limit showing that the upper bound approaches one?
You can integrate over (0,1) as the point {1} is of zero measure
i was thinking same
William Churcher Well yes, but it’s a bit weird answering the question that way as our friend has probably not seen any measure theory yet.
Using the Riemann definition of the integral, you can show that removing (or adding) any finite number of points from your domain of integration does not change the value of the integral. The formal language in which this sort of thing is discussed is called measure theory, if you want to look into it a bit more :)
Will you make series for all topics of undergraduate mathematics in future ?
Mind-spinning, mesmerizingly enchanting
ou quelque chose comme ça
Great video.
You are ammazing math
You should go on Penn and Teller's Fool Us with this wonderful magic trick!
I found your problem is very good ☺️
I have seen more proofs of this remarkable identity, but if you like using a bit of tough and solid calculus, then this is the one you will like!
Amazing!
Great maths approach
You are too much Sir
Amazing proof!
Very nice to see that it indeed was probable with nothing but stuff from early calculus course. Or course I'm not sure any students would be so confident and succeed on their own as it was quite an undertaking. 😂
very interesting ! all with elementary calculus !!
Is this your result, or just a less commonly known result?
that’s fantastic
do you have a dictionary of those "tools" ?
I think this is the length y proof of the Basel problem.Any how the proof is interesting with many clever tricks.
This approach is quite elegant and more simple than Euler's one. But I thing that you need to know the goal to build this ! Isn't it ?
@ No ! It 'is what I want to say. The Euler's method is more complicated but he began from ... nothing, and of course without computer. It is easier (hum!) to build elegant demonstrations when you know the goal.
Cool "Dune" T-shirt :>
Well, that is Your best video, as far, as I'm concerned - in so far as I was able to follow - and wonder what happened to 1/(1+x)
Does anyone know how you could prove the sum of the reciprocal of odd numbers is 3x the sum of the reciprocal of even numbers without resorting to the proof here i.e without showing that the the sum of the reciprocal of evens is 1/4 of the the sum of all reciprocals? The reason I ask is because I was quite astounded as it does not seem immediately obvious it is 3x. I guess it is important because it may help to solve other problems like figuring out other sums like 1/3^3 + 1/7^3 + 1/11^3 + 1/15^3 which would help to solve Apery's number.
Very nice
This is fuckiing amazing
Lovely!
The best method of them all. Who did it first, I wonder?