9:46 very sneaky, erasing the -2 in the cut

At first after seeing those integrals, I was like "no way this is elementary", but as it turns out it was completely comprehendible. Nicely explained Prof. Penn.

This is a very clever proof, but it relies on two magic integrals, and I don't see the motivation. How would someone come up with An and Bn?

A note: for the bit where you prove that sin(x) >= 2/pi * x on [0,pi/2], a more rigorous way to prove that--whilst still drawing the same picture--is to note that the second derivative of sin(x) is -sin(x), which is non-positive on [0,pi/2]. Hence sin(x) is concave on [0,pi/2], and thus lies above the secant line.

Would be nice to hear a little about how you would figure out what An and Bn should be in order to prove this.

All the steps were straightforward, but imagine the foresight needed to see the path ahead. Very, very nice.

Fun fact: this was actually the final question in the 2010 Australian HSC Mathematics Extension 2 paper, you can search up the video where Eddie Woo talks about it

Way easier to understand compared to other proofs that I’ve seen on this problem.

Probably the Fourier transform on a finite interval is the simplest way to solve the Basel problem (provided you have the machinery in place...) Just using that the delta function is the sum of all exponentials and that integrating twice is dividing Fourier coefficients by n^2.

Did Doner give motivation for An and Bn? From a student's perspective, your presentation, while very understandable, doesn't tell me how I might have come to solve such a problem. It seems like magic furthering the view that math is just a bunch of tricks that certain people just know.

This is incredible! Thank you

great work, very comprehensive and clear. :D

Absolutely amazing.

Thank you for the video. I've been stuck at various points on your previous videos. And one of the things that definitely stuck out to me is that that I need to get better at trig. I've definitely been watching them though! Just haven't finished them, so I'll be going back to them. Also I'm not sure that I agree this is necessarily all that easy, but it is an interesting use of analysis. Thank you for finding the problem.

I almost didn't believe that the proof could be so simple so I was waiting for some super complicated step to show up. I finally realized that wasn't happening once you showed the telescoping! The whole proof seems like black magic.

Great work! Thanks for sharing this problem... I'm interested to learn about its use by Daners.

love your videos!

Nice work!

This was so cool, thanks!

Here's a fun fact -- the asymptotic density of squareless integers (that is to say, numbers that cannot be simplified under the radical sign because they have no squares as divisors) is 6/pi^2, or the reciprocal of the basel number. Makes sense in an intuitive way: one is the limit of the sum of the reciprocal of squares, the other is, basically what you get when you have a "limit" that's not horizontal of all integers that have no reciprocal basil numbers as divisors. 6/pi^2 shows up somewhere else that's interesting. I think Wolfram's site has it.

Fourier series are hidden behind this. Integrating something against cos^n(x) and against cos(nx) is not that different.

This way is probably one of the most "easy" to follow, but also one of the less elegant, because the setting comes from nowhere. Thanks !

Hermoso resultado!

Easy? Clever, yes. And easy to understand, once derived.

Wow. One of your finest.

Love your t-shirt! It's my favorite formula

Very cool!!! Does this "elementary" method have analogues/generalizations to work for the other even values of the Riemann Zeta function? Maybe they get too impractical too quick but it'd still be interesting to see!

multiplying the sum by the object at 14:20 had me confused. It's just multiplying by one though. A = B + C + D 1 = (B + C + D)/A. Its just a very fancy version of one.

Wonderfull!!👊

excellent proof, very interesting

17:39

Wow. Amazing how such contrived power function integrals of Trig functions can be used to determine the sum of the inverse squares.

Simply beautifull !!!

I had seen the beginning of the video and then I tried to solve the integral by parts to determine the relationship between An, Bn and Bn-2. The result was not the same as shown on the board (😆). At 9:46 the fix appeared (🤩)...Excellent video lesson and way to solve this classic problem. Sorry my english...

I'm glad this was an "easy solution" to the Basel problem !😄

oh lord. now it began to dawn on me what "some elementary" calculations mean. Very nice. Thank you very much to lead me through this jungle of integrals and trigsfunctions. I looked for a long time for this result pi^2 over 6. it is Riemann's zeta function evaluated for zeta(2). ❤

I don't know the intuition behind An and Bn, but the proof looks like something which is closely related to Parseval's identity.

Cannot we use both Riemann Zeta function and define a Fourier series by f(x) = x when -π

1:47 from 0 to π/2

it was very beautiful math

There is a quicker way to prove this , using the integral of xcos(nx) between 0 and pi , equal to ( (-1)^n + 1 )/ n^2 (riemann 's lemma in a simple case )

Nice

i know it's unrelated to the video here but i'd just like to share with you guys that i had the fermat's little theorem proved today at my abstract algebra classes which was exciting

i d wanna know how did they think of using this method? the math is easy to understand after the first 3 hints, but how do you reach the notion of using these integrals?

Is there a general way to construct an An,Bn to make these kind of series telescopic?

the result is the same as the 2 * average of f(x)=x^2 between x=0 and x=pi/2. amazing

Wait not only you found the answer for N tending to inifinty but that was the answer for any N(I mean the part where he used telescoping)

Very satisfying and clever proof, thanks Prof. When Prof. was calculating lim n --> inf, B_n/A_n , I was thinking of L'Hopital's rule. It probably wouldn't do the trick though.

I wouldnt say my fav basel problem proof.

Lovely proof.

there is a quicker way to prove this using the integral of xcos(nx) between 0 and pi .(equal to ( (-1)^n + 1 ) / n^2 .

Might just be the bias of knowing about power series from Calc II, but I’d say the factored maclaurin series argument is more “elementary”

Thank you, professor. That was long, but clear.

And what’s the link with the sum of the reciprocals of n squared ?

Omg this is so surprising 🙀

Quite easy to follow but I don't see why we're using 2n instead of n?

Yesterday I looked through a STEP 3 past paper from 2018, question 7. Which was a way of solving the Basel problem. I don't know if it's a well known method but it is another simple method.

Can we build up An and Bn for sinx instead of cosx and what do we get from there ?

An is a walis integral for even numbers?

I don't see how this is "easy". The series A, B come out of nowhere. I think the simplest proof (I've seen) is using Fourier series, for example for function x^2 on (-pi, pi) (and then plug in x = pi), or for function sgn(x)*(pi - abs(x))/2 and using Parseval's theorem (Fourier series is sin(nx)/n, so Parseval's theorem will lead to the sum of 1/n^2). At least in case of the Fourier transforms, the motivation is clear: either construct a series whose coefficients are 1/n so when squared, we have the desired sum from Parseval's theorem, or straight up the series with coefficients equal to 1/n^2 and then just plug in a specific value of x.

Given that a binary sequence is considered "friendly" if every digit in the sequence neighbours a 1. Eg:0,1,1,0,1,1,1,0,1,1,0 Then find the smallest n such that the number of friendly binary sequences of size n is greater than 100. (Source: IOQM 2022)

Hola

U can just equal sin taylor series expansion and u got the sum of the roots equal bassel sum just 2 lines for god sake

Easiest one is Euler's one, even if it is not really a proof.

I think the word "approximation" is more accurate. Don't you think?

Interesting proof. Blackpenredpen also has an easy solution

Yes, but Asubn and Bsubn, there’s absolutely NOTHING intuitive about them! Pretty easy to get somewhere when you aren’t starting from scratch.

Credit worthy is problem.

I like proofs that give you a feeling of why the result should be the case rather then just grinding through the equations. This wasn't one of those.

Thank you for the proof. But I'm curious about your T-shirt. Do you know of any interpretation of this equality?

Intelligent but no way I could think that

Yeah, that was short.

Gotta correct you on one thing, Michael, if you don't mind. Basel is pronounced B-"ah"-sel. Greetings from Switzerland.

Easy ? Just consider the Fourier series of abs(x) period 2.pi and you get the result in 2 minutes if slow in algebra. By the way, this is a Riemann series power 2. All the Riemann series of even power are 'pseudo-rational', equal to pi at equal power times the Bernoulli number : pi^2/6 for n=2, pi^4/90 for n=4, pi^6/945 for n=6, etc. Proof by Fourier series not so hard, especially if you know the Parseval theorem... My regretted colleague Roger Apery proved in 1977, published the year after, that the Riemann series of odd power are irrationnal except for n=3. Here this is really high level algebra. I still do not unsterstand all of his arguments. But his paper was proved correct afterwards by numerical simulation. We are not all equal in terms of neuronal efficience... I just realized scrolling down the comments flow that other people are aware of Fourier series. This is the easy way to evaluate Riemann series of even power. The one in this video is just a funny trick to find the result for n=2.

Oh come on, at 9:47 you sneakily changed the second claim. I lost ten minutes trying to prove that on my own...

Zum Glück war es eine einfache Lösung...

could you do x^x = Gamma(x) and hence show that one of the two functions is always going to be larger than the other after a certain point

Interesting approach, though, like other comments, the choice of A and B is a mystery. Also, those interested in Basel might also enjoy the video by 3Blue1Brown on the subject

asnwer=1 isit

Have you ever tried visualizing it and using Geometric Constructions to view pi-squared as six times the solution?

not a good method

an easy solution takes 18 minutes?....yeah, this is AMAT 413 all over again....

This proof was actually developed for an examination for highschool students here in Australia with aid from Sydney Uni. You can find it as the last question in this paper: www.boardofstudies.nsw.edu.au/hsc_exams/hsc2010exams/pdf_doc/2010-hsc-exam-mathematics-extension-2.pdf I think it's kinda cool because the exam question was set in 2010 while D Daners only officially published the proof in 2012 in Math magazine

Greetings, it remember me Chebyshev polynomials, thanks for your work and for aharing it, I appreciate it :peace_symbol: