An easy solution to the Basel problem
HTML-код
- Опубликовано: 14 окт 2024
- 🌟Support the channel🌟
Patreon: / michaelpennmath
Merch: teespring.com/...
My amazon shop: www.amazon.com...
🟢 Discord: / discord
🌟my other channels🌟
Course videos: / @mathmajor
non-math podcast: / @thepennpavpodcast7878
🌟My Links🌟
Personal Website: www.michael-pen...
Instagram: / melp2718
Randolph College Math: www.randolphcol...
Research Gate profile: www.researchga...
Google Scholar profile: scholar.google...
🌟How I make Thumbnails🌟
Canva: partner.canva....
Color Pallet: coolors.co/?re...
🌟Suggest a problem🌟
forms.gle/ea7P...
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 what I came here to say 😂
How is it at all comprehe Sibley why anyone would.thinknof this to.begin woth..and what are An amd Bn..some sort of constants or parameters..I don't think he said...
9:46 very sneaky, erasing the -2 in the cut
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.
I was gonna comment the same thing. I love the simplicity of graphs but they don't always scream rigor !
Is it really needed to prove such obvious things which can be confirmed without a proof?
@@Cjendjsidj When the given proof would be that simple, might as well do it.
@@Cjendjsidjit depends on the audience is the real answer, but it is good practice to prove things that “seem” obvious but may not be to those with a lower level understanding than you. Further, it will prevent you from getting into the habit of assuming things that seem “obviously” true that actually turn out not to be true at all.
Would be nice to hear a little about how you would figure out what An and Bn should be in order to prove this.
You sell your soul to the devil and he will reveal those integrals to you
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
That doesn’t surprise me, that question in different years has also done proofs of the irrationality of e.
Extra fun fact: Daniel Daners also writes the Australian HSC Maths Ext 2 papers and this proof (published 2012) makes a slight improvement on the HSC question.
Edit: I'm not exactly sure if he wrote this question specifically, but I would not be surprised if he did.
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.
Integrating the Taylor expansion of log(2 cos x) = ix + Log(1+e^(-2ix)) from 0 to pi/2 also gives the result pretty painlessly.
Or the contour integral around all positive real axis poles of cot(pi x)/x^2 (computed with a few tricks like deforming the contour to a big half circle. The residues on the poles give the sum).
@@JosBergervoet I think you can do the contour as a square (with side length R -> infinity) which is easier maybe
And that inequality at around 10:00 comes out of nowhere..I don't see why anyone would.think of thst right? I would do it using McClaren expa sion of cosine series and then sine expa suon maybe and setting x equal to pi I think that would work out?. .or something like that or Eulers method maybe..
@@leif1075 That inequality sin x >= (2/pi)*x for 0
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?
I would belive if someone likes those integrals for other reasons and stumbled upon the identety proof by chance (serindipity). It happens a lot when you do research; one of my teachers told me he took identities that arise from comparing coeficients of polynomials in complex number theory and later put them in calc I tests for first years to prove them by induction. Some trigonometry identities are discovered by serindipity (arctan(x) + arctan(1/x) = pi/2)
maybe they were playing with finding variance of cos^2n and a famous approach is reduction formula so it's only logical to try to find a recurrence for it. tangling the two recurrence relationship together im not sure what the motivation is for that but afterwards, noticing the n^2 term gives motivation for the limit tool he proved third.
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.
All the steps were straightforward, but imagine the foresight needed to see the path ahead. Very, very nice.
great work, very comprehensive and clear. :D
Absolutely amazing.
Way easier to understand compared to other proofs that I’ve seen on this problem.
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.
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.
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 !
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!
This is incredible! Thank you
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.
The Basil problem is a pretty hard problem to solve if you just sit down and stare at it without already knowing the answer. The typical method of using the Taylor series of sin(x)/x is (in my opinion) the easiest to motivate, since all it involves is matching coefficients and using a well-known series expansion, both of which are “tricks” that every undergraduate has seen by the time they take real analysis. If you’ve never seen the Taylor series proof, it’s the first one given on the Wikipedia page for the Basel problem. Once you know the answer, it’s not so hard to come up with other ways of reaching it, and as a result we get nice proofs like this!
@@heyitsmedave2354 Fine, assume the result, even present the Taylor series approach. How do you go from that to coming up with An and Bn? That's really the crucial aspect in teaching and learning mathematics - to learn how to think like a mathematician
Referencing an old analogy, you've shown a beautiful building, but you took down and discarded the scaffolding. But the scaffolding has enormous value.
Typically what happens is someone comes up with an intuitive but very long and laborious proof that uses a lot of heavy machinery. Then over time, people find ways to simplify the proof bit by bit, cutting parts out, taking short cuts, and using more elementary machinery. Eventually, you end up with a dead simple proof like this that seems like black magic that fell from the sky.
Fourier series are hidden behind this. Integrating something against cos^n(x) and against cos(nx) is not that different.
Yeah, I thought so as well. Have you ever seen a proof using fourier series of f(x) = x over the interval [0,1]?
Easy? Clever, yes. And easy to understand, once derived.
Great work! Thanks for sharing this problem... I'm interested to learn about its use by Daners.
Wow. One of your finest.
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). ❤
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.
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...
Wow. Amazing how such contrived power function integrals of Trig functions can be used to determine the sum of the inverse squares.
Love your t-shirt! It's my favorite formula
Cannot we use both Riemann Zeta function and define a Fourier series by
f(x) = x when -π
Nice work!
I'm glad this was an "easy solution" to the Basel problem !😄
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?
This was so cool, thanks!
love your videos!
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
The sum of squares theorem?
@@Noam_.Menashe no, the fermat's little theorem
@@kkanden yes, but doesn't it say that the sum of squares if integers is either 0 or 1 mod 4?
@@Noam_.Menashe I think he's referring to a^p=a (mod p)
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 )
This is called Fourier series.
excellent proof, very interesting
17:39
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)
Is 010 considered a friendly sequence or not in this convention?
@@Mod_on_exp no
Let f(n) be the number of friendly sequences of length n. Then f(n) = f(n-1) + f(n-3) + f(n-4) for n > 4.
Proof: Partition the set of friendly sequences of length n based on the first four digits: (a) {0110}; (b) {1100, 1101}; (c) {0111, 1110, 1111}. (All other four-digit starting strings are invalid.) The recursion follows, because these subsets are in bijection with the friendly sequences of lengths n-4, n-3, n-1, respectively. (a) and (b) are clear, while for (c), the bijection is to insert/remove the 1 as the second digit.
The rest is easy; with the initial values f(1) = 0, f(2) = 1, f(3) = 3, f(4) = 4, we find f(11) = 105 is the first value > 100.
I'll start just looking at the number friendly sequences ending with 1 (which is equivalent to ending in 11), call that s(n). We can make a longer sequence by just appending '1' (does not add a new 0) or by appending '011' (adds a new 0). This then implies a recursive formula s(n) = s(n-1) + s(n+3).
Friendly sequences ending in 0 are in a 1-to-1 relation with one bit shorter friendly sequences ending in 1. Let f(n) be the number of all friendly sequences of length n then f(n) = s(n) + s(n-1) = (s(n-1)+s(n-3)) = (s(n-2)+s(n-4)) = (s(n-1)+s(n-2)) = (s(n-3)+s(n-4)) = f(n-1) + f(n-3).
By direct calculation f(2)=1, f(3)=3, and f(4)=4. Then by the recursion the sequence continues 1, 3, 4, 5, 8, 12, 17, 25, 37, 54, 79, 116.... 116=f(13), so the answer is 13.
@@johnchessant3012 I'm having trouble with 1100 in set (b). Your argument seems to suggest that 1100 can be front appended to a valid sequence like 011. But that result is 1100011, which is not a friendly sequence.
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.
the result is the same as the 2 * average of f(x)=x^2 between x=0 and x=pi/2. amazing
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.
To apply hopital you need to differentiate with respect to n, not x. I dont see how you can do it unless swapping integrals and derivatives is a thing
@@bernat8331 It was just a fleeting thought that I had, it wouldn't be of any practical use, however it is true that A_n , B_n -->0 is it not ?
I don't know the intuition behind An and Bn, but the proof looks like something which is closely related to Parseval's identity.
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.
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 .
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)
it was very beautiful math
Lovely proof.
Hermoso resultado!
And what’s the link with the sum of the reciprocals of n squared ?
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”
Can we build up An and Bn for sinx instead of cosx and what do we get from there ?
I achieved lim(Bn/An)=(pi^2)/4 and the bound is from 0 to pi if we do sinx instead of cosx
I wouldnt say my fav basel problem proof.
Simply beautifull !!!
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.
Quite easy to follow but I don't see why we're using 2n instead of n?
1:47 from 0 to π/2
Is there a general way to construct an An,Bn to make these kind of series telescopic?
Thank you, professor.
That was long, but clear.
Yes, but Asubn and Bsubn, there’s absolutely NOTHING intuitive about them! Pretty easy to get somewhere when you aren’t starting from scratch.
Interesting proof. Blackpenredpen also has an easy solution
Wonderfull!!👊
An is a walis integral for even numbers?
I think the word "approximation" is more accurate. Don't you think?
Nice
Thank you for the proof. But I'm curious about your T-shirt. Do you know of any interpretation of this equality?
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
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
Omg this is so surprising 🙀
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.
Intelligent but no way I could think that
Easiest one is Euler's one, even if it is not really a proof.
Gotta correct you on one thing, Michael, if you don't mind. Basel is pronounced B-"ah"-sel. Greetings from Switzerland.
Yeah, that was short.
Credit worthy is problem.
Oh come on, at 9:47 you sneakily changed the second claim. I lost ten minutes trying to prove that on my own...
Hola
Zum Glück war es eine einfache Lösung...
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
Have you ever tried visualizing it and using Geometric Constructions to view pi-squared as six times the solution?
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
asnwer=1 isit
not a good method
an easy solution takes 18 minutes?....yeah, this is AMAT 413 all over again....
Greetings, it remember me Chebyshev polynomials, thanks for your work and for aharing it, I appreciate it :peace_symbol: