Some geometry behind the Basel problem 

An ordinary simple magic sequence of "And then I'm going to do something weird with this sum right here" Michael Penn's style steps.

The whole time I’m imagining what Euler would have thought if he had seen this. I think he’d have found it incredible

It’s quite amazing how much gymnastics can be done to turn this thing first into something where the geometric series gets rid of the summation and then use much more general tricks to turn it into a path integral and finally just a simple integral. Now it would be nice to see the whole proof in reverse to see the the motivation behind all the seemingly bizarre rewriting of this.

15:42 argument from necessity: We know the first term has to sum to zero because it is a wholly imaginary number, and the sum of the reciprocals of the squares is obviously a real number with no imaginary part.

This is a very beautiful argument, but a rather unnatural way of presenting it. The main idea here is that the inscribed angle theorem implies arg(1 + exp(iθ)) = θ/2 (where 0 < θ < π). Integrate both sides from 0 to π, then recognize that the left-hand side is the imaginary part of log(1 + exp(iθ)). Expand this as sum((-1)^(n+1)*exp(inθ)/n) and integrate term by term with respect to θ. We're now left with a sum of 1/n^2 but only the odd terms, so we just need to massage it into the Basel sum. Nowhere was it required to make any steps that would leave the reader scratching their head wondering where we could possibly be going next.

The Basel problem is a favourite of mine. I've never seen this approach before. I never would have thought of it, but it's delightful!

Amazing!🎉🎉🎉 This less than 20 minutes video took me more than one hour to fully understand! And I could follow all the steps thanks to what I learned in the past year. I felt so happy while I was following it. Thank you very much for bringing this to us!

BOOM! My favorite video of yours. The only way it could have been better is having a back flip in the middle, but that might be expecting too much.

Wow, that's a real breathtaking adventure!

Как по мне, это лучше решение проблемы, из тех, что я когда-либо видел. Конечно, справедливость перестановки предельных операций ещё предстоит доказать, но в остальном, всё идёт просто по маслу!

wow that was cool, its so interesting to see all the different methods of solving the basel problem, keep them coming!

Wow! It’s an amazing result!

what did I just saw it was amazing thanks for delivering such quality content

Alternatively, we can see that arg(1+e^(i*θ))) is θ/2 by plotting e^(i*θ), then adding 1, which makes a rhombus. Thus the argument of 1+e^(i*θ) is the argument of the diagonal of this rhombus. But we know from high school geometry that the diagonal of a rhombus is an angle bisector, and thus we have θ/2 for its argument.

Well that was impressive and delightful. There's no way in the universe of universes that I would have come up with anything like that, but I did thoroughly enjoy it.

Adding zero and multiplying by one are so useful. If you can guess which of the infinite forms of zero or one you need

Thanks

OMG that's just over the top. Thanks 🙏

I never thought about it , nice one , keep going .

I'm curious about the justification behind letting the integral of z go from 0 to e^iθ so easily. Isn't that a line integral in the complex plane? I feel like that needs some more definition, or you should've talked about how z will travel that path in the complex plane, no? Am I overthinking it?

Wow! This is flabbergasting! Absolutely stupefying!

This approach is just the reverse of the proof using the Fourier series of F(x) = x^2. It uses almost the same path but backwards, until it arrives in the integral part.

Nice to see a circle while solving this problem.

This is a great video. I am of the biggest critics of this channel and, in all truth, I was impressed. Congratulations!

Nice! Can you do this to find the Apery's constant?

Method which Euler used can easily be generalized to sums of reciprocals of even powers

I swear this "zero-th integral" idea is your favourite thing in the world

Name sounded familiar: Indeed, I started at University of Copenhagen at the same time as Dr. Brink. Small world

That was super duper cool !!!!!

I liked this one very much!

What this video, like so many others, glosses over is the fact that it would be almost impossible to come up with these steps, in a reasonable amount of time, without knowing the method beforehand. So we see a series of "tricks" which probably do not help us to develop a reasoned way of solving a wider class of problems.

That was a long ride. And great stuff.

How does this method break down if you try to use it to find a closed form for zeta(3) in terms of pi?

That derivation was sexy af Mr. P!

Makes me think of gravitational force components adding up on the surface of a spherical mass of uniform density: g at position r=xi^+yj^+zk^=|r|r^=(R,θ,φ) is: g(r)=μΣ(1/|r-r'|²)(r-r')^=(μ*/R²)(-r^) Where: (GM=)μ*=40.02×10¹³ (SI units: Nm²/kg), R=6400km, μ=μ*/N N= number of component masses. We can go a little further: μΣ(1/|r-r'|²)(r-r')^=(μ/|r|)Σ(1/|r^-r'/|r||²)(r-r')^ r'=|r'|(r')^ is the variable for a particular r=|r|r^. Note that: |r'|≠|r|=R or |r'|=R, such that: 0≤|r'|≤2R. Why does |g|≈π² [m/s²]❔🤔

You chose the number in arg(1+z) equal to theta/2, which is the same as the integral variable. I would have chosen it as phi/2 since it should have nothing to do with the theta in the integral.

3Blue1Brown has a beautiful visualization of this topic.

e^(i*Theta) has a length of 1. Does the 1/(1-z) have a complex radius of convergence? Does the single value of 1 act similar to an improper integral and did you hand wave past that point? Please excuse my ignorance.

Please, can you explain the details you left? Why 1 + e^{i*theta} is equal to 2*cos(tehta/2}? Pleaseeee

19:30

just wondering how do we know mod Z is less than 1 so that infinite geometric series sum can be applied.

Just wow!!!

Zeta(2) = Pi^2 / 6

15:33 it would [2cos(x/2)]²

It's a good one.

Amazing!

Beautiful

Wow, awesome.

Wow!

Amazing!!!

As nice as this is (and I did enjoy it!), it's to me a bit of false advertising. What we have here is a lot of algebra / calculus crunching that at one single point makes use of a geometrical insight to simplify an integrand. When I hear "some geometry behind the Basel problem", I'm expecting some way of showing why summing an infinity of (reciprocal) squares leads to something that has to do with a circle (pi squared) - in a geometrical way. We get nothing of that sort. In terms of the expression we are simplifying, I have no geometrical intuition why this appears, why it's being integrated, what the factor 2/3 is doing there, etc.

wow. beautiful

At first when I saw the video title I thought this would be similar to the 3Blue1Brown video on the Basel Problem where he shows geometrically how lighthouses surrounding a circular lake have an intensity that, as the lake's radius increases to infinity, approaches the Basel Problem solution. It's a pretty neat graphical solution if you haven't seen it yet. 🙂 ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-d-o3eB9sfls.htmlsi=xIg2HcvqUkHdeiDh

Thank you for mentioning the complex logarithm function, but those switches between sum and integrals and that complex integral you write without specifying its path should be explained and justified by you and not the viewer. It is extremely delicate to change the order of sum and integrals in series that do not converge absolutely in a neighborhood of the region you are interested in. Bad things can happen when you do this in conditionally convergent series, so more care should be taken to explain these changes in integration. Also, I realize you need to have a 1/(1+z) to make your picture beautiful, but if you had not put that (-1)^(n+1) there you would have obtained -Log(1-z), which gives you a similar picture to the one you did. The geometry is the important part, and that is the nice part of this video. If only you were more clear about the switch in integration and sum and the paths of integration, this would be a jewel to learn from.

As excellent as your presentation is, the problem is unmotivated in the sense that you already know the answer. A priori, one would never think of a thing as equalling to 4/3 of the thing minus 1/3 of the thing, UNLESS one already knew the answer and was in the mood to Symbol Mash.... Still very cool thumbnail with circle and angles to motivate the miraculous appearance of PI in the answer!

Error: the lower bound of the z-integral should be 1.

🤯

7:42 whay (-1)**n+1

Wow

0:02 the is pronounced "thi" when after "the" is letter like a, o, e, i or u

Do you want a deeper answer?

waouh !

𝔸 𝕞𝕒𝕘𝕚𝕔𝕚𝕒𝕟!

PLEASE NATURAL PLEASE N A T U R A 10

That was scam. It's in no way a geometric view of the problem; it was just a trick to evaluate the argument. Nice calculus approach though!

Beautiful