Some geometry behind the Basel problem
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- เผยแพร่เมื่อ 13 ก.ย. 2024
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The whole time I’m imagining what Euler would have thought if he had seen this. I think he’d have found it incredible
He’d probably been simultaneously astonished by achievements of science and modern technology and shocked by the abundance of stupidity.
Holy smokes. I was thinking this exact same thing. Haha. Euler would be amazed how far math has come.
@@woody442 can you give some examples of such stupidity?
@@woody442 How do we know he didn't think of this?
@@woody442 You of course mean denial of science and history ... brought to you by the supporters of DJT.
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.
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.
agreed, especially when Michael pulls the "0th integral" trick it's very likely that the proof makes much more sense in reverse order. In this case we really should be starting with the inscribed angle theorem to get 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. Much smoother in my opinion.
@@johnchessant3012 You also more or less work backwards through this proof if you solve the integral from 0 to pi/2 of ln(sinx) dx by replacing sinx with (e^ix)(1-e^(-2ix))/2i
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.
I don't see where the mistake is, but I can very easily tell that the imaginary component integral does not evaluate to zero.
@@minamagdy4126it is 0
@@minamagdy4126 There wasn't a mistake, his argument why the integral is 0 is just needlessly complicated. The integral ln |1+exp(i*theta)| dTheta is a real number that then gets multiplied by i, resulting in a purely imaginary number. Likewise the integral over arg(1+exp(i*theta)) dTheta will be a real number since arg() is a real-valued function.
Since we now have a purely imaginary number plus a real number equal to a real number (the original sum), we know that the imaginary number and hence the integral over ln|1+exp(i*theta)| must be 0, so you can skip the evaluation entirely.
@jay_sensz the thing is, you can skip the evaluation, or you can do it. If you do it, it should yield the same result. My work shows me that this is not the case (integral of cos(t/2) dt from 0 to pi is 2, not 0). My question is, where is the mismatch coming from? Is it a mistake on my part? Michael's? Is there some other theory where the non-zero imaginary result makes sense, as wild as that is?
@@minamagdy4126 He's referring to the argument of the ln function, not the whole expression under the integral because (according to Wolfram Alpha) abs(1+exp(i*theta)) = abs(2*cos(theta/2)) = 2*cos(theta/2) for theta in [0,pi]. However, the resulting integral is still very complicated and the closed form Wolfram Alpha spits out for the indefinite integral is an absolute monstrosity.
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.
So basically going from the other direction?
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!
Adding zero and multiplying by one are so useful. If you can guess which of the infinite forms of zero or one you need
Wow! It’s an amazing result!
Yes but there were some amazing steps along the way, too.
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.
Mathoma has appeared from beyond the pale
(you don't know me but I'm a big fan)
How do you get a rhombus from e^(i*θ)+1?
@@galoomba5559 Draw it out. We know e^(i*θ) is on the unit circle in the complex plane, since it has a modulus of 1. Using the parallelogram method of adding complex numbers, e^(i*θ)+1 falls one unit to the right. Note that all sides of this parallelogram have length 1. Thus, it's a rhombus.
@@Math_oma Ah, i get it now. That's very nice
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?
it's not said explicitly in the video, but if the integrand is a holomorphic function (and 1/(1+z) is besides z = -1) then the value of an integral depends only the endpoints. so in this case, while it's not standard, the notation makes sense.
@@kkanden Is that because the Cauchy-Reimann equations essentially make any polya vector field conservative? I think I remember seeing something like that? That a polya vector field doesn’t curl, making it conservative
@@hydropage2855 you got me here because i don't know the linear algebraic technicalities, i am much better versed in the raw theory ;) but from what i know, all the nice theorems and properties of holomorphic functioncs come down to complex derivatives being a "nice" linear transformation so it may very well be what you're saying
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.
wow that was cool, its so interesting to see all the different methods of solving the basel problem, keep them coming!
Thanks
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.
OMG that's just over the top. Thanks 🙏
I never thought about it , nice one , keep going .
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.
Nice to see a circle while solving this problem.
Method which Euler used can easily be generalized to sums of reciprocals of even powers
Nice! Can you do this to find the Apery's constant?
Surely not
@@Nolord_ Why not? Just use the squine and cosquine functions from the previous video! 🤔
@@RGAstrofotografia Go ahead and you'll become famous
Name sounded familiar: Indeed, I started at University of Copenhagen at the same time as Dr. Brink. Small world
This is a great video. I am of the biggest critics of this channel and, in all truth, I was impressed. Congratulations!
Why would you criticise this channel?
That was super duper cool !!!!!
Wow! This is flabbergasting! Absolutely stupefying!
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.
You have a point. Michael should have chosen a different title for this video. I enjoyed it all the same.
19:30
Are you his wife or something? That's my guess.
The penn is mightier than the sword
15:33 it would [2cos(x/2)]²
That derivation was sexy af Mr. P!
I liked this one very much!
That was a long ride. And great stuff.
Zeta(2) = Pi^2 / 6
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.
Just wow!!!
Amazing!
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]❔🤔
just wondering how do we know mod Z is less than 1 so that infinite geometric series sum can be applied.
was thinking the same
Please, can you explain the details you left? Why 1 + e^{i*theta} is equal to 2*cos(tehta/2}? Pleaseeee
That's simple geometry. 1+exp(iϴ) is the diagonal of the rhombus spanned by 1 and exp(iϴ). Hopefully, you are familiar with the definition of sine and cosine.
3Blue1Brown has a beautiful visualization of this topic.
It's different
@@Nolord_ It's still "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.
He definitely hand waved past that point. The series representation of 1/(1+z) has radius of (uniform) convergence equal to 1, meaning that you can normally swap the series representation only if the parameter is strictly less than 1 in absolute value. In the paper mentioned in the video (you can find it online for free), they explicitly avoid Abel's theorem (which could sometimes be used to deal with these kinds of issues) and take a different approach by appealing to the Mercator series. It is pretty tricky to get around that issue at the boundary of convergence.
7:42 whay (-1)**n+1
It's a good one.
Wow, awesome.
Amazing!!!
wow. beautiful
Beautiful
Wow!
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.
This is what i don't like too. In the original paper they do a way better job. This video is like the equivalent of "proving" that sqrt(2 + sqrt(2 + ...)) = 2, using an argument where one takes convergence for granted; it's just incomplete. It's especially frustrating that Michael puts it as some kind of homework, like it is the easy part which you could work out in a couple of minutes.
@@joelganesh8920 Would you tell me the title of the original paper? I was not able to find it.
@@erikysilvagomes5496 The title is "A Solution to the Basel Problem that Uses Euclid’s Inscribed Angle Theorem" (by D. Brink, 2014 as stated in the video). It is available for free on ResearchGate.
I will say that they do not work out the details, but they are at least way more clear on how to work them out if you wanted to.
🤯
Error: the lower bound of the z-integral should be 1.
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!
Yeah I'm unsure why he took this route. The original paper even motivates it the "correct" way, starting at the Mercator series, working its way to the solution to the Basel problem.
0:02 the is pronounced "thi" when after "the" is letter like a, o, e, i or u
WTH are you talking about?
@@godfreypigott if you have -9 brain sels you won't get it
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. 🙂
th-cam.com/video/d-o3eB9sfls/w-d-xo.htmlsi=xIg2HcvqUkHdeiDh
Wow
𝔸 𝕞𝕒𝕘𝕚𝕔𝕚𝕒𝕟!
Do you want a deeper answer?
Not from a SHALLOW mind.
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!
Everything is geometric if you squint hard enough.
The title says "some geometry". There was "some geometry" -it's literally the thumbnail- it doesn't say a geometric way of solving the problem.
waouh !
PLEASE NATURAL PLEASE N A T U R A 10
Beautiful