There's always a moment in these videos where I suddenly realise "oh, I can see where this is heading now", and I immediately grasp how a bunch of mathematical ideas I thought had nothing to do with each other are actually working closely together. That moment never fails to bring a broad smile to my face.
Even though the argument given here is new, the Wallis product has been known a long time, and there are other arguments for it than the one given here. For example, just as our previous video on 1 + ¼ + 1/9 + … was based on a paper by Johan Wästlund clearly showing the connection between that sum and circles, there is also a beautiful paper by Wästlund showing a connection between the Wallis product and circles via a different approach than we’ve taken here, which you may find interesting. Donald Knuth has also put out descriptions building off this work by Wästlund. You can check both of those out in the links below. And of course there’s Wallis’s original 17th century argument, based on analysis of certain integrals, though this can make the connection to circles hard to see directly. But, naturally, we’re fondest of the proof we ended up giving here, for its simplicity, for the directions in which it generalizes, and, hell, for the opportunity to re-use our lighthouse animations. And we hope you enjoyed it too. *Edit*: It looks like some people are asking about why the segment at 12:33 is okay, given that it feels like taking 0/0. Keep in mind, the actual goal at that spot is to find a polynomial whose roots are L_1, L_2, ... L_{N-1}, so the concrete result being stated is that (x - L_1)(x - L_2)...(x - L_{N-1}) will expand out to become 1+x+x^2+....x^{N-1}. No division by zero issues there. Sure, plugging in x=1 to (x^N - 1)/(x - 1) is undefined (at least before explicitly stating the intention to extend the function via a limit), but the reason for doing that polynomial division was just to see how (x - L_1)(x - L_2)...(x - L_{N-1}) would expand. All that division is asking is (x - 1)(...what?...) = (x^N - 1). Here, to give a really simple example, it's like saying x^2 - 1 has roots at 1 and -1, so dividing it by (x - 1) gives a polynomial with just a root at -1, namely (x^2 - 1) / (x - 1) = x + 1. "But wait!", someone could say, "you can't plug x = 1 into that fraction!". For sure for sure dude, but that doesn't change the fact that x + 1 is legitimately a polynomial which just has -1 as a root. Maybe you justify that division by saying something about limits, or about analytic continuation, or just by reframing to say what you care about is the question (x - 1)(...what?...) = x^2 - 1, but that's all kind of beside the point. Also, many of you are asking "Isn't the 'distance is proportional to angle' approximation only valid for lighthouses near the observers? What about all the lighthouses on the far end of the circle?". The key is that the product we are ultimately interested in is made up of the asymptotic contributions of each particular lighthouse (in the sense of, e.g., "The 53rd lighthouse after the keeper"), in the limit as N goes to infinity. Whatever particular lighthouse you are looking at, in that limit as N goes to infinity, it will be bunched right next to the observers, and so distances will be proportional to angles for computing its asymptotic contribution. As noted in the section on formalities, Dominated Convergence then rigorously assures us that it's ok to equate "The product of each particular lighthouse's asymptotic limit contribution" (which is the product we're interested in: the Wallis product, or sine product more generally) with "The asymptotic limit of the product of the contributions from each particular lighthouse" (which is the asymptotic limit of the products we have an easy time calculating: the distance-products our lemmas directly address). For more technical details on this use of Dominated Convergence, see the supplemental blogpost. Our supplemental blogpost: www.3blue1brown.com/sridhars-corner/2018/4/17/wallis-product-supplement-dominated-convergence Another cool way of approaching the Wallis product: www.math.chalmers.se/~wastlund/monthly.pdf apetresc.wordpress.com/2010/12/28/knuths-why-pi-talk-at-stanford-part-1/
Your result is true when you sum over positive integer. He sums over both positive and negative. You can find your result if you combine k and -k in the same factor.
The sheer quality of this content is quickly surpassing not only everything else but itself as well. When I thought 3b1b couldn't get any better it does twofold yet again.
Your videos are the most creative representations of mathematic principals I’ve ever seen, it’s always such a joy sitting down to watch one of these. You’ve turned math back into the beautiful and elegant process that I lost touch with, please never stop doing what you do
Sridhar Ramesh! That guy's writing on Quora is pretty awesome, good to see he's joined the 3B1B team. Great proof, and I'm proud that I caught the technicalities myself; was going to write a comment about them, but as usual, you have addressed them yourself!
The Wallis product looks very musical to me. 2 is going up an octave, 2/3 is going down a perfect fifth, 4/3 up a fourth, and so on. I don't know if it's possible, but it would be interesting if there's some kind of proof from this direction. Basically, the Wallis product gives a series of musical intervals that converges to a "note" that is pi/2 above the starting note, which works out to about 782 cents, a rather flattened minor sixth.
Hé Lê (@Science4All) tu as vu la dernière de Mathologer (ruclips.net/video/yk6wbvNPZW0/видео.html ) elle aussi elle est pas mal même si visuellement elle n'atteint pas le niveau de 3Blue1Brown
I have had similar thoughts. It seems, as an observation on the large-scale structure of mathematics, that it is a number-line emphasis on the "line" and so much of "mathematical weirdness" comes from forcing a line to bend or a circle to be straight. These are different metrics and live in different worlds, they do not really want to talk to one another unless forced to. Lines really only want to be a grid!
Every time I watch one of your videos I feel the urge to tell you how awesome your channel is. This connection between the roots of unity and the product expansions of sine is something I didn't know yet, explained in such an intuitive gemetric way! Seriously. Thanks for doing all this!
I, as a 10th year student, haven't understood anything, but that give me more interest in studying maths! So, great job! It means that your videos are really interesting and that you're encouraging more young people like me to study the beauties of maths!😀
I love the little introductions before each video. It's interesting to hear some background on why exactly you made it. It's always an amazing feeling when you stumbled upon something new like in this one!
You're a genius educator, I mean it. Every single one of these videos is so carefully made. Even though I'm already familiar with most of the concepts I find myself learning. It really blows my mind. I just wanted to say that ^^ Congratulations for the proof!
I was looking for a way to "generate" pi out of this product from scratch, so to speak, but this is the next best thing I could have possibly asked for. Your use of infinitely large circular lakes with lighthouses, observers, and light-reception-metrics to prove facts about infinite products and sums is by a significant margin the most beautiful mathematics I have ever (and probably will ever) encounter. I would love to see more of this kind of thing, even if its another video explaining another set of previously known results.
Yeah relatable Ngl highschool math is a joke compared to following his videos in real time, if i want to underatand it i would have paused, and replay the video probably 5 times
Mathologer and 3Blue1Brown are honestly legends, revolutionaries. You guys change the world with every video. Absolutely amazing communicators, a skill sadly rare in higher education and complex topics. decades from now, you guys will be like the Feynman of math education. Keep up the amazing work
I always have to rewind and rewatch your videos so that I’m sure I have really grasped everything that is going on. I’ve been watching your videos for a few years now and I love them!
The level that I love this channel, cannot be expressed with real or imaginary numbers. I am 40 and a 3rd try 1st year Calculus student but the amazing thing, to me at least. Is that there are small snippets of videos like this, that I understand or look "familiar" to me in some way. I view that as my own personal mathematical enlightenment developing. Honestly, that makes me love maths even more; even though I struggle horribly with it. I have a brilliant calculus teacher who thankfully seems to have patience with me as well as the host of tutors that help me every week. I hope one day, I can look back on these videos and either expound upon them, or say "Ahh, yes. It is that way" and actually understand why. Thank you Grant and everyone that contributes to these visual dialog, and that includes people in the comments who push questions with more questions.
This is really cool, and that is not even my favorite video of yours. As you mention in the beginning, a big part of the value in your videos is attributable to the presentation and the communication of the result and I would like to say that you do a fantastic work in this matter. The representations you propose are always very insightful on top of beeing beautiful. In essence, thank you for the hard work and keep doing such intersesting an wonderful videos ! (Sorry for the approximative english, hope you still get the message =) )
I love how you lead me to an understanding. When you teach, I grok. The way you build up clues for me to start piecing things together is akin to being lead through the plot of a great mystery novel. You first help one to construct an intuition and then you reinforce it. This wonderfully developed skill-set you weild shows the beauty of your mind. Because of you (and a few other brilliant minds) I am able now to learn anything mathematical if I just think about it geometrically/trigonometrically. Array manipulation perceived as translation and rotation through 'N' Dimensional space has changed the way I see the world. I love you for this. Thank you for sharing your understandings in such a beautiful way!
Amazing doesn´t describe this accurately, this project is far beyond that, keep it up! At 12:51 you can also show that the left side of this equation exists, when you substitute x with the value x=1 by using L´Hospitals rule, so lim x-->1 of (x^n-1)/(x-1) = lim x-->1 of (n*x^(n-1))/1 = n. Absolutely brilliant stuff in my eyes!
What you brought up about how rearranging the factors in the infinite product can change the product reminds me a lot of the Riemann Rearrangement Theorem. Mathologer has a wonderful video about the Riemann Rearrangement Theorem where he mentions that there are 3 key ingredients. 1. The series as a whole converges 2. The sum of the positive terms diverges to ∞ 3. The sum of the negative terms diverges to −∞ Let's assume we have an infinite product in which all factors are positive. Modifying these 3 key ingredients for products, we have the following 3 "key" ingredients for the infinite product you mentioned 1. The infinite product as a whole converges 2. The product of the factors greater than 1 diverges to ∞ 3. The product of the factors less than 1 converges to 0 Intuitively, following the same logic for the Riemann Rearrangement Theorem for series, with these three key ingredients, you should be able to prove the "Riemann Rearrangement Theorem for Products" - that for any positive real number M, there is a way that you can rearrange the factors of your infinite product to get it to converge to M. (Again, assuming all factors are positive.) For now, I need to get to bed, but I definitely want to go through that argument to see if it actually works or if there is any hiccup. Although the thought occurs to me right now that I may not have to do much work at all, by making use of the isomorphism between the additive group of real numbers and the multiplicative group of positive real numbers. I'll definitely look into both of these this week. It gets me wondering: is there a notion of a "conditionally convergent" and "absolutely convergent" product? If so, what are the definitions? I'll have to look into this :)
I decided to go with the isomorphism route. Suppose that you have the infinite product a₁∙a₂∙a₃∙... where all of the factors are positive and satisfying the 3 key ingredients I mentioned for products. Let M be a positive real number. The infinite product is defined to be lim(n→∞) (a₁∙a₂∙a₃∙...∙aₙ). Since the natural log is a continuous function, ln(lim(n→∞) (a₁∙a₂∙a₃∙...∙aₙ)) = lim(n→∞) ln(a₁∙a₂∙a₃∙...∙aₙ) = lim(n→∞) [ln(a₁)+ln(a₂)+ln(a₃)+...+ln(aₙ)] So, ln(a₁∙a₂∙a₃∙...) = ln(a₁)+ln(a₂)+ln(a₃)+... So the natural log converts infinite products into infinite series. Now, since a₁∙a₂∙a₃∙... converges, we have P = a₁∙a₂∙a₃∙... for some positive real number P. Thus, ln(P) = ln(a₁∙a₂∙a₃∙...) = ln(a₁)+ln(a₂)+ln(a₃)+..., which gives that the infinite series ln(a₁)+ln(a₂)+ln(a₃)+... is convergent. List the factors of the infinite product which are greater than 1 as p₁, p₂, p₃,.... Now, ln(p) > 0 if and only if p > 1. So ln(p₁), ln(p₂), ln(p₃), ... are precisely the positive terms of this infinite series. Since lim(n→∞) (p₁∙p₂∙p₃∙...∙pₙ) = ∞, by the continuity of natural log, lim(n→∞) [ln(p₁)+ln(p₂)+ln(p₃)+...+ln(pₙ)] = lim(n→∞) ln(p₁∙p₂∙p₃∙...∙pₙ) = lim(x→∞) ln(x) = ∞ So the sum of the positive terms of the infinite series diverges to ∞. Similarly, list the factors of the infinite product which are less than 1 as q₁, q₂, q₃,.... Now, ln(q) < 0 if and only if 0 < q < 1. So ln(q₁), ln(q₂), ln(q₃), ... are precisely the negative terms of this infinite series. Since lim(n→∞) (q₁∙q₂∙q₃∙...∙qₙ) = 0 and since each factor is positive, by the continuity of natural log, lim(n→∞) [ln(q₁)+ln(q₂)+ln(q₃)+...+ln(qₙ)] = lim(n→∞) ln(q₁∙q₂∙q₃∙...∙qₙ) = lim(x→0+) ln(x) = −∞ So the sum of the negative terms of the infinite series diverges to −∞. Therefore, this infinite series satisfies the three key ingredients for the Riemann Rearrangement Theorem. Therefore, there exists a rearrangement of the terms of ln(a₁)+ln(a₂)+ln(a₃)+... which converges to ln(M). But rearranging those terms is the same as rearranging a₁, a₂, a₃,... Let b₁, b₂, b₃,... be the rearrangement of a₁, a₂, a₃,... so that ln(b₁)+ln(b₂)+ln(b₃)+... = ln(M). But by a similar argument to the continuity argument we used at the beginning (technically we need the converse: that if an infinite series converges, then the corresponding infinite product converges where you exponentiate with e instead of taking the natural log), ln(b₁)+ln(b₂)+ln(b₃)+... = ln(b₁∙b₂∙b₃∙...). Thus, ln(b₁∙b₂∙b₃∙...) = ln(M). After exponentiating both sides, you get that there is a rearrangement of the factors of a₁∙a₂∙a₃∙... whose product is M. :) Of course, this technique of converting infinite products to infinite series and vice versa can give a whole lot more information too. Such as: if you have an infinite product which converges, then the sequence of factors must converge to 1. This follows from the corresponding fact about the sequence of terms converging to 0 for an infinite convergent series. And then using the fact I just stated above, you can then go through the same logic as the actual proof of the Riemann Rearrangement Theorem. Whenever your partial product is below M, keep multiplying by factors above 1 until you get above M. Whenever your partial product is above M, keep multiplying by factors below 1 until you get below M. And this will converge to M since the sequence of factors must converge to 1. Pretty fun!
This is so good I feel equal parts joy and sadness. Joy over how great it is and sadness over how poorly math was communicated throughout all my years of studying it.
I appreciate the comments about convergence of a product. I'm not an analyst, but I felt like you'd done something a bit naughty in the final steps of the proof.
Such a wonderful presentation of this concept, something that is so abstract is explained in such a lucid, pleasant, logical and visual manner ! Way to go ! I am a big fan. Binge watching math videos first time in life :D
I'm glad I watched enough mathologer videos to instinctively question the commuting of limits and interweaving of infinite products. While, I can't determine when those are possible, it's still nice to know I learned something.
Great video as always. The fact, that I did not understand the "keeper/sailor" idea is my fault of lack of brain mass. A very complicate proof. The one shown in a Mathologer video is more intuitive and easy. But anyway, I like the lighthouse concept a lot! Brought much insight to me.
This video is a lot harder to digest than your previous content. But that means I get to rewatch this video multiple times until I think I get it. Great animations though :-)
I always had a hard time convincing to myself the infinite product representation of sin z in complex analysis class. It all make sense to me now. Thank you so much!
Many ideas in this video that may appear convoluted to a layman is actually very "natural" to someone with a solid higher education in a math related field. You spent years playing with complex numbers and polynomials and what not, eventually they become part of your intuition.
Casually gonna demostrate de Wallis product of pi and the sin formula Mindblowing dude, just mindblowing. Don't be scared at all to explain more, and more slowly as well. I'm always afraid of losing details. I would even recommend you to make 2-parts videos, at least I ould recommend you consider doing it. I mean, you don't make such a beautiful proof everyday...
if you trace Wallis product step-by-step you would get impression that Pi/2 is a rational number, but that can't be because Pi is a irrational number. The result of Rational Number times 2 can not be an irrational number.
I also notice that, taken in pairs, this product is prod[ n^2 / (n^2 -1) ] for n even, and is also 2 * prod[(n^2 -1) / n^2 ] for n odd. It seems that those forms should provide a relationship to the Basel problem and thus also to sines. The most interesting thing to me about Euler's solution to the Basel problem is that it only uses the coefficient of one of the powers of x in the expression for the sine.
I see that subtle easter eggy use of tau instead of pi to record the complex number's phase angle around the unit circle when talking about roots of unity. ;) Makes it so much easier to communicate radians relative to a complete turn of the unit circle~
Pi doesn't have to die. There's areas of math where it's more useful than tau. ;) I personally find myself on the side of preferring mathematical notation that is also intuitive. Using tau over pi does a lot of that where appropriate. But tau doesn't always work better than pi everywhere.
"tau doesn't always work better than pi everywhere" This. There are equations and infinite sequences that, in whatever way, simplify down to something based on pi. Say, pi^2/6 or pi/2. But then there's using rcis(theta) on the Argand plane, or simple harmonic motion in physics which makes better use of tau I'd say both should be in frequent use for wherever they make the most sense
At 11:223, why can one replace the bracket signs with modulo signs showing distance - as the subtraction of two complex numbers doesn't give you a completely real number...
Math totally aside, damn your graphics are well done. The foggy effect around those LED-palette lighthouses against a night background really pulls me in for some reason. Your graphics are very good on your other vids too, but this one made me think to pause and comment on it.
I just saw the video … what I found hand-wavy is the fact that the distance is compared to the angle … it works for small angles and large N but what about near pi angles … yes, these would be large lighthouses indices but still in the product. It is as if you only consider small angles and, with N large enough, they contribute large integer numbers. Thanks to clarify.
Am I missing something? At 12:33 when O = 1, the fraction on the left becomes 0/0, so how can you still conclude that it equals a partial sum of the geometric series?
That is a completely reasonable question. Well, I would say there are several ways you can be convinced about that fact. The first is there is no problem about the denominator being ''0'' because that was an algebraic identity, this is, a formal identity, no matter about evaluating at specific values. You need basically the fact that (I denote by P a ''pi'' letter of product) P (x-j) (with j taking values of complex n-th roots of unity distinct to 1) is equal to x^(n-1)+...+x+1. He somehow deduces this identity by dividing the whole product P (x-j) (this time including the n-th root 1) by x-1 (the factor you don't want) to get x^n-1/x-1 which is in fact the same as before. As I have mentioned, this can be just understood ad ''formal'' equalities not being worried about evaluating on a specific value. Recall that this is true even for a more general field with unity 1. When we say x^n-1/x-1 = x^(n-1)+...+x+1 ''in a formal way'' we are actually saying that x^n-1=(x^(n-1)+...+x+1)(x-1), this last identity is true even considering evaluations for complex x. This is also an equality of polynomials. Since x^n-1 = (x-1) * P (x-j) (product over j distinct to 1) we can deduce (x^(n-1)+...+x+1)(x-1)= (x-1) * P (x-j) (product over j distinct to 1) . And then just delete x-1 on both sides. Does it make sense to delete x-1 on that equation? What if x=1? Well, they are just polynomials, x-1 is an element of a ring (ring C[x] of complex polynomials) which is also an integral domain (this means the ring has unity, is commutative and, the most important property in this case, it does not have zero divisors, this is, if a and b are nonzero then ab is, too, and this is precisely the property that allows us to delete nonzero numbers on a equation of the above type, for example, if az=bz and z is nonzero then z(a-b)=0 and then a-b=0 because there aren't zero divisors, and this is the same as deleting z on the equation ;) ) and as I was saying x-1 is nonzero so that it can be cancelled on the equation. Once we have (x^(n-1)+...+x+1)=P (x-j) (product over j distinct to 1) making j=1 we get n=the desired product. One last comment, you can also understand the discussed fact by observing the smoothness of the equation. I mean, you would agree that the equation written on the video with O-1 on the denominator is true whenever O is not 1. Well, do not evaluate at O=1, but since you are tallking about polynomials then make the limit O--->1, and by the continuity of the involved functions this will behave in a good way, and so what do you get? The desired equation anyway. I hope to have been clear enough :)
The RHS is a polynomial that, when multiplied by x-1, gives x^N-1. There is only one polynomial like that, namely, x^{N-1}+x^{N-2}+\dots+x+1. Thus, they must be equal… for all values of x, including x=1.
Integrating the limits sequentially one at a time makes senses because for any given distance to a point, there is a corresponding mirrored point (with a negative sign for example) which should be integrated before moving to the next distance. For example: If you move away from the Sailor on the X axis, towards the origin and beyond, you're going to pass lighthouses in groups of two (a positive A and negative B lighthouse), passing one just ever so slightly before the other. Having started your journey from the Keeper's perspective, you'll pass the two lighthouses simultaneously each time. So it seems logical that the limits should be integrated in the order you used (A1*B1*A2*B2), and not something different like A1*B1*B2*A2.
Thanks ♥ for you efforts and for these amazing videos, Please we want another episodes about deep learning and Machine learning algorithms (RNN, K-means, Logistic regression, SVM/SVR, ....) ♥
I also had to go back and rewatch this to understand what he meant. He's taking the limit of the series of numbers in each column, which will always tend to 1 for large enough N since there's only one 7 in the column. His point is that if you take the product of each column's limit of index N as N -> inf, it's not equal to the limit of the product of each column's index N (i.e. row N) as N -> inf, so the product of limits isn't equal to the limit of products.
@@briangronberg6507 he’s not taking a limit of a sum, he’s taking the limit of a series of numbers. In the first case, that series is the product of row i from i=1 to inf, and in the second case it’s simply the numbers in a given column. In the first case, the limit of those products is 7 since every number in the series is 7, and in the second case the product of the limits of those series is 1, since each column’s series tends to 1 for large enough N.
At 12:25, instead of the little bit of algebra to get the resulting sum, I applied L'Hopital's rule for the limit as O approaches 1, which quickly yields n. Other than that my approach was the same, save for superficial differences. I began by writing an expression for distance product in the natural(?) way (write the points in their exponential form, shift it all over by 1, product of the absolute values) but this isn't really needed. I guess my initial thought was just to write the damn thing down. 😂 In any case, Lemma #2 is such a pretty little thing, I felt the need to play with it. This isn't expected at all, it provides a beautiful example of solving a geometric problem in the complex plane, and it even has a nice hook in the form making an initial check of easy cases: I first convinced myself ... that I convinced myself, that n=0 does not fit the problem, as an observer is assumed. The trivial case of n = 1 yields no line segments, corresponding to the empty product of 1. The case n = 2 gives one line segment of length 2, which is also the product of "all the lines." I skipped over n = 3 because it felt like I'd need to write stuff down for that one. My favorite easy, but not _trivial_ (in the usual sense) case is when n = 4. We have the line segment of length 2 that goes across the circle, and the other two are hypoteni(?) of right triangles. We get 2*sqrt(2)*sqrt(2) = 4 = n. Overall a 10/10 adorable tidbit, I'm surprised this isn't a popular cool math problem, at least among math majors.
To write a limit formaly correct, it should be like for example: lim( sum of 1/2^n) for n->infinity, which is 2. How does the correct form of this look like? What goes to infinity here exactly?
Frederik Huber The limits taken vertically are not based on the products of the columns (notice the lack of dots between numbers in the vertical direction as opposed to the dots in the horizontal rows). The vertical limit is simply the value that the sequence approaches, i.e., the limit of the nth term (not the limit of partial products). Since each sequence eventually only has 1's, the limit of the nth term is 1. i.e., the rows are lim(Product of n terms) and the columns are lim(nth term) as n --> ∞.
Aha! Yeah that makes a lot more sense. Rewatching this I noticed I overlooked that the limit of 7, 7, 7, ... = 7, which wouldn't make sense if it was a product. Thanks!
You can view it as a sequence of functions, fn = 1*1*1.... n times * x (with x = 7 in the particular example). When n goes to infinity you "lose" the x... This is an example of a sequence of functions that converges pointwise but not uniformly because, as you said so yourself, at any finite n there's still a 7 there.
Hi, I found an alternative, but easy visual way to find the distance product of Observer replacing one of the lighthouses at 12:00 First, calculate the distance product of the Observer but only using lighthouses 1, 3, 5, 7. Then the Observer is 1/2 of the way between lighthouse 7 and 1, so the distance product is 2. You then still have to multiply this by the distance product of the Observer and the remaining lighthouses (2,4,6). Here you do the same thing, you calculate the distance product of the Observer with lighthouses 2 and 6, which is equal to 2. Then only lighthouse 4 remains and the distance product of the Observer with lighthouse 4 is equal to 2 as well. You then multiply these 3 distance product to get the answer, 8.
Apparently, that does only work for N = 2^k with some natural number k. Nice finding, anyway. And for the proof of the Wallis product in the video (better to say: sketch of a proof) it is sufficient to have _some_ large numbers N, which may well be powers of 2.
I used to think imaginary and complex numbers where something stupid. How wrong I was. Now I know that even the most abstract ideas in math can allow us to find very deep and interesting relations between things
You could prove Lemma 2 using the fact that the guy is e^{pi i / n} in the complex plane, and we want to find the magnitude of the product from k = 1 to n - 1 of w^k - e^{pi i / n} where omega = L_1 = e^{2 pi i / n} so this whole thing becomes, since we have magnitude, we can flip sign to get |product(e^{pi i / n} - w^k)| = |(e^{pi i / n})^n - 1| = |-1 - 1| = |-2| = 2. If it helps, note that w^k = L_k.
I had to present a proof of the stirling formula last semester and the main part of it was proving the Wallis product (after you have this the rest is more or less trivial). I also had an geometrical approach based on a paper I found but it wasn't as visual as this here of course. I must say that I didn't like this as much as the Basel problem video because it was more algebraicly playing around than geometrical intuition. I also think think that this approach is a bit harder to make rigorous for a non-mathmatician audience. If I don't have a rigorous proof then I can't be sure that my intuition is correct. So this proof here is not ideal for my taste but I really like that you try to make "nice" proofs for already proven things :) EDIT: I just saw that the the sorce I used for my proof was the one by Johan Wästlund in the description. I of course rephrased a lot of stuff he did to make it easier to understand. I think his method is easier for everyone and could even be done in schools but the main problem is the proof is pretty boring in the middle when deriving a lot of side stuff algebraicly.
The boring parts is where you say, "Exercise!" And then follow it with, "Justkidding. You can read the boring derivation in the book." Then everyone feels like they've been sold short, but the lesson is not too uninteresting.
David Herrera The thing is that I did not do that at all. Like I said my proof was rigorous, no matter how small I proved every single step. I even needed an inequality with e for the final proof that is a common exercise in analysis and I sat there for quite a while trying to figure out how to make this complete but also understandable for people without a math background. As I tested it with friends I knew that my proof was pretty much that, the most difficult thing was knowing what the number e is. I personally hate leaving stuff out, I even had to reformulate the last steps because I didn‘t want to use rules about asymptotic approximations ~ that I wouldn‘t prove so I only used the definition of the ~ symbol. Like I said in the end the proof was a bit boring cause in the middle part were a lot of lemmas that I proved which made everything easier to follow but not really exciting. At the end however when I showed the geometry, how to interprete the results as circles and area and so on went cool again. If Inwould make it again I would try to think of a way to make the middle part better
3Blue1Brown: explains an exclusively arithmetic complex problem with a very specific solution Also 3Blue1Brown: so you're probably wondering how this relates to geometry
I wish a more standard mathematical proof was also provided alongside with the long paragraphs of discussion in the blog. Its good to provides insights on how the proofs were constructed, but there are still other technicalities underlying the proof. I know you guys defintely are capable of doing so. For example, another issue I am considering is that although all of the ratios are converging to its numerical limits, the speed of which they are converging is slower and slower going down the sequence, i.e. terms are not uniform convergence. I suspect that causes the interchange of the orders generate a different ''value'' for the limit, rather than "absolute convergence" you mentioned in the blog post.
Great video again! Just one small suggestion. I found 9:00 quite confusing at first, as I thought the distance is not simply (O-L_i) but |O-L_i|, and the magnitude of a multiplication of the former is not a multiplication of the latter. Then I did search around and found the property for complex numbers that |xy|=|x||y|. Maybe it would be helpful to note this property somewhere for people who are not that familiar w/ complex numbers.
The property |xy| = |x||y| has always been true for the real numbers so he probably thought that it wouldn't cause that much confusion when he also used it for complex numbers
I recently had to prove these things in a home assignment. You start with the Fourier series of cos(tx), then use it to show that the integral from 0 to xpi of [cot(t) - 1/t] is the infinite sum of log[1 - x^2/n^2] But doing standard integration you also get log(sinx/x) So that means log(sinx/x) = sum of log(1 - x^2/(npi)^2) = log of the infinite product (1 - x^2/(npi)^2); You get rid of the log in both sides and substitute x = pi/2... voila
I think it wouldn't hurt to quickly mention that the absolute value of a product is the product of absolute values also when talking about complex numbers
Yes I agree, although the proof of this is so easy both to state and to visualise that maybe it slipped their mind! (write z1 = n*e^(i*a), z2 = m*e^(ib), where n, m >= 0 are real magnitudes. Then z1*z2 has a magnitude of n*m -- that's half of the definition of complex multipation. the other half being the new angle of (a+b) mod_tau. Or you can manually prove it algebraically z1 = a_ib, z2 = c + id, but this doesn't seem very natural to me.)
This video is part of math history now. In math books of my school-time there was "interesting fact" text insertions to make pupil more interested in a subject. In the math books of future they will give link to this video.
Man, I was so freaked out after doing the re-upload and publishing. "WHY IS IT 360p?!?!, Did I do the export wrong?!?". Silly RUclips, easing it's way into HD.
Trying to explain what you’ve seen to a friend: Intricate mathematical interrelationships explained in a novel and aesthetically alluring way by means of ingenious animations, an authoritative yet affable (saganesque) voice, and background music perceptible enough to ensure a rhythm, yes subtle enough to avoid distraction. Anyone who says this isn’t art...
As stated in this year's Pi Day video, all of the Greek letters in Euler's day were just variables; it was just some one mathematician who decided that pi should be taken as a constant, based on a recurring statement by Euler: "Let pi be the ratio of the semicircumference to the radius of the given circle."
Im just doing my thesis on abelian number fields, hence lots of cyclotomic fields stuff is in play (thanks to Kronecker-Weber). I wouldn't be surprised if this was burried somewhere in the bast literature, but it is cool as heck. I would love to see more.
This magnum opus is the epitome of human creation. Each day my thoughts wander to your impeccable and uniquely inspiring content. The subtle messages conveyed in each of your pieces are hallucinatory and bring us, the human race, glorious euphoria. The kind of euphoria that ignites woodlands like an Einstein-Rosen Bridge collision. I'm deeply engaged in the metaphor of sloths, one of the seven deadly sins, which is easily comparable to the ancient greek myth, pandora's box, juxtapositional acquiesce to the current state of affairs of this tumultuous society we find ourselves in today. Before the wondrous discovery of this series, I felt chap-fallen, woebegone, and forlorn. Instantaneously, these fabrications took the quimp out of my day and lit the fire in my eyes that is the motivation to achieve one's passions and become the best version of yourself. As Galileo once said, "Répondre intelligemment même à un traitement inintelligent". I'll abandon you on this inscription, wishing you the best of all good fortune.
At around 6:30, you talk about the magnitude and the angle. You should mention that it has to do with the polar coordinate form (maybe not HOW it does, but just mention it, so people who don't already know about the polar coordinate form can look it up)
Gotta love how the commutativity of fractions doesn't always hold for infinite products. If you write them as one fraction with one vinculum, i.e. 2*2*4*4*6*6.../1*3*3*5*5*7..., you might simplify it and remove the one in the denominator. Then you end up with (2/3)^2 * (4/5)^2 * (6/7)^2 ... * (2n/(2n+1))^2, which converges to zero. Even if you account for the discrepancy in the number of numerators vs. the number of denominators and append " * (2n+2)" to the end, it would amount to multiplying infinity by zero at the point of convergence. Yet writing it as 2 times the rest of the fraction brings it back to a product of pi/2! It's like there's a whole new kind of place value.
![BLACK BAG - Official Trailer [HD] - Only in Theaters March 14](http://i.ytimg.com/vi/Du0Xp8WX_7I/mqdefault.jpg)
There's always a moment in these videos where I suddenly realise "oh, I can see where this is heading now", and I immediately grasp how a bunch of mathematical ideas I thought had nothing to do with each other are actually working closely together.
That moment never fails to bring a broad smile to my face.
suuuuuuuper nice :-)
like meeting an old friend in an unexpected place
Yes.
why did you say "broad"? wasn't that overly emotional?
7:30 is that moment for me
I like to call that a math-gasm.
What's better than a 3Blue1Brown video?
A 3Blue1Brown video with an original proof by the creators
tony nixon Have you ever heard of Mathologer? If not, you'll like what happens there, the guy does that all the time :)
Béranger Seguin i love mathologer long time ago but when i saw his video proving that numberphile is wrong, i was like mathologer sucks.
what is better than a 3blue1brown video? it is 2 3blue1brown videos
Béranger Seguin yes. I do watch mathologer videos. Very inspiring
So, you went from loving his videos to thinking he sucks when he *proved* that someone is wrong. Very reasonable.
Novel mathematical proofs being presented on RUclips is proof to me that we live in the future.
Well, you said that twelve hours ago so... Yes, now we do. ;)
Putting it that way certainly does make it clear how far we've come
We live beside the future.
seems to me that the more future, the harder it is to be novel. I think the novelty means we're still finishing up the. past
Seán O'Nilbud in, beside, underneath, above... the future is our climbing frame.
A mathematical proof first shown on youtube by the best mathematical youtube channel ... What a time to be alive !!
I fucking love your videos!!
Even though the argument given here is new, the Wallis product has been known a long time, and there are other arguments for it than the one given here. For example, just as our previous video on 1 + ¼ + 1/9 + … was based on a paper by Johan Wästlund clearly showing the connection between that sum and circles, there is also a beautiful paper by Wästlund showing a connection between the Wallis product and circles via a different approach than we’ve taken here, which you may find interesting. Donald Knuth has also put out descriptions building off this work by Wästlund. You can check both of those out in the links below. And of course there’s Wallis’s original 17th century argument, based on analysis of certain integrals, though this can make the connection to circles hard to see directly.
But, naturally, we’re fondest of the proof we ended up giving here, for its simplicity, for the directions in which it generalizes, and, hell, for the opportunity to re-use our lighthouse animations. And we hope you enjoyed it too.
*Edit*: It looks like some people are asking about why the segment at 12:33 is okay, given that it feels like taking 0/0. Keep in mind, the actual goal at that spot is to find a polynomial whose roots are L_1, L_2, ... L_{N-1}, so the concrete result being stated is that (x - L_1)(x - L_2)...(x - L_{N-1}) will expand out to become 1+x+x^2+....x^{N-1}. No division by zero issues there. Sure, plugging in x=1 to (x^N - 1)/(x - 1) is undefined (at least before explicitly stating the intention to extend the function via a limit), but the reason for doing that polynomial division was just to see how (x - L_1)(x - L_2)...(x - L_{N-1}) would expand. All that division is asking is (x - 1)(...what?...) = (x^N - 1).
Here, to give a really simple example, it's like saying x^2 - 1 has roots at 1 and -1, so dividing it by (x - 1) gives a polynomial with just a root at -1, namely (x^2 - 1) / (x - 1) = x + 1. "But wait!", someone could say, "you can't plug x = 1 into that fraction!". For sure for sure dude, but that doesn't change the fact that x + 1 is legitimately a polynomial which just has -1 as a root. Maybe you justify that division by saying something about limits, or about analytic continuation, or just by reframing to say what you care about is the question (x - 1)(...what?...) = x^2 - 1, but that's all kind of beside the point.
Also, many of you are asking "Isn't the 'distance is proportional to angle' approximation only valid for lighthouses near the observers? What about all the lighthouses on the far end of the circle?". The key is that the product we are ultimately interested in is made up of the asymptotic contributions of each particular lighthouse (in the sense of, e.g., "The 53rd lighthouse after the keeper"), in the limit as N goes to infinity. Whatever particular lighthouse you are looking at, in that limit as N goes to infinity, it will be bunched right next to the observers, and so distances will be proportional to angles for computing its asymptotic contribution.
As noted in the section on formalities, Dominated Convergence then rigorously assures us that it's ok to equate "The product of each particular lighthouse's asymptotic limit contribution" (which is the product we're interested in: the Wallis product, or sine product more generally) with "The asymptotic limit of the product of the contributions from each particular lighthouse" (which is the asymptotic limit of the products we have an easy time calculating: the distance-products our lemmas directly address). For more technical details on this use of Dominated Convergence, see the supplemental blogpost.
Our supplemental blogpost:
www.3blue1brown.com/sridhars-corner/2018/4/17/wallis-product-supplement-dominated-convergence
Another cool way of approaching the Wallis product:
www.math.chalmers.se/~wastlund/monthly.pdf
apetresc.wordpress.com/2010/12/28/knuths-why-pi-talk-at-stanford-part-1/
3Blue1Brown why did you reupload?
Amazing video both times I saw it.
ArpholomuleNutt He made a small mistake with the original.
3Blue1Brown
sin(fX)/fX=product of {1-(X/N)^2)}
But your result is different!!!
Richard Reynolds what was it?
Your result is true when you sum over positive integer. He sums over both positive and negative. You can find your result if you combine k and -k in the same factor.
You're a freaking math communication genius dude
Ditto!
The sheer quality of this content is quickly surpassing not only everything else but itself as well. When I thought 3b1b couldn't get any better it does twofold yet again.
112BALAGE112 Don't fucking jinx it!
Your videos are the most creative representations of mathematic principals I’ve ever seen, it’s always such a joy sitting down to watch one of these. You’ve turned math back into the beautiful and elegant process that I lost touch with, please never stop doing what you do
And he hasn’t stopped
Sridhar Ramesh! That guy's writing on Quora is pretty awesome, good to see he's joined the 3B1B team. Great proof, and I'm proud that I caught the technicalities myself; was going to write a comment about them, but as usual, you have addressed them yourself!
The Wallis product looks very musical to me. 2 is going up an octave, 2/3 is going down a perfect fifth, 4/3 up a fourth, and so on. I don't know if it's possible, but it would be interesting if there's some kind of proof from this direction. Basically, the Wallis product gives a series of musical intervals that converges to a "note" that is pi/2 above the starting note, which works out to about 782 cents, a rather flattened minor sixth.
Woooow!! Congrats for the beautiful proof! This is both extremely cool and hugely impressive!
S4A? You here? Ah, I see you're a man of culture as well :')
Thanks dude!
Oh hello there, comment ça va? xD
No one ever misses 3B1B videos x)
Hé Lê (@Science4All) tu as vu la dernière de Mathologer (ruclips.net/video/yk6wbvNPZW0/видео.html ) elle aussi elle est pas mal même si visuellement elle n'atteint pas le niveau de 3Blue1Brown
I keep noticing, how many interesting ideas come out when you connect square-related and circle-related concepts!
I have had similar thoughts. It seems, as an observation on the large-scale structure of mathematics, that it is a number-line emphasis on the "line" and so much of "mathematical weirdness" comes from forcing a line to bend or a circle to be straight. These are different metrics and live in different worlds, they do not really want to talk to one another unless forced to. Lines really only want to be a grid!
The Poincaré conjecture is a case in point, I suppose.
if you think about it it really is squares... here is what I mean:
2/1*2/3*4/3*4/5*...
= 4/3 * 9/8 * 16/15 * 25/24 * 36/35 * 49/48 * 64/63 * ...
@@wrog7616 isn't it 4/3 * 16/15 * 36/35 * 64/63 * ... ?
@@pietervannes4476 Yeah. Thanks. lol
Every time I watch one of your videos I feel the urge to tell you how awesome your channel is. This connection between the roots of unity and the product expansions of sine is something I didn't know yet, explained in such an intuitive gemetric way! Seriously. Thanks for doing all this!
I, as a 10th year student, haven't understood anything, but that give me more interest in studying maths! So, great job! It means that your videos are really interesting and that you're encouraging more young people like me to study the beauties of maths!😀
I am 10 too and I understand 20%😢
I love the little introductions before each video. It's interesting to hear some background on why exactly you made it. It's always an amazing feeling when you stumbled upon something new like in this one!
TAU SPOTTED
YES! Finally. The Tauist revolution has begun!
I was wondering about the taus, by the way.
Tau! may i redirect any pi-ists to tauday.com ?
π is dead. Long live the vastly superior τ.
Euler used Pi like Theta.
I personally believe Tau is simply the limit of pi.
Limits are awesome.
You're a genius educator, I mean it. Every single one of these videos is so carefully made. Even though I'm already familiar with most of the concepts I find myself learning. It really blows my mind.
I just wanted to say that ^^ Congratulations for the proof!
I was looking for a way to "generate" pi out of this product from scratch, so to speak, but this is the next best thing I could have possibly asked for. Your use of infinitely large circular lakes with lighthouses, observers, and light-reception-metrics to prove facts about infinite products and sums is by a significant margin the most beautiful mathematics I have ever (and probably will ever) encounter. I would love to see more of this kind of thing, even if its another video explaining another set of previously known results.
Who else had no idea what he's actually talking about but can't stop watching his videos
Yeah relatable
Ngl highschool math is a joke compared to following his videos in real time, if i want to underatand it i would have paused, and replay the video probably 5 times
I'm blown away with every new video. I said it before and I say it again : this channel is pure youtube gold !
Thank you, I genuinely thank you.
Mathologer and 3Blue1Brown are honestly legends, revolutionaries. You guys change the world with every video. Absolutely amazing communicators, a skill sadly rare in higher education and complex topics. decades from now, you guys will be like the Feynman of math education. Keep up the amazing work
Yeah
"local mathematicians"
i don't think my town has any
@@lanye2708 primary school too ?
@@lanye2708 most school level math teachers majored in econ or finance
If you don't know any local mathematicians, that's because you're the local mathematician.
@@Pablo360able my self-esteem is boosted
Noo! That poor Pi creature at 3:03 became a skeleton!
Rather, it got petrified! This is some Medusa-level sorcery and I shall NOT tolerate it
#PrayForPi
I always have to rewind and rewatch your videos so that I’m sure I have really grasped everything that is going on. I’ve been watching your videos for a few years now and I love them!
The level that I love this channel, cannot be expressed with real or imaginary numbers.
I am 40 and a 3rd try 1st year Calculus student but the amazing thing, to me at least. Is that there are small snippets of videos like this, that I understand or look "familiar" to me in some way. I view that as my own personal mathematical enlightenment developing. Honestly, that makes me love maths even more; even though I struggle horribly with it. I have a brilliant calculus teacher who thankfully seems to have patience with me as well as the host of tutors that help me every week. I hope one day, I can look back on these videos and either expound upon them, or say "Ahh, yes. It is that way" and actually understand why.
Thank you Grant and everyone that contributes to these visual dialog, and that includes people in the comments who push questions with more questions.
Did you watch his essence of calc series
I have, several times. Its amazing.
Check Wikipedia about inaacessible cardinals if you want bigger numbers than complex numbers.
I love your voice. It helps me fall asleep. You’re not boring but rather quite calming
This is really cool, and that is not even my favorite video of yours. As you mention in the beginning, a big part of the value in your videos is attributable to the presentation and the communication of the result and I would like to say that you do a fantastic work in this matter. The representations you propose are always very insightful on top of beeing beautiful.
In essence, thank you for the hard work and keep doing such intersesting an wonderful videos !
(Sorry for the approximative english, hope you still get the message =) )
I’m a simple man, I see 3B1B, I click like.
I'm a simpler man. I see a video I like, I click like.
Same. No matter what.
I think if you were a simple man, you wouldn't be interested in math channels
My mind has never not been blown by this channel!
everytime i watch a 3b1b video i always manage to get lost, then completely enlightened at the end. insane quality vids
Great that you talked about the issue with limits and infinite products !
I love how you lead me to an understanding.
When you teach, I grok.
The way you build up clues for me to start piecing things together is akin to being lead through the plot of a great mystery novel. You first help one to construct an intuition and then you reinforce it. This wonderfully developed skill-set you weild shows the beauty of your mind.
Because of you (and a few other brilliant minds) I am able now to learn anything mathematical if I just think about it geometrically/trigonometrically.
Array manipulation perceived as translation and rotation through 'N' Dimensional space has changed the way I see the world. I love you for this.
Thank you for sharing your understandings in such a beautiful way!
This is the best channel on youtube!
Amazing doesn´t describe this accurately, this project is far beyond that, keep it up!
At 12:51 you can also show that the left side of this equation exists, when you substitute x with the value x=1 by using L´Hospitals rule, so lim x-->1 of (x^n-1)/(x-1) = lim x-->1 of (n*x^(n-1))/1 = n. Absolutely brilliant stuff in my eyes!
3:03 when an extra gets a part in the movie
What you brought up about how rearranging the factors in the infinite product can change the product reminds me a lot of the Riemann Rearrangement Theorem.
Mathologer has a wonderful video about the Riemann Rearrangement Theorem where he mentions that there are 3 key ingredients.
1. The series as a whole converges
2. The sum of the positive terms diverges to ∞
3. The sum of the negative terms diverges to −∞
Let's assume we have an infinite product in which all factors are positive.
Modifying these 3 key ingredients for products, we have the following 3 "key" ingredients for the infinite product you mentioned
1. The infinite product as a whole converges
2. The product of the factors greater than 1 diverges to ∞
3. The product of the factors less than 1 converges to 0
Intuitively, following the same logic for the Riemann Rearrangement Theorem for series, with these three key ingredients, you should be able to prove the "Riemann Rearrangement Theorem for Products" - that for any positive real number M, there is a way that you can rearrange the factors of your infinite product to get it to converge to M. (Again, assuming all factors are positive.)
For now, I need to get to bed, but I definitely want to go through that argument to see if it actually works or if there is any hiccup. Although the thought occurs to me right now that I may not have to do much work at all, by making use of the isomorphism between the additive group of real numbers and the multiplicative group of positive real numbers. I'll definitely look into both of these this week.
It gets me wondering: is there a notion of a "conditionally convergent" and "absolutely convergent" product? If so, what are the definitions? I'll have to look into this :)
I decided to go with the isomorphism route.
Suppose that you have the infinite product a₁∙a₂∙a₃∙... where all of the factors are positive and satisfying the 3 key ingredients I mentioned for products. Let M be a positive real number.
The infinite product is defined to be lim(n→∞) (a₁∙a₂∙a₃∙...∙aₙ).
Since the natural log is a continuous function,
ln(lim(n→∞) (a₁∙a₂∙a₃∙...∙aₙ))
= lim(n→∞) ln(a₁∙a₂∙a₃∙...∙aₙ)
= lim(n→∞) [ln(a₁)+ln(a₂)+ln(a₃)+...+ln(aₙ)]
So, ln(a₁∙a₂∙a₃∙...) = ln(a₁)+ln(a₂)+ln(a₃)+...
So the natural log converts infinite products into infinite series.
Now, since a₁∙a₂∙a₃∙... converges, we have P = a₁∙a₂∙a₃∙... for some positive real number P. Thus,
ln(P) = ln(a₁∙a₂∙a₃∙...) = ln(a₁)+ln(a₂)+ln(a₃)+...,
which gives that the infinite series ln(a₁)+ln(a₂)+ln(a₃)+... is convergent.
List the factors of the infinite product which are greater than 1 as p₁, p₂, p₃,.... Now, ln(p) > 0 if and only if p > 1. So ln(p₁), ln(p₂), ln(p₃), ... are precisely the positive terms of this infinite series. Since lim(n→∞) (p₁∙p₂∙p₃∙...∙pₙ) = ∞, by the continuity of natural log,
lim(n→∞) [ln(p₁)+ln(p₂)+ln(p₃)+...+ln(pₙ)]
= lim(n→∞) ln(p₁∙p₂∙p₃∙...∙pₙ)
= lim(x→∞) ln(x) = ∞
So the sum of the positive terms of the infinite series diverges to ∞.
Similarly, list the factors of the infinite product which are less than 1 as q₁, q₂, q₃,.... Now, ln(q) < 0 if and only if 0 < q < 1. So ln(q₁), ln(q₂), ln(q₃), ... are precisely the negative terms of this infinite series. Since lim(n→∞) (q₁∙q₂∙q₃∙...∙qₙ) = 0 and since each factor is positive, by the continuity of natural log,
lim(n→∞) [ln(q₁)+ln(q₂)+ln(q₃)+...+ln(qₙ)]
= lim(n→∞) ln(q₁∙q₂∙q₃∙...∙qₙ)
= lim(x→0+) ln(x) = −∞
So the sum of the negative terms of the infinite series diverges to −∞.
Therefore, this infinite series satisfies the three key ingredients for the Riemann Rearrangement Theorem.
Therefore, there exists a rearrangement of the terms of ln(a₁)+ln(a₂)+ln(a₃)+... which converges to ln(M). But rearranging those terms is the same as rearranging a₁, a₂, a₃,... Let b₁, b₂, b₃,... be the rearrangement of a₁, a₂, a₃,... so that ln(b₁)+ln(b₂)+ln(b₃)+... = ln(M). But by a similar argument to the continuity argument we used at the beginning (technically we need the converse: that if an infinite series converges, then the corresponding infinite product converges where you exponentiate with e instead of taking the natural log), ln(b₁)+ln(b₂)+ln(b₃)+... = ln(b₁∙b₂∙b₃∙...). Thus, ln(b₁∙b₂∙b₃∙...) = ln(M).
After exponentiating both sides, you get that there is a rearrangement of the factors of a₁∙a₂∙a₃∙... whose product is M. :)
Of course, this technique of converting infinite products to infinite series and vice versa can give a whole lot more information too. Such as: if you have an infinite product which converges, then the sequence of factors must converge to 1. This follows from the corresponding fact about the sequence of terms converging to 0 for an infinite convergent series.
And then using the fact I just stated above, you can then go through the same logic as the actual proof of the Riemann Rearrangement Theorem. Whenever your partial product is below M, keep multiplying by factors above 1 until you get above M. Whenever your partial product is above M, keep multiplying by factors below 1 until you get below M. And this will converge to M since the sequence of factors must converge to 1.
Pretty fun!
This is so good I feel equal parts joy and sadness. Joy over how great it is and sadness over how poorly math was communicated throughout all my years of studying it.
It's amazing how your videos keep getting better and better!
I appreciate the comments about convergence of a product. I'm not an analyst, but I felt like you'd done something a bit naughty in the final steps of the proof.
Davy Ker
i know, right? before he mentioned the subtleties i had assumed that the mathematics mafia would show up or something to "take care of" him
Such a wonderful presentation of this concept, something that is so abstract is explained in such a lucid, pleasant, logical and visual manner ! Way to go ! I am a big fan. Binge watching math videos first time in life :D
Everything is linked. That's awesome! Euler would have loved this video:)
Yes he would
Drinking game: bingewatch 3blue1brown videos and drink every time there is a use of complex numbers
I'm glad I watched enough mathologer videos to instinctively question the commuting of limits and interweaving of infinite products. While, I can't determine when those are possible, it's still nice to know I learned something.
Great video as always. The fact, that I did not understand the "keeper/sailor" idea is my fault of lack of brain mass. A very complicate proof. The one shown in a Mathologer video is more intuitive and easy. But anyway, I like the lighthouse concept a lot! Brought much insight to me.
8:14
The tau rebellion will never die
There comes a time when I'm not actually able to get what are you talkin' 'bout but still I watch it further..and I realize that I'm digesting it now!
This video is a lot harder to digest than your previous content. But that means I get to rewatch this video multiple times until I think I get it.
Great animations though :-)
I always had a hard time convincing to myself the infinite product representation of sin z in complex analysis class. It all make sense to me now. Thank you so much!
24:47 "=" sign is missing in 1/1^2 + 1/2^2 + ... = pi^2/6
You are infinite years old now
I'm happy that firstly I had the time to watch the whole video at once and secondly it was that long. Keep it up. I keep recommending your channel.
This is just.... hell, how did you get such ideas?! They're so convoluted I don't see a way for someone to 'notice them.'
that's why it's called a trained mathematician
Many ideas in this video that may appear convoluted to a layman is actually very "natural" to someone with a solid higher education in a math related field. You spent years playing with complex numbers and polynomials and what not, eventually they become part of your intuition.
5:10 is 1 second for a golf backswing(observer to --> point A) + 1 second for a golf swing && follow through (point A to --> point B) == 2 seconds.
How can you dislike this guy?
A new 3b1b video on 4/20? You never fail us
It is one of the most understandibale and captivatng explanation I've ever seen. Thank you!
The vibes of the start of your videos are amazing! Always encourage us to be curious about mathematics.
Casually gonna demostrate de Wallis product of pi and the sin formula
Mindblowing dude, just mindblowing. Don't be scared at all to explain more, and more slowly as well. I'm always afraid of losing details. I would even recommend you to make 2-parts videos, at least I ould recommend you consider doing it. I mean, you don't make such a beautiful proof everyday...
if you trace Wallis product step-by-step you would get impression that Pi/2 is a rational number, but that can't be because Pi is a irrational number.
The result of Rational Number times 2 can not be an irrational number.
I also notice that, taken in pairs, this product is prod[ n^2 / (n^2 -1) ] for n even, and is also 2 * prod[(n^2 -1) / n^2 ] for n odd. It seems that those forms should provide a relationship to the Basel problem and thus also to sines.
The most interesting thing to me about Euler's solution to the Basel problem is that it only uses the coefficient of one of the powers of x in the expression for the sine.
I see that subtle easter eggy use of tau instead of pi to record the complex number's phase angle around the unit circle when talking about roots of unity. ;) Makes it so much easier to communicate radians relative to a complete turn of the unit circle~
B...but pi looks more beautiful. Me don’t like pi dead
Pi doesn't have to die. There's areas of math where it's more useful than tau. ;) I personally find myself on the side of preferring mathematical notation that is also intuitive. Using tau over pi does a lot of that where appropriate. But tau doesn't always work better than pi everywhere.
AHHHHH HERESY!!!!!
"tau doesn't always work better than pi everywhere"
This. There are equations and infinite sequences that, in whatever way, simplify down to something based on pi. Say, pi^2/6 or pi/2. But then there's using rcis(theta) on the Argand plane, or simple harmonic motion in physics which makes better use of tau
I'd say both should be in frequent use for wherever they make the most sense
Agreed 100% wundrweapon :)
At 11:223, why can one replace the bracket signs with modulo signs showing distance - as the subtraction of two complex numbers doesn't give you a completely real number...
Math totally aside, damn your graphics are well done. The foggy effect around those LED-palette lighthouses against a night background really pulls me in for some reason. Your graphics are very good on your other vids too, but this one made me think to pause and comment on it.
I just saw the video … what I found hand-wavy is the fact that the distance is compared to the angle … it works for small angles and large N but what about near pi angles … yes, these would be large lighthouses indices but still in the product. It is as if you only consider small angles and, with N large enough, they contribute large integer numbers. Thanks to clarify.
Am I missing something? At 12:33 when O = 1, the fraction on the left becomes 0/0, so how can you still conclude that it equals a partial sum of the geometric series?
Use calculus and its limits.
Just like x/x -> 1 when x->0. When you have a 0/0 limit, you just rephrase it in different way to get the solution.
Just wrote a little thing in the pinned comment to address that. Hope it helps!
That is a completely reasonable question. Well, I would say there are several ways you can be convinced about that fact. The first is there is no problem about the denominator being ''0'' because that was an algebraic identity, this is, a formal identity, no matter about evaluating at specific values. You need basically the fact that (I denote by P a ''pi'' letter of product) P (x-j) (with j taking values of complex n-th roots of unity distinct to 1) is equal to x^(n-1)+...+x+1. He somehow deduces this identity by dividing the whole product P (x-j) (this time including the n-th root 1) by x-1 (the factor you don't want) to get x^n-1/x-1 which is in fact the same as before. As I have mentioned, this can be just understood ad ''formal'' equalities not being worried about evaluating on a specific value. Recall that this is true even for a more general field with unity 1. When we say x^n-1/x-1 = x^(n-1)+...+x+1 ''in a formal way'' we are actually saying that x^n-1=(x^(n-1)+...+x+1)(x-1), this last identity is true even considering evaluations for complex x. This is also an equality of polynomials. Since x^n-1 = (x-1) * P (x-j) (product over j distinct to 1) we can deduce (x^(n-1)+...+x+1)(x-1)= (x-1) * P (x-j) (product over j distinct to 1) . And then just delete x-1 on both sides. Does it make sense to delete x-1 on that equation? What if x=1? Well, they are just polynomials, x-1 is an element of a ring (ring C[x] of complex polynomials) which is also an integral domain (this means the ring has unity, is commutative and, the most important property in this case, it does not have zero divisors, this is, if a and b are nonzero then ab is, too, and this is precisely the property that allows us to delete nonzero numbers on a equation of the above type, for example, if az=bz and z is nonzero then z(a-b)=0 and then a-b=0 because there aren't zero divisors, and this is the same as deleting z on the equation ;) ) and as I was saying x-1 is nonzero so that it can be cancelled on the equation. Once we have (x^(n-1)+...+x+1)=P (x-j) (product over j distinct to 1) making j=1 we get n=the desired product. One last comment, you can also understand the discussed fact by observing the smoothness of the equation. I mean, you would agree that the equation written on the video with O-1 on the denominator is true whenever O is not 1. Well, do not evaluate at O=1, but since you are tallking about polynomials then make the limit O--->1, and by the continuity of the involved functions this will behave in a good way, and so what do you get? The desired equation anyway. I hope to have been clear enough :)
The RHS is a polynomial that, when multiplied by x-1, gives x^N-1. There is only one polynomial like that, namely, x^{N-1}+x^{N-2}+\dots+x+1. Thus, they must be equal… for all values of x, including x=1.
You are the most underrated channel on RUclips. I’m sad that you don’t get millions of views despite making interesting videos.
11:59, a fancy way of dividing by zero without the universe ending. :)
Phew
Integrating the limits sequentially one at a time makes senses because for any given distance to a point, there is a corresponding mirrored point (with a negative sign for example) which should be integrated before moving to the next distance.
For example: If you move away from the Sailor on the X axis, towards the origin and beyond, you're going to pass lighthouses in groups of two (a positive A and negative B lighthouse), passing one just ever so slightly before the other. Having started your journey from the Keeper's perspective, you'll pass the two lighthouses simultaneously each time. So it seems logical that the limits should be integrated in the order you used (A1*B1*A2*B2), and not something different like A1*B1*B2*A2.
Thanks ♥ for you efforts and for these amazing videos, Please we want another episodes about deep learning and Machine learning algorithms (RNN, K-means, Logistic regression, SVM/SVR, ....) ♥
How dare you make something so complicated so clear and simple using clever techniques and beautiful animations. Outrageous. Who do you think you are?
19:55 I didn't get that, why is the limit of the collumns 1? :/
I also had to go back and rewatch this to understand what he meant. He's taking the limit of the series of numbers in each column, which will always tend to 1 for large enough N since there's only one 7 in the column. His point is that if you take the product of each column's limit of index N as N -> inf, it's not equal to the limit of the product of each column's index N (i.e. row N) as N -> inf, so the product of limits isn't equal to the limit of products.
@@PresAhmadinejad the limit of the infinite sum of the columns tends to infinity, not 1. I’m not sure what I’m missing.
@@briangronberg6507 he’s not taking a limit of a sum, he’s taking the limit of a series of numbers. In the first case, that series is the product of row i from i=1 to inf, and in the second case it’s simply the numbers in a given column. In the first case, the limit of those products is 7 since every number in the series is 7, and in the second case the product of the limits of those series is 1, since each column’s series tends to 1 for large enough N.
@@PresAhmadinejad Thank you. That’s much clearer!
@@briangronberg6507 you’re welcome!
At 12:25, instead of the little bit of algebra to get the resulting sum, I applied L'Hopital's rule for the limit as O approaches 1, which quickly yields n.
Other than that my approach was the same, save for superficial differences. I began by writing an expression for distance product in the natural(?) way (write the points in their exponential form, shift it all over by 1, product of the absolute values) but this isn't really needed. I guess my initial thought was just to write the damn thing down. 😂
In any case, Lemma #2 is such a pretty little thing, I felt the need to play with it. This isn't expected at all, it provides a beautiful example of solving a geometric problem in the complex plane, and it even has a nice hook in the form making an initial check of easy cases:
I first convinced myself ... that I convinced myself, that n=0 does not fit the problem, as an observer is assumed.
The trivial case of n = 1 yields no line segments, corresponding to the empty product of 1.
The case n = 2 gives one line segment of length 2, which is also the product of "all the lines."
I skipped over n = 3 because it felt like I'd need to write stuff down for that one.
My favorite easy, but not _trivial_ (in the usual sense) case is when n = 4. We have the line segment of length 2 that goes across the circle, and the other two are hypoteni(?) of right triangles. We get 2*sqrt(2)*sqrt(2) = 4 = n.
Overall a 10/10 adorable tidbit, I'm surprised this isn't a popular cool math problem, at least among math majors.
Why is (at 19:52) the limit of 7*1*1*1*1...=1?
That doesn't make sense to me.
Frederik Huber because for each sequence, at a certain point the sequence will be all 1's.
To write a limit formaly correct, it should be like for example: lim( sum of 1/2^n) for n->infinity, which is 2.
How does the correct form of this look like?
What goes to infinity here exactly?
Frederik Huber The limits taken vertically are not based on the products of the columns (notice the lack of dots between numbers in the vertical direction as opposed to the dots in the horizontal rows). The vertical limit is simply the value that the sequence approaches, i.e., the limit of the nth term (not the limit of partial products). Since each sequence eventually only has 1's, the limit of the nth term is 1.
i.e., the rows are lim(Product of n terms) and the columns are lim(nth term) as n --> ∞.
Aha! Yeah that makes a lot more sense.
Rewatching this I noticed I overlooked that the limit of 7, 7, 7, ... = 7, which wouldn't make sense if it was a product.
Thanks!
You can view it as a sequence of functions, fn = 1*1*1.... n times * x (with x = 7 in the particular example). When n goes to infinity you "lose" the x...
This is an example of a sequence of functions that converges pointwise but not uniformly because, as you said so yourself, at any finite n there's still a 7 there.
Also 24:09 resembles the gamma function reflection formula, gamma(f)*gamma(1-f)=pi/sin(pi*f)
Your use of colour in your videos is always beautiful. Keep it up :)
Hi, I found an alternative, but easy visual way to find the distance product of Observer replacing one of the lighthouses at 12:00
First, calculate the distance product of the Observer but only using lighthouses 1, 3, 5, 7. Then the Observer is 1/2 of the way between lighthouse 7 and 1, so the distance product is 2. You then still have to multiply this by the distance product of the Observer and the remaining lighthouses (2,4,6). Here you do the same thing, you calculate the distance product of the Observer with lighthouses 2 and 6, which is equal to 2. Then only lighthouse 4 remains and the distance product of the Observer with lighthouse 4 is equal to 2 as well. You then multiply these 3 distance product to get the answer, 8.
Apparently, that does only work for N = 2^k with some natural number k. Nice finding, anyway.
And for the proof of the Wallis product in the video (better to say: sketch of a proof) it is sufficient to have _some_ large numbers N, which may well be powers of 2.
I used to think imaginary and complex numbers where something stupid. How wrong I was. Now I know that even the most abstract ideas in math can allow us to find very deep and interesting relations between things
You could prove Lemma 2 using the fact that the guy is e^{pi i / n} in the complex plane, and we want to find the magnitude of the product from k = 1 to n - 1 of w^k - e^{pi i / n} where omega = L_1 = e^{2 pi i / n} so this whole thing becomes, since we have magnitude, we can flip sign to get |product(e^{pi i / n} - w^k)| = |(e^{pi i / n})^n - 1| = |-1 - 1| = |-2| = 2. If it helps, note that w^k = L_k.
I had to present a proof of the stirling formula last semester and the main part of it was proving the Wallis product (after you have this the rest is more or less trivial). I also had an geometrical approach based on a paper I found but it wasn't as visual as this here of course. I must say that I didn't like this as much as the Basel problem video because it was more algebraicly playing around than geometrical intuition. I also think think that this approach is a bit harder to make rigorous for a non-mathmatician audience. If I don't have a rigorous proof then I can't be sure that my intuition is correct. So this proof here is not ideal for my taste but I really like that you try to make "nice" proofs for already proven things :)
EDIT: I just saw that the the sorce I used for my proof was the one by Johan Wästlund in the description. I of course rephrased a lot of stuff he did to make it easier to understand. I think his method is easier for everyone and could even be done in schools but the main problem is the proof is pretty boring in the middle when deriving a lot of side stuff algebraicly.
The boring parts is where you say, "Exercise!"
And then follow it with, "Justkidding. You can read the boring derivation in the book."
Then everyone feels like they've been sold short, but the lesson is not too uninteresting.
David Herrera The thing is that I did not do that at all. Like I said my proof was rigorous, no matter how small I proved every single step. I even needed an inequality with e for the final proof that is a common exercise in analysis and I sat there for quite a while trying to figure out how to make this complete but also understandable for people without a math background. As I tested it with friends I knew that my proof was pretty much that, the most difficult thing was knowing what the number e is. I personally hate leaving stuff out, I even had to reformulate the last steps because I didn‘t want to use rules about asymptotic approximations ~ that I wouldn‘t prove so I only used the definition of the ~ symbol. Like I said in the end the proof was a bit boring cause in the middle part were a lot of lemmas that I proved which made everything easier to follow but not really exciting. At the end however when I showed the geometry, how to interprete the results as circles and area and so on went cool again. If Inwould make it again I would try to think of a way to make the middle part better
I wasn't saying that you did that. It was a recommendation made in jest.
The supplemental blog post does not seem to exist any more...
3Blue1Brown: explains an exclusively arithmetic complex problem with a very specific solution
Also 3Blue1Brown: so you're probably wondering how this relates to geometry
I wish a more standard mathematical proof was also provided alongside with the long paragraphs of discussion in the blog. Its good to provides insights on how the proofs were constructed, but there are still other technicalities underlying the proof. I know you guys defintely are capable of doing so.
For example, another issue I am considering is that although all of the ratios are converging to its numerical limits, the speed of which they are converging is slower and slower going down the sequence, i.e. terms are not uniform convergence. I suspect that causes the interchange of the orders generate a different ''value'' for the limit, rather than "absolute convergence" you mentioned in the blog post.
Great video again!
Just one small suggestion. I found 9:00 quite confusing at first, as I thought the distance is not simply (O-L_i) but |O-L_i|, and the magnitude of a multiplication of the former is not a multiplication of the latter. Then I did search around and found the property for complex numbers that |xy|=|x||y|. Maybe it would be helpful to note this property somewhere for people who are not that familiar w/ complex numbers.
The property |xy| = |x||y| has always been true for the real numbers so he probably thought that it wouldn't cause that much confusion when he also used it for complex numbers
Thank you!
I recently had to prove these things in a home assignment.
You start with the Fourier series of cos(tx), then use it to show that the integral from 0 to xpi of [cot(t) - 1/t] is the infinite sum of log[1 - x^2/n^2]
But doing standard integration you also get log(sinx/x)
So that means log(sinx/x) = sum of log(1 - x^2/(npi)^2) = log of the infinite product (1 - x^2/(npi)^2);
You get rid of the log in both sides and substitute x = pi/2... voila
OH YES. Another 3b1b Video.
The excellent animations and graphics are critical for us visual learners.
I like how some people dislike the video when it’s only been up for 2 minutes and the video is 25 minutes long...
UCrafter5000 not fully true.
On original non reuploaded video it was 235 likes and 0 dislikes. Now it has 3 dislikes already.
In all the videos I just think and at the end I realize how elegant the solution is.
I think it wouldn't hurt to quickly mention that the absolute value of a product is the product of absolute values also when talking about complex numbers
Yes I agree, although the proof of this is so easy both to state and to visualise that maybe it slipped their mind! (write z1 = n*e^(i*a), z2 = m*e^(ib), where n, m >= 0 are real magnitudes. Then z1*z2 has a magnitude of n*m -- that's half of the definition of complex multipation. the other half being the new angle of (a+b) mod_tau. Or you can manually prove it algebraically z1 = a_ib, z2 = c + id, but this doesn't seem very natural to me.)
This video is part of math history now. In math books of my school-time there was "interesting fact" text insertions to make pupil more interested in a subject. In the math books of future they will give link to this video.
Man this sounds interesting, but your animations are too good to be watched in 360p, so brb.
Man, I was so freaked out after doing the re-upload and publishing. "WHY IS IT 360p?!?!, Did I do the export wrong?!?". Silly RUclips, easing it's way into HD.
Matter of patience :)
Trying to explain what you’ve seen to a friend: Intricate mathematical interrelationships explained in a novel and aesthetically alluring way by means of ingenious animations, an authoritative yet affable (saganesque) voice, and background music perceptible enough to ensure a rhythm, yes subtle enough to avoid distraction.
Anyone who says this isn’t art...
8:18 LONG LIVE TAU!
And at 6:33!
Dont be conceited if tau appears, idiots.
As stated in this year's Pi Day video, all of the Greek letters in Euler's day were just variables; it was just some one mathematician who decided that pi should be taken as a constant, based on a recurring statement by Euler: "Let pi be the ratio of the semicircumference to the radius of the given circle."
This video is literally a giant proof for a pi/2 identity. Where's your tau nau?
Im just doing my thesis on abelian number fields, hence lots of cyclotomic fields stuff is in play (thanks to Kronecker-Weber). I wouldn't be surprised if this was burried somewhere in the bast literature, but it is cool as heck. I would love to see more.
This magnum opus is the epitome of human creation. Each day my thoughts wander to your impeccable and uniquely inspiring content. The subtle messages conveyed in each of your pieces are hallucinatory and bring us, the human race, glorious euphoria. The kind of euphoria that ignites woodlands like an Einstein-Rosen Bridge collision. I'm deeply engaged in the metaphor of sloths, one of the seven deadly sins, which is easily comparable to the ancient greek myth, pandora's box, juxtapositional acquiesce to the current state of affairs of this tumultuous society we find ourselves in today. Before the wondrous discovery of this series, I felt chap-fallen, woebegone, and forlorn. Instantaneously, these fabrications took the quimp out of my day and lit the fire in my eyes that is the motivation to achieve one's passions and become the best version of yourself. As Galileo once said, "Répondre intelligemment même à un traitement inintelligent". I'll abandon you on this inscription, wishing you the best of all good fortune.
This is the only science channel I watch and don't understand.
Isn't discovering a proof like this worth an award these days?
Chvocht - iff you're in the club
I would suspect it’s too simple of a problem.
not even close
screwhal underhill Ok
well it already had proofs so huh
At around 6:30, you talk about the magnitude and the angle. You should mention that it has to do with the polar coordinate form (maybe not HOW it does, but just mention it, so people who don't already know about the polar coordinate form can look it up)
Great job guys :)
I can't wait to see what you guys come up with next
Around 9:36 you say "seven" instead of "N". Excellent video, as always!
Notification squad here. Been waiting for the video to come out :)
Gotta love how the commutativity of fractions doesn't always hold for infinite products. If you write them as one fraction with one vinculum, i.e. 2*2*4*4*6*6.../1*3*3*5*5*7..., you might simplify it and remove the one in the denominator. Then you end up with (2/3)^2 * (4/5)^2 * (6/7)^2 ... * (2n/(2n+1))^2, which converges to zero. Even if you account for the discrepancy in the number of numerators vs. the number of denominators and append " * (2n+2)" to the end, it would amount to multiplying infinity by zero at the point of convergence. Yet writing it as 2 times the rest of the fraction brings it back to a product of pi/2! It's like there's a whole new kind of place value.