The simplest reason why π/4 = 1 - 1/3 + 1/5 -...

[@mathemaniac]

Like the video - it really helps the channel! | Visit brilliant.org/Mathemaniac/ to get started learning STEM for free, and the first 200 people will get 20% off their annual premium subscription. Signing up is also a great way to support the channel! At the end, essentially during the argument, both sides of the sum are just Cesàro sums, just that the LHS can be interpreted as a regular sum, and so we proved the identity. For an even more rigorous treatment, read the description. By the way, on second thought, this looks similar to Fourier coefficients, at least the time average bit - though I can't see whether this is the same proof as the one using Fourier series of sgn(x). It feels very connected, but also very different in the sense that we don't need to use the Cesàro sums of the Fourier series. Please let me know if you have any ideas regarding this.

[@strikeemblem2886]

Regarding the missing link to the Fourier series of sgn(x), here is the connection: . At the very heart of the argument, you have constructed a function f(x) that is the same as the square wave, as it has the same fourier coefficients. In particular, this construction is for f(1), because of the setup used on the radius, and the "minute hand", see 10:50. . Jordan's criterion tells us that the square wave (which is of bounded variation), has its Fourier series S_N f(1) converging to 1/2(f(1+) + f(1-)). This is 0 by definition of the square wave. This should be compared with the RHS of the eqn in 14:45. . So why is it no coincidence that the method of Cesaro summation also gives us 0? The keyword is "good kernels". In the proof of Jordan's criterion, we will find the use of something called a "Dirichlet kernel" D_N(t). While this a kernel, it is not a "good" kernel. But we may use it to define the "Fejer kernel" F_N(t) = 1/n sum_k D_k (t), which is a "good kernel". . The Fejer kernel is closely related to Cesaro summation: If we consider the Cesaro summation of successive Fourier series, i.e. consider [cesaro f_n](x) = 1/n (S_0 f(x) + ... + S_N-1 f(x)), then this equals the convolution of the Fejer kernel with f at x, i.e. (F_N * f)(x) !!! . The precise result that connects everything is this: The Fejer kernel is a family of "good kernels". As a consequence we have, if f is L1(torus) + bdd + the limits f(x-) and f(x+) exist, then (F_N * f)(x) -> 1/2(f(x+) + f(x-)). . For our case x=1, this says that [cesaro f_n] (1) -> 1/2(f(1+) + f(1-)) = 0.

[@archibaldhaddock3485]

Brilliant idea. You are so underrated. I request to bring more series or courses just like before, this will make us like you more.

[@mathemaniac]

I do notice that viewers tend to like those series or courses more, but if I have to be honest, I really like these standalone videos where I just share the bits of maths that are under-appreciated. I just had a lot more fun doing these (and actually truer to the much earlier state of the channel), BUT there might be some series later on. Stay tuned!

[@comuniunecuosho-campulbudi7611]

New Calculus channel is the most underrated because it's the most clear and revolutionary

[@johnchessant3012]

Great video! By the same method averaging from 0:20 to 0:40, I found that 1 - 1/2 + 1/4 - 1/5 + 1/7 - 1/8 + 1/10 - 1/11 + ... = pi/(3*sqrt(3)). Try doing that with calculus!

[@renka-chan9213]

w o w w h a t

[@mathemaniac]

Really nice! I think it should be doable with calculus (just wouldn't be that obvious), but this is so nice!

[@laxminarayanbhandari855]

Here you go. This is pretty actually easy to do. I hope you know LaTeX. It's quite painful, but understandable, that KZbin doesn't support LaTeX. We have the infinite sum S=\sum_{k=0}^\infty\frac1{3k+1}-\frac1{3k+2} We start with the power rule of integration. \int_0^1 x^n\,dx =\frac1{n+1} Setting n=3k and n=3k+1, we get S=\sum_{k=0}^\infty\int_0^1 x^{3k}-x^{3k+1}\, dx Interchanging the order of summation and integration and then using the geometric series, S=\int_0^1\frac{dx}{x^2+x+1} This integral trivially evaluates to S=\frac\pi{3\sqrt3}

[@caspermadlener4191]

My proof: The amount of solutions to N=x²-xy+y² is equal to 6(divisers of 3k+1 - divisers of 3k+2), which tells us that the average amount of solutions is equal to 6N(1/1-1/2+1/4-1/5+1/7-1/8+...). The average amount of solutions to N=x²-xy+y², is also equal to N*2*pi/sqrt(3), because we have x²-xy+y²

[@josephmathmusic]

It is a Dirichlet L series! You can write this, for j = cube root of unity, [(j + j^2/2 + j^3/3 + ...) - (j^2 + j^4/2 + j^6/3 + ...)]/(j-j^2) = ( - ln ( 1 - j) + ln (1 - j^2) )/(j-j^2) = i (-Arg(1 - j) + Arg(1 - j^2))/ ( i sqrt(3)) = i(pi/6 + pi/6)/i sqrt(3) = pi/(3 sqrt 3)

[@thelosttomato4020]

My favorite way to prove this is by using the expansion for ln(1+x) ln(1+x)= x- x^2/2+x^3/3-x^4/4.... plugging in ix instead of x, we get ln(1+ix)= ix +x^2/2-ix^3/3-x^4/4.... etc separating out the real and imaginary parts, we get, ln(1+ix)=i(x-x^3/3+x^5/5-x^7/7.... inf) + (x^2/2-x^4/4 + x^6/6... inf) plugging in x=1, we notice that the imaginary part on the right is our desired sum, call it S ln(1+i)= S*i + (x^2/2-x^4/4 + x^6/6... inf) ln(1+i) can be written as ln(sqrt(2)*e^(ipi/4))= 1/2 ln(2)+i*pi/4 equating imaginary parts, we get S=pi/4 It's one of the coolest things I've ever done in math , and it just randomly clicked when I was sitting in class lol

[@yinq5384]

Great proof as always! 3:05 The average position is the center of gravity of the highlighted semi-circle.

Yes, but I'm not quite sure how you figure that one out elementarily? I know how to figure that one out with analysis. I suspect there might be some elegant argument from Euclidean geometry? In any case, the argument used in the videos was great! (A tiny bit sloppy though: our host did not mention that the transformation that matches velocity with position and vice versa is a linear one. That way it commutes with taking the average, which is also a linear operation.)

[@yinq5384]

@ I don't know any elementary way actually. And I suspect the same. :P BTW, do you reckon the argument in this video an "elementary" one?

[@ajreukgjdi94]

Me: mom, can we get calculus Mom: we have calculus at home The calculus at home: "you have to be careful when adding up infinitely many things" This just feels like calculus with more steps. It's just that your dx (or rather dt) steps are of size 1/2 hour. The only step you're missing is to let dt approach zero and boom, calculus. You just don't add any resolution after 1/2 hour

[@iaroslavragel2387]

Very nice proof. But still it requires calculus. Even to rigorously define an average of something over a period of time one needs integration. So rigorously speaking this "proof" does not use calculus because it's not a proof, but only a sketch, an explanation. A roadmap for a rigorous proof, if you want. And when one starts making this proof rigorous he finds out he needs calculus. And this should not be a surprise because the statement of the problem itself is already made in terms of calculus. Infinite series summation is already calculus in essence.

[@mathemaniac]

Yep. This is exactly why I deliberately used the word "see" rather than "prove" in the title. Though it has to be said that it is only because of my two most recent videos that I learn that the word "calculus" also means limits and sequences and series and stuff like that (in my view, it only covers differentiation and integration, and the other stuff are grouped under "analysis").

[@DDranks]

I have hard time imagining what proof of pi would _not_ involve calculus, as it is not a constructible number or even an algebraic number, so there's no hope of representing it but as a limit over a series of tightening approximations - something that screams "calculus" to me. Heck, even it's actual definition - half of the arc length of the unit circle - screams "calculus".

[@schizoframia4874]

Nothing in life is free

[@johnchessant3012]

(This definitely isn't rigorous, but) you can use geometric series to say that ... + exp(-2ix) + exp(-ix) + 1 + exp(ix) + exp(2ix) + ... = 1/(1-exp(ix)) + 1/(1-exp(-ix)) - 1, which indeed simplifies to 0. In fact, you can even put z = exp(ix) and get that the sum over all integer powers of z is 0. With z = 2, this leads to the whole 1 + 2 + 4 + 8 + ... = -1 business which leads us to 2-adics...

[@captainsnake8515]

It’s amazing how many beautiful “ah ha” moments there are in this proof. Finding d using the average velocity, finding the sum of all the tips with a rotation, etc. I loved it.

[@matrix8163]

Hi Trevor! I'm a high school student. I've a question which came in my mind a long time ago but I'm unable to solve it. The problem is: Take N number of random points on the circumference of the circle. Join them to form a polygon. What's the probability that the polygon so formed contains the centre of the circle?

[@victorribera5796]

I have to admit that the one from 3b1b is more apealing to me, but this one is really beautiful too, great job

[@Robert-jy9jm]

Strongly disagree, but 3b1b is also great! :)

[@victorribera5796]

@@Robert-jy9jm That's why I said it is to me, I liked this one a lot too, and understand why some people may prefer this one. But in my case, I prefer 3b1b's

[@Christopher._M]

I agree but this is also great

[@laxminarayanbhandari855]

I still prefer standard calculus methods. I'm much more familiar with them and use them on a daily basis.

[@caspermadlener4191]

This argument might be useful to proof Dirichlet's theorem (the one about primes in arithmetic progressions)! This is not only an easy way to have moduli, but addition and multiplication and addition both work amazing! Also, I wouldn't be surprised if this would result in the 69420th proof of quadratic reciprocity :)

[@Infinium]

Lovely visualised proof, thanks for sharing!

[@mathemaniac]

Glad you enjoyed it!

[@PASHKULI]

you lost me at 3:00 placing the yellow dot on a dashed line and not on the intersection of dashed with circle.

[@two697]

You have to take the average of the horizontal and vertical components of displacement separately. Think of the green arc as a wire and you are trying to find its centre of mass. Obviously, the centre of mass can't line on the wire. It has to lie somewhere of the dashed line

[@PASHKULI]

@@two697 got it, thanks, so you decide to take "half an hour", makes sense now, thanks!

[@dopplershift4767]

Dude, I love the style of your videos! Do you base it off of 3blue1brown's? Love your content!

[@medexamtoolscom]

My favorite one is the sum of the natural numbers to the -2 power adding to pi^2/6 which can be proved with the coefficients of a fourier series for a squarewave.

[@kabirsethi2608]

Beautiful video used some basic trig and ofc calculus to derive something special i guess. root(3)pi/6=1-1/5+1/7-1/11+1/13... and so on

[@derinates9927]

Wow thank you so much for brightening our perspective about the problemm!!

[@happyjohn1656]

I couldn't remember the word "counterclockwise" and you using "anticlockwise" confused me so much 😂 10:47 PM 5/4/2022

[@thankyou3634]

As soon as these clocks came up I was 100% sure that there would be some section on the relation to the Fourier Series of a square wave...

[@mathemaniac]

Nope - I thought about it as well - but it is very different from a proof by Fourier series. This observation was said in the pinned comment.

[@raulyazbeck7425]

Thank you for this video! Now watching

[@Mutual_Information]

Just discovered this channel. Very impressed!

[@0willow0]

I'm an adult with 10th grade precalculus math (working on it), so I'm out of my element for sure, but I am able to grasp how this clock analogy is simulating the series, why the evens don't contribute, why the odds oscillate between negative and positive and contribute smaller and smaller values. What has confused me is where the 4k+1 and 4k+3 comes from 8:45? I think I understand why it's 1/4(4k+1 or 3), it's expressing where on the clock you'll start, but what is k to begin with?

[@linoarenz548]

How? How can this argument be so wonderful? To use the velocity instead of the position to find d... Just wow... If you see it it looks so simple but I would never thought about it that way Just wow

[@thomaskn1012]

Wow, this is so much more convoluted than all of the other ways of figuring out pi/4.

[@hybmnzz2658]

Convoluted? All arguments were simple applications of linearity of expectation and geometry of vectors. You can just integrate 1/1+x^2 I know, but then showing the arctan power series expansion using elementary methods would feel more convoluted.

[@laxminarayanbhandari855]

@@hybmnzz2658 exactly. and that's the real beauty.

[@chriszachtian]

Sorry, this time you lost me at 11:00... But I will continue to keep up

[@alexpotts6520]

Am I right in thinking that this is basically doing Fourier analysis by stealth?

[@lidorshimoni5470]

Turn with timing or number theory. both interpretations are great for understanding. Thank u. Great video!

[@imperiumgraecum9126]

Several years ago, I was building a silly pet project in C, trying to compare various methods of calculating π as to their accuracy, and I clearly remember this sum's convergence speed was atrociously low.

[@Felipe-sw8wp]

This didn`t hit so hard as the series for sine and cosine video... Not complaining though, very interesting! It was a new concept to me. Keep it up.

[@PaulMurrayCanberra]

But, the expression isn't the average location of all the points, it's the average of all the points taken from 0:15 to 0:45. You can't just "add theta" to that, because it's no longer within that time interval.

[@Enderman-en3dv]

Nah, adding theta is sort of an independent argument. To solve the problem of finding the average center, notice that hypothetically rotating the clock by theta only reshuffles the points, so the average stays the same.

[@mathemaniac]

I was not talking about the average at that part of the video - I want to sum the positions, THEN take the average. Think of the diagram as just a snapshot in time, then we just sum the positions, which from the symmetry argument, has to be 0.

[@gutwrenchingdeaths]

this channel is really cool

[@mathemaniac]

Thanks!

[@inverse_of_zero]

This was really enjoyable. Thank yoU!

[@avz1865]

This proof deserves to go in the Book

[@BangkokBubonaglia]

You completely lost me @11:30. You talk about the position having moved 2 units to the left (i.e. horizontal) and then somehow use that to calculate the vertical average position??? I'm not a stupid guy, but I have to admit I simply could not follow at all what you were trying to do here. I get that you are calculating the average of sin(x) over the range [0,π], and with calculus this would be trivial (it is 2/π), but whatever you are doing here with vectors left me scratching my head. I was hoping to show this to a group of advanced 7th graders, but I can't use this. Even I couldn't follow it.

[@mathemaniac]

Basically at each point during that time period, we can obtain the position vector by simply doing two things to the velocity vector: (1) rotating it anticlockwise by 90 degrees, and (2) dividing the speed by 2 pi. So to calculate the average position, we can simply do exactly these two things, i.e. rotate and shrink the average velocity vector to get the average position vector (because at each point in time, it is exactly what we do).

[@BangkokBubonaglia]

@@mathemaniac Thank you. I believe I see it now. I had to let that simmer for a while before it became apparent what you were saying. I appreciate your thoughtful response.

[@Czeckie]

this is cool. Because of the Cesaro summation I'm 100% sure this is a Fourier series argument in disguise

[@mrshodz]

Hello. I have a naive question. How can you use the same variable d for the average position of different hands? All these hands rotate at a different position there their averages would be different. Can you please explain this to me🙏

[@mrshodz]

I see. No need to explain.

[@xaviersoenen4460]

J'ai une idée qui me ferait préférer le système en base 8: pi =6 arcsinus(1/2) Je développe (1+x²)^(-1/2) qui est la dérivée d'arcsin x, j'intègre ce développement comme vous l'avez fait pour arctan x et je remplace x par 1/2.

[@spazmoidectomorf6209]

For mathsemaniac, or anyone else with insight, I'm very keen to study calculus at a higher level, I've completed things such as differential equations involving damping and all that stuff, or for anyone in the uk, the furthermaths syllabus. I would like suggestions of good books to learn along with practice questions of concepts like jacobian calculus and greens differential equation methods. Could someone recommend any good math work books, or any other further books I can look into.

[@sumdumbmick]

whoa... you didn't awkwardly retcon it to yield pi. it's almost like you have a shred of integrity.

[@veenasv954]

Great one!! Ur videos are really good.

[@mathemaniac]

Glad you like them!

[@RomanNumural9]

A cool way to shoehorn the chesaro sums into the video, but since your construction involves taking an average to begin with, it should be enough that as n trends to infinity, 1/n trends to 0. I.e. Cauchy criterion for convergent sequences/series

[@mathemaniac]

Of course the LHS does not need Cesàro sums, but the RHS requires it - the position of each hand has length 1, so actually the sum of the positions at each point in time does NOT converge in the Cauchy sense, and we really need the Cesàro sums to justify that.

[@chair547]

Calc proof doesn't make my brain hurt like the number theory one or this one.

[@koleso1v]

The proof of the total sum to be zero is actually quite weak. It is based on rotation, in particular, for every point that used to be at n*phi there is a point at (n+1)*phi, so that all points go to the next neighbours. However, the phase is wrapped around a circle, i.e. all values are modular 2*pi. Thus, the distribution of the points can possibly be inhomogeneous around the circle and that will violate all the proof.

[@mathemaniac]

The distribution of the points is irrelevant - the argument has not depended on that. Not sure what you mean by "inhomogenous" - it is entirely possible that there are just two possible positions that the tips could take (namely at half past, there would just be up and down), BUT the same argument would still work. If we rotate by pi, then as you said, all the points go to the next neighbours, and because we are considering an infinite number of these, we are considering the exact same points anyway. The only time that this argument would fail is every o'clock. But the argument failed there NOT because of the inhomogeneity of the distribution, but because during the rotation argument, we have not rotated at all (or only rotated an integer number of revolutions around the clock)! So of course, after the "rotation", the sum would stay the same! So we can't deduce anything from this argument. This time point is deliberately omitted in the half-hour period from quarter past to quarter to, so during that half-hour period, this argument is valid. And EVEN IF this argument is really dependent on "homogeneity of the distribution" (which isn't as reiterated above), assuming what you mean is that the points are not dense on the circle, the only times when this is violated would be a rational multiple of pi, where there will be a finite number of points on the circle (again, the argument still works if that is the case, as stated above, only saying that IF it somehow doesn't work), is negligible when taking the time average, AND the "sum" would be bounded, and so we can ignore those points of time!

[@racpa5]

Awesome video!

[@user-jm3xl7rg5k]

Really brilliant!

[@carly09et]

you sussumed analysis as an axiom... other wise a nice demonstration.

[@meofamily4]

I understood this explanation just fine, until the author proposed to make the proof rigorous. At that point his statements were, for me, impossible to follow. Namely @15:53. (an infinite sum of positions doesn't converge? Eh?) The sum of positions of all the points on the circle's perimeter surely equals the center.

[@JojiThomas7431]

Beautiful video

[@mathemaniac]

Thank you!

[@sunritpal9596]

Very interesting 👌

[@mathemaniac]

Glad you think so!

[@SimonDoer]

me: ''I also know a proof that involves a Fourier series, but that one also uses integration.'' Mathemaniac: Basically uses Fourier series for the proof. me: surprised Pikachu face

[@EPMTUNES]

Great video.

[@mathemaniac]

Thanks!

[@NoNameAtAll2]

"clock with infinite hands" oh no, it's Fourier transform, isn't it?

[@mathemaniac]

Not really - I also have the same queries while making the video - but the argument is very different from using a Fourier series.

[@thomasmorinaga4540]

虽然看到一半就懵了，但我大受震撼👍

[@alexcwagner]

Maybe it's just me, but I think it'd be easier to just learn calculus.

[@evanchae2657]

so great

[@siegfriedbarfuss9379]

First explain the idea top down and then maybe follow through bottom up.

[@user-jc2lz6jb2e]

So basically a hidden Fourier transform?

[@swamiaman7708]

Thumbs up if you don't understand anything even after watching the whole video 😂.

[@Hexor1211]

Yeah, right, after you learn Fourier analysis then this is much simpler method than other proofs. :-)

[@josephmathmusic]

The most elementary proof I know!

[@omargaber3122]

amazing❤❤❤

[@mathemaniac]

Thank you!

[@user-iw1qn3mt7e]

Дякую!

[@kosttavmalhotra5899]

who invented this proof

[@DeepFriedLiver]

Neat proof. I have to disagree that this is simpler than other methods. I don’t think this is at all accessible to someone without calculus. Even though it’s not relied on heavily. Also Some of the explanation borrowed on familiarity with some very complicated ideas like Hilbert’s Hotel. Which is fine, but it’s not simple. All that said. The core idea’s are great and the whole thing does provide nice insights for those who take the time to digest it.

[@gauravbharwan6377]

So you are making video again . Take breaks but keep animating maths

[@mathemaniac]

Honestly I have never taken breaks - just that KZbin is clearly not my full-time focus. I am only making these videos when I have the time to do so.

[@gauravbharwan6377]

@@mathemaniac so you tube is hobby, good .I just want mathematics to have more channels like 3blue 1brown

[@muazkoc8279]

👌

[@satyadas4519]

Good imagination.

[@behzadfarhadi]

زیر نویس فارسی نداره😔

[@shiv9582]

English is Not my Mother Toungue But i would recommend you to learn English it will do help you for surving in today's world

[@non-inertialobserver946]

Sheeesh

[@tolkienfan1972]

It's not really a proof. Some steps are unjustified. No definition of average point was given. It's probably difficult to define average point without calculus. A very interesting video nonetheless

[@SuperYoonHo]

:)

[@panchovilla5400]

A lot of mumbo jumbo. You say your explanation won´t involve number theory or differenciation. Well, I´m sorry to say but your explanation turned out to be more convoluted and harder to understand than the other explanations!!!!

[@mohamedfaresessid8855]

worst explication ever

[@grzechu9751]

beautiful, we need more those types of proofs!

[@mathemaniac]

Exactly!

[@JehTweenGong]

In pi. Proof of the Goldbach's Conjecture and more is available at tienzengong.files.wordpress.com/2020/04/zeta-2.pdf

[@krzysztofciuba271]

"Perfect" (almost)-only in Euclidean space. In the real, non-fictional space, i.e. Minkovskian space-time", 2X"Pi"^2=I^2 ; hence/ Pi/= I, I= square of "-1"(the mystical unit ("one") )accord to Minkowski) !- it is calculated from SR/GR@Doppler Lorentz formula@experiments, by Ive-Stilwell, AD 1938(the first one, confirmed later with more precise devices)@Sagnac effect; this calculation is a v.simple(!),but .....if Satan, a liar (J8:44, indirect in Aristotle on evil: privation of good,@A.Tarski's, formal Definition of Aristotle's Truth(sentence), AD 1933, prevents you to know it (the proof) ask me for the Revelation(from H.Spirit(J3:8).Ps. in the Bible, 1 Kings. 7:2, 23, this "Pi"= 3 (according to the unit measure at that time) and that is OK according to mathematics: a sample standard deviation@Jacobian transformation's formula. "

[@dinocoder1281]

when the schizophrenia

[@brendawilliams8062]

I say Stein Paradox plus Sphere Packing plus this informative video would make an interesting dance.

[@fibbooo1123]

Great video!

[@mathemaniac]

Thanks!