Euler's real identity NOT e to the i pi = -1

[@SmileyMPV]

There is a law, called Stigler's law, which states that no scientific discovery is named after its original discoverer. This law was of course discovered by Robert Merton.

[@mathsmasters]

:) Yes, all very sad :)

[@lethargogpeterson4083]

I love it!

[@anand.suralkar]

Hahaha

[@JK-jt3lr]

:-D

[@sumdumbmick]

nah, son. it was discovered by all the poor brown people that science stole knowledge from and then claimed to have discovered.

[@Mathologer]

I am a huge fan of Euler and had been wanting to to make this video for a long time. Pretty nice how it did come together I think. One of the things I like best about making these videos is how much I end up learning myself. In this particular instance the highlights were actually calculating those other sums I mention myself using Euler’s idea (the Riemann Zeta function evaluated at even numbers) as well as learning about this alternate way to derive the Leibniz formula using the zeros of 1-sin(x). Oh, and one more thing. Euler’s idea of writing sin(x) in terms of its zeros may seem a bit crazy, but there is actually a theorem that tells us exactly what is possible in this respect. It’s called the Weierstrass factorization theorem. As usual if you'd like to help me please consider contributing translations of the English title, captions and description into other languages. Enjoy :) Today's t-shirt I got from here: shirt.woot.com/offers/pi-rate?ref=cnt_ctlg_dgn_1

[@naimulhaq9626]

Please make a video on Madhava, I long to see his complete exposition. Did he really know about the Zeta function or Euler's sin function?

[@tabularasa0606]

That is quite an ingenious proof.

[@Mathologer]

+Naimul Haq Madhava did not know about the Zeta function, but he certainly did some amazing work on infinite series. Definitely one of the (relatively) unsung heroes of mathematics. I've had a close look at the results that are attributed to him and his school. Would be great to make a video about him. What I am struggling with a little bit is to figure out how exactly he arrived at his results :)

[@enzogiannotta]

Yeah it would be awesome to know how Madhava arrived to his results, probably his proofs where not rigorous enough, as I suspect that he lacked of enough rigorous math tools.

[@sansamman4619]

Mathologer, it is weird how smart people have existed and found out cool things and no one cared.tho we all know how mathematicians do maths for their amusement and dont care about fame so it is funny xd

[@DavidB5501]

Roger Cotes isn't entirely forgotten, at least to anyone with a good edition of Newton's Principia. Cotes collaborated with Newton on the second edition, and wrote a preface to it. When Cotes died tragically young, Newton reportedly said "if he had lived, we would have known something".

[@atharvas4399]

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

[@GiantKush]

I think it would be wrong to not mention Numberphile and maybe Mind Your Decisions ( Presh Talwalkar )

[@gurri_8]

also Michael Penn

[@discreet_boson]

How to start a war in the comments about who is the best teacher

[@takyc7883]

And blackpenredpen

[@AZTECMAN]

Flammable Maths also

[@charlesaugustus5508]

Nice pi-rate shirt.

[@Brooke-rw8rc]

The pi rate is one half turn per second.

[@devd_rx]

@@Brooke-rw8rc y/whoosh

[@marktilley7222]

C/(2π)!!

[@phyngineer]

1:17 nice one: Jacob Bernoulli, Johann Bernoulli, Leibniz, John Wallis, Mathologer

[@Mathologer]

Couldn't resist :)

[@completeandunabridged.4606]

Mathologer I like your pi-rate t-shirt.

[@judychurley6623]

from above discussions, seems some would rather it be a tau-shirt...

[@blackpenredpen]

Nice!

[@completeandunabridged.4606]

blackpenredpen Hey!

[@gerryg6439]

I am a new math major who was very depressed lately. I can not help but think that I am not fit for math(since there is someone super good at math in my house, and constantly impose that peer pressuring). After watching your video, I felt much better by learning something and regained my interest in math. Thank you!

[@viktorramstrom3744]

Believe in yourself. You can do anything if you put your mind to it.

[@JorgeGarcia-jt4kq]

Mathologer, I really love your videos. I'm still in junior high and I probably haven't seen any math topics you explain at school, But I found your videoS very easy to understand and my passion for math has grown because of you. Keep up with the videos and WITH THE GREAT CONTENT. I follow many math channels, but none are as good and as detailed as yours is. Really really really THANK YOU VERY MUCH.

[@Mathologer]

That's great, mission accomplished :)

[@itismethatguy]

That’s what i feel in class 10…what is junior high though

[@star_ms]

@@itismethatguy 10th is sophomore, 11th is junior high

[@johncowart9536]

I was just reading about this identity, and there were steps I didn't quite get how Euler made the leap (wasn't well explained, and it's hard to extract the steps from just reading), so thank you sooooooo very much for posting this. Love your videos and shirts :)

[@onlyonecjb001]

I have just stumbled upon this video and this is an incredible proof. Bravo.

[@AzrgExplorers]

At 6:06 "How did Euler manage to prove his identity?" With his driver's license, of course!

[@ifroad33]

Do you mean his horse and carriage license?

[@raynmanshorts9275]

You bastard! I was just about to use that incredibly lame joke!

[@megauser8512]

lol

[@themaverick1891]

Such a stupid comment.

[@bilaalahmed8923]

He identified as π²/6

[@trelosyiaellinika]

I have watched many a solution to the Basel problem, some quite interesting, with Fourier series, double integrals etc. But this explanation of yours, and all your posts in general, stand out at least in one aspect that others lack: the beautiful animations, which I absolutely like! Also, the additional details, that maybe do not contribute directly to the solution chain but explain how the exploring mind works, e.g. establishing an upper limit of

[@whatever5575]

the beauty of math is it seems to strive to condense complex ideas into simple and elegant formulas. thanks for the video.

[@thomaspickin9376]

I can't believe you managed to do this video without saying the words "basel problem" once.

[@Mathologer]

Yep, decided against mentioning it at some point :) (It's in the keywords though)

[@RaveScratch]

Currently in my final year of getting my undergrad for "pure math," and I feel this video highlights the bittersweet reality of getting a formal education in math. When going through the mountain of videos about the Basel Problem, I must have seen over a dozen beautiful proofs, or at least outlining of proofs, that get me really excited to learn more. However, we recently went over this problem in my Math Analysis II course, we were shown using Perseval's Theorem with f(x)=x which just... gives the answer. No nice intuitions, no incites, nothing. I know it is important to have these tools under your belt, but the way I'm learning them, at least to me, makes them seem like no more than useful algebraic magic.

[@diegocastillo6470]

I can spend a whole day watching everyone of your videos, they're all amazing, really really amazing. I already loved maths, but you're starting to make me into a math addict. Excellent work

[@MichaelRothwell1]

Excellent video on Euler's solution to the Basel problem. Thanks for pointing the way to finding the sum of the reciprocal fourth (and higher even) powers. Here's my solution for finding the sum of the reciprocal fourth powers. We start with the equation suggested at 12:12 : x-x^3/3!+x^5/5!-... = x(1-x^2/π^2)(1-x^2/(2π)^2)(1-x^2/(3π)^2)... The idea is to use Mathologer's clue (at 14:00) to finding the sum of the reciprocal fourth powers (which is π^4/90), which is to equate coefficients of x^5. To make things clearer, we'll make a couple of changes to this equation: Divide by x: 1-x^2/3!+x^4/5!-... = (1-x^2/π^2)(1-x^2/(2π)^2)(1-x^2/(3π)^2)... Replace x^2 by y: 1-y/3!+y^2/5!-... = (1-y/π^2)(1-y/(2π)^2)(1-y/(3π)^2)... Now we wish to equate the coefficients of the y^2 terms. So that the patterns will be clearer, let's write the right hand side as (1-αy)(1-βy)(1-γy)... where α = 1/π^2, β = 1/(2π)^2, γ = 1/(3π)^2, ... and note that what we really want to calculate is α^2+β^2+c^2+... = 1/π^4 + 1/(2π)^4 + 1/(3π)^4 + ..., or rather π^4 times this. Now let's multiply out the first few terms of (1-αy)(1-βy)(1-γy)...: (1-αy)(1-βy) = 1 - (α+β)y + αβy^2 (1-αy)(1-βy)(1-γy) = 1 - (α+β+γ)y + (αβ+βγ+γα)y^2 - αβγy^3 We see that the coefficient of y^2 is the sum of the products of the numbers α, β, γ taken two at a time, and in fact this pattern continues to hold as the number of terms increases. Equating coefficients of y^2, we get 1/5! = αβ+βγ+γα+... Now how do we calculate α^2+β^2+γ^2+...? Taking the case with three terms, we see: (α+β+γ)^2 = α^2+β^2+γ^2+2αβ+2βγ+2γα So α^2+β^2+γ^2 = (α+β+γ)^2 - (2αβ+2βγ+2γα) Again, this pattern continues to hold with more terms So we can write: α^2+β^2+c^2 + ... = (α+β+c+...)^2 - (2αβ+2βc+2cα+...) =(1/6)^2 - 2/120 =1/36-1/60 =1/12(1/3-1/5) =1/12×2/15 =1/90 As α^2+β^2+c^2 + ... = 1/π^4 + 1/(2π)^4 + 1/(3π)^4 + ... we have 1/π^4 + 1/(2π)^4 + 1/(3π)^4 + ... = 1/90 so 1 + 1/2^4 + 1/3^4 + ... = π^4/90 QED! Those who are familiar with Viète's formulae for the coefficients of polynomials in terms of the roots will notice that it is these formulae that crop up here, but that instead of working from the highest power of x downward, to analyse a power series it is more convenient to start with the lowest power (the constant term x^0) upwards, and instead of the roots we consider the reciprocals of the roots, but otherwise the formulae are exactly the same. This means that we can use Newton's sums (or identities - see artofproblemsolving.com/wiki/index.php?title=Newton%27s_Sums) for the sum of the nth powers of the numbers α, β, γ etc (which will give us the sum of the 2nth reciprocal powers of the positive integers) in terms of the coefficients (which are essentially the elementary symmetrical polynomials) to churn out the formulae for the sum of the reciprocal even powers quite easily. Newton's sums (or identities), adapted to reverse polynomials (of degree N) and reciprocal roots tell us that if P(x) = a0 + a1x + a2x^2 + ... + aNx^N and if Pn is the sum of the nth powers of the reciprocals of the roots then a0P1 + a1 = 0 . . . (1) a0P2 + a1P1 + 2a2 = 0 . . . (2) a0P3 + a1P2 + a2P1 + 3a3 = 0 . . . (3) etc. Since these are true for any degree N, they will also be true for our power series if we let N tend to infinity. Notice in our case the sum of the nth powers of the reciprocals of the roots is 1/π^(2n) times the sum of the reciprocal 2nth powers of the positive integers, so once we know Pn, we find the the sum of the reciprocal 2nth powers of the positive integers as π^(2n)×Pn. Substituting in successive formulae, we get: Equation (1): 1×P1 - 1/3! = 0 P1 = 1/6 ∴ Sum of reciprocal squares = π^2/6 Equation (2): 1×P2 - 1/3!×P1 + 2×1/5! = 0 1×P2 - 1/6×1/6 + 2×1/120 = 0 P2 = 1/36 - 1/60 = 1/90 ∴ Sum of reciprocal fourth powers = π^4/90 (as before) Equation (3): 1×P3 - 1/3!×P2 + 1/5!×P1 - 3/7! = 0 1×P3 - 1/6×1/90 + 1/120×1/6 - 3/5040 = 0 P3 = 1/6×1/90 - 1/120×1/6 + 3/5040 P3 = 1/540 - 1/720 + 1/1680 P3 = 1/945 ∴ Sum of reciprocal sixth powers = π^6/945 (This agrees with the value on the video at 2:11!) and so on...

[@Mathologer]

Welcome back! Very nice solution. It's also quite rewarding to consider what identities you get when you compare the coefficients of higher order terms for the 1-sin(x) setup that I mention at the end. (Setting things up from scratch for cos(x) is also fun :)

[@davidrheault7896]

I know the zeta function from physics here is how I do any of these even integers. Use the Fourier series of a quadratic and integrate according to parceval identity there you go pi^4/90 = zeta(4) QED zeta(8) use Fourier series of x^4 and hit it with parceval you end up with pi^8/9450

[@anand.suralkar]

What is this

[@Jared7873]

@@anand.suralkar 🤔🤓🤷‍♂️

[@anand.suralkar]

@@davidrheault7896 i means what is this hit with Percival

[@matthewlind3102]

Thank you! I love learning about math, even if I don't fully understand what you are explaining, getting a glimpse is rewarding in itself.

[@gairikbanerjee4439]

To calculate C at 11: 06, differentiate C(Pi-x)x(pi+x) = sinx on both sides WRT x and then substitute x=0, to get C=1/pi squared

[@ripansharma5259]

Thanks man

[@bosorot]

c = Weierstrass Products of Sine = product from n to infinite of ( -1/n^2*pi^2) . simple differentiate C(Pi-x)x(pi+x) = sinx will not work because C(Pi-x)x(pi+x) is not equal to sin x

[@benjaminbrady2385]

This is extremely well made and edited! Great job as always Mathologer!

[@enzogiannotta]

I loved your introduction to power series. I really enjoy this topic because power series are so versatile that are very handy for a lot of problems, on of them is arriving to closed formulas for infinite sums (and that for me is really cool).

[@Tim3.14]

I really enjoyed this video! For the formula for 1-sin(x), I think it should be: (1-(2x)/(1pi))^2 times (1+(2x)/(3pi))^2 times (1-(2x)/(5pi))^2 times (1+(2x)/(7pi))^2 times ..., so that the zeros are at x = pi/2, -3pi/2, 5pi/2, -7pi/2, etc. [This is the same as what's shown in the video at 16:23, except half the factors have a plus sign instead of a minus.] Expanding each of the squared terms we have: (1 - 4x/pi + 4x^2/pi^2)(1 + 4x/(3pi) + 4x^2/(3pi)^2)(1 - 4x/(5pi) + 4x^2/(5pi)^2)(1 + 4x/(7pi) + 4x^2/(7pi)^2)... Mulitplying this out and grouping together powers of x, we have: 1 - (4/pi)(1 - 1/3 + 1/5 - 1/7 + ...)x + higher powers of x. By comparison with the McLauren series (1 - x + higher powers), we can see that -1 = -(4/pi)(1 - 1/3 + 1/5 - 1/7 + ...), which gives us pi/4 = 1 - 1/3 + 1/5 - 1/7 + ...

[@richardschreier3866]

Another glorious eye-opener. It astounds me that something so beautiful and accessible to only the world's top mathematicians in the 1700s can now be understood and marvelled at by millions. Thanks for another fine addition to the Mathologer Magic Show.

[@utkarshrastogi2482]

Really appreciate your mention of indian mathematician Madhava 👍👍. Sign of a true enthusiast.

[@carollaw21]

6:07 "How did Euler manage to prove his identity?" Did he have a birth certificate?

[@AMIT_OP]

Ghanta lele

[@rezumavuj483]

Ok I understand what u mean😂

[@joevogl3431]

I watched this a year or so ago and loved every minute of it. I watched it again and am still astounded at Euler’s creativity and Dr. Polster’s brilliant ability to communicate it.

[@ahoj7720]

I strongly recommend reading Euler's works, especially his "Introductio in Analysin Infinitorum". Various editions in Latin are available on line and are easy and fun to read (yes, even in Latin!) I worked in the theory of partitions of integers and Euler's seminal work on the subject would make great videos! Thank you for this beautiful video!

[@kimothefungenuis]

Where can I find them

[@ny6u]

this is perhaps one of the most beautifully rewarding videos you have made 👏🏻👏🏻👏🏻👏🏻

[@mpcc2022]

This is so beautiful. Euler and Ramanujan are easily some of the most aesthetically tasteful Mathematicians.

[@_polargaming]

In my opinion this is the best video you've done so far. Good job, keep it up!

[@mariomario-ih6mn]

You can get better and better approximations by stopping it at finite points and solving the polynomials.

[@wallaceOne5900]

This video is one that is both commendable and unforgettable.Keep it up

[@Alex-mq1yd]

Awesome video, also on an unrelated note: is your t-shirt a pi-rate?

[@Mathologer]

Absolutely :)

[@davidh1958]

Thank you so much for explaining this wonderful identity in such a straight forward and delightful manner.

[@Mathologer]

:)

[@maxnullifidian]

It boggles my mind how mathematicians could figure out so much cool stuff while having to calculate everything by hand! That, to me, is what made them truly great.

[@ancamg]

I am aware it's a three years old message, but calculating by hand it's not that difficult, if you get the hang on it. In my school we were not allowed calculators. Everything went easy until we did nuclear/atomic physics Math itself was easy, but working with never ending numbers was not funny

[@musicalBurr]

This is the best video you've done yet. Great work - it's getting better and better.

[@ThinkTwiceLtu]

Great video, once again:)

[@Mathologer]

I am also slowly making my way through all your videos. Really nice stuff :)

[@ThinkTwiceLtu]

Mathologer glad you enjoy them, even though your videos are on a completely different level than mine.

[@doudsbass]

It's very poetic and noble, I think, to address these sophisticated videos to youngsters that wouldn't know what a derivative is. Thanks a lot for the upload 😊

[@MrSigmaSharp]

I follow many math channels and Mathologer is by faaaar the best. Please do a video on boolean calculus and/or non decimal base calculations. Thanks for the good work.

[@Mathologer]

Glad you think so :)

[@sansamman4619]

Sigmad, to be honest mathologer should be in the entertainment category.for example when you watch khan academy, you are like: aha.ye.ok. When you watch mathologer: wat THATS AMAZIIINNGGG I LOVE MATH!!!

[@hypercent]

@Sigmad: Do you follow 3blue1brown? Just wondering because you said "by faaaar". I think 3blue1brown is also excellent and certainly not far (if at all) behind Mathologer. Watching this Mathologer video I was actually reminded of 3blue1brown's video "Pi hiding in prime regularities" which was similarly mindblowing.

[@MrSigmaSharp]

hypercent You are right it's a great channel too but that I think is beside the point. I like Mathologer better because I beleive it's well oriented. Again, there is no scale to judge math channels by.

[@hypercent]

@Sigmad. Okay. If there is no scale, then "by far" makes no sense though. Also, you said "Mathologer >> every other math channel I know", I asked "does that include 3b1b? Because I think thats on a similar level.": What is beside the point here? Anyway, I have no problem with you liking Mathologer better. Just your wording ("by faaaar") gave me the impression you might not even know 3b1b. But you do, so all is fine. :)

[@jamesfleming1155]

I can't handle how awesome this is. Great explanation. So thankful for your channel.

[@AdityaShirolkar]

Thanks for reference to Indian Mathematician "Madhava of Sangamagrama" in the end. We definitely care. The series is also called as Madhava-Leibniz series. link: en.wikipedia.org/wiki/Leibniz_formula_for_%CF%80

[@Chris_Bonner]

Very much appreciated.

[@varadarajcuram2238]

Thanks for mentioning Madhava towards end of video.

[@wilhelmsarosen4735]

1:30 Aww, putting your face and handle among the mathematicians is very nice. :)

[@Mathologer]

Could not resist :)

[@Momotaroization]

I'm rewatching this video after a while, and it really clicked for me. One viewing is not enough, but don't change anything! It was great!

[@alhobbel]

I first read about this exposition in Havil's book about the gamma constant and it blew my mind. This is such a treasure of daring mathematics and an example of the brilliance of Euler. It wasn't rigorous but blimey it was beautiful.

[@Mathologer]

In my opinion all of Havil's books are fantastic. Usually when I watch mathematical KZbin videos or read sort of popular expositions of math there is hardly anything I have not seen already somewhere else. In Havil's books I often find bits and pieces that are new to me :)

[@eliaskhoury1484]

evey minute in your videos can be extremelly helpful , thank you , and keep the hard work

[@SeleniumGlow]

I swear... This man has the most interesting T-shirts that I've ever seen.

[@ianslater2936]

Can't claim to understand it all but your presentation is terrific. Maths is definitely beautiful.

[@cphVlwYa]

Amazing video as always! I find it hilarious that this is how I take breaks from studying math hahahaha.

[@Mathologer]

Yes, I usually also unwind from doing math by doing different sorts of math :)

[@rgarciaf071]

This has been a nice surprise to find on KZbin thanks for sharing !

[@wendolinmendoza517]

You can find a good exposition of Euler's development on this and other subjects on the book 'Euler, the master of us all', by William Dunham. The author adds comments which really help understand how they used and thought of math; It's really worth :)

[@danielribastandeitnik9550]

What a coincidence! In the last 2 weeks I focused on learning infinite, Taylor and McLauren series to begin solving differential equations using series. It was nice to watch this video since I'm fresh with all this knowledge! Great video.

[@omislash7135]

Mathematicians always like to trace their advisor's heritage. I find it endearing.

[@AlphaGeminorum1]

I'm about 5 weeks into a rapid fire learning experience regarding ontological maths. Between these extremely succinct and highly digestible vids and a ton of books I've purchased on the subject this artist is becoming familiar with an entire world - indeed the entire universe - of the foundation of our reality. A reality that wasn't at all properly taught to me 5 decades ago. Way to go, Mathologer!

[@JumperAce]

i hit my confusion limit when he starting making the polynomial coincide with the sine wave. just seems, so bizarre

[@nathanisbored]

i mean, thats basically what a sine wave is, an infinite polynomial. think of a power series as the "decimal approximation" of a function. decimal expansions are to numbers as power series are to functions. just like you can get closer and closer to pi by adding more digits, you can get closer and closer to sinx by putting in more terms. and you can do this with any (well-behaved) function. and just like rational numbers are the only numbers whose expansion terminates, polynomials are the only functions whose power series terminates.

[@frankschneider6156]

It's called "fitting". If you do this in scientific resarch by gathering more and more data until you have at least some points fitting your hypothesized curve its called "faking" (or junk research tobe more polite). And if done in economics it is called "cooking the books". So mathematicians are the only ones to get away with this.

[@JumperAce]

nathanisbored that was... amazingly descriptive, ok, less confused. I guess my hangup is how the polynomials stretch into positive and negative infinity in the y axis, whereas the sine wave is contained at + and -1. I understand as you expand the polynomial you sort of, mitigate this, but even in an infinite sense i still feel the polynomial will always be unbound

[@fabibi_ha]

@Mike S. When i first heard of infinite series i also thought like you: "In the end the graph of this sum just has to rapidly drift out of the "box" where the usual sin(x) is in. It has to break out of the boundary [-1,1] at some point. And with what we learned about polynomials, the higher the exponent, the stronger/faster the graph will rise or fall. With each term we add the fitting polynomial gets a little longer, but the at its breakoff point is also increasing." That's where the magic about infinite series comes in. In the exact moment, where you transfer from finite series to infinite series, suddenly you get new properties. Each graph where you believe it has to break out of the boundary is just not finished. In fact: If you start your thinking process with adding more and more terms, you'll never finish. It's like in limit problems: the closer you get to your limit, your numbers usually get nastier and more complicated. But then you just take a big leap to "the end" and end up with a really nice number :D At school i imagined infinity just as a very huge number, like 10^9 or if it was needed in context 10^100 maybe. This worked out for things like e=lim h->inf (1+1/h)^h But as i learned more and more infinity became more like: "doing a process for eternity". What do you do at the moment in terms of learning math? Are you in school or do you visit college or sth like that?

[@hectarsavoie8166]

@Frank Schneider All of scientific research is bases on curve fitting to some extent, the trick is for your mathematical model to be within the margin of error of your instruments.

[@danielcookman3971]

Wonderful video! I also came across this sum whilst learning about complex analysis and (in particular) contour integration, though what you showed helps to explain why pi appears in it!

[@richardtheimer7959]

At 11:00 the constant c is turned into (1/pi)(1/pi) with the justification that it's not hard to see how this makes the best fit to the sin function. I don't see it though. Can someone explain this?

[@ethanyap8680]

That's a little bit of trickery I believe he skipped. I think he covers it in another of his videos

[@supernovaw39]

It's so ingenious! Great video

[@bardofvoid174]

Damn, I've searched the internet pretty deeply (not very deep, actually, but more than most would look-) a few times for the actual *formula* function of sin, and you're the only place I've found it- t h a n k y o u.

[@Metalhammer1993]

wow you took 5min to explain mac Lauren series. my maths prof took two ours. and your explanation how to do them was way more useful. thanks man

[@math2693]

Unbelievable 🤯🤯🤯🤯!!! What a smart proooooof Just one question,,, how did he derive that C = 1/pi^(2)?

[@kingofspain1234]

It comes from the same reasoning that was used to derive Taylor/Maclaurin series: matching values of nth derivatives. The first derivative of sin evaluated at 0 is 1, so we want the first derivative of our sin product to be equal to 1 at 0 as well.

[@cwaller1151]

Just to write it out for anyone searching the comments, because it took me an embarrassingly long time to get it: y=C(pi+x)x(pi-x)=C*x(pi+x)(pi-x)=C*x(pi^2-x^2)=C*pi^2*x-C*x^3 dy/dx=C*pi^2-3Cx^2. dy/dx @ 0 = C*pi^2-3C(0)^2=C*pi^2. Since we want dy/dx to be 1 instead of pi^2, C=1/pi^2.

[@ny6u]

Stunningly beautiful presentation 👏🏻👏🏻👏🏻👏🏻👏🏻👏🏻👍🏻👍🏻👍🏻

[@Tomaplen]

I found another way to prove this (I don't know if someone else mentioned this or found it already) 1) Write the fourier series for f(x)=x^2 in the interval [-2,2] 2) evaluate that series in x=2 3) Voila!

[@randomaccessfemale]

Euler has left the chat.

[@gurkiratsingh7tha993]

We could calculate the value of pi from the maclaurian series of sinx by using newton-ramphson method

[@stefanfink2023]

I'm struggling with the product formula for 1-sin(x): How did you find (1 - 2 x/Pi)^2 (1 + 2 x/(3 Pi))^2 (1 - 2 x/(5 Pi))^2 (1 + 2 x/(9 Pi))^2...? Especially the squares seem be coming from nowhere. Thanks for the great video.

[@jamesgrove8791]

Wow, some of these mathematicians were insanely smart.

[@hagaimichel7027]

can't you get the sum of odd powers by approximating cos(x) as Euler did for sin(x)?

[@Mathologer]

Give it a try :)

[@Kazetomosuki]

Hagai Michel had the same thought. Did U already try? :D

[@sparshsingh3991]

guess that means NO!! Obviously, Hagai Michel its the first intuition; but that that won't do. Moreover, we don't have to worry as it must have been tried out. ;)

[@Kazetomosuki]

Sparsh Singh hehe ... guess it's like the egg of Columbus: U never know unless U try it yourself

[@Mathologer]

Might as well tell you. Cos also gives nice identities. The first one is pi^2/8= 1/1^2+1/3^2+1/5^2+... (so the sum of the reciprocals of the odd squares is pi^2/8 :) Nothing really new though because this identity is only two easy steps away from the identity this video is about: pi^2/6= (1/1^2+1/3^2+1/5^2+...)+(1/2^2+1/4^2+1/6^2)= = (1/1^2+1/3^2+1/5^2+...)+1/4(pi^2/6). Therefore, 1/1^2+1/3^2+1/5^2+...=pi^2/6(3/4)=pi^2/8 :)

[@thangerlek5071]

You know it's going to be a good video when you're less than a minute in and already thinking about the zeta function!

[@himanshumallick2269]

Zeta(3) is known, and is famously called the apery's constant. And the maclaurin series for arctan(x) yields the leibnitz formula for pi/4 by setting x=1 Pi can also be calculated through ramanujan's formula (which converges rapidly) or chudnovsky brothers' formula. Plz make a video for the riemann hypothesis.

[@Mathologer]

Actually, it's not a problem to calculate approximations of zeta(z) for any z. And so, yes, zeta(3) is known in the sense that you can approximate it and it's named after Apery because he proved that is is an irrational number. However, the question is whether its precise value can be expressed in terms of any other well-known mathematical constant like pi or e and as I said in the video, nobody has a clue :)

[@SG1guru]

what he means i think, it's not known whether zeta(3) has a "nice" representation like zeta(2) does. Of course you can calculate any values of zeta function numerically to any precision

[@davidrheault7896]

Mathologer how about pi^3/26? Recently it has been calculated to great precision in the Riemann hypothesis and the value is closed to pi^3/25.8....loads of decimals

[@timh.6872]

David Rheault Sounds like you should try doing a rigorous proof!

[@philippenachtergal6077]

I guess you could call Apery's constant "nice" if it shows up in various other places. Sadly, I have no idea if it does.

[@djconnel]

Wow -- the John Wallis product is particularly shocking, since the 1 is redundant in the product, so it can be written: (π/2) = (2/3)·(2/3)·(4/5)·(4/5)·(6/7)·(6/7) which is of course incorrect since each of the terms on the right < 1, so the product must be < 1. But that's a matter of grouping. You can get any answer you want if you pick the right grouping.

[@Mathologer]

Actually, believe it or not but the 1 is not redundant. Have a look at what it means for an infinite product to converge. If we remove the 1 the product does not converge any more :)

[@djconnel]

@@Mathologer That's what's really weird. Surely there's a lemma that multiplying by 1 is an identity operator. This example suggests otherwise.

[@stevethecatcouch6532]

Pausing at 7:09 to approximate pi. If we call pi/1 the 0th partial sum, then the nth partial sum is a polynomial in pi of degree 2n+1. pi=0 is interestingly a root of each,a fact of little use at the moment. Being confident that pi is not equal to 0, I would divide by pi to get rid of that root. That leaves us with a polynomial of degree 2n. (x^2n)/(2n+1)! + .... +(x^8)/9! -(x^6)/7! +(x^4)/120 -(x^2)/6 + 1 = 0 Solve for x. I used Wolfram Alpha, but the series converges fast enough that a patient person, such as Leonhard Euler, could solve for the relevant roots by hand. n ....one value for x 1 .... 2.4495 2 .... ? 3 .... 3.0786 4 ....3.1487 5 ... 3.14115 6 ....3.14161 In addition to this sequence, the roots generate other sequences of real numbers, one of which, after flailing about a bit, appears to be converging to 2*pi. Edited to add: which makes sense on reflection because sin(2*pi) is also 0.

[@Mathologer]

That's what I had in mind :)

[@drewduncan5774]

No mention of how to solve for x. Not a straightforward thing to do when the degree of the polynomial is high...

[@zecheng3771]

Maybe solving root of polynomial by Newton's method

[@maxsilvester1327]

@@zecheng3771 I've tried this some time ago and you can get a many digits of pi by using newtons method using Newton's method with initial value x_0 = 3, a calculator, and the Taylor polynomials of sin and cos truncated to just 10 terms. I got the same value as the built-in pi constant using just 3 iterations.

[@granttherock9121]

Best explanation of this proof I have seen so far Much appreciated

[@omarsamraxyz]

But wait, the constant "e" was discovered by Euler or what... So how come Roger Cotes wrote about it, and a basic thing in it has not been discovered?? (correct me if I'm wrong)

[@yodaadoy2863]

Its a misconception i guess

[@DanSchrimpsher]

The constant that Euler called "e" was in fact found by Jacob Bernoulli in 1683 (forgive me for spelling). Cotes was working with Newton on Calculus and dividing circles into n parts or some such.

[@elnurhajiyev2477]

your presentations are always lovely, sir. I am learning quite a bit of math from these videos.

[@xDR1TeK]

Please can we have some history and explanation behind bessel's equations?

[@xiomaragonzalez6584]

Thanks a lot for your videos! I understood already a lot of things thanks to them that I never found in other parts of the internet

[@aaronleperspicace1704]

The infinite series for sine of x was actually discovered by Madhava of Sangamagrama from India :( Not any european mathematician.

[@dekippiesip]

If discovered independently(very likely) I don't think you can assign blame to anyone. Many results in maths have been discovered prior in China or India only to be rediscovered independently in Europe(often with gaps of more than 1000 years in between).

[@peterb9481]

Good video. Well presented, good content and animations. Good to see credit also given to the original discoverers.

[@jpnesseth]

Am I the only one who was hoping for more than a "neat trick" (4:30)?

[@mariomario-ih6mn]

No

[@bahraam19]

Good job, love the channel!

[@andrewmirror4611]

Oh, god, graduated the school two months ago, so miss the Math. Ohh, it's like a dose

[@franzluggin398]

You could always, you know, study math...

[@andrewmirror4611]

No, I couldn't. First month is for exams (And math is first), second is for preparing to the university, third is for getting visa and other things. And I am a chemist not mathematic. P.s. I am Russian and I am going to get graduated in China

[@DonCDXX]

I haven't been in a class room for years, but I still find time for a little recreational math. I definitely suggest finding time for some recreational math.

[@luigiionascu9797]

.. and i am prof of chemistry. .....the enigmatic is inccorupt body at sains ..!..in chemistry cllassic aminoacid travel in amine if go radical carboxil -cooh .., fatt travel in acido fatt and glicerol and autooxidating and glucid glucoza and fructoza etc go mollecule more simple and oxidating ...!..at sains corps body is not this procces ..!..it s enigmatic and divin ..!..

[@siddheshrane]

coincidentally at this point in time I too have graduated college two months ago

[@acerovalderas]

Beautiful and clear! I love your videos.

[@drewduncan5774]

Numerical analysts will immediately remember Cotes from the Newton-Cotes formulae.

[@ingGS]

Drew Duncan well, in the Spanish speaking world they are sometimes called simply “Newton formulae”, it was just later when I moved to the U.S. that I came to know them as Newton-Cotes, and I make an effort to carry that naming back home every time I can.

[@justinlink1616]

This is a video that warrants more than one thumbs up from me. Well done.

[@mr.incognito9100]

At 11:07 he says that you can see that the constant is 1/pi^2. How do you figure this out?

[@Mathologer]

For any polynomial the derivative (slope) at zero is just the coefficient of the x term :)

[@chedagoz7145]

You want the derivative of Sine and the derivative of the polinomial to be the same at 0. The derivative of sinx is cosx, de derivative of the polinomial is c(pi^2-3x^2). So we have cosx=c(pi^2-3x^2) which at 0 is 1=c(pi^2) so c=1/pi^2

[@evanfarhood8649]

Mathologer I believe you made an error when you factored in the 1/(pi^2) into the other therms. In the video, @ 11:08, you factor in what appears to be the 1/pi term, not 1/(pi^2).

[@DineshMishra-jh5ow]

@@Mathologer I am trying to verify the same for myself using five points and I got C = 1/pi^4. Is that what it is supposed to be? Can't get to the expression you have at 12.02 from there.

[@ganesansrinivasan6063]

Absolutely fantastic! And so brilliantly done!

[@nathanielglover65]

But why 1/pi^2

[@jayashrishobna]

I was wondering, too, but I managed to work it out - you get the 1/π^2 from equating the slopes of the two curves at x=0. The two curves we're comparing are: y = sin x y = C(x)(π+x)(π-x). As he mentioned in the video, there are infinitely many curves of the form C(x)(π+x)(π-x), so we need to find the C that fits the best. Since my math sucks, I expanded [(x)(π+x)(π-x)] fully: (x)(π+x)(π-x) = x(π^2 - x^2) = xπ^2 - x^3 So we have the following equation: sin x = C[xπ^2 - x^3] To make the right side be a good approximate for the left side, we need the slopes of both sides to be the same at x = 0. Differentiate both sides, using product rule "left d right, right d left": cos x = C*d/dx[xπ^2 - x^3] + [xπ^2 - x^3]*d/dx(C) = C*[π^2 - 3x^2] + [xπ^2 - x^3](0). Since we want to set the slopes equal at x=0, we evaluate this at x=0: cos (0) = C*[π^2 - 3(0)^2] + 0 1 = C*π^2 + 0 Rearranging, we get: C = 1/π^2.

[@nathanielglover65]

Thanks!

[@jayashrishobna]

np. Edited to reflect that d/dx(C) = 0, not 1 as I previously typed!

[@stridentdust597]

Just for clarification, you don’t need to use the product rule. When you expand C[(Pi + x)x(Pi - x)] you get: x(Pi^2 - x^2) = C[xPi^2 - x^3] When you differentiate this you can use the power rule (because C and Pi are constants - not functions) Therefore once you’ve set this to sine(x) you and differentiated you get: Cox(x) = CPi^2 - 3x^2 - let x = 0 1 = CPi^2 Therefore C = 1/(Pi^2) That should help if the previous comment was a bit hard to read.

[@fatmanurgurbuz1829]

@@jayashrishobna that was very helpful.thanks

[@andreachiarini4685]

Non riesco a fare a meno di cliccare i tuoi video, sono veramente fantastici ed interessanti.

[@TheFunnyMTM]

9:18 at infinity they coinSINe 😂😂

[@megauser8512]

lol

[@SnabbKassa]

I was hoping you were going to reveal that Euler's real identity was Bruce Wayne

[@andrescolomarcedeno9952]

Xd

[@aroei9103]

16:58 How very sad. (Proceeds to laugh) xDDD

[@AdhiNarayananYR]

Really sad

[@dashnarayana]

Tons of thanks for remembering Madhaba of Sangram gram

[@mariomario-ih6mn]

7:07 tell us in the comments. No one answers.

[@vaasudevsuryahimawari953]

No words , I got goosebumps. 🙏🙏🙏. Thank you sir, love you❤.

[@ben_clifford]

13:55 The payoff of this video was very satisfying

[@wiler5002]

One of your best videos

[@mguieb1988]

Would it be possible to do one on the Monstrous Moonshine? I just love how you introduce and teach math.

[@morelelfrancel6603]

Damn I love this guy. Dude you're funny, and your videos are as beautiful informative. Keep up!

[@glum_hippo]

16:55 Magavar of sangamaagramma? Who what when?!

[@Mathologer]

As usual a quick google search "Madhava math" will get you the basic info :)

[@boltez6507]

Ancient Indian mathematians are truly underrated