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

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.

How was this video not titled "Identity Theft"?

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 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

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

Nice pi-rate shirt.

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".

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!

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

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

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.

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

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

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.

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 :)

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).

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.

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

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

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.

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...

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 + ...

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

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

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.

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

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

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

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!

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

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!

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.

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

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

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!

Thanks for mentioning Madhava towards end of video.

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

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.

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

This has been a nice surprise to find on RU-vid thanks for sharing !

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

Great video, once again:)

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

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

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.

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.

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 😊

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

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 :)

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

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

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

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.

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.

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

Some of this is still, unfortunately, over my head, but I'm working on it - & I DO know enough to see its beauty! Thanks! tavi.

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.

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

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

absolutly brilliant! Thanks for explaining this so understandably! Very nice indeed.

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

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?

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!

Absolutely fantastic! And so brilliantly done!

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!

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

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

Beautiful and clear! I love your videos.

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

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

Really nice, intuitive explanation of the limit described at 5:30 being equal to 2!

Thank you for the video! All of you friends are super awesome! Oh, moments in this video are sad.

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

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

this is one of the very simple yet way out of the box thinking that elegantly and enjoyably blows your mind.

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

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

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

Love maths, Zeta function and many more Thanks for making the understanding very simple.

It's so ingenious! Great video

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)

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

I just watched it again for the first time in a long time and it is amazing. Even more fun the second time around.

Good job, love the channel!

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

9:18 at infinity they coinSINe 😂😂

Great video. I love your video’s. For the question posed at 7:05. I believe you can find a series which you can multiply the series for Sin(x) to produce x (Inverse Sine function). With the new series you can input 1 into it and should get pi/2. Although there are easier ways of deriving the inverse Sine Function series. I hope this helps.

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

13:55 The payoff of this video was very satisfying

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

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

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

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.

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

Really an other amazing video! Thank you :)

Thanks for the great video! I would like to see more about that zeta of 3 series:) Sounds strange that nobody knows if it converges or not.

But why 1/pi^2

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

Nice video. I remember from my mathematics study in university that Euler's equation could be proven by calculating a complex integral of a function with poles at 1/(n*n) for integer n, but it is nice to see an argument that uses simple calculus.

Impressive how much you can cram into a seventeen minute video, prof P!