Not much Mathologer action lately, sorry. Just insanely busy at work at the moment. Like pretty much everywhere else, my uni here in Melbourne has moved to teaching online this semester and that means crazy hours for at least another six weeks in my case. Happy to at least get this video out. Not the one I planned to do (sign of permutations) but since I had some of the animations lying around much more doable than one from scratch. Anyway, hope you are all staying safe.
High schools online too. I have been making youtube videos for each for one of my classes. I only I had have made them last year, then this would have been easy!
I just wanted to say thanks so much for your videos. I'm an engineer with only a engineer's background in math, so I can only follow a reasonable percentage of what you talk about. But I had always loved math, and your videos keep my interest in math fresh and fun.
4:45 "On the right, what do we have. Oh, well, whatever that is, it should be infinity. HMMM" "On the left, we are dividing by zero. Double HMMMM" DEAD
You remove all the tedium, replacing it with slick graphics, perfect music and your happy personality. Your love of math is the "answer" for us. Thank you!
It is amazing, maths is like intellectual freedom. I think if we were truly free, we'd just see maths alone. Freedom is associated with anarchy (t-shirt!), which just makes things better. Nerd Power!
Maths.... is the most beautiful thing ever. So subtle, so abstract and so transcendental... it always existed and will always exist. What we know is a tiny fraction, the most will always be beyond our comprehension.
Beautiful presentation. Euler's original book "Introductio in Analysin Infinitorum" is a treasure. It's easily readable, even if written in Latin! (there are translations, of course.) It's exactly the same spirit as your presentation.
Very much worth doing but really tricky stuff to get right. I am actually regularly feature Ramanujan's fast converging Pi series exams for one of the courses that I teach at uni. So definitely on the radar ...
Very much worth doing but really tricky stuff to get right. I am actually regularly feature Ramanujan's fast converging Pi series exams for one of the courses that I teach at uni. So definitely on the radar ...
@@Mathologer I said, the comment you posted was duplicated, and this error may be due to a connection problem, as I have witnessed such problems on youtube before
The general formula for the sum of 1/x^(2n) is hidden in here too, and only requires a few extra steps. Start from the chapter 2 formula and take logs and derivatives to get the formula at 10:40: cot(x) = 1/x + 1/(x-pi) + 1/(x+pi) + 1/(x-2pi) + 1/(x+2pi) + ... Move the 1/x term to the left side, and take 2n-1 more derivatives. The k'th derivative of 1/x is (-1)^k * k!/x^(k+1), so when k = 2n-1 this is -(2n-1)!/x^(2n). So we have (d/dx)^(2n-1) (cot(x) - 1/x) = -(2n-1)! * (1/(x-pi)^(2n) + 1/(x+pi)^(2n) + ...) Now divide by -(2n-1)! and take the limit as x -> 0. 1/(2n-1)! * lim x -> 0 [(d/dx)^(2n-1) (1/x - cot(x))] = 1/(-pi)^(2n) + 1/pi^(2n) + 1/(-2pi)^(2n) + 1/(2pi)^(2n) + ... On the right side, all the negatives are squared away, and we end up with 2 copies of each term. So multiply by pi^(2n)/2, and we get this: pi^(2n)/2 * 1/(2n-1)! * lim x -> 0 [(d/dx)^(2n-1) (1/x - cot(x))] = 1 + 1/2^(2n) + 1/3^(2n) + ... = zeta(2n) This formula looks messy, but there's a trick: notice that on the left, we have something of the form 1/k! * k'th derivative of f(x) at x=0. These are just taylor series coefficients! The left side is really just pi^(2n)/2 times the coefficient of x^(2n-1) in the taylor series of 1/x - cot(x). There are a few other things we can do to make the formula easier to read. We can multiply the function by x to make the powers line up nicely (otherwise the 1/k^8 sum will be related to the coefficient of x^7, instead of x^8). This gives: zeta(2n) = pi^(2n)/2 * coefficient of x^(2n) in the taylor series of 1 - x cot(x) The next thing we can do is move the pi^(2n) "inside" the taylor series, by replacing x with pi x. We can also move the factor of 1/2 into the function. Then we get: zeta(2n) = coefficient of x^(2n) in the taylor series of (1 - pi x cot(pi x))/2, or equivalently, (1 - pi x cot(pi x))/2 = sum n=1..inf, zeta(2n)x^(2n) And indeed, if you ask wolframalpha to compute the taylor series of (1 - pi x cot(pi x))/2, you get pi^2/6 x^2 + pi^4/90 x^4 + pi^6/945 x^6 + pi^8/9450 x^8 + ... Finally, comparing this series to the standard taylor series for cot in terms of Bernoulli numbers gives Euler's general formula for zeta(2n)
Wow! A way to generate a sum series of the Euler pi expressions. I wonder if any complex analysis (or higher dimensions) can allow formulas & solutions for the odd powers of pi? (I noticed the general "super" formula/expression by Euler would result in imaginary #s (i, or (-1)^(1/2)) when trying to achieve the odd powers.)
@@danielreed540 there is a slightly different formula that can be proved in basically the same way, but for the sum of zeta(n) x^n for n>=2. specifically this sum is equal to -x(gamma + polygamma(1-x)) where gamma is the euler mascheroni constant, and polygamma is the log derivative of the gamma function. then the formula says that the derivatives of this function (call it f) satisfy f^(n)(0)/n! = zeta(n). using the reflection formula polygamma(x) + polygamma(1-x) = pi cot(pi x) allows you to recover the formula from my first comment by computing (f(x) + f(-x))/2, but of course replacing x by -x and adding them will wipe out all of the odd terms of the series.
@@darkseid856 π^2 being close to g is not a coincidence. "g" depends on the unit system, and the meter was chosen (or it was at least one of the definitions) so that a meter-long pendulum had a period of exactly two seconds (the second was chosen a long time ago as 1/86400 of a day). If this is exactly true, that gives L=g/π^2=1 (in SI units). It is 99.3 cm actually with the standard g, of course it changes a bit by the place on Earth. edit: not saying that you don't know this. but pretty sure at least some people in the comments don't. I found this out many years after learning about "π^2=g"
@@blizzbee No. Euler's conjecture is a variation on Fermat's Last Theorem and was actually proven false in the 20th century and a counterexample was found. However, he didn't claim it was necessarily true, he simply conjectured it...so I'm not sure what you mean by statements.
The crisis has kept me busier than ever, and so I am even that much *more* grateful that you posted a video so I could take a good mental health break. PS: Although one can find patterns in anything, it's fun that 28:57 is part of the continuing fraction of 7.
To get the sum[1/s**2n, s=1 to s=infinity], just take derivatives of the function (x*cot(x)) 2n times, then multiply by -pi**2n*1/2. I was literally working on this the whole past week.
Another great video from prof. Burkard. I love it! Thank you and stay safe professor. Note: I'm amazed that even Euler is Mathologer's Patreon Supporter.
Despite the lockdown, the comforts I appreciate in life have really shown themselves to me and I feel grateful for what I have. These maths animations are among that!
The thing about Euler's solution to the Basel problem is that the maths in it isn't actually very hard to follow, it's sixth-form-level stuff really. But the reason the Basel problem stumped the likes of Fermat and Bernoulli was because the solution requires thinking outside the box, and skating on thin ice with regards to mathematical rigour. Euler took a few steps that more experienced mathematicians might have disregarded as nonsense - but he got the right answer, didn't he? It was Euler's creativity that was what really unlocked the problem.
Exactly. And now, unfortunately, we live at a time when academic authorities try to prevent students to enjoy mathematics like Euler did, even mocking the ones that dare to do that, because of "lack of rigour". Such awful people.
I'm amazed by how useful are the addition (0) and multiplication (1) identities, two of the five members of Euler's identity, btw... Nice video! Thanks a lot for this great channel.
I am a student of class 11. But I love mathematics. I love to learn and think. But the teachers I know always teaches the things which is important for our exam. But I am not satisfied with that. That's why Internet is the only source where I can learn. I think this channel is the best. Lots of love for you sir for making these kind of videos.
Woww...an amazing insight into how Euler's mind worked. I sooo much enjoyed this video. It involves taking notes and a bit of thought but it is worth all the effort in the end.
just wow ... i am a maths teacher to high schoolers but let me tell u ... i feel like i am a happy little kid with his toy when i watch your videos . you are amazing . thank you
Leibniz-Madhava series converges really slow but we can apply Euler's series transformation to get convergence rate about one bit per iteration (approximately three new decimal places per 10 iterations)
General note: there is a general product equivalent of the taylor series using something called product calculus. Very much worth checking out and useful in statistics as well!
Another delightful video illuminating the connections between various infinite sums (and an infinite product) and powers of pi by using the product formula for sine. Glorious. In response to your query about what is missing from the Nike argument, two things stand out for me. If you start with the product formula for sine as a postulate, you need to show that a) the product converges and that b) the leading coefficient needs to be one. I expect there is a theorem that allows you to then conclude that this infinite product which has the same zeros and scale factor as the sine function is indeed the sine function. If you are looking for more potential material, I think the Gamma function, its various identities (the reflection formula, the Legendre duplication formula) and its use in evaluating infinite products is pretty cool.
Hey @mathologer , if you're doing a future video on quintic equations ; it would be fascinating to see how even the greatest mathematicians fell short on trying to find a formula and their insights- especially euler ( i can't seem to find his attempt on the quintic).and George Jerrard who was reluctant to accept the quintic was unsolvable by radicals etc . I'm sure there also must've been interesting attempts by ferrari and cardano . Tschirnhaus transformations....... (Fingers crossed) this is the content of your next video
That got me stumped for a bit, too, but I think I have found a way to explain it: The sum we want is every single term (1/i²)(1/j²) for every combination of i and j from 1 to infinity. So, we have (1/1²)(1/1²) + (1/1²)(1/2²) and so on when the first term is (1/1²). This is (1/1²)(1/1² + 1/2² + 1/3² + ...) Then the same for (1/2²): (1/2²)(1/1²) + (1/2²)(1/2²) + (1/2²)(1/3²) + ... giving (1/2²)(1/1² + 1/2² + ...) Grouping those together, we get (1/1² + 1/2²)(1/1² + 1/2² + 1/3² + ...) And so on for every 1/i² in the first term, giving (1/1² + 1/2² + 1/3²...)(1/1² + 1/2² + 1/3² + ...) = (1/1² + 1/2² + 1/3² + ...)² I hope that helped.
You can skip a lot of calculating by adding up answers from the Leibnitz series like you would for a Pascal's Triangle. And if you start with four and us it for the numerator you will converge on π/1 instead of π/4. It just makes the formula much more pleasing to work with. You can also use different numbers for the numerator to get different multiples of pi. It depends on what multiple you are looking for. Something that I just figured out recently.
At 4:32 an expression for [( sin x ) / x ] is found, making me wonder if an integral expression can be found from the right hand side, bypassing both Feynman’s technique for finding that integral and using contour integration over the complex plane to find the result.
Your videos are great review and sometimes brand new stuff. I had seen this at some other point in time in history of mathematics, but my teacher must have skipped a few steps or I was half asleep.
There are lots of interesting infinite fractions for Pi, aren’t there? Is it a characteristic of a transcendental number that there are an infinite number of continued fraction representations? (While for strictly irrational algebraic numbers there is only one unique family of continued fractions?)
Thank you Lord Mathologer. I would love to see how the Pi continued fraction defined by b's(3 , 6, 6, 6 ...) and a's(1^3, 1^3 + 2^3, 1^3 + 2^3 + 3^3, ...) is derived. Its neat because Pi - 3 is that sum of cubes fraction which is also equivalent to an alternating sum involving (n, k). I would also be supremely interested to know how Nilankantha accelerated the 'leibniz' arctan(1) by multiplying by the square root of 3 on one side and dividing by powers of three in the sum. I just read Pi Unleashed but not able to find answers to these questions.
Music (all from the free audio library that RUclips provides to creators): ruclips.net/user/audiolibrarymusic?nv=1 Take me to the Depth Fresh fallen snow English country garden Morning mandolin
did you know that if you square the Fourier series of f(x)=x between [-T,T] and equate its infinite coefficients with the coefficients of the Fourier series of f(x)=x^2 in the same interval, one gets all the identities of even powers of pi at once?
Oh, I just noticed that a few of the mobius strip geometry videos are private now when I wanted to show a friend of mine (the hypertwist video is a bit above their weight). Are they coming back? (maybe on the second channel if they're not fitting of this one?)
Euler formula, Taylor series, and of course the mysterious pi! Thanks for the video, very inspiring. once I can not follow the auto formula, I can easily go to meditation mode, thanks for the music.
Why does 1/(2×3×4) − 1/(4×5×6) + 1/(6×7×8) − ··· = (π−3)/4 ? (One of the "beautiful equation" candidates.) That π/4 looks like arctan(1) from Leibniz's formula, and cf. the generalised continued fraction in Brounker's formula! PS: Also, Viète's formula is rather cute: 2/π = cos(π/4) × cos(π/8) × cos(π/16) × cos(π/32) × ··· Thanks for the channel.
How much did they do graph theory? I don't remember hearing any results by either of them. Then again, I'm not sure whether graph theory really existed back then...
It would be fun to see your interpretation of the BBP hexadecimal digit of Pi extraction formulas, which seems (to me) conceptually impossible for an irrational number and also seems to hint at a subtle relationship between Pi and the number 16. Also, the basic BBP formula for Pi seems to generate better than 1 decimal digit of precision per iteration which seems amazingly fast even if each step requires quite a lot of computation. I enjoy your content, thanks.
If anyone is curious to learn more about the Weierstrass product formula which Mathologer mentions at the end of this video, you will probably want to check out the product formula for the Gamma function, which is related to the Sine product formula by the expression: Gamma(x) * Gamma(1 -x) = pi / sin(pi*x). You can learn more about the product formula for the Gamma function from the brand-new video I just posted on my channel
Hans Enedahl it turns out to be ok, but you're quite right in questioning this seemingly obvious step. My guess is that a rigorous justification would involve (1) clearly identifying the domain of validity of the two expressions (where each is well-defined either as a finite expression or as a limit), (2) showing that the two definitions coincide over this domain, (3) showing that both expressions are differentiable over this domain. If I'm not mistaken, these properties should already force the derivatives to be equal (either as finite expressions or as limits).
Great video :) Most interesting I find the numbers on the left side in the denominator: 6, 90, 945, ... where do they come from? Is there a formula for them?
I think one problem with the derivation is that we need to know if the infinite product converge before using it and that if we write sin(x) in function of its zeros, how we know that it doesnt have complex zeros that we need to add. Great video by the way.
![BLACK BAG - Official Trailer [HD] - Only in Theaters March 14](http://i.ytimg.com/vi/Du0Xp8WX_7I/mqdefault.jpg)
Not much Mathologer action lately, sorry. Just insanely busy at work at the moment. Like pretty much everywhere else, my uni here in Melbourne has moved to teaching online this semester and that means crazy hours for at least another six weeks in my case. Happy to at least get this video out. Not the one I planned to do (sign of permutations) but since I had some of the animations lying around much more doable than one from scratch. Anyway, hope you are all staying safe.
No problem!
@@2false637 pun of the week award.
High schools online too. I have been making youtube videos for each for one of my classes. I only I had have made them last year, then this would have been easy!
I just wanted to say thanks so much for your videos. I'm an engineer with only a engineer's background in math, so I can only follow a reasonable percentage of what you talk about. But I had always loved math, and your videos keep my interest in math fresh and fun.
no problem but how do you do these animations ?
4:45
"On the right, what do we have. Oh, well, whatever that is, it should be infinity. HMMM"
"On the left, we are dividing by zero. Double HMMMM"
DEAD
Well given he had infinity on one side its a good thing the other side is dividing by 0. 😄
0/0=1 ?
Well , its limits here . The calculus limits
You remove all the tedium, replacing it with slick graphics, perfect music and your happy personality. Your love of math is the "answer" for us. Thank you!
I had my worst two weeks lately, and this 30ish minutes was the first I was feeling joyful. Thank you!
It is amazing, maths is like intellectual freedom. I think if we were truly free, we'd just see maths alone. Freedom is associated with anarchy (t-shirt!), which just makes things better.
Nerd Power!
So do you think the North should be represented by positive reals? (referring to your current Mandelbrot Set profile image)
Maybe...
I am just glad I don't have to meet stupid people and stay undisturbed for Physics and Mathematics
@@LittleBishop001 I just like this rotation better ;)
Maths.... is the most beautiful thing ever. So subtle, so abstract and so transcendental... it always existed and will always exist. What we know is a tiny fraction, the most will always be beyond our comprehension.
Earlier I wondered: "When is mathologer posting his next video?"
Couldn't have gotten a better answer :)
I first read you comment as "Euler wondered..."
Great video
Flammable Maths hey Papa
Great job flammable maths
This is like a grand unified theory for pi formulas. Amazing!
Beautiful presentation. Euler's original book "Introductio in Analysin Infinitorum" is a treasure. It's easily readable, even if written in Latin! (there are translations, of course.) It's exactly the same spirit as your presentation.
Glad you are back! I was concerned after the channel was hijacked. You are a tier above the other math youtubers in my opinion! :)
The world is always a bit better with a video from the great Mathologer. I am glad you put this together, thank you Mathologer!
Would love to see about Ramanujan's fast converging Pi series. There are plenty of interesting theorems by Ramanujan. Please demonstrate those.
Very much worth doing but really tricky stuff to get right. I am actually regularly feature Ramanujan's fast converging Pi series exams for one of the courses that I teach at uni. So definitely on the radar ...
Very much worth doing but really tricky stuff to get right. I am actually regularly feature Ramanujan's fast converging Pi series exams for one of the courses that I teach at uni. So definitely on the radar ...
@@Mathologer connection problem? This was posted twice :)
@@kimmalyncleaveway2907 ??
@@Mathologer I said, the comment you posted was duplicated, and this error may be due to a connection problem, as I have witnessed such problems on youtube before
The Auto Algebra Song™ makes me so happy.
Well, I do need this happiness badly, because I understand nothing.
Same here. It’s like the File Select Theme for ”Super Mario 64”. 🙂
Beautiful. I usually do not comment on channels, but i had to on this one. Amazing. Never stop making these videos.👏👏👏
Every time you release a new video it just happens to be exactly what I needed to see! Thank you!
The general formula for the sum of 1/x^(2n) is hidden in here too, and only requires a few extra steps. Start from the chapter 2 formula and take logs and derivatives to get the formula at 10:40:
cot(x) = 1/x + 1/(x-pi) + 1/(x+pi) + 1/(x-2pi) + 1/(x+2pi) + ...
Move the 1/x term to the left side, and take 2n-1 more derivatives. The k'th derivative of 1/x is (-1)^k * k!/x^(k+1), so when k = 2n-1 this is -(2n-1)!/x^(2n). So we have
(d/dx)^(2n-1) (cot(x) - 1/x) = -(2n-1)! * (1/(x-pi)^(2n) + 1/(x+pi)^(2n) + ...)
Now divide by -(2n-1)! and take the limit as x -> 0.
1/(2n-1)! * lim x -> 0 [(d/dx)^(2n-1) (1/x - cot(x))] = 1/(-pi)^(2n) + 1/pi^(2n) + 1/(-2pi)^(2n) + 1/(2pi)^(2n) + ...
On the right side, all the negatives are squared away, and we end up with 2 copies of each term. So multiply by pi^(2n)/2, and we get this:
pi^(2n)/2 * 1/(2n-1)! * lim x -> 0 [(d/dx)^(2n-1) (1/x - cot(x))] = 1 + 1/2^(2n) + 1/3^(2n) + ... = zeta(2n)
This formula looks messy, but there's a trick: notice that on the left, we have something of the form 1/k! * k'th derivative of f(x) at x=0. These are just taylor series coefficients! The left side is really just pi^(2n)/2 times the coefficient of x^(2n-1) in the taylor series of 1/x - cot(x).
There are a few other things we can do to make the formula easier to read. We can multiply the function by x to make the powers line up nicely (otherwise the 1/k^8 sum will be related to the coefficient of x^7, instead of x^8). This gives:
zeta(2n) = pi^(2n)/2 * coefficient of x^(2n) in the taylor series of 1 - x cot(x)
The next thing we can do is move the pi^(2n) "inside" the taylor series, by replacing x with pi x. We can also move the factor of 1/2 into the function. Then we get:
zeta(2n) = coefficient of x^(2n) in the taylor series of (1 - pi x cot(pi x))/2, or equivalently,
(1 - pi x cot(pi x))/2 = sum n=1..inf, zeta(2n)x^(2n)
And indeed, if you ask wolframalpha to compute the taylor series of (1 - pi x cot(pi x))/2, you get pi^2/6 x^2 + pi^4/90 x^4 + pi^6/945 x^6 + pi^8/9450 x^8 + ...
Finally, comparing this series to the standard taylor series for cot in terms of Bernoulli numbers gives Euler's general formula for zeta(2n)
Got pretty close to including a variation of this in the video on the Bernoulli numbers and the Maclaurin summation formula.
Nice insight
Wow! A way to generate a sum series of the Euler pi expressions. I wonder if any complex analysis (or higher dimensions) can allow formulas & solutions for the odd powers of pi? (I noticed the general "super" formula/expression by Euler would result in imaginary #s (i, or (-1)^(1/2)) when trying to achieve the odd powers.)
@@danielreed540 there is a slightly different formula that can be proved in basically the same way, but for the sum of zeta(n) x^n for n>=2. specifically this sum is equal to -x(gamma + polygamma(1-x)) where gamma is the euler mascheroni constant, and polygamma is the log derivative of the gamma function. then the formula says that the derivatives of this function (call it f) satisfy f^(n)(0)/n! = zeta(n). using the reflection formula polygamma(x) + polygamma(1-x) = pi cot(pi x) allows you to recover the formula from my first comment by computing (f(x) + f(-x))/2, but of course replacing x by -x and adding them will wipe out all of the odd terms of the series.
Nice
Who all agree that this is the best mathologer video ever made? Thanks a lot Sir for this wonderful video.
Engineers: Hold my Taylor Series
No need, sin(x) = x
@@henryhayton8784 cos(θ)=1
π^2 = e^2 = g
This one: ruclips.net/channel/UC7590VTWe6m0kq3gJcgLINg
@@darkseid856 π^2 being close to g is not a coincidence. "g" depends on the unit system, and the meter was chosen (or it was at least one of the definitions) so that a meter-long pendulum had a period of exactly two seconds (the second was chosen a long time ago as 1/86400 of a day). If this is exactly true, that gives L=g/π^2=1 (in SI units). It is 99.3 cm actually with the standard g, of course it changes a bit by the place on Earth.
edit: not saying that you don't know this. but pretty sure at least some people in the comments don't. I found this out many years after learning about "π^2=g"
I’ve been waiting for a good lesson on this for many years. Thank you!
I usually never comment on videos but this video is a masterpiece. I never knew this could be conceptualized so easily. You are a legend.
A mathologer video 😊😊😊😊😊 I keep waiting for the great videos like it. Just love watching mathologer videos.
Euler is quickly rising in the ranks of "potentially the smartest human ever lived" in my estimation.
Historians estimate that 40% of all mathematical papers written during the years Euler did his maths, were written by Euler.
@@alexpotts6520 I want a citation for that. It's amazing.
Didn't watch video yet... Did Euler prove every math statements he invent? Just asking.
@@blizzbee No. Euler's conjecture is a variation on Fermat's Last Theorem and was actually proven false in the 20th century and a counterexample was found. However, he didn't claim it was necessarily true, he simply conjectured it...so I'm not sure what you mean by statements.
@@pinchus2714 axioms, lemmas, theorems, conjectures.. Duh
Absolutely charming video, it's amazing how products and sums are linked and how one is able to derive the known formulae so effortlessly.
I just went half way the video and WOW! Euler was amazing, but you are amazing too, real quality stuff! Thanks!
The crisis has kept me busier than ever, and so I am even that much *more* grateful that you posted a video so I could take a good mental health break. PS: Although one can find patterns in anything, it's fun that 28:57 is part of the continuing fraction of 7.
Just watched an old Video and now there is a new Video! Da steckt wirklich großartige Arbeit in den Videos, vielen Dank
Ooh ooh I've been working on polylogarithm functions and Dirichlet series recently and what you have shown is very exciting to me.
To get the sum[1/s**2n, s=1 to s=infinity], just take derivatives of the function (x*cot(x)) 2n times, then multiply by -pi**2n*1/2.
I was literally working on this the whole past week.
Another great video from prof. Burkard. I love it!
Thank you and stay safe professor.
Note: I'm amazed that even Euler is Mathologer's Patreon Supporter.
Always having a great time watching Mathologer !
Despite the lockdown, the comforts I appreciate in life have really shown themselves to me and I feel grateful for what I have. These maths animations are among that!
Your videos are AMAZING....and the explanation is always very brilliant!!!
The thing about Euler's solution to the Basel problem is that the maths in it isn't actually very hard to follow, it's sixth-form-level stuff really. But the reason the Basel problem stumped the likes of Fermat and Bernoulli was because the solution requires thinking outside the box, and skating on thin ice with regards to mathematical rigour. Euler took a few steps that more experienced mathematicians might have disregarded as nonsense - but he got the right answer, didn't he? It was Euler's creativity that was what really unlocked the problem.
Exactly. And now, unfortunately, we live at a time when academic authorities try to prevent students to enjoy mathematics like Euler did, even mocking the ones that dare to do that, because of "lack of rigour". Such awful people.
Thank you sir.. the contents of your videos are truly wonderful... !! Absolutely one of the bests on youtube...! : )
I'm amazed by how useful are the addition (0) and multiplication (1) identities, two of the five members of Euler's identity, btw... Nice video! Thanks a lot for this great channel.
I am a student of class 11. But I love mathematics. I love to learn and think. But the teachers I know always teaches the things which is important for our exam. But I am not satisfied with that. That's why Internet is the only source where I can learn. I think this channel is the best. Lots of love for you sir for making these kind of videos.
Breath taking, awe inspiring and spellbinding. Thank you Euler/Mathologer.
Unfortunately im a big Euler fan and i already knew everything in this video, but i love to see it over and over again.
There is never enough Euler.
Enjoyed it. Great insights. Good to see. Stay safe Mathologer.
Great video! As I'm wrapping up Calc II, this was a really great way to extend what I've seen of infinite series.
Glad you enjoyed it!
Anyone try to work out Sum_i 1/i^6? Using the method in the video I ended up with
Sum_i 1/i^6 = sum_{i,j,k} 1/(i^2j^2k^2) - 2\sum_{i
"Just tell our brains to shut up." one of the nicest phrases concerning the understanding of math problems.
smooth, elegant and graceful always.... thanks Mathologer
28:11 Hey! Euler is your Patreon patron!
28:25 And so is Mandelbrot!
Yes, great, isn't it :)
Oh god
Great presentation. You made it looked easy.
Woww...an amazing insight into how Euler's mind worked. I sooo much enjoyed this video. It involves taking notes and a bit of thought but it is worth all the effort in the end.
That's the most enlightening math video I have watched recently.
just wow ... i am a maths teacher to high schoolers but let me tell u ... i feel like i am a happy little kid with his toy when i watch your videos . you are amazing . thank you
This reminds me of my high school Math teacher, he used to tell us " and this is the art of mathematics"
25:45 When the class nerd reminds the teacher to give some homework:
Leibniz-Madhava series converges really slow but we can apply Euler's series transformation
to get convergence rate about one bit per iteration (approximately three new decimal places per 10 iterations)
*Euler* again !
He was really a badass his work is everywhere メ!
General note: there is a general product equivalent of the taylor series using something called product calculus. Very much worth checking out and useful in statistics as well!
Another delightful video illuminating the connections between various infinite sums (and an infinite product) and powers of pi by using the product formula for sine. Glorious.
In response to your query about what is missing from the Nike argument, two things stand out for me. If you start with the product formula for sine as a postulate, you need to show that a) the product converges and that b) the leading coefficient needs to be one. I expect there is a theorem that allows you to then conclude that this infinite product which has the same zeros and scale factor as the sine function is indeed the sine function.
If you are looking for more potential material, I think the Gamma function, its various identities (the reflection formula, the Legendre duplication formula) and its use in evaluating infinite products is pretty cool.
Wow the logarithmic trick was amazing!
I'm a french high school student and i don't understand a lot, but what a pleasure to listen to !
Hey @mathologer , if you're doing a future video on quintic equations ; it would be fascinating to see how even the greatest mathematicians fell short on trying to find a formula and their insights- especially euler ( i can't seem to find his attempt on the quintic).and George Jerrard who was reluctant to accept the quintic was unsolvable by radicals etc . I'm sure there also must've been interesting attempts by ferrari and cardano .
Tschirnhaus transformations.......
(Fingers crossed) this is the content of your next video
Please keep this good work going forever!
Damn Euler! The greatest mathematician of all time. True superhero.
Isn't Newton or Gauss are greatest
@@pardeepgarg2640 you may be right, but I'll stick with Euler
@@newlaty72 me with Ramanujan :D
Everyone has its own Taste :D
This is great, I wish we would go even deeper into the maths
Wheeeeere do you want to go???
24:21 I still don't understand how the orange sum is the square of (π^2)/6
That got me stumped for a bit, too, but I think I have found a way to explain it:
The sum we want is every single term (1/i²)(1/j²) for every combination of i and j from 1 to infinity. So, we have (1/1²)(1/1²) + (1/1²)(1/2²) and so on when the first term is (1/1²). This is (1/1²)(1/1² + 1/2² + 1/3² + ...)
Then the same for (1/2²): (1/2²)(1/1²) + (1/2²)(1/2²) + (1/2²)(1/3²) + ... giving (1/2²)(1/1² + 1/2² + ...)
Grouping those together, we get (1/1² + 1/2²)(1/1² + 1/2² + 1/3² + ...)
And so on for every 1/i² in the first term, giving (1/1² + 1/2² + 1/3²...)(1/1² + 1/2² + 1/3² + ...) = (1/1² + 1/2² + 1/3² + ...)²
I hope that helped.
@@davidgould9431 thank you👍👍👍. It was helpful.
Thanks for yet another enjoyable math insight.
Very high quality Mathologer
Awesome video, as always. What is the music at 1:48 please ? Thanks
The legend is back!
Love the pattern in trig to explain this! Worth trying them out for sure!
You can skip a lot of calculating by adding up answers from the Leibnitz series like you would for a Pascal's Triangle. And if you start with four and us it for the numerator you will converge on π/1 instead of π/4. It just makes the formula much more pleasing to work with. You can also use different numbers for the numerator to get different multiples of pi. It depends on what multiple you are looking for. Something that I just figured out recently.
At 4:32 an expression for [( sin x ) / x ] is found, making me wonder if an integral expression can be found from the right hand side, bypassing both Feynman’s technique for finding that integral and using contour integration over the complex plane to find the result.
Quando vedo i tuoi video, devo spegnere la TV, e spegnere la radio.... Devo concentrarmi senza un minimo rumore
Your videos are great review and sometimes brand new stuff. I had seen this at some other point in time in history of mathematics, but my teacher must have skipped a few steps or I was half asleep.
Some of the things I show in this video you won't find anywhere else :)
There are lots of interesting infinite fractions for Pi, aren’t there? Is it a characteristic of a transcendental number that there are an infinite number of continued fraction representations? (While for strictly irrational algebraic numbers there is only one unique family of continued fractions?)
Watched 28 minutes video, but only thought: "why so short? I wanted more pi formulas!"
Well there is definitely more material for part 2, 3, 4, ... :)
Thank you Lord Mathologer. I would love to see how the Pi continued fraction defined by b's(3 , 6, 6, 6 ...) and a's(1^3, 1^3 + 2^3, 1^3 + 2^3 + 3^3, ...) is derived. Its neat because Pi - 3 is that sum of cubes fraction which is also equivalent to an alternating sum involving (n, k). I would also be supremely interested to know how Nilankantha accelerated the 'leibniz' arctan(1) by multiplying by the square root of 3 on one side and dividing by powers of three in the sum. I just read Pi Unleashed but not able to find answers to these questions.
Pi unleashed, great book. I'll keep returning to pi in the future. Lots more amazing stuff to talk about :)
Each term of the continued fraction adds exactly one more term of the sum (via Euler's continued fraction formula)
@Mathologer : Thumbnail is mispelled, it says "infiNTE" instead of "infiNITE"
i would like to know the names of your music choices, they are nice
Music (all from the free audio library that RUclips provides to creators):
ruclips.net/user/audiolibrarymusic?nv=1
Take me to the Depth
Fresh fallen snow
English country garden
Morning mandolin
@@Mathologer Thank you very much! I will check it out.
O wow! I didn’t know Euler was one of your patrons!
did you know that if you square the Fourier series of f(x)=x between [-T,T] and equate its infinite coefficients with the coefficients of the Fourier series of f(x)=x^2 in the same interval, one gets all the identities of even powers of pi at once?
It's a special case of Parseval's identity,
This was a really great journey.
Oh, I just noticed that a few of the mobius strip geometry videos are private now when I wanted to show a friend of mine (the hypertwist video is a bit above their weight). Are they coming back? (maybe on the second channel if they're not fitting of this one?)
Euler formula, Taylor series, and of course the mysterious pi! Thanks for the video, very inspiring. once I can not follow the auto formula, I can easily go to meditation mode, thanks for the music.
Why does 1/(2×3×4) − 1/(4×5×6) + 1/(6×7×8) − ··· = (π−3)/4 ?
(One of the "beautiful equation" candidates.)
That π/4 looks like arctan(1) from Leibniz's formula, and cf. the generalised continued fraction in Brounker's formula!
PS: Also, Viète's formula is rather cute:
2/π = cos(π/4) × cos(π/8) × cos(π/16) × cos(π/32) × ···
Thanks for the channel.
Dis is pure beauty ♥️♥️😍😍 much love math mathologer ♥️♥️♥️♥️
Could you please do a video on Pade approximants?
I'll actually do something related to Pade approximants in my next video ;)
GORGEOUS VIDEO 👏🏻👏🏻👏🏻👏🏻👏🏻
Try to find a mathematical branch where you don't find a speck of either Euler's it Gauss' contribution. I'll wait.
How much did they do graph theory? I don't remember hearing any results by either of them. Then again, I'm not sure whether graph theory really existed back then...
Grade 0 math.
@@tetraedri_1834 Euler wrote what is now considered to be the first paper on graph theory and introduced the notion of Eulerian paths and circuits.
It would be fun to see your interpretation of the BBP hexadecimal digit of Pi extraction formulas, which seems (to me) conceptually impossible for an irrational number and also seems to hint at a subtle relationship between Pi and the number 16. Also, the basic BBP formula for Pi seems to generate better than 1 decimal digit of precision per iteration which seems amazingly fast even if each step requires quite a lot of computation. I enjoy your content, thanks.
So nice math video. Thank you so much
Wow, the equality between the partial sums and the partial fractions kinds blows my mind.
omg the music you use to introduce the chapters always remind me of That Chapter, so I expect people to die or to disappear... Nice video though!
Perhaps you could do some videos on statistics? Like Student's t-test, or Bayesian inference.
A class 11th student in India with basics clear and with bunch of curiosity can understand this.
That anarchist A on your math shirt is fire. 😄
If anyone is curious to learn more about the Weierstrass product formula which Mathologer mentions at the end of this video, you will probably want to check out the product formula for the Gamma function, which is related to the Sine product formula by the expression: Gamma(x) * Gamma(1 -x) = pi / sin(pi*x). You can learn more about the product formula for the Gamma function from the brand-new video I just posted on my channel
Question: Is it ok to differntiate and still keep the equal-sign?
Hans Enedahl it turns out to be ok, but you're quite right in questioning this seemingly obvious step. My guess is that a rigorous justification would involve (1) clearly identifying the domain of validity of the two expressions (where each is well-defined either as a finite expression or as a limit), (2) showing that the two definitions coincide over this domain, (3) showing that both expressions are differentiable over this domain. If I'm not mistaken, these properties should already force the derivatives to be equal (either as finite expressions or as limits).
Great video :) Most interesting I find the numbers on the left side in the denominator: 6, 90, 945, ... where do they come from? Is there a formula for them?
At 13:24 in the video after removing the bracket shouldn't the negative sign be distributed over the entire term what was between the brackets
Have another close look I'd say. You seem to have blinked at the wrong moment :)
Sorry, I got it later when I rewatched it.
Your videos are amazing.
Please do a video on Apéry's constant and positive odd zeta values!
What happened to the last video
I think one problem with the derivation is that we need to know if the infinite product converge before using it and that if we write sin(x) in function of its zeros, how we know that it doesnt have complex zeros that we need to add.
Great video by the way.
sir you are really awesome.How can I support you?If there is a way I shall definately do