Ramanujan's easiest hard infinity monster (Mathologer Masterclass)
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- čas přidán 8. 06. 2024
- In this masterclass video we'll dive into the mind of the mathematical genius Srinivasa Ramanujan. The focus will be on rediscovering one of his most beautiful identities.
00:00 Intro
02:48 How did his mind work?
09:12 What IS this?
15:11 Fantastic fraction
18:12 Impossible identity
23:38 Thanks!
This video was inspired by two 2020 blog posts by John Baez:
math.ucr.edu/home/baez/ramanu...
Here are some links to selected Mathologer videos dealing with Ramanujan's mathematics:
Numberphile v. Math: the truth about 1+2+3+...=-1/12: • Numberphile v. Math: t...
How did Ramanujan solve the STRAND puzzle? • How did Ramanujan solv...
Ramanujan's infinite root and its crazy cousins: • Ramanujan's infinite r...
Check out the article "Inequalities related to the error function" by Omran Kouba for the nitty gritties about Ramanujan's infinite fraction: arxiv.org/abs/math/0607694v1
Further discussion of the error function: tinyurl.com/mu5vywsz
Another interesting stack exchange discussion: math.stackexchange.com/questi...
Survey of the problems that Ramanujan submitted to the Journal of the Indian Mathematical Society. See page 29 for a discussion of the identity that we talk about in this video. Also of interest in the problem discussed on page 30: faculty.math.illinois.edu/~be...
This is the letter that Ramanujan sent to Hardy. Identity VII 6 is closely related to what we are talking about in this video www.qedcat.com/misc/ramanujan...
The answer to Ramanujan's challenge appeared in the February 1916 issue of the Indian Mathematical society (vol. VIII, no. 1, pp. 17-20) "Answer to Problem 541 by K.B. Madhava".
A couple of links and remarks about the "square root of the Wallis product":
Wiki page for the Wallis product: en.wikipedia.org/wiki/Wallis_... (among other things check out the discussion on the value of the derivative of the Riemann zeta function at 0 at the end of this page).
Mathologer video "Euler's infinite pi formula generator" has a proof for the Wallis product • Euler's infinite pi fo...
Discussion on stackexchange of the asymptotic behaviour of the "square root" tinyurl.com/3yxyhjmp
Also check out the discussion in A. De Morgan, "On the summation of divergent series", The Assurance Magazine, and the Journal of the Institute of Actuaries, 12 (1865), pp. 245--252.
Here is a connection to the discussion of ways of associating meaningful values to certain divergent series in the Mathologer videos on 1+2+3+ "=" -1/12:
log (the product) "=" - log 1 + log 2 - log 3 + log 4 - log 5 + log 6 - ... = log 2 - log 3 + log 4 - log 5 + ....
and the last divergent series is known to have Cesaro sum log (pi/2)^(1/2). (essentially due to Euler, I think). See also exercise 207, page 515, in Knopp's book "Theorie and Anwendung der unendlichen Reihen", 2nd edition, Springer, 1924.
Obviously, in the last part of the video, when we plug x=0 into the infinite fraction, we just go for it a la Nike: "Just do it", (or a la Ramanujan: 1+.2+3+...=-1/12). Having said that, us ending up with root pi over 2 which is exactly what we want, is really too weird a number to pop up by coincidence. As I said in the video, to really pin down why our manipulations give the right answers is tricky. For example, we need a justification for the way I arrive at the 1/1/2/3/4... fraction in the first place. Usually infinite fractions are evaluated by first turning them into a sequence of partial fractions. Then the value of the infinite fraction, if it exists, is the limit of this sequence. The partial fractions result by truncating the infinite fraction at the plus signs. For many infinite fractions you get a different sequence having the the same limit by truncating at the fractions bars instead.
A very good book on infinite fractions featuring, among many other things, the Wallis product and the error function: Sergey Khrushchev, Orthogonal polynomials and continued fractions, Cambridge University press, 2008 (p.198, has a high-level proof for our infinite fraction in x rep. of the error function.)
Bug report:
1. At some point I copied and pasted the warm-up infinite series instead of Ramanujan's infinite series.22:42
2. Almost invisible: An "(x)" is hiding in Ramanujan's hair :) at • Ramanujan's easiest ha...
T-shirt: I bought today's t-shirt many years ago. When I just looked online I could not find it anymore. However, there are many similar designs available. Just google "Paranormal distribution".
Music: Down the Valley by Muted
The infinity sign turning into two question marks animation is based on an illustration entitled "Infinitely Many Questions" by Roberto Fernandez. See page 76 of my book Eye Twisters.
Enjoy!
Burkard
Ramanujan is my favorite. He saw things in a way nobody else ever has. Maybe someday we will have another like him. I love his work it seems so natural yet is so deep.
vvhöö säß he´$ -8önn-? ?
his indiän mäFF titchäiR xD
make jökes the löck släyce v v
idk about natural 😅
i am but a simple man
i see video about Ramanujan I click
That's definitely the way to go.
Your user name reaffirms this.
My man you're old !! So ancient!
I am just amazed by how Ramanujan's mind worked.
There are people in history (and a few undoubtedly alive today) whose minds work at a freakishly different level: Archimedes, Euler, Gauss, Ramanujan, Escher, Tesla, …
@@walterrutherford8321 Indeed. Also, Stephen Hawking.
Wow just the first 2 minutes of introduction is just amazing, what a great mind Ramanujan had, and beautifully explained professor!
True that I'm figuring stuff out now and it's just like whoa I mean I get to say that it's part of my generation it's just like whoa you understand this guy figured this stuff out just one day I mean I did the same thing but but he already did it so it's not quite the same because maybe part of the collective conscious help me do it but the collective conscious wasn't there yet and this guy just fucking I don't know just pulled it out of the fucking chaos and made it order for us which is amazing and transcended and awesome and beautiful and everything that I love about the human species that I hope I absolutely hope we figure out how to save we figure out how to get along on this beautiful Earth if you sweep away the grime and the dirt the beautiful Earth is under there the green below and the blue skies above it's all there and we can just sweep it away you understand ✨
The mathologer said it exactly right this man is one of the smartest people that ever lived you could put them up with sir Isaac Newton. Two the smartest people that have ever probably existed on this earth at least that we know of My mom was smarter than me and my sister was the math genius of the family. I wish she was still here because who knows what she could have done by now she would have been 39.
No offense but she probably would have had a CZcams channel she was one of the first people that showed me how smartphone worked and how you could have CZcams on a mobile we were at a beach and we were watching afroman videos but like she had the G2 the one with the flip out keyboard she was so smart I miss her so much and I guarantee you if you think I've come up with anything my sister I walked in one time when she was doing homework as a senior in high school she was in calculus and she was doing it in her head figuring out the answers and then showing her work afterwards and I'm the only person in this world that knows about this and someday I will tell her kids how damn smart their mom was. Even if that is my life purpose I hope they seek me out. Brenden and Trenton find me someday and I'm going to tell you how awesome your mom was I promise you I'm the only person that knows your grandfather is going to tell you a bunch of bullshit I promise you my sister your mother was sublime and God damn she was smart smarter than I ever was hell she might have been able to hang with the mathologer without having to Google a whole bunch of stuff like I do. ✨
Ramanujan had many identities with exponential functions, burying Euler's famous identity.
Absolutely amazing. I did not want the video to end!
Thank you very much :)
8:30
You can still use the exponential function in the following way:
Let y = ce^f(x)
y' = f'(x)ce^f(x)
Therefore f'(x) = x and f(x) = ½x² + const
So in the case of y' = xy we have y = ce^½x²
Those three chords played at the start sound EXACTLY like the first few seconds of Babooshka by Kate Bush.
@@simonmultiverse6349Wth are you talking about? Well a few seconds later I've heard what you were talking about. Good catch!
I also like to comment like you. Sometimes, he shows very immature style but truly genius. Some methods he shows is just copy paste. I never see his own style. Undoubtedly, Ramanujan developed his own method to solve series problem. That is why it was pretty difficult to capture the then mathematician.
Am I the only one to notice the infinite sum on the last slide is wrong? It has the product of natural numbers in the denominator instead of just odd numbers. And it includes even powers of x as well.
It depends on if you're counting spaces or integers
i noticed that too
I had worked on this problem in my pre-university days. Though I eventually had to see the solution out (couldn’t solve it), it was my first deep exposure to Ramanujam’s mind and mathematical thinking. (The first time mathematical connect was obviously the introduction to limits concept).
That was when I truly understood why he is called “the man who knew infinity”. No wonder, one of the greatest son of Indian soil 🙇
You make complicated-looking maths problems sound so easy! Please keep up the great work!
Continued fractions are truly amazing and, for most people, mysterious mathematical objects. This is really a pity, because just as the natural base for logarithms is e rather than 10, and the natural measure of an angle is radians rather than degrees, the most natural representation of a real number, in a sense, is a continued fraction.
i don't think so. real numbers are not meant to be written down by their construction. they are incountably infinite, so humanity can only ever be able to write down countably many real numbers. but in application, humans don't care about the exact value. and the scientific notation does approximate numbers perfectly.
if we're going that deep, arguably the natural measure of an angle might be its cosine
@@toniokettner4821 With all due respect, I do think that you're missing the point. phi (the golden ratio) for example can be written in continued fraction form with just one number: 1. Admittedly, that has another perfectly good precise representation.
@@RobinHillyard I think that's just one example you can provide for that side of the argument. Looking at the overall picture, the scientific decimal notation is more useful as well as aesthetic to look at, in most cases, compared to continued fraction representation.
For example, multiplication using decimal notation is arguably simpler than using continued fraction representation...
@@RobinHillyard only very special numbers have a nice continued fracrion representation. mainly roots of integers
This is the most mind boggling video ever made on this equation. and definitely a masterclass. How Ramanujan thought this is simply impossible to fathom. It appears that when it comes to Maths first comes god and then comes Ramanujan.
@sarcastic_math343Neumann cannot even come close to Ramanujan
This is one of the clearest mathematical expositions I have ever heard.
Encore une vidéo prouvant le génie de Ramanujan.
Démonstration hallucinante !!!!
Bravo, comme d'habitude. Les vidéos mathématiques les plus géniales du net.
I'm just stoked that I immediately recognized the intro as Ramanujan. Not because I recognized the identity, or am familiar with the math, but simply because any time Mathologer breaks out the continued fractions, it's a one-way ticket to Ramanujan-town.
Your combination of complex theory, with a simple start, great graphics and robust thought was a true work of art here, very well done. I really appreciated the disclaimer section at the end as the needed warning of not being too glib.
Best one Ramanujan's explanation I have never seen before . Awesome MATOHLOGER .
It has been nearly a decade since I sat in a maths classroom and it wasn't until I began to watch your videos that I realized how much I miss it.
Im a math failure. I'm here for your t-shirts 😅
No one is a math failure, only people who were not taught some small but critical math rules or ideas
@@mrboombastic_69420😂bro you said 'only people...' i get u didn't mean that but 😂
If your definition of a math failure is failing in a math test then I'm too(failed in my mid term, 8th grade) , but I've graduated with math and now will be going for further studies because it's a really really good subject
On a different note, the person who commented above is at the least partially correct. Because it's always either we weren't taught the right way or our own fault for ignoring it or our studies. It's not a big deal, it's common we do sometimes neglect our studies unless you're a complete nerd. So, you were never a failure, it's just you not seeing the other way around (meaning you don't feel or think that you can turn it around)
@@1stlullaby484 What’s funny about ”only people…”? What am I missing, here? 🤔😅
You know what is really magical?
You have made your instruction and animation line up, to almost create a gamified experience. This was wonderfully engaging and explantory, well done!
Mathematical Tetris :)
It would be nice to see a follow up filling in the details that were left on the table at the end of the video. A video on the connection between continued fractions, cutting sequences, and trajectories of billiard tables could also be a fun "spiritual successor". ;)
Thank you for another amazing exposition! I had seen this identity a long time ago and always wondered how it could be derived. Totally agree that it is just a beautiful identity to look at!
Ps. I noticed that the yellow integral identity holds for negative x as well. Fascinating that the error function has this sort of continued fraction expansion.
Glad you enjoyed this explanation :)
Dear Professor, another shining explanation about a brilliant gem. Thanks a lot. Again: this is my favourite channel EVER!
Another truly amazing video! Ramanujan was amazing, and so are you!
Whoa, getting name-checked in a Mathologer video! Achievement unlocked!
Congratulations on coming up with that nice way of getting the fraction (and on unlocking another achievement level :)
For the challenge near the end related to the Wallis product, if you simply eliminate the 1 in the denominator then each factor is < 1 converging to 1, meaning the overall product is finite.
It's finite, yes, but it's not what we want! The first factor is 2/3, which is already smaller than sqrt(pi/2), and it can only get smaller from there. In fact, my experiments suggest that it converges to 0.
You could do the same for the other product too, but it apparently does not converge anyway. Funny how adding 1s changes nothing even though it seems like it should. It all comes down to how to write the product, what is the general term? That determines if the 1s should stay or not, removing them is not allowed, it changes the terms of the product, and hence the overall value.
I love how some of the transitions make the image of Ramanujan smile.
that was so corny though
21:40 Decompose both expressions as products of fractions by pairing each term in the numerator with the term below it in the numerator. On the right, you have that each fraction is > 1, so their product will always grow. On the left, however, they alternate between > 1 and < 1, so it's at least possible for it to converge.
That's it :)
If you group them in pairs though you get n*n/[(n-1)(n+1)] = n^2/(n^2-1), which is always > 1. Why does this not suggest the left fraction grows infinitely?
@@dylan7476 Exactly! I too had the same question. Can someone answer this?
@@Sameer_S_Kulkarni You can bound (2n)^2/((2n)^2 - 1) by 1+1/n^2 which can be further bounded by e^(1/n^2). This reduces the product into a sum, and since the sum of reciprocals of squares converges, you're happy :3
For the other product, you can write that as (1+1/1)(1+1/3)(1+1/5)... and now it's easy to see that 1/1+1/3+1/5... is a lower bound.
@@Sameer_S_Kulkarni there's always a mathologer video for it ;)
But I'm not sure which particular one. Have to look it up..
Fantastic video! So incredibly clear how this all is derived, step by step, from some simpler warm up expressions. Great job!!!!
So many nice reminders of maths I haven't thought about in a while. Great pedagogical approach.
Is anybody else more excited about Mathologer videos than Hollywood videos.
I can’t explain the excitement I feel when I get the notification.
Beautiful video!
What an amazing identity. A really great video as well, breaking it all down in a very digestible way, thank you!
E ceva fenomenal , unii oameni se nasc geniali , a fost ceva deosebit acest film extraordinar -Multumesc mult Maestre -Romania!
On Ramanujan's channel this video is 20 seconds long and the explanation consists of him saying "I saw that this identity must be true"
Love me some mathologer masterclasses... still waiting on that Kurosawa length Galois theory video :)
Fantastic! Amazing. The last bit left unanswered questions, as intended. Everything else was clear.
Wow, I love this mix of algebra, calculus and clever manipulations ♥️
For me the most beautiful part of the derivation was 1/1/2/3/4/...=sqrt(pi/2)
Thank you for your work 👌
Thank you so much for your wonderful and inspiring videos!
Your videos are greatly awaited and they are always worth waiting for. Your videos always generate love for Mathematics. I wish I had a teacher like you in the school days. Lots of love and respect to you. Always ❤
For real. Great way to show so many topics in this one. The diff eq. in here was great and so much more accessible. Now I get what it can do a lot more clearly.
Glad you are enjoying these videos so much :)
Wow fascinated to see an incredible math art work by the god mathematician Ramanujan, explained by another great mathematician who simplifies all math to easy understanding.
Wow, that knocked my socks off Professor Polster! Beautiful beyond belief! How can I get back to work now with this spinning in my head! What I need is some tea and some just sitting stupefied, savoring the aftertaste.
ANOTHER MASTERPIECE LETS GO!
what an incredible journey of revelations. I truly appreciate your work in showcase these amazing feats.
Beautiful 😊 what a mind Ramanujan had
The animations of the equations are so perfect to illustrate things
Amazing video mathologer as usual. I really liked the tiny bits of sneak manipulations with calculus in the video.
Glad you liked it!
Beautiful as always
One of the channels for which I will always give a thumbs up, even before watching 😅
Thanks Mathologer.. what a mind Ramanujan had..
Absolutely loved the video.
I've missed your videos for so long, good to see the king talk about another king of math..
Wow, thanks
Brilliant!! I am hooked on your math.
Please make more of such contents
Wie immer super!
17:31 Slight mistake, the sum is supposed to be x/1 + x^3/1.3 + x^5/1.3.5 and so on.
The cut-and-paste autopilot demon strikes again :(
Ramanujan is my favorite. I simply cannot begin to comprehend how his mind worked.
Wow 😮
Beautiful and very ingenious ❤
You are awesome. Your explanations are always very good. 😄
Marvelous & lovely!
It was Douglas R. Hofstadter's GEB that introduced me (and I guess many of us) to this fascinating man from India, while the intriguing man from Germany quasi introduced himself, through these nonpareil YT videos of his.
Thank you for delivering so nice content, which helps more people experience the beauty of mathematics.
Ramanujan was an amazing genius. Love from Sweden💛💙
Yes, smarter than Euler
Great video. Although wasn't this fraction in particular discovered by Laplace and proved by Jacobi. Of course, Ramanujan os a genius to have rediscovered it all by himself, but I was really hoping we'd get more about similar fractions. Digging deeper I found a book by S Khrushchev which discusses a whole theory of continued fractions like these along with great and largely unknown work done by Euler in this field. I think it can be found online as a pdf.
Send me link please I would like to read as well
@@1stlullaby484 Sorry can't provide the link, but if you search orthogonal polynomials form Euler's point of view pdf, I think you'll find it online. If not, I'll try adding the link.
The book by Khushchev is great. Also have a look at my notes in the description of this video :)
This is so astonishingly beautiful 😍
This video is a great journey 👍
Another wonderful video.
Easily one of the top three math shirts of the channel right here.
Nice video.
Ramanujan is my favourite mathematician.
Beautiful, Beautiful, Beatiful !!!!!
Definitely another interesting video. Not only maths part was interesting, even the music at the end was very soothing too.
Glad you enjoyed it!
you are the best math youtuber keep going pls
I love your videos, thank you!!!
Another great video! I always tell the class about your channel and Ramanujan
Thanks for sharing!!
@@Mathologer you're most welcome!
Pretty cool how something this crazy is understandable with just a little bit of calculus and a couple of prior results.
Honestly it blows my mind when I see these mathematicians from India generated some of the most popular integrations, limits and whatnot with sheer simplicity yet mysteriously and seamlessly embedded them either in a poem or a mantra or a prose so not only the willing one is able to synthesize the literature written but able to practically implement the math encrypted in it. Even more interesting is that these mathematicians always dedicated their discoveries to God and let the discovery have an open access for all irrespective of their background without claiming to be the founder of the said discovery; precisely why I am convinced to believe why several of their discoveries that we today are studying/ using don't bear their names.
Omg i just realized,,,uve been posting in the exact same style for 8 yrs👏👏
Too late to watch tonight. Will finish in the morning.
trulyyyy magic, brilliant!!
first time I got confused on these videos, heh
warning about baby calculus wasn't strong enough to include baby differential equations as well
I wasn't ready
always great to have your mind blown.
Great video. The most amazing thing is that the root of pi divided by 2 can be represented as the sum of two numbers. An amazing result considering it is obtained from a normal distribution. Thank you, something to think about. Thanks again for the video :).
11:00 Also; √2 is also a component, in the left side of Ramanujan’s identity; it’s just in the denominator. The left side is, basically, just: (√π*√e)/√2. 🤔
At 20.40 a leading factor 1 in the numerator has just been suppressed. Pop it back in and the product inside the red box becomes (1/1)(2/3)(4/5)…..Each term is less than 1 so the product is convergent.
That's definitely an important observation. In particular, in the Wallis product it is very important to include the seemingly superfluous 1 at the bottom to get the pairing right. However, in terms of making sense of why there is a root pi/2 hiding that new product, putting the 1s in or leaving them out does not get us to core of the matter. Have you had a chance to watch this ? czcams.com/video/YuIIjLr6vUA/video.html
Yes. A great presentation. Where I get hung up on a philosophical hook is this. Start with the sum of a geometric series with constant ratio r, -1
@@Mathologer I have been reflecting on the root pi/2 point, which is far from intuitively obvious.
The Wallis product is a consequence of Euler's formula for sin(x) expressed as an infinite product.
Euler gives sin(x) = x (1 - (x/pi)^2)(1 - (x/2pi)^2)(1 - (x/3pi)^2)........ and if x< pi this expression must certainly converge
Setting x equal to pi/2 we have sin(pi/2) = 1 = (pi/2)(1-(1/2)^2)(1-(1/4)^2)(1-(1/6)^2)....... = (pi/2) (1-(1/2))(1+(1/2)) (1-(1/4))(1+(1/4)) (1-(1/6))(1+(1/6)) .........
And this is convergent.
That simplifies to 1 = (pi/2) * (1/2)(3/2) (3/4)(5/4) (5/6)(7/6) .......
Still convergent (but the existence of terms greater than 1 means you cannot automatically conclude from this expression on its own that it necessarily converges).
It is now possible to see that the even numbers all appear twice in the denominators and the odd numbers all appear twice in the numerators (except 1 which only appears once but we can deem that solitary 1 to be 1^2 without affecting the result).
Taking square roots we have 1 = sqrt(pi/2) * (1.3.5.7........)/(2.4.6.8...)
And that is why there is a root pi/2 lurking in this infinite product.
@@MathologerI need to go back to maths more.. I have even forgotten my chain rules and product rules in calculus.
But in this Wallace equation I keep thinking you can just add 1* to the numerator or denominator as many times as you like.. and would mess up the evaluation done in pairings.,
Don't you think Ramanujan solved the two simple related functions first. Then he saw that they they could be added together to give an even more fascinating result.
So brilliant~
Always good to see a new Mathologer video! Nice shoutout to my old friend John Baez.
I've been following his blog for years :)
I think this is one of your best videos!
It was quite a drama filled with lots of "aha!" moments, but also making sure to watch out for any sneaky moves (knowing that the Wallis Product is "conditionally convergent" primed me for the big reveal that things weren't quite what they seemed with taking the square root of it).
I would say that the fantastic fractions segment was my favorite part. :)
Have not heard from you for a while :) Glad you enjoyed this video and thank you very much for your continuing help with answering questions. I also think this video worked out very well. By the looks of it, not a video that will be hugely popular. Still very much worth doing.
@@Mathologer It's unfortunate that it's not super popular!
And yes, this past academic year was quite busy. But hopefully this next year will be easier (the lie we all tell ourselves every year, right?)
You got me into continued fractions Mathologer. Now I have published work on folding continued fractions.
Folding continuous fractions. That sounds interesting :)
Long waited 😊
Just amazone, more on ramanujan pls
It's so beautiful.
What really amazes me about this, is that we found the solution working backwards having already been given the answer. How Ramanujan found this from scratch I will never be able to understand
y'know, the warm up puzzle i actually once thought of in like 8th grade in geography class cause i was bored, but i had no knowledge of calculus, so lets just say it stumped me for a while (until i aproximated and than guessed e-1 cause you know, e is pretty famous), so seeing it as a warm up puzzle in this video made me feel a bit nostalgic, so thanks i guess
Never imagined that there might be a relationship between calculus and continuous fractions🤯
Cool!
Even without baby calculus, I have watched you enough, @Mathologer, to be able to keep up with how this works -- Danke sehr für das Video!
After some bit long time ❤ waiting jusst for vedio
Nice content
This is a great piece and exposition of calculus, differential equations and continued fractions. The only thing lacking is a reason for guessing that the great Ramanujan approached the problem this way himself. Do we have even a hint of a reason that this is even remotely his own approach?
Mind blowing