Ramanujan's easiest hard infinity monster (Mathologer Masterclass)

[@jppagetoo]

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.

[@cv507]

vvhöö säß he´$ -8önn-? ? his indiän mäFF titchäiR xD make jökes the löck släyce v v

[@warguy6474]

idk about natural 😅

[@aayushkc5824]

Wow just the first 2 minutes of introduction is just amazing, what a great mind Ramanujan had, and beautifully explained professor!

[@thej3799]

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 KZbin channel she was one of the first people that showed me how smartphone worked and how you could have KZbin 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. ✨

[@sonarbangla8711]

Ramanujan had many identities with exponential functions, burying Euler's famous identity.

[@GlutesEnjoyer]

i am but a simple man i see video about Ramanujan I click

[@Mathologer]

That's definitely the way to go.

[@Hi-Phi]

Your user name reaffirms this.

[@1stlullaby484]

My man you're old !! So ancient!

[@paperclips1306]

​​@@1stlullaby484yeah. When ancient people like glutes appreciate ramanujan we like it.

[@risingsun9064]

I am just amazed by how Ramanujan's mind worked.

[@walterrutherford8321]

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, …

[@PC_Simo]

@@walterrutherford8321 Indeed. Also, Stephen Hawking.

[@harshavandu]

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 🙇

[@eliyasne9695]

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

[@simonmultiverse6349]

Those three chords played at the start sound EXACTLY like the first few seconds of Babooshka by Kate Bush.

[@maxprofane]

​@@simonmultiverse6349Wth are you talking about? Well a few seconds later I've heard what you were talking about. Good catch!

[@robaxhossain5653]

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.

[@TheMsksk]

Absolutely amazing. I did not want the video to end!

[@Mathologer]

Thank you very much :)

[@popa42]

You make complicated-looking maths problems sound so easy! Please keep up the great work!

[@guruprasadr6743]

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.

[@imtiazmohammad9548]

​@sarcastic_math343Neumann cannot even come close to Ramanujan

[@volodymyrborodin4001]

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.

[@thej3799]

It depends on if you're counting spaces or integers

[@whyre69]

i noticed that too

[@denisdeffunt2974]

Encore une vidéo prouvant le génie de Ramanujan. Démonstration hallucinante !!!! Bravo, comme d'habitude. Les vidéos mathématiques les plus géniales du net.

[@bestsolutionlaid]

Best one Ramanujan's explanation I have never seen before . Awesome MATOHLOGER .

[@1AdityaSingh]

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.

[@toniokettner4821]

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.

[@lexyeevee]

if we're going that deep, arguably the natural measure of an angle might be its cosine

[@RobinHillyard]

@@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.

[@ಭಾರತೀಯ_ನಾಗರಿಕ]

@@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...

[@toniokettner4821]

@@RobinHillyard only very special numbers have a nice continued fracrion representation. mainly roots of integers

[@jamesmstern]

This is one of the clearest mathematical expositions I have ever heard.

[@PrairieWolff]

Im a math failure. I'm here for your t-shirts 😅

[@mrboombastic_69420]

No one is a math failure, only people who were not taught some small but critical math rules or ideas

[@1stlullaby484]

​@@mrboombastic_69420😂bro you said 'only people...' i get u didn't mean that but 😂

[@1stlullaby484]

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

[@1stlullaby484]

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)

[@PC_Simo]

⁠@@1stlullaby484 What’s funny about ”only people…”? What am I missing, here? 🤔😅

[@naipaulojar7761]

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.

[@bobSeigar]

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 :)

[@idrisbalavakos]

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.

[@tracyh5751]

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". ;)

[@DrAvery-lc6bs]

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.

[@QuantumHistorian]

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.

[@Mathologer]

That's it :)

[@dylan7476]

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?

[@ಭಾರತೀಯ_ನಾಗರಿಕ]

@@dylan7476 Exactly! I too had the same question. Can someone answer this?

[@det1729]

@@ಭಾರತೀಯ_ನಾಗರಿಕ 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.

[@shoam2103]

@@ಭಾರತೀಯ_ನಾಗರಿಕ there's always a mathologer video for it ;) But I'm not sure which particular one. Have to look it up..

[@amkamath]

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.

[@Mathologer]

Glad you enjoyed this explanation :)

[@alexandrewashington6447]

Dear Professor, another shining explanation about a brilliant gem. Thanks a lot. Again: this is my favourite channel EVER!

[@phenixorbitall3917]

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 👌

[@jaredjbarnes]

Is anybody else more excited about Mathologer videos than Hollywood videos. I can’t explain the excitement I feel when I get the notification.

[@WolfdogMusic]

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"

[@frankjohnson123]

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.

[@gcewing]

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.

[@Ranoake]

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.

[@Maddy.lotus11]

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.

[@dcterr1]

Another truly amazing video! Ramanujan was amazing, and so are you!

[@LeoStein]

Whoa, getting name-checked in a Mathologer video! Achievement unlocked!

[@Mathologer]

Congratulations on coming up with that nice way of getting the fraction (and on unlocking another achievement level :)

[@m4mathematix381]

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 ❤

[@wonderbars36]

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.

[@Mathologer]

Glad you are enjoying these videos so much :)

[@grimreaper7854]

I love how some of the transitions make the image of Ramanujan smile.

[@Mathuyius]

that was so corny though

[@whobalobalo]

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.

[@160p2GHz]

So many nice reminders of maths I haven't thought about in a while. Great pedagogical approach.

[@oldschoolsoldier1634]

Love me some mathologer masterclasses... still waiting on that Kurosawa length Galois theory video :)

[@PC_Simo]

22:00 Wild guess: Wallis’s product does not explode, because every second factor is less than 1 (2/3, 4/5, and so on); whereas, in our infinite product, all factors are greater than 1 (2/1, 4/3, and so on). So, it’s a matter of alignment; just like, in the Numberphile introduction to 1+2+3+… ”=” -1/12.

[@21nck93]

17:31 Slight mistake, the sum is supposed to be x/1 + x^3/1.3 + x^5/1.3.5 and so on.

[@Mathologer]

The cut-and-paste autopilot demon strikes again :(

[@dakkumar]

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.

[@leonhardeuler7647]

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.

[@1stlullaby484]

Send me link please I would like to read as well

[@leonhardeuler7647]

@@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.

[@Mathologer]

The book by Khushchev is great. Also have a look at my notes in the description of this video :)

[@colinpountney333]

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.

[@Mathologer]

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 ? kzbin.info/www/bejne/j6asep2Cp5upi6M

[@colinpountney333]

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

[@colinpountney333]

@@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.

[@patinho5589]

@@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.,

[@MihaiDumitrumath]

E ceva fenomenal , unii oameni se nasc geniali , a fost ceva deosebit acest film extraordinar -Multumesc mult Maestre -Romania!

[@henridelagardere264]

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.

[@reynalindstrom2496]

Ramanujan was an amazing genius. Love from Sweden💛💙

[@imtiazmohammad9548]

Yes, smarter than Euler

[@darwinboor1300]

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.

[@Crazy_mathematics]

(a+xb)/(xa-1)= 1 gives x = (a+1)/(a-1) Substitute x in eqn gives a(a-1)+b(a+1)=a(a+1)-(a-b) If b=1,that implies a(a-1)+(a+1)=a(a+1)-(a-1) Therefore 998×999+1000=999×1000-998 Or √2(√2-1)+(√2+1)=√2(√2+1)-(√2-1) Or π(π-1)+(π+1)=π(π+1)-(π+1) ❤️from🇮🇳

[@panPetr0ff]

Typo at the end: (pi - 1)

[@sophiophile]

The most impressive thing about Ramanujan is that his problems are solvable using incredibly simple techniques. I can't imagine what would have happened if he had access to the breadth of knowledge that is available today.

[@Mathologer]

Actually very few of his results can be proved using relatively simple mathematics like in this video :)

[@sophiophile]

@@Mathologer I guess I have an exposure/confirmation bias, because the only ones that I can follow are impressive and (comparatively) simple- and I just skip over/am unaware of the ones that are out of my leave. A big part of it (at least in the videos you have done re: his work) is how beautifully you present it, too. I'm sure if I was staring at this on a piece of paper it wouldn't seem as simple as here.

[@prakashpanangaden1373]

Fantastic! Amazing. The last bit left unanswered questions, as intended. Everything else was clear.

[@deepasadish]

Ramanujan is my favorite. I simply cannot begin to comprehend how his mind worked.

[@Syntax753]

Thank you so much for your wonderful and inspiring videos!

[@Sajatzsiraf]

what an incredible journey of revelations. I truly appreciate your work in showcase these amazing feats.

[@ronbally2312]

One of the channels for which I will always give a thumbs up, even before watching 😅

[@replattus]

What an amazing identity. A really great video as well, breaking it all down in a very digestible way, thank you!

[@secretsorcerer]

You are awesome. Your explanations are always very good. 😄

[@Formal_Geography_Channel.]

(IMO) Mathloger is 1 of the best channels to exist for a idea, Note: Its just My own Opinion on the suggestion, Advice; "Feel free to exchange eachothers own Opinion even mine* to eachother".

[@ModusTollendoTollens]

Hello, I like to do the homework you leave in videos. I'm no differrential eq. connoisseur, but you can brute force the solution for y'= 1 + xy. Asuming I'm not dividing by 0 and that we apply fundamental theorem of calculus (FTC) the right way (i dont put constants while calculating because they will be 0 by the way we use the FTC): y' - xy = 1 y(y'/y - x) = 1 factor y y'/y - x = 1/y not dividing by 0 y'/y = x + 1/y ln(y) = (x^2)/2 + int_0^x (dt/y ) integrating, forming the ugly integral (from 0 since ln is defined in positive numbers) y = e^(x^2 /2) e^(ugly integral) get rid of ln and do sums into products. If we derive this last expression, by Leibnitz rule we get y' = (e^(ugly integral))' e^(x^2 /2) + x e^(x^2 /2) e^(ugly integral) , but we know y' = 1 + xy , i.e. y' = 1 + x e^(x^2 /2) e^(ugly integral) = (e^(ugly integral))' e^(x^2 /2) + x e^(x^2 /2) e^(ugly integral) implies 1= (e^(ugly integral))' e^(x^2 /2) then (e^(ugly integral))' = e^(-x^2 /2) integrating e^(ugly integral) = int_0^x (dt e^(-t^2 /2)) so the solution is y(x) = e^(x^2 /2) int_0^x (dt e^(-t^2 /2)).

[@Mathologer]

That's it :)

[@josephsilver9162]

Although I'd be exaggerating if I said I understood this without reviewing some portions of the explanation, I believe the explanation was extremely well done! These videos are, for the most part, very satisfying mathematically. As some have already suggested perhaps we could one day find another human with the skill set of Ramanujan - but I'm not holding my breath!

[@dhoyt902]

You got me into continued fractions Mathologer. Now I have published work on folding continued fractions.

[@Mathologer]

Folding continuous fractions. That sounds interesting :)

[@OllyJ70]

Beautiful 😊 what a mind Ramanujan had

[@zachb.4429]

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

[@spoxx1802]

I had never heard of Ramanujan before this video. What an absolute genius. His unfortunate early death set humanity back decades. Imagine how much more he could have done with even just 20 or 30 more years on his planet.

[@capteagle9]

Omg i just realized,,,uve been posting in the exact same style for 8 yrs👏👏

[@GoldenAgeMath]

Beautiful video!

[@nothinginteresting1662]

Never imagined that there might be a relationship between calculus and continuous fractions🤯

[@jamesbond_007]

Fantastic video! So incredibly clear how this all is derived, step by step, from some simpler warm up expressions. Great job!!!!

[@davidhand9721]

There's a guy on KZbin that thinks continued fractions like this explain all of physics, Gavin Wince. The numerology is strong with that one.

[@username-ur6dq]

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

[@yurisich]

Easily one of the top three math shirts of the channel right here.

[@sohampinemath1086]

I've missed your videos for so long, good to see the king talk about another king of math..

[@Mathologer]

Wow, thanks

[@soheilshirmohamadi3449]

Please make more of such contents

[@petrospaulos7736]

Great video! Thank you. I believe that there is a mistake at 23:00 . The denominators should be 1*3*5*7 and not 1*2*3....

[@drgatsis]

Another great video! I always tell the class about your channel and Ramanujan

[@Mathologer]

Thanks for sharing!!

[@drgatsis]

@@Mathologer you're most welcome!

[@dmitryramonov8902]

Very nice example of divergent series!! -1/2 ln(pi/2) = ln1 - ln2 + ln3 - ln4 +... = zeta'(0) - 2ln2 zeta(0).

[@kecskemetib]

Great video, Ramanujan never fails to amaze. Do you plan to at some point cover Ramanujan's constant exp(pi*sqrt(163)) ?

[@hymnodyhands]

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!

[@MrTiti]

bah i soooo much love your German influenced way of speaking. Makes it soooö much easier for me to understand English :)

[@DrEnzyme]

As an armchair maths fan I sometimes wonder what the world today might look like had Ramanujan lived longer. He seems like another Euler.

[@Mathologer]

We definitely missed out on lots of amazing mathematics. For example, the formula that this video is about would most likely never have been discovered if not for Ramanujan :)

[@Mod_on_exp]

Quite naturally, that comparison was made by the OG mentor Hardy himself.

[@heliy_25]

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 :).

[@umpatte0]

The animations of the equations are so perfect to illustrate things

[@manitoublack]

Too late to watch tonight. Will finish in the morning.

[@L13832]

Amazing video mathologer as usual. I really liked the tiny bits of sneak manipulations with calculus in the video.

[@Mathologer]

Glad you liked it!

[@MuffinsAPlenty]

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. :)

[@Mathologer]

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.

[@MuffinsAPlenty]

@@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?)

[@bravemortal]

At 21:32, if you put a one in the beginning. You can do (1/1)(2/3)(4/5)..... which indeed converges.

[@mrautistic2580]

11:57 “Let’s switch to Genius Mode…”. - love that quote!! 😅😅

[@PC_Simo]

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. 🤔

[@PC_Simo]

Thank You, for the heart (❤), professor Polster 😌. Also; you can, of course, also write the left side, as: √(πe)/√2. I just took it apart, as much, as possible, for the sake of clarity. 😌

[@RamanujanPi]

After Euler he his the next Hero for math's

[@stevenvanhulle7242]

On the positive side: I understand it the way Burkard explains it. Not very easy, but comprehensible. 🙂 On the negative side: I would never be able to come up with this myself 😞

[@dougr.2398]

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?

[@Tehom1]

Always good to see a new Mathologer video! Nice shoutout to my old friend John Baez.

[@Mathologer]

I've been following his blog for years :)

[@quantumgaming9180]

ANOTHER MASTERPIECE LETS GO!

[@NoNameAtAll2]

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

[@yinq5384]

Great video as always! 8:26 It's a separable equation if 1 wasn't there. dy/y=xdx and we have y=C[e^(x^2/2)-1]. 21:42 The Wallis product W = (4/3)(16/15)(36/35)...>1, at the same time W = 2(8/9)(24/25)(48/49)...

[@shohamsen8986]

for the sq root of wallis product which is the ramanujan's identity. Its clear that a subsequence converges. If you treat the product as (2/1)*(4/3)*(6/5)... then it might explod as each term is greater than 1. But Now treat it like the Wallis product that it is. 1*(2/3)*(4/5)*(6/7)... Its a product of fractions less than 1. Thus it too must be less than 1. The convergence of the product is subsequence dependent. Now you should argue which subsequence makes more sense from the integral expression it is being equated to.

[@Mathologer]

That's all true and important. However, neither way of bracketing our new product is THE correct one. Instead, what's important here is HOW things explode/implode to infinity/0. For x > 0 the fraction converges to the correct value of the expression featuring the integral. However it does so slower and slower. A close inspection of how things collapse at x=0 then shows that the correct value at this point should be root pi over 2.

[@shohamsen8986]

@@Mathologer thanks for the answer.

[@kianushmaleki]

Magical. How did Ramanujan think? How did he come up with crazy identities like this?

[@advaykumar9726]

The fact he was majorly self taught makes it more mond boggling

[@Mathologer]

... and explains why a lot of his mathematics is so different from that of other mathematicians :)

[@shashankjaiswal2266]

Definitely another interesting video. Not only maths part was interesting, even the music at the end was very soothing too.

[@Mathologer]

Glad you enjoyed it!

[@jaredgreen2363]

21:55 simple. The numerators are always larger than the denominators in the second, but smaller half the time in the first. The divergence would be removed if one extra denominator was included at each step. The product becomes equal if the square root of the next denominator is included in the full denominator.

[@Mathologer]

Neither way of bracketing is the CORRECT one. Maybe check my notes in the description of this video :)

[@My_Food_Opinion.]

(IMO) this is a amazing video to watch for a idea, Disclaimer: Its just My own Opinion on the suggestion, Advice; "Be very proud of having a Opinion on something that counts for eachother including *me.”

[@BrooksMoses]

Beautiful! With your explanation, I have to wonder if what happened was that Ramanujan was working on the continued fraction, came out with the square-root-and-integral result, and had a "aha" moment of remembering that there was an infinite sum that filled in the other half of the integral and so the sum of the two would be pleasantly simple.

[@davidwilkie9551]

Ramanujan had a particular belief in an invisible helper, teacher, "goddess", which anyone can recognise in themselves as who they talk to when solving problems, ..he was considerably more intense and definite about putting a name on the access to universal memory we share as sense-in-common. So if you see Euler's symbolic representation of imaginary real-time relative-timing functionality in a specific Singularity-point projection-drawing that you can reconstruct reverse quantization processes to demonstrate how logarithmic condensation self defines. A Unit Circle derivivation example of the e-Pi-i infinitesimal shaping function projected via Singularity-point Eternity-now 0-1-2-ness Entanglement Fusion-Fission in a picture-plane perspective of resonant unity here-now-forever vanishing-into-no-thing @.dt 1-0-infinity instantaneous trancendental cross-sectional composition of this Holographic Universe. A reorientation to the Singularity-point Apature, i-reflection orthogonality time-timing or log-antilog nodal-vibrational emitter-receiver relative-timing quantization-bonding reciprocation-recirculation potential crystallisation.., ect, etc, something like another aspect-version of consultation with a personal access to conscious awareness, sense-in-common.., and so on. Have to come back later to this..

[@lunafoxfire]

Pretty cool how something this crazy is understandable with just a little bit of calculus and a couple of prior results.

[@johndoyle2347]

Reposting and slight editing of recent mathematical ideas into one post: Split-complex numbers relate to the diagonality (like how it's expressed on Anakin's lightsaber) of ring/cylindrical singularities and to why the 6 corner/cusp singularities in dark matter must alternate. The so-called triplex numbers deal with how energy is transferred between particles and bodies and how an increase in energy also increases the apparent mass. Dual numbers relate to Euler's Identity, where the thin mass is cancelling most of the attractive and repulsive forces. The imaginary number is mass in stable particles of any conformation. In Big Bounce physics, dual numbers relate to how the attractive and repulsive forces work together to turn the matter that we normally think of into dark matter. The natural logarithm of the imaginary number is pi divided by 2 radians times i. This means that, at whatever point of stable matter other than at a singularity, the attractive or repulsive force being emitted is perpendicular to the "plane" of mass. In Big Bounce physics, this corresponds to how particles "crystalize" into stacks where a central particle is greatly pressured to break/degenerate by another particle that is in front, another behind, another to the left, another to the right, another on top, and another below. Dark matter is formed quickly afterwards. i to the i power: the "Big Bang mass", somewhat reminiscent of Swiss cheese, has dark matter flaking off, exerting a spin that mostly cancels out, leaving potential energy, and necessarily in a tangential fashion. This is closely related to what the natural logarithm of the imaginary number represents. Mediants are important to understanding the Big Crunch side of a Big Bounce event. Matter has locked up, with particles surrounding and pressuring each other. The matter gets broken up into fractions of what it was and then gets added together to form the dark matter known from our Inflationary Epoch. Sectrices are inversely related, as they deal with all stable conformations of matter being broken up, not added like the implosive "shrapnel" of mediants. Ford circles relate to mediants. Tangential circles, tethered to a line. Sectrices: the families of curves deal with black holes and dark matter. (The Fibonacci spiral deals with how dark matter is degenerated/broken up and with supernovae. The Golden spiral deals with how the normal matter, that we usually think of, degenerates, forming black holes.) The Archimedean spiral deals with dark matter spinning too fast and breaking into primordial black holes, smaller dark matter, and regular matter. The Dinostratus quadratrix deals with the laminar flow of dark matter being broken up by lingering black holes. Delanges sectrices (family of curves): dark matter has its "bubbles" force a rapid flaking off - the main driving force of the Big Bang. Ceva sectrices (family of curves): spun up dark matter breaks into primordial black holes and smaller, galactic-sized dark matter and other, typically thought of matter. Maclaurin sectrices (family of curves): older, lingering black holes, late to the party, impact and break up dark matter into galaxies. Dark matter, on the stellar scale, are broken up by supernovae. Our solar system was seeded with the heavier elements from a supernova. I'm happily surprised to figure out sectrices. Trisectrices are another thing. More complex and I don't know if I have all the curves available to use in analyzing them. But, I can see Fibonacci and Golden spirals relating to the trisectrices. The Clausen function of order 2: dark matter flakes off, impacting the Big Bang mass directly and shocking the opposite side, somewhat like concussions happen. While a spin on that central mass is exerted, all the spins from all the flaking dark matter largely cancel out. I suspect that primordial black holes are formed by this, as well. Those black holes and older black holes, that came late to the Big Bounce, work together to break up dark matter. Belows method (similar to Sylvester's Link Fan) relates to dark matter flaking off during a Big Bang event. Repetitious bisection relates to dark matter spinning so violently that it breaks, leaving smaller dark matter, primordial black holes, and other matter. Neusis construction relates to how dark matter is broken up near one of its singularities by an older black hole and to how black holes have their singularites sheared off during a Big Crunch. General relativity: 8 shapes, as dictated by the equation? 4 general shapes, but with a variation of membranous or a filament? Dark matter mostly flat, with its 6 alternating corner/cusp edge singularities. Neutrons like if a balloon had two ends, for blowing it up. Protons with aligned singularities, and electrons with just a lone cylindrical singularity? Prime numbers in polar coordinates: note the missing arms and the missing radials. Matter spiraling in, degenerating? Matter radiating out - the laminar flow of dark matter in an Inflationary Epoch? Connection to Big Bounce theory? "Operation -- Annihilate!", from the first season of the original Star Trek: was that all about dark matter and the cosmic microwave background radiation? Anakin Skywalker connection?