Ramanujan's infinite root and its crazy cousins

In this video I'll talk about Ramanujan's infinite roots problem, give the solution to my infinite continued fraction puzzle from a couple of week's ago, and let you in on the tricks of the trade when it comes to making sense of all those crazy infinite expressions. Featuring guest appearances by Vihart's infinite Wau fraction, the golden ratio and the Mandelbrot set.
Here is a link to a screenshot of Ramanujan’s original note about his infinite nested radical puzzle: www.qedcat.com/misc/ram_incomp...
Check out the following videos referred to in this video:
• Ramanujan: Making sens... Mathologer video on Ramanujan and 1+2+3+...=-1/12. This one also features an extended discussion of assigning values to infinite series in the standard and a couple of non-standard ways
• Infinite fractions and... Mathologer video on infinite fractions and the most irrational of all irrational numbers.
• The dark side of the M... Mathologer video on the Mandelbrot set. The second part of this one is all about a supernice way of visualising the infinite expression at the heart of this superstar.
• Wau: The Most Amazing,... Vi Hart's video on the mysterious number Wau, a must-see :)
Enjoy :)

[@LurkerPatrol5]

Can I just say that I greatly appreciate how well you pronounce Srinivasa Ramanujan's name correctly.

[@statinskill]

LP Jones I'm German. If I pronounce any foreign name like I would in Germany, more often than not I get it right. Mathologer sounds like he is one of us, and if he's not then he's Austrian. Pretty much the same difference, unless of course he's Swiss-German.

[@jaybajrangbaligamingyt7486]

श्री निवास रामानुजन

[@pleappleappleap]

Actually, his emphasis is on the wrong syllable.

[@tinu5779]

@@jaybajrangbaligamingyt7486 Thank you, that makes it easier for me to read his name correctly :)

[@kumarupendra]

kzbin.info/www/bejne/mqSkfnuFlpySbLc

[@LucasFreitas-zy2ly]

Ramanujan is a god, a guy with no graduation at all, figures out so much, and its good to remember, that he lived a very short life, imagine his legacy if he had lived more.

[@addityasinghal897]

Yeah. I wish I could meet him and learn from him

[@stevenvanhulle7242]

Additya Si - That might have been a disappointment. Lacking a rigorous mathematical background Ramanujan often made errors in his proofs, if he had a proof at all. He was all about intuition, which is why he didn't care much about rigorous proof. The most surprising thing is that his intuition was most of the time correct. Not always, though, and it's a bit strange that his errors didn't harm his reputation, rather the contrary. Probably because for a god this all might have been obvious, but his errors showed that he wasn't a god, but just the most gifted mathematician ever. Probably.

[@anushka_nd]

Agreed. He used to get dreams of solutions to the toughest of his problems, and when he’d wake up and try the method from his dream it would solve the sum! Truly magical

[@annaisabanana6848]

I think his mind completely understood mathematics, much better than pretty much anyone ever. I wouldn't be surprised if he would have had an intuition as to whether collatz conjecture was true or not, but just couldn't rigorously prove it, if he had lived till then.

[@parthrenwa]

Thinking that he is God is a material thinking. He is obviously and inarguably better than us, but the is not God. He said that he got Every thing from God and said himself as devotee of God.

[@jwm239]

..."The Man Who Knew Infinity." Get the book, see the movie!

[@Mathologer]

Got the book and seen the movie. Loved the book, hated the movie :)

[@arnavanand8037]

Great

[@clprackers7093]

I loved the movie and hated the book. :)

[@legyengeza4768]

Didn't read the book, but the movie was accurate in some ways. Like here with the infinite square roots, he always knew, always felt that his answer is right, he just could't prove it.

[@hewhomustnotbenamed5912]

Haven't read the book, have seen the movie. The movie was trash, I guess I should look into getting digital version of book.

[@dougr.2398]

Thank GOD someone can pronounce his name correctly!!

[@mishthiexplores3732]

Yeah.... Its S. RAMANIJAM. As indian names are so difficult to pronounce as hindi/sanskrit language produces maximum vibrations whichare resultee by the same vibrations in our throat to tongue...

[@NuisanceMan]

@@mishthiexplores3732 I'm surprised all Indians don't end up with sore throats!

[@aaronleperspicace1704]

Here's some extra interesting info about his name. His first name, Srinivasa, is a Sanskrit epithet of the Hindu God Vishnu. It can be split into: Sri and Nivasa. Sri is another name of Lakshmi, the Hindu goddess of fortune and the wife of Vishnu. It is a belief in Hinduism that goddess Lakshmi (Sri) resides in Vishnu's heart. Nivasa in Sanskrit means "abode", so Sri-nivasa means, "the abode of Lakshmi", ie, Vishnu. Rāmānuja can be split into: Rāma, anu and ja. Rāma is the name of the hero of the Hindu epic Ramayana, the warrior prince of Ayodhya. Anuja in Sanskrit means little brother as it is composed of anu and ja. Anu means "subsequently" and "ja" means born. Thus anuja means "subsequently born", ie, a younger brother. So Rāmānuja means "the younger brother of Rāma", ie, Lakshmana. It is a practice in South India (where Ramanujan was from) to add an "n" after Sanskrit words ending with a. Hence it became Ramanujan. Ignore the dolt who said Indian names produce vibrations and whatever.

[@go9565]

@@aaronleperspicace1704 very nicely explained. And lol at that last comment about the dolt!

[@pupperemeritus9189]

the r is not a french kind of r.other than that the pronunciation here is pretty accurate. the i is a little bit more emphasized in srinivasa. and the ni is short. va is again long. sa can be long . if r is a bit softer and its just about perfect

[@robertgumpi7235]

Wow. Very good. I learned of Ramanujan as a math-interested boy and he immediately amazed me. Now getting older I understand even more and more but it amazes me still more and more.

[@MrSupernova111]

I really love this video. I am a math enthusiasts but by no means a mathematician. I've seen lots of great youtube videos, documentaries, and films about famous math subjects but they rarely dive into the process of solving problems. You got a new sub. Thanks for interesting content in your channel!

[@Mathologer]

That's great. It's mainly for people like you who are interested in some in depth explanations that I am making these videos :)

[@veerabhadraswamy-vbs5607]

MrSupernova111

[@bongo990]

Mathologer your work is now being showcased in primary schools here in India! You have reached a huge audience.

[@santiagocalvo]

same case here! im subbing right now

[@raghavdodla1376]

Bro, then plz plz work on the infinte paradox

[@ulilulable]

Thanks for Ramanujan's actual solution in 11:26! It made thing much clearer! (And strict!) I love how what would otherwise take hundreds,or thousands of words ad-hoc can be condensed into such a clear, concise solution mathematically.

[@Mathologer]

Yes, his reasoning is very neat. However, as I said in the video, as an argument showing that the infinite nested radical is actually equal to 3 his argument is not complete :)

[@infoeducardo6915]

@@Mathologer late to the party, is the solution to unsolved infinite root question 4? n(n+3)=n√(n+5+(n+1)(n+4))

[@samueldeandrade8535]

​@@Mathologer I think it is kinda arr0gant, at least dismissive, when any person makes "rigorous observations" about statements of mathematicians from the past. So many people overrate rigor, when actually a lot of Math exists because Euler, Ramanujan and others didn't restrict themselves that much to rigorous thinking.

[@Fire_Axus]

your feelings are irrational

[@GMPranav]

He lived just 33 years, the world wasn't ready to take the math he would have found if would have lived more. It didn't deserve it.

[@gamefun404]

He just lived only 32 yrs not 33 😭

[@Fire_Axus]

@@gamefun404 stop getting so emotional

[@localtitans4166]

Salute to one of the greatest mathematicians of all time 👍👍👍

[@harrymills2770]

When I was generalizing Cauchy-Schwarz into Hölder's, I had a small Ramanujan moment, with page after page of sums. Luckily, I'm no genius, so it passed.

[@somniad]

I've got to say, compared to, for instance, Numberphile (although that's still an awesome channel), you do a really excellent job of explaining the concepts you talk about, and getting into the maths behind it deeply. Definitely one of the best channels KZbin has to offer! Keep it up ^^

[@soupisfornoobs4081]

I'd say numberphile is more about the surface of the problems, with how short their videos usually are, while mathologer talks about technique and goes deeper, with longer videos. Love them both equally, though

[@615bla]

just found this channel yesterday and watching these videos since, im so happy stuff like this exist. thank you for your work here!

[@Mathologer]

Glad all this works for you and thank you very much for saying so :)

[@rajendrasankpal3024]

@@Mathologerñq sun by 2 Malik my no fully

[@StarGazer-0807]

Ramanujan is a ideal of many people including me also.... He got funs in infinite series Hardy understood his grey matter or knowledge and took him out of india to improve his educational knoledge Many great mathematicians are not gone to neer of him He really knew the INFINITY We ar unlucky that the great Genius man could not live many days in the world😢😢😢😢

[@AnilKumar-xl2te]

Ramanujan never dies. Ramanujan lives for infinity.

[@david-melekh-ysroel]

Which is -1/12

[@chilewong1324]

Srinivasa Ramanujan died on 266h April,1920 R.I.P. What's a pity!!!!

[@yvesnyfelerph.d.8297]

What?

[@stevejob.417]

I don't know what's a pity

[@wishamahmad2719]

WHAT!!!

[@Tuberex]

Ah yes, 266h April, my birthday

[@OmarLakkis]

Find 6.

[@micagasan9814]

Everytime I watch video on theme like this one, I fall in love with Maths over and over again.

[@Mathologer]

That's great, mission accomplished than as far as the Mathologer is concerned :)

[@nathanmajaw7436]

Me too ... And also it excites me . And i want to learn more... And more

[@hsreenivas5551]

Ramanujan was amazing talented person

[@rky6756]

ramanujam the great .............

[@chatt31]

My humble thanks to Mathologer for clear and concise stepwise formulation. Much appreciated.

[@maxguichard4337]

SPOILER ALERT FOR CHALLENGE (11:55) Inspired by Ramanujan's own solution here is my mine in a similar vain. The nested radical is √(6 + 2√(7+3√(8+4√... Ramanujan decided to construct a function satisfying the equation f(n) = n√(1+f(n+1)) to solve the problem. I will ignore the initial 6 in the nested radical and just look at 2√(7+3√(8+4√... Taking 2 to be the initial n, we need a function satisfying: f(n) = n√((n+5)+f(n+1)) TBH I wasn't sure of a good approach to solving this, so I just did a bit of guessing. Squaring both sides and expanding yields: f(n)^2 = n^2(n+5)+n^2(f(n+1)) I noticed that if f(n) was a polynomial of degree 2, then the maximal degrees would be the same on both sides. The minimal degree on the RHS will be n^2, so f(n) couldn't have a constant.Therefore I guessed: f(n) = An^2 + Bn Evaluating the LHS gives: f(n)^2 = (An^2 + Bn)^2 = A^2n^4 + 2ABn^3 + B^2n^2 = n^2(An^2 + 2ABn + B^2) The RHS gives: f(n+1) = A(n+1)^2 + B(n+1) = A(n^2+2n+1) + B(n+1) = An^2 + n(2A + B) + (A+B) n^2(n+5)+n^2(f(n+1)) = n^2(n+5+An^2+n(2A + B) + (A+B)) = n^2(An^2 + n(2A + B + 1) + (A+B+5)) Finally since these are equal we can pattern math to find the coefficients A and B: A^2 = A, so A = 1 2AB = 2A + B + 1 => 2B = 3 + B => B = 3 And we can verify that B^2 = 9 = (A + B + 5) = (1 + 5 + 3) = 9 So the function f(n) = n^2 + 3n satisfies the relationship we needed. So f(n) = n√((n+5) + (n+1)√((n+6) + (n+2)√... = n(n+3) Setting n = 2: 2*5 = 10 = 2√(7 + 3√(8 + 4√... Going back to the original we have : √(6 + 2√(7+3√(8+4√... = √(6 + 10) = √16 = 4 Let me know if there are easier ways to achieve this result, I'd be interested to hear!

[@BruhGamer05]

This is such an interesting solution Max.

[@simonthor7593]

I love this video! It explains infinite series so well and I also now feel really fascinated about mathematics!

[@Magicsquarepuzzles]

I have loved Srinivasa Ramanujan early. But I love him more. What a beautiful identities and very well explained.

[@reubenemmanuel3287]

I really like your channel. Keep uploading such fascinating stuff.

[@LuisManuelLealDias]

e to the i to the e i o equals e to the wau to the tau wau wau

[@abadlydrawnsnowman1648]

is it a vi hart reference?

[@semiawesomatic6064]

Abiyyi Ramadhan yeah 😂

[@chorthithian]

Luis Dias LOOOOOOOOL

[@et496]

6 a a66662927726

[@CarelessMiss]

Abiyyi Ramadhan yes.....

[@mhdsahil3515]

India is blessed to have such a personality..❤️

[@ernestdecsi5913]

Very enjoyable performance. I say this to a pensioner living in Slovakia whose mother tongue is Hungarian. If I had listened to such lectures when I was young, we could have been colleagues. Mathematics had one big hurdle for me: the English language. As much as I loved math, I hated English so much. I also write this through a compiler.

[@jamesclark6864]

"the world only makes sense if you force it to." - Batman (Batman V Superman)

[@imbatman6519]

🙏🏿

[@imbatman6519]

You know it was very good movie

[@legendhero-eu1lc]

Thank you for the video! All of you friends are super awesome!

[@janherfs3063]

12:24 Should put a Jumpscare warning before showing that power monster :D

[@deusexmaximum8930]

2:41 "...tower of power." "Oh, that's a good band."

[@u.v.s.5583]

UNLIMITED POWER!!! (Mr. Sheev Palpatine)

[@JohnPZiagos]

Love this! Keep up the great work!

[@stephenpyons3994]

an amazing self auto-teaching, he's incredibly fantastic!

[@0621098045]

def infiniteRoot(num): if num == 100: return 0 return sqrt(1+(num+1)*infiniteRoot(num+1)) print(infiniteRoot(1)) this recursion function calculate "infinite root" and result is 3.0

[@kilimanjarobottomup1852]

awesome

[@animator897]

Ramanujan started a revolution in mathematics history.and he find multiple theorm and solve multiple unbelievable mystery of mathematics he gives many unresolved theory proof. help of ramanujans theory and formulas now we know many secret of mathematics. He was the God of mathematics. I am proud of my country which has given great people like ramanujan to this world. I am proud to be an Indian🇮🇳🇮🇳

[@fabylizzoad8869]

Ishiih

[@brucefrizzell4221]

Thank you for the subtitles .

[@davidwilkie9551]

The "Mathemagian" reveals the hidden trick of a simple sequential selection methodology, sanity returns to the discussion, thank you.

[@amberheard2869]

I am still learning mathematics to be a mathematician wish I will also do great contributions in math.

[@modern_genghis_khan0393]

Are you still learning Math, yet ?

[@EugenioDeHoyos]

"And then, maybe finally one of my favorite equations: x^(x^(x^(...))) = 8 Solve for x... Have fun." ROFL

[@shivamarya2334]

Eugenio De Hoyos 8^(1/8)

[@wolfie6175]

I got the same answer. Try putting it in the calculator . You won't get 8

[@rohitchourasia8500]

e^(log 8 /8)

[@Cherry-xc9dr]

8^(1÷8)

[@Destroier534]

I also got e^(ln(8)/8) = 8^(1/8), but the series is really finicky and won't converge to 8 if you start the series with x = 8^(1/8), converging instead to 1.4625. The equation is an unstable equilibrium point for the series, so if you want to calculate it you have to start with 8: x1 = 8 x2 = (8^(1/8))^x1 x3 = (8^(1/8))^x2 etc. In a computer this will eventually diverge and go back to 1.4625 due to floating point imprecision, but the more precise the math (more bits) the longer it will stay at 8. If you do it by hand you can see that it should stay at 8 forever, because (8^(1/8))^8 = 8.

[@AtmosMr]

Great videos and explanations. Thank you from an interested lay person. I seem to understand every stage of your video as you go through it but when I stop the video and try it out myself it all goes horribly wrong! Reminds me of my school days. Really enjoy the calm presentation.

[@Adriano0net]

Man to begin to understand each video he makes, I shall need to watch 3 or more, and so on to an infinite growing need of knowledge, if I get lucky and don't die in the process... lol Great work by the way! Continue doing this that I promise to continue to strive on to understand it! lol²

[@austinchesnut3449]

4:52 Mind blown.

[@deepakchawda5233]

From india, i am big fan of Ramanujan ji

[@abhishekmahapatra9069]

We cannot express Mr. Srinivasa Ramanujan in words he is an exceptional creation of GOD. Whether it's Ramanujan's paradox or his infinite series expression they all are exceptionally wonderful creations of Ramanujan. Not only for him but also for us, also for me MATHEMATICS is like an art unto itself. And the video is quite a good one. The way he is explaining it's wonderful. Thank you sir for such an excellent explanation. Thanks a lot. Regards

[@munendrathakur9864]

Proud To Be A Citizen Of The Land Where Ramanujan Lived And Made All That Great Things 🙏🏻❤

[@assortedtea902]

😒

[@Cjendjsidj]

😒

[@sunsathiya66]

Can you make an another infinite series of Ramanujan which is 1^3+2^3+3^3 ...... = 1/120 And nice video

[@TheFrewah]

That would be nice since I have only seen the solution which is equal to the negative area if you make a continous plot. But how did he do it?

[@quinn7894]

0:09 Amazing pronunciation

[@adrianhdz138]

Kuin Firipusu XDXDXD hahahahhahah too funy

[@thermotronica]

Cool video. That was a great explanation

[@ahmedamraniakdi2143]

Well played with the Wau number, well played.

[@matthewschad6649]

*POWER* *TOWER*

[@MegaMGstudios]

Sounds like it would be an exercise routine

[@Sauspreme]

I feel like the only way to fix these 3=4 and 1=2 situations, we'd just have to accept that we cannot make infinite expressions themselves "equal" to something. they are infinite. instead maybe we should use a new symbol in place of "equals" symbol that instead means something like, "is possibly". Maybe not "is possibly" symbol, but just something along those lines. Either way, Ramanujan's ideas are amazing!

[@learningisecstatic9348]

Thank God i found your channel. Happy to subscribe. You are great.

[@davidpaganin3361]

Many thanks, as always!

[@Borthralla]

x^x^x^x^x^x^x^x... converges only for 1/e < x < e^(1/e). A reason for this is because, of the numbers y the function converges to, the solution is x=y^(1/y), whose maximum occurs at y = e. Past e=2.71, including 8, there is no x that will converge to it. Logically, since the maximum occurred at y = e, 8^(1/8) is Less than e^(1/e) and so it makes sense that it wouldn't converge to something greater than it after infinite exponentiation.

[@kemcorpvirtualenterprises3776]

Man, where the HELL am I...!?

[@vitakyo982]

I don't know .... Somewhere ? (Tell us if not )

[@mheermance]

Yay, I made it to the end of the video and at no point did my head hurt. That's an achievement when any mathematics goes near infinity. At least for me.

[@dbell95008]

I love all of these lectures! And anything Ramanujan, of course... When I first saw the expansion and solution of the infinite series, 1+1/2+1/4+1/8 ..., while I appreciated and understood the solution process, I saw a ridiculously simple alternate: Represent the series not as decimal fractions, but in binary. Clearly, 1.1111111111111... sums to 2. Or, divide the series (again in binary notation) by 2, and you have 0.1111111111... or 1

[@firstnamelastname8684]

for anyone wondering the value at 11:57 is (i think) 4

[@firefly618]

√[6 + 2·√[7 + 3·√[8 + 4·√[9 + ... ]]] (solution below) √[6 + 2·√[7 + 3·√[8 + 4·√[9 + ... ]]] = 4 because: 4 = √16 = √[6 + 2⨉5] then take the 5 at the end and do the same: 5 = √25 = √[7 + 3⨉6] then the 6: 6 = √36 = √[8 + 4⨉7] and so on. At every step you're replacing: x = √[x²] = √[x+2 + x²−x−2] = √[(x+2) + (x−2)(x+1)]

[@jagdishramanathan4620]

i like it

[@achalpatel8655]

Done right!

[@utsavlakhera1043]

extremely well way of teaching

[@sam111880]

great video there is also a infinite continued fraction form of the talyor series so you can come up with almost anything you can represent in talyor infinite series form in a infinite continued fraction form which is a nice way to prove irrationality because a irrational number has a infinite continued fraction.

[@birb1686]

Staying up past my bed time to watch maths videos

[@gunnarbehl5237]

me too

[@shridhariyer8370]

Here you have started with 3 and then grown it into an infinite series. Can you prove it is as equal to 3 by starting from the infinite radical?

[@peterbentleyhk]

Amazing - wonderful !

[@jokerestaurant8427]

Hi Mr. Mathologer. I have a question. When you were discussing the puzzle fraction at 10:04 in the video, technically, wouldn't this be a solution too. We can not use the second method of chopping up the fraction since the blank space under the 2 can be seen as "0", hence giving 2/0, then 2/3-(2/0), and so on and so forth?

[@Hwd405]

Very informative video. It's pretty nice to know that I was totally right about the "1=2" infinite continued fraction thing ;)

[@sillysad3198]

1=2 only IF you assert TRANSITIVITY to the relation "=" that you have established between the numbers and the infinit fraction in question.

[@Hwd405]

+Silly Sad er, I wasn't saying 1=2. I was just making it clear what I was referring to.

[@TheAgamemnon911]

I actually watched the Wau-Video that was linked and now I feel like I have been trolled... hard.

[@havewissmart9602]

Loooool

[@greg55666]

That was really great.

[@Deathlupus]

Amazing... I'm just amazed. Wow.

[@francomiranda706]

pterofractals are spectating in approval

[@fatimajn5033]

I didnt expect that Ramanujan could do that also👏That's perfect😍

[@melody_florum]

What math needs is a symbol like the sigma sum symbol but for recursive/iterative functions. It’d work the same way with the initialization on bottom and limit on top but instead of adding or multiplying the result, you just insert it back into itself. This kind of notation could be useful for example expressing the Mandelbrot set as an inequality with one of these recursive functions being less than infinity

[@tanimahmed8095]

Awesome video on Mathematics... Thanks

[@BigDBrian]

x^x^x^x... = 8 x^(x^x^x^x...) = 8 x^8 = 8 x = 8^(1/8) = 1.297 roughly. However if you try partial sums it converges to about 1.4625 So let's try again x^x^x^x... = 8 (x^x^x^x...)^x = 8 8^x = 8 x = 1 obviously this raising one to itself will only end up with 1, no matter how many times you do it. So let's try again... Damnit mathologer you're not making this easy for me!

[@Mathologer]

This one is not an easy one :)

[@BigDBrian]

by just plugging in numbers using wolframalpha it seems like no value of x converges to 8. All the convergent values end up plenty below, and the divergent will tend to infinity, of course.

[@littlebigphil]

"(x^x^x^x...)^x = 8" That's invalid because (a^b)^c ≠ a^(b^c)

[@BigDBrian]

littlebigphil an attempt was made. Also, the infinite exponent is poorly defined, I just used two different ways to interpret it!

[@littlebigphil]

mrBorkD It's just that, if it were to mean that, it probably would have been written as x^(x*x*...) = 8

[@leocherry]

10:46 Да кто мы такие, чтобы судить Рамануджана! Just who we are to judge Ramanujan?

[@myxail0]

знаю, что спустя 3 года, ноо.. В математике это и есть важнее всего, истина. И критика) Тут мысль и логика намного важнее чистого авторитета

[@leocherry]

@@myxail0 я теперь так и смотрю на свой коммент :)

[@Piffsnow]

Very interesting. Thanks !

[@robertgumpi7235]

Hi and thank you for your good videos. Can you explain a bit how you make them? How do you create the graphics? How do you mix yourself into the graphics? etc.

[@moatl6945]

At 15:48 I had to re-hear it several times, if he said »they're« or »der«…

[@gometoful]

The solution to Srinivasa's second infinite square series you asked us to try is 4. I have the solution but it's long. I'm convinced that's the answer.

[@Mathologer]

Good work :)

[@dark808bb8]

I look forward to your videos on fixed point iteration

[@traso56]

i came before and after knowing how the mandelbrot set is generated wow it really helps

[@chrisg3030]

" Ramanujan rewrites [ √9 ] like this" points to √(1+(2√16)), "well, square root of 16 I can rewrite like that" points to √(1+(2√(225/4))). In that case shouldn't we simply replace the √16 in the former expression with the whole of the latter expression to get √(1+(2(√(1+(2√(225/4)))))? Which equals 3. I suggest that at whatever computable place you discontinue the expression it always equals the number you started with. 3 = √(1+(2*√(1+(3*√(1+(4*√(1+(5*7)))))))))). If we start with 2 and go 2=√4, 4=1+3, 3=1*3, 3=√9, and continue as per Ramanujan, then we have 2=√(1+(1*√(1+(2*√(1+(3*√(1+(4*√(1+(5*7)))))))))). If we start with 4 then we get 4= √(1+(3*√(1+(4*√(1+(5*7)))))). We're adding something to the beginning or left hand end, so maybe that's where we should be putting those dots.

[@prakhar9473]

In this sea of mathematics I am a nomath.

[@mariasoduzai5751]

Same here 😂😂😂

[@giovannigrassi2707]

I love your videos!

[@Mathologer]

:)

[@lalitparashar6657]

Genius ramanujan ❤

[@theo7371]

x^x^x^......=P(x) P(1)=1 If we looked at the sequence: x, x^x, (x^x)^x, ((x^x)^x)^x, ...., it converges to 1 when 0

[@ZonkoKongo]

👍🏿

[@theo7371]

+Raghu Raman Ravi Yeah, I know the second series is the right one because this is how you do it by default (without the parentheses). P(0) is undefined but the limit of the function as x gets close to zero is zero. I'll edit it out anyway since it is not necessary.

[@ZonkoKongo]

+theo konstantellos lim n->0 n^n^n... = 0 for n not being complex

[@Kosekans]

12:23 x^x^x^... = 8 ==> x^8 = 8 ==> x = 1.2968.... Can you do it like that?

[@xamzx9281]

Kosekans this doesn't work, it has no answer

[@jyotishr6093]

Wau That was one hell of a troll! ;)

[@MathZoneKH]

wow! great video!

[@infosneakr]

3:52 thank you for saying that. It was what i was thinking at that time of the video.

[@spaceshipable]

I noticed that you can pull the same trick with 1 + 1/2 + 1/4 ... by doing this: x = 1 + 1/2 + 1/4 ... x = 1 + x/2 x = 2

[@tusharphogat78]

ME : WOW YOU FOUND IT FBI: WHY DONT WE FIND GUYS LIKE THEM

[@djbj1993]

Did you just troll us into watching 5 minutes of properties of the number 1 in Vihart's video? Wau

[@bobsoup2319]

Rumanujan’s second infinite square root answer is 4 right. Because each term of value N is broken down into sqrt(n^2)=sqrt[(n+2) + (n-2)(n+1)]. And then the first term for that would then be n=4 bc n+2=6. So the whole thing = 4.

[@jamalsalameh4105]

Best channel keep up the good work. You should do a video with numberphile

[@samouhamza]

Is the fraction F in the video equal to 1 ?

[@yosuakrisnando]

I don't know. If it's yes, you have to pay respect... 😂😂😂

[@dabossbabie3605]

at 7:57, why did he want to find the sums...? Why could not he have done this? : S = 1 + 1/2 + 1/4 + 1/8 + . . . S/2 = (1 + 1/2 + 1/4 + 1/8 + . . .)/2 = (1)/2 + (1/2)/2 + (1/4)/2 + (1/8)/2 + . . . (optional step) = 1/2 + 1/4 + 1/8 + 1/16 + . . . Thus, S - 1 = S/2 S/2 = 1 S = 2 I feel that this is a more algebric way to do this. This proves that S = 2, and you don't just see the "limit" as the partial sums go to 2. Please give me a reason why his way is right.

[@corpsiecorpsie_the_original]

What are the algebraic rules for infinite sums?

[@rivareck5875]

Actually, the mathematically correct definition of the "value" of an infinite sum is the limit of the value of the consecutive sums. So he is just following the mathematical definition. The problem with saying "let 1+2+3+4+5+6+...=S" is that S here is not really defined mathematically... You do not know whether the serie converges, so S may very well not exist or be the infinite. In your case, you would first have to prove that 1+1/2+1/4+... does converge (which it does, you can prove it, it is one of Riemann's series). Then, doing your calculation may start to make sense. But if you were to come across it in maths study most of the time you would detail the calculation with more precise explanation through the definition of the limit of the sum.

[@helenabegum500]

1=1/2 + 1/2^2+1/2^3+1/2^4 So it is 2

[@subhasmitadash5064]

Great mathematician 🙏🙏🙏

[@getmeoutoftheyoutubeservers]

concerning the infinite root: i evaluated 4 using the same process as was done for the original and found it equal to sqrt(1+3sqrt(1+4sqrt(1+5sqrt(...)))). i found this interesting as it looks the same as the 3 version without the outermost (sqrt(1+2x)) layer. this works for 2 by surrounding the original with sqrt(1+x) and i presume the same as with 4 applies for higher natural numbers.

[@FlyingSavannahs]

"Let's just call it r for 'rude.'"

[@andrejunior2484]

I laughed a lot at 3:53 ... . Ofcourse I know exactly what to do... Yes ofcourse I do... yeah... (Sounds of me crying and leaving the chat) ....

[@mathunt1130]

The way forward is to use a recurrence relation. By defining a recurrence relation you can prove convergence and uniqueness of limit. That will then allow you to use whatever method you like to find the limit.

[@avanishr]

love how ramanujan's answer starts with assume f(n) = n(n+2)