Ramanujan's infinite root and its crazy cousins

[@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

[@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.

[@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.

[@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

[@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

[@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😢😢😢😢

[@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

[@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.

[@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

[@localtitans4166]

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

[@hsreenivas5551]

Ramanujan was amazing talented person

[@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.

[@Magicsquarepuzzles]

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

[@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.

[@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.

[@rky6756]

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

[@mhdsahil3515]

India is blessed to have such a personality..❤️

[@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

[@AnilKumar-xl2te]

Ramanujan never dies. Ramanujan lives for infinity.

[@david-melekh-ysroel]

Which is -1/12

[@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.

[@k0pstl939]

Couldnt it be complex? Also my naive answer was to rewrite as x^8=8, ln on both sides, take power down and divide, get lnx=ln8/8(or3ln2/8) so x=8*e^1/8 or 2*e^3/8(same number i believe)

[@janherfs3063]

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

[@simonthor7593]

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

[@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?

[@joelkositsky4263]

I think that your "power tower" needs a clarifying definition: It could be the limit of 2 different sequences: 1) Term 1 = x, Term 2 = x^x, and Term N+1 is x^(Term N). or 2) Term 1 = x, Term 2 = x^x, and Term N+1 is (Term N)^x. with the value of the power tower being the limit as N goes to Infinity. Of course if x>1, the second sequence blows up to +Infinity very quickly, while the first sequence may have a finite limit. In your example of the power tower = 8.... x = 8^(1/8) is a solution. Not sure if a complex root of this answer (e.g. [8^(1/8)*e^(i pi/4)] would also solve your tower power equation.

[@joelkositsky4263]

On further consideration, my answer posted a few minutes ago to the power tower=8 problem is incorrect. If the power tower = 2 then the correct answer is indeed x=sqrt(2), and the simple trick of replacing all the power tower, except for the first x, works. But 8^(1/8) < sqrt(2), so that can't possibly be the answer to your problem. In fact, the power tower = any real number greater than about 2.7 seems to have no solution. The simple replacement trick doesn't seem to work for this general problem. Any help out there???

[@joelkositsky4263]

After more thought I realized that the power tower does will blow up for values of x > e^(1/e). (This fact is Left as an exercise for the prof) So the maximum value the power tower can have is 'e' Am I getting there?

[@Mathologer]

definitely getting there :)

[@Vaibhavjha91]

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

[@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.

[@jamesclark6864]

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

[@imbatman6519]

🙏🏿

[@imbatman6519]

You know it was very good movie

[@darlingdarling2943]

For the …((c^2 + c)^2 + c)… equation I found a formula that seems to work for real numbers in the domain -0.75

[@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

[@chatt31]

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

[@munendrathakur9864]

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

[@assortedtea902]

😒

[@Cjendjsidj]

😒

[@deusexmaximum8930]

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

[@u.v.s.5583]

UNLIMITED POWER!!! (Mr. Sheev Palpatine)

[@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 ?

[@peyoje]

According to Mathworld when the following series converges: a^(1/a)^(1/a)^(1/a)... it converges to 1/(a^a). Changing a to 1/x you can show that if x^x^x^...=8 then x=-1/8

[@qwertyuioph]

x^x^...=8 -> x^8=8 -> x= 8^1/8

[@austinchesnut3449]

4:52 Mind blown.

[@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.

[@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!

[@kurzackd]

11:00 -- ok, so we've PROVEN Ramanujan's association with 3 the CORRECT way... However, you initially also demonstrated that the same thing can be done with 4 (sqrt[16]). So, how / what way do you CHOP the roots to eventually achieve FOUR ??

[@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.....

[@davidwilkie9551]

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

[@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.

[@hemantjoshi825]

Ur comment that the infinite series is equal to 4, i don't think is proper, coz the last number in the series will turn out to be a fraction While sir ramanujan only dealt with natural numbers. So i suggest you to check once

[@quinn7894]

0:09 Amazing pronunciation

[@adrianhdz138]

Kuin Firipusu XDXDXD hahahahhahah too funy

[@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

[@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?

[@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

[@deepakchawda5233]

From india, i am big fan of Ramanujan ji

[@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.

[@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.

[@michaeldiakakis4581]

I do not think you are being fair with Ramanujan or his argument. There is a V. strong argument in the (implicit) fact that his own numbers form a sharp progression (the natural number sequence), , so his soln is constrained, which is not the case with yours (lots of freewheeling, with that huge number in the nested radical), or so it seems to me, and I am probably about to be corrected.

[@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

[@maheshwariravichandran6866]

For the first sum if you make a program to calculate upto n terms and data type of sum is long then for n>57,sum=3.00

[@fatimajn5033]

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

[@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

[@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!

[@alexandertsiolkovski5848]

@ 1:98, but isn't there a critical difference between the red and the blue portions of the two equations? Red is always an integer but blue is fractional

[@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

[@stephenpyons3994]

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

[@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.

[@aee220phmunirabad]

Ramanujan = King of infinite series! Nobody ever worked so hard like Ramanujan till date. Mind blowing equations and solutions. World still not knows how he creates these equations!

[@moatl6945]

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

[@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.

[@matthewschad6649]

*POWER* *TOWER*

[@MegaMGstudios]

Sounds like it would be an exercise routine

[@pierreabbat6157]

In the 2-adic numbers, 2*2*2*2*... converges to 0. 2/(3-2/(3-...)) is 1 in the reals, but in the 2-adics it's 2, and in other p-adic systems it doesn't converge. To see this, take the sequence 0/1, 2/3, 6/7, 14/15, ... and subtract each from 2. You get 2/1, 4/3, 8/7, 16/15, ..., increasing powers of 2 divided by odd numbers, which sequence converges to 0 in the 2-adics.

[@f5673-t1h]

Pierre Abbat nice

[@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

[@mihininja]

wouldn't x^(x^(... = 8 not have any real solution because the x power tower can only have a solution if y (in this case =8) is in the range e^-1

[@birb1686]

Staying up past my bed time to watch maths videos

[@gunnarbehl5237]

me too

[@tomkerruish2982]

Looking back at this video (and having learned a bit more math, thanks to you and others), it struck me that 2×2×2×2ו•• does converge to 0 in the 2-adic number system.

[@kemcorpvirtualenterprises3776]

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

[@vitakyo982]

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

[@manudude02]

For the 2/(3-(2/3-(...., if we set that to equal x, then we have x=2/(3-x), then assuming x!=3, we can rearrange that to get x^2-3x+2=0 which means x is both 1 and 2.

[@firstnamelastname8684]

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

[@karangupta1825]

When I tried to sum1 + 1/2 + 1/4 + 1/6 + ..... using Cesaro method It ended me at 0 I also founded that : 1 +1/2 +1/4 +1/6 + ..... = 1 + 1/3 + 1/5 + 1/7 + 1/9 + ..... = 1 + 1/2 + 1/3 + 1/4 + ..... = 0 All the three series are equal and converge at 0 I can be wrong as well The cool thing is that I am just 13 and a big fan of your channel

[@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 :)

[@stelladavis1798]

Wouldn't the 2/(3-2/(3-2/(3-...))) converge to one with the second method? The limit as the number of times goes to infinity would be 2/2 which is one. Right?

[@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 :)

[@germanmoshes4657]

For that Ramanujan's puzzle, I think if the radical is brought to a general form of f(x)= sqrt(ax+(1+a)^2+x*sqrt(a(x+1)+(1+a)^2+(x+1)*sqrt.., it is very easy to show that f^2 (x) = ax+(1+a)^2+x*f(x+1), it can be satisfied by f(x) = a+x+1. Initially we had a=0 and x=2, which gave us 3. In the second case, a=1 and x=2 satisfies all the values in the radical, so f(2) = 4

[@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

[@fmakofmako]

wow I learned a lot between this video and last. You replied to my comment previously saying it was a great answer and I was waiting restlessly for validation/clarification. I'm really happy with the insight that it matters how you do it. I replied to someone else the Ramanujan puzzle answer (4) and did a little playing on my own to come up with a general solution to those types of set ups: let w = (a-2c)n+a^2-ac+c^2 then (n+a)^2=(n+c)(n+a+c)+w so f(n)=n(n+a)=n√(w+(n+c)(n+a+c))=n√(w+f(n+c))=n√((a-2c)n+(a^2-ac+c^2)+(n+c)√((a-2c)(n+c)+(a^2-ac+c^2)+(n+2c)√...)) On the importance of how you set up the series it should be noted for your x^x^x^...=8 problem that if you have a series f(1)=x, f(2) = x^x, f(3) = x^x^x then f(n+1)=x^f(n) so if it converges to y then x=y^(1/y) which in this case is 2^(3/8). I immediately thought that was the answer then started second guessing myself when I started messing up the order of operations, so if anyone else does the same kind of testing you have to be careful to not do (x^x)^x instead of x^(x^x) because then you'll get a divergent series of the form f(n+1) = f(n)^x. Thanks again.

[@TheAgamemnon911]

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

[@havewissmart9602]

Loooool

[@getmeoutoftheyoutubeservers]

an interesting property i noticed: at 2:59 the mathologer introduces a continued fraction, which ramanujan declares equivalent to 1. this causes the denominator to have to be equal to 2. solving for 3-x=2, we get 1 and find the same continued fraction. at about 3:18 the mathologer declares that 2 is also equal to the fraction. a similar thing applies - the denominator must be 1, 3-x=1 gives x=2, same fraction. for 3, however, something else emerges. the required denominator is 2/3 and when solving as above, the result is 2+1/3. doing the same gives 2+1/7. recursing this gives 2+1/(2^x) for a number of recurses x. the same process applied to 4 gave similar results, with the recursed pattern being 2+1/(1.5(2^x)-1). higher values and numerical extensions to the original numbers assumed to be equivalent to the continued fraction (integers, real numbers, complex numbers?, quaternions???) are left as an exercise to the reader and may be posted as responses to this comment.

[@cezarybotta1610]

x^x^x^... = 8 x^8 = 8 x = 8^(1/8). This means that if the equation x^x^x^... = 8 has a solution, then it is equal to 8^(1/8). Let's assume that the number 8^(1/8) is indeed a solution. Therefore the sequence: a(1) = 8^(1/8) a(n+1) = (8^(1/8))^(a(n)) (for n >= 1) converges to 8. Let's consider the function f(x) = x^8 - 8^x. It is obviously continuous. We have f(1) < 0 and f(7) > 0. Using the intermediate value theorem we get that there exists such a number p from the interval [1, 7] that satisfies the equation f(p) = 0. Therefore: 8^p = p^8 p^(1/p) = 8^(1/8). This means that we have a(1) = p^(1/p) < p and also for any integer k >=1 satisfying a(k) < p: a(k+1) = (8^(1/8))^(a(k)) = (p^(1/p))^(a(k)) < (p^(1/p))^p = p. Thus for any positive integer n we have a(n) < p. Hence if the limit of a(n) exists, then it is not greater that p and therefore not equal to 8. Contradiction. This finally means that the equation x^x^x^... = 8 has no solutions. Sorry for my bad English.

[@danildmitriev5884]

DISCLAIMER: this comment features a lot of words. It essentially constitutes my stream of consciousness, which was inspired by reading the comment above. I'm sorry for my complete lack of brevity :( If you want to read the most important part (from my perspective), please go to the last two paragraphs. And then, if you feel interested, work your way through the whole text. Dear TheCB, your comment inspired me to do some experimental research, and a lot of interesting things popped up. I am really grateful to you for writing this argument, it was very interesting and inspiring! I decided to write a little follow-up to it, which may serve as an inspiration to someone else to explain some odd things which I have found during my "research", and which I wasn't able to explain. At first, I read your argument, saw that it was completely true (and very neat). Then I decided to check experimentally what this magical "p" would be in case of 8^(1/8), to understand more thoroughly what is happening. It turned out to be around 1.4625, but that's beyond the point now. I decided to run similar experiments for other natural numbers, such as 2, 3, 4, and so on. Interestingly, I found that the ONLY integer number (out of first eight, to be honest) for which the sequence x^(x^(x^...))) converges to the "correct" value is just 2. For 3, 4, 5, 6, and 7, just as in case of 8, the corresponding sequences stop growing at some point, not even reaching 3. Speaking in terms of your argument, it is also quite easy to show why this fact is true. Suppose that we are solving an equation x^(x^(x^...)))=k, where k is some natural number larger than or equal to 2. Clearly, we can do the same logic as you did and presume that the only possible solution to the equation can be x=k^(1/k). Then we consider the function f(x)=x^k-k^x, and show that for k>2 we have: f(1) = 1-k < 0, while f(3) = 3^k-k^3 > 0 (can be shown by induction, with base k=3). After that your argument flows well. However, the only number for which this argument does not work is 2, since there we cannot claim that f(k-1)>0. And further, the function f(x)=x^2-2^x is less than 0 for 1

[@cezarybotta1610]

You can prove fact 2) this way: If the limit is equal to some number a, then of course x = a^(1/a). This means that x 0 our sequence converges. Therefore we have x = k^(1/k) for some number k =1 we have a(n) < k (where a(n) is defined by the equations a(1) = x and a(n+1) = x^(a(n))). Hence a(n) < k

[@danildmitriev5884]

Yeah, this does make sense. Thank you once again, then :)

[@Lightn0x]

x^8 = 8 actually has 8 different solutions. One real and 7 complex. The fact that you are only considering the real solution makes the proof incomplete, but you are right that there are no solutions :)

[@maciejp7829]

@@avraham4497 LOL noob go back to school

[@sanjaydoshi7968]

S. Ramanujam's works are airh the Institute of Mathematical Sciences, Chennai. To what use are his discoveries, theorems,. Formulas being put to ?

[@djbj1993]

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

[@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.

[@prakhar9473]

In this sea of mathematics I am a nomath.

[@mariasoduzai5751]

Same here 😂😂😂

[@yahccs1]

Another fascinating one... ad infinitum...! However, I did find this a bit absurd: The two square root expressions for different numbers starting the same way, taking out 1's and the same factors of the next number seems a bit contrived, since the bit at the end (inside the last square root sign) is always different as long as it's a finite sum, and we can't see how they differ as it goes on and on making each step look the same on both! I guess the difference is they take differing numbers of steps to reach the final value - or you might say one infinity must be bigger than the other!?! At any stage the result is impossible to compute unless you know what's in the last square root sign, and that was worked out from the original amount anyway. I struggle to see how you can raise x to infinitely many powers of x and make any sense of that, so I thought what happens with a finite number of x powers going up. If I think a = x with N powers going up and a^x = x with N+1 powers going up which is the same as x^a since it is x to the power of x with N powers going up, so then we could have a^x = x^a or x ln a = a ln x or ln (a) / a = ln(x) / x perhaps this might give an idea how x^x^x... changes as you add one more power of x - the answer looks the same (a = x seems to be the solution)? so if (x^...) = (x^...)^x then x must be 1 but how can it work? If x^x = x then x ln(x) = ln(x) which only works at x=1 and ln(x) = 0. Then 1^1^1^1.... = 1 how can it be any other value? This is really mind-boggling. Any real number raised to power 1 is itself and to power of 0 is 1 but any other power would make it go to infinity if x>1 and to 1 if x

[@mcol3]

So why is Ramanujan's sum not 4? Where's the mistake in that?

[@lukapopovic5802]

For philosophers it is 3 and 4 in the same time, matematicians when doesn't know what to they invite rules (that doesn't exist in nature)

[@fotina45]

axioms

[@lukapopovic5802]

No, axioms are different things, axioms they are not only true, they represent the only true solution. This is in the video is one of the results that are agreed by mathematicians to be the true (one and only true) solutions. Axioms are BY NATURE only true solutions, not agreed by humans

[@fotina45]

nope axioms are assertion we assume to be true for proving some other function...axioms can be non existent in real physics

[@lukapopovic5802]

We intuitively know that they are true, which is not the case with the infinite series in this video

[@lugyd1xdone195]

With the convergent series substitution also works. 1/2 + 1/4 + ... is the same as writing r/2. That gives us 1+ r/2 = r. With simple arithmetics it turns out r=2

[@tubebrocoli]

at p = 2*p, there's more numbers that solve this equation! there's 0, and any kind of infinite pretty much =)

[@tubebrocoli]

why not?

[@tubebrocoli]

Well, you definitely can't *subtract* any kind of positive infinity from both sides of an equation, that's true. This doesn't mean that inf = 2*inf isn't true though.

[@franzluggin398]

Infinity is not a number. But that doesn't mean you cannot make equations with infinity. The equation inf = inf + 1 is actually sometimes defined to be true (or rather, infinity is defined with this property in mind). inf = 2 * inf doesn't mean 1=2, because dividing by infinity is not an equivalence relation. Or is 1*0 = 2*0 false just because 1=/= 2?

[@LtLabcoat]

@Mandelbrot I think it would be more correct to say that infinity = infinity + 1 (and similarly, infinity = infitiy*2) is true, but infinity - infinity = 0 is not.

[@Icenri]

Seeing Mandelbrot arguing with brocoli was just too fractal.

[@nonexistence5135]

I don't have much of a mathematical background so I'm not sure (I just started my first year of high school), but I am pretty sure the Wau one (2/3+1/3+1...) you get it equals 2. My argument for this is that 3+1 equals four. With this in mind, you divide 3 by four and 1 by four, which gives you 3/4+1/4, which gives you one. With this in mind, you divide 2 by 1 infinitely (2/1/1/1/1/1...) which equals two. I cannot think of another way to break this up and still get a finite answer, but if I made a mistake or that can be broken up differently, please correct me.

[@FlyingSavannahs]

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

[@davidb2885]

12:29 Actually it should be ^8√8... But spontaneously you would say that for x>1 this goes instantly towards infinity. Why is that not the case?

[@flamingpaper7751]

x^x^x^x^...=8 has no solution since infinite tetration (x^x^x^...) has a range from 1/e^e (0.065...) to e (2.718...) and 8 doesn't fall in that range. Therefore no solution

[@irrelevant_noob]

Umm... What do you mean the range is limited by e? Clearly it needs to include infinity, doesn't it? o.O

[@flamingpaper7751]

@@irrelevant_noob Any input larger than e^1/e diverges to infinity, even if slowly

[@nicholasthesilly]

You're confusing domain and range.

[@avanishr]

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

[@toxications]

p=2*2*2*2*2... -> p=2*p -> p= -∞ and p= 0 and p= ∞ because 2*∞ =∞ Or am I doing something wrong here?

[@ld8336]

I think toxications is right. In the last part of the video you are invited to consider what is the building process on the expression: the function which is iterated. Here, starting from x, you take 2x an repeat, again and again, infinitely. There are 3 possible limits then: - if you start from x=0, the limit is 0. - if you start from a positive number x>0, the limit is +infinity. - if you start from a negative number x

[@EwieEnde]

Infinite and 0 are equal both are singularities

[@Chalisque]

I imagine a more formal way of describing the point: If we have a recurrence relation x_n+1 = f(x_n), which has only one stable fixed point, whose basin of attraction is the entire domain of f, except for a countable set, then that stable fixed point is the only sensible value to give to the recurrence relation itself. Something like that. Essentially with the x = 2/(3-x) thing there are two values which don't end up at 1: 2 and 3. 3 leads to x being undefined, and 2 leads to x being 2, but all other roads, as if built by the Romans, lead to 1.