Infinite tetration of rad2 = 2 [tetration]

[@a_random_girl_on_the_internet]

I’ve only found this channel an hour ago but he’s making me love math more and more every minutes

[@JUSSTLORE]

same, yet i always hated math and found it uninteresting. hes the only person that can explain math that is interesting to me.

[@JUSSTLORE]

@kevinclark3086 oh math and music are both great :D

[@mariasultana6465]

Same here 😅

[@jensberling2341]

This presentation is fantastic. Not a dull moment, and it is performed with humor. I have never seen such a beauty in demonstrating a mathematical truth. Thank you from my whole heart.

[@DrR0BERT]

For all those asking why 4 isn't the solution, when it it a solution to √2^x = x, it can be shown that each term of the sequence {a_n} that defines the infinite super power is less than 2 with a simple induction argument. To see this, define g(x) = √2^x. Note that a_(n+1) = g(a_n). The induction proof begins with a_1 = √2 < 2. Now suppose a_n < 2. We need to show that a_(n+1) < 2. First notice that a_(n+1) = √2 ^ (a_n). Also let me remind you that g(x) = √2^x is an increasing exponential function. With a_n < 2, then g(a_n) < g(2). That is, a_(n+1) < √2^2 = 2. Note, one can also show that {a_n} is increasing as an induction proof, which I will omit here as it is a straight forward proof using the fact that g(x) is an increasing function. So {a_n} is a monotonic increasing function bounded by 2. Therefore it converges. Its limit must be ≤ 2. This would rule out 4 as a possible solution. I get why he didn't include all this, as he wanted to streamline the presentation over being very meticulous an losing a good portion of the audience.

[@tomctutor]

Indeed but our host has shown that x=2 is "a" solution to the problem, not that there may be "other" solutions. We might call it the _principle_ solution if you want to be semantic. Just like the √ 4 =2 , is usually what my students write, but we all know that √ 4 ={2, -2}. BTW the first step in you inductive proof was a_1 = √2 < 2, could easily have started a_1 = √2 < 4?

[@nigerianprinceajani]

@@tomctutor sqrt is a well-defined function that is the inverse of x -> x² for x > 0 so in fact sqrt(4)=2, not sqrt(4)={2,-2}. If you teach your students that sqrt(x) has multiple solutions for a real x, you're teaching them something wrong. It is correct though that {-2,2} is the set of solutions to the equation x²-4=0. This doesn't mean that sqrt isn't well-defined, though.

[@tomctutor]

@@nigerianprinceajani sqrt(4) =/= -2 then?

[@nigerianprinceajani]

@@tomctutor Indeed.

[@tomctutor]

@@nigerianprinceajani I know you are talking about Reals, but lets extend this to complex numbers for a moment, whats the √1? =1^(1/2)=e^i(0+2nπ)/2=e^i(nπ)={-1,1} √4= 2e^i(2nπ)/2={-2,2} I know what you are saying though, if we interpret the radical √, meaning the positive root then you would be correct. But, unless you are wanting to abandon √4 ☰ 4^(1/2) and it's extension into complex field, then you can't ignore the -ve branch of the root. All that said there is nothing wrong with asserting √x ≥ 0 ( only the principle branch) as also there is nothing wrong with choosing a ≤√x ≤ b for some arbitrary choice of real interval [a,b], thus a restriction on your domain, which you need to do in some geometrical problems.

[@michelbaba3002]

your channel is very underrated Mr. Prime. I respect you more than I do my actual math teacher

[@PrimeNewtons]

Your math teacher deserves all the respect. I appreciate the compliment, though.

[@kitcutting]

The flat cap, the mesmerizing accent, the theatrics, the MATH, this guy is perfect

[@afjelidfjssaf]

I just discovered your channel from your popular tetration video! Thank you for making all these videos, I'll definetly be using them in the future for learning math

[@jimbalmer5267]

2/20/24 I have a 69 year old brain and I'm having some concerning brain issues. This topic intrigues me, I know nothing about tetration. I never saw the concept. Your video shows that you are a teacher. With that, I am certain. Thank you for this, I'll watch it over and over as well as watching related vidoes. I will understand this. I plan to beat down cognitive issues using leaning events. Thank you for pushing me toward this path. It's a good one

[@obamabingchilling]

After experimenting with tetration, the e root of e seems to be the limit of infinite tetration. Any number higher than this infinitely tetrated will grow till infinity, whereas the e root of e infinitely tetrated settles at the constant e. Very interesting stuff. Thank you Mr. Prime!

[@meraldlag4336]

The other bound is 1/(e^e) before things get unstable :)

[@Scratch3.0Tutorials]

yes, this is due to the derivative of the graph x^x or x^-x I can't remember which

[@mari-with-a-gun]

I don’t have a formal proof but I think the infinite tetration of the nth root of n always converges to n (and the max of n root n is e root of e)

[@cwldoc4958]

@mari-with-a-gun It was fairly straight forward to show that what you say (that is, that the tetration converges to n) is true for 1e) there are two solutions to the equation x^(1/x)=n^(1/n), and there exists 1

[@cwldoc4958]

@@mari-with-a-gun I believe your statement holds for 1

[@fullfungo]

For simplicity let’s denote root(2)=r. Now, by r^r^r^r^… we mean the value that the following sequence converges to: a_1 = r a_2 = r^r a_3 = r^r^r … a_{n+1} = r^(a_n). First of all, this sequence is monotonically increasing and bounded from above by 2. a_1 = r < 2. a_n < 2, so a_{n+1}=r^(a_n)

[@m.h.6470]

WAY to complicated a solution for such a simple problem. Just do the following: √2 = 2^(1/2) √2^√2 = 2^(1/2 * 2)^(1/2) = 2^1^(1/2) But 1^x is always 1, therefore √2^√2 = 2^1 = 2 It doesn't matter, how many √2 you stack on top of each other. Anything beyond the second √2 can simply be ignored, as it is just an exponent for 1.

[@eiseks3410]

That's the correct solution

[@fullfungo]

@@m.h.6470 did you seriously just say that 1/2•√2 = 1??? Please check your math before you call my method “WAY to [sic] complicated”.

[@m.h.6470]

@@fullfungo Where exactly did I say that? Please check the math presented before you say something that simply isn't true. I replaced ALL √2 with 2^(1/2), so there is no 1/2*√2 anywhere in my calculations. There is only 1/2 * 2 = 1, which you should agree is correct.

[@fullfungo]

@@m.h.6470 as you said √2 = 2^1/2 So √2^√2 = 2^(1/2•√2) which is *NOT* 2^1. For some weird reason you concluded that 1/2•√2 = (1/2•2)^1/2, but this is not correct. 1/2•√2 = 1/2•(2^1/2) which is completely different. You would know this if you weren’t so smug and just used a calculator √2^√2 = 1.6325…

[@levistepanian5341]

is the idea that the powers are raised from the right to left? if so this makes sense. Imagine this: the rightmost sqrt2 is raised to the sqrt2, then sqrt2^(sqrt2^sqrt2), keep in mind that every time, you end up raising a sqrt2 to a power less than 2, I say this because sqrt2^sqrt2 < 2 because sqrt2^n where n is less than 2 is less than 2, just logic, and we repeat this process: sqrt2^(sqrt2^(sqrt2^sqrt2)) keep in mind that the inner bracket are less than 2, and as such the outer brackets are less than 2, thus this pattern continues, and the power approaches 2 as the reiteration approaches infinity. This is beautiful.

[@marc-andredesrosiers523]

Would have been worthwhile to check that there is a limit and that it is finite. Producing one upper bound would have been valuable, as the process produces an increasing sequence. An upper bound of 16 should be producible by induction.

[@liams9047]

Earlier I made a comment because I made the classic order-of-operations-mistake. Rather than deleting I edited it to say what I am saying now in case anyone read the comment earlier.

[@magefreak9356]

Why is this the case? I see what you mean by having an upper bound and using induction, but if I solve the same way for the infinite power tower of (3)^(1/3), I get 3 as the answer. However if I plug my answer into a calculator, I don't get 3. I know that working with infinity, weird things can happen, but where in the steps does this break down and not work?

[@syamantagogoi]

An exceptional articulation on the stuff. The way it is being presented it's really awsome.Though I am a retired person (Petroleum Exploration & Production Business) with Chemistry background I have become regular viewer of this channel ever since I discovered it in KZbin.By viewing this ,I recall all those good memories of doing these sums with great efforts while in our college days.Thanks to Prime Newtons.... and we do hope that it will keep on educating us

[@panagiotisvlachos6114]

You are the reason that I'm thinking to love Mathematics!! One of your best videos!! Congratulations!!

[@tharteon1866]

The bob ross of maths. Calling Phi a flower is just poetic.

[@husseinhussam6014]

The equation (sqrt(2))^x = x has two solutions which are 2 and 4. Then why does the infinite tetration of the square root of two does not equal 4?

[@bhaskarporey3768]

In right hand side of the equation the number is 2. So according to the question we have to show the value of x which satisfy the equation. We are not here to find the possible value of that equation. I think that's what he tried to prove here.

[@PrimeNewtons]

When I make videos. I try to make it as easy to understand as possible. I know the tetration converges to 2. I just needed to show the only valid answer. It is like solving for the sides of a triangle. If you get +2 and -3, you only keep the valid answer, in this case +2. Hope this clarifies it.

[@antonvakhitov9477]

I also found out that sqrt(2) ^ 4=4 holds and I was wondering where it a step which is technically illegal. Turned out that this is the second from the end step. If W-function of something equals to W-function of something else, these somethings aren't necessarily equal. (W(x) = W(y) does not mean x=y)

[@rainerzufall42]

As I said above, W delivers 2 values in that range! W_0(y) = - ln 2 and W_1(y) = - ln 4. This implies the correct results 2 and 4. You may call 2 a principal branch result, just as for W() function, but I'd say, it's just not clever to use the W() function for real values < 0. The only value < 0, that doesn't have zero or two results, is W(- 1/e) = -1.

[@yurenchu]

​@@antonvakhitov9477 Actually, it's the other way around. W(x) = W(y) does mean x = y , but u*(e^u) = v*(e^v) does not automatically mean u = v . As a consequence, -ln(2)*e^[-ln(2)] = -ln(4)*e^[-ln(4)] does not automatically mean -ln(2) = -ln(4) Note: the lefthandside is -ln(2) * e^[-ln(2)] = = -ln(2) / e^[ln(2)] = -ln(2) / 2 while the righthandside is -ln(4) * e^[-ln(4)] = = -ln(4) / e^[ln(4)] = -ln(4) / 4 = -ln(2^2) / 4 = -2 * ln(2) / 4 = -ln(2) / 2 which matches the lefthandside.

[@Marcel-yu2fw]

Hm, but why can't x=4? It is definitely a solution to the equation "rad2^x=x" and you lost that solution, when using the W-"function". (You seem to treat it as the inverse to the function f(x)=x*e^x which isnt bijective..) If you look at the graphs y=x and y=rad2^x they clearly intersect in exactly 2 points: x=2 and x=4.To determine which solution for x is correct you'd first have to define what you mean by infinite tetration. For example, you could define it as the limit of rad2^rad2^...^rad2 (n times) as n goes to infinity. In that case 2 seems to be the correct solution (you'd have to prove that though), but there might be other ways to define it and get 4 as an answer.

[@Bradley2016_]

he also treated lnx as the inverse of e^x, negating all solutions which require ln(x) = ln(x+2πn) but thats not what he was proving, he was proving that √2 ^^ inf = 2, not 4

[@Marcel-yu2fw]

@@Bradley2016_ Well I dont think we need complex numbers here, and over the real numbers ln(x) IS the inverse to e^x, so I think thats fine. But his prove that it equals 2 just ignores 4 as a possible solution, he doesnt show, that it can't be 4. What he would have to do is show that rad2^rad2^...^rad2 (n times) approaches 2 as n goes to infinity (which seems to be the case). All his approach can show is that the answer is either 2 or 4, or indeed infinity, i forgot that possibility in my original comment: If x=rad2^^inf=inf then the equation "rad2^x=x" also holds.

[@Bradley2016_]

@@Marcel-yu2fw no, im not disagreeing with you, that IS a solution, 3b1b has demonstrated it himself, but my point is that there are infinitely many solutions, hes trying to prove ONE of which, not two, not 786.5, ONE

[@blizzardgm2797]

Why does Arsene Lupin have a KZbin channel now

[@sebastianmaccaroni6859]

@@Bradley2016_i was thinking the same, you’re able to make a divergent series appear to be a lot of different solutions with various manipulations

[@Sigma.Infinity]

Putting the W(known constant) result into a form which can be evaluated "by hand" rather than using LambertW lookups is a new idea to me. I like it!

[@HoudiniHamster]

J'aime beaucoup votre style (surtout la casquette), votre dynamisme. Et cerise sur le gâteau, votre diction est très claire et facile à suivre pour un non anglophone.

[@cwldoc4958]

For the purpose of calculating the infinite tetration for any positive real number, let a > 0, and define the sequence S(n) by S(1) = a, and S(n) = a^S(n-1) for n > 1. It turns out (to be proved below) that for 0 < a < (1/e)^(e), the sequence diverges and for a > e^(1/e), the sequence diverges to infinity. For (1/e)^(e) 0, that is, f is strictly increasing (2) f(0) = a and f(1) = a^a (3) f has one point of inflection, at xo = [ln(ln(1/a))]/ln(1/a)

[@marcolima89]

I give a like to all your videos even before I see everything, cuz I know how good it's gonna be. In this one, I understood the math, but I still opened an excel sheet to test it out and check :D

[@bonity3605]

I love ur videos!! As a high school student, I would appreciate it sm if u could do a video on Calculus introduction :) I wish all my maths lessons were like urs haha

[@childrenofkoris]

this is just crazy 😂😂😂 i learned so much from your channel, just beyond physics understanding that we can actually calculate any expression with infinite series 😂😂😂 i still cant believe this

[@alexandermorgenstern4585]

Wow! I just stumbled across your videos yesterday and i don't know why but they reminded me immediately on "Bob Ross and the Joy of Painting". I adore the way you present the topics with passion and joy. Just fascinating! You should go to high schools and teach young people maths: They'll start loving it :-). Greetings from Germany. Alex.

[@ShivamBajpai219]

I understand the use of W function. But for the youngsters who possess basic knowledge of log functions, there's an easier way. We have x ln (root 2)= ln(x) => x ln(2)^0.5=ln(x) =>x/2*ln(2)=ln(x) =>x ln(2)=2ln(x) This is a one on one function unless x itself is a function. So by observation x=2 Or rearrange it like X/ln(x) =2/ln(2) And we get the same result. Discovered this channel 5 minutes ago, but am hooked with your energy man. Great work 👌🏻

[@Notthatkindofdr]

But your final equation also has the solution x=4. That is, 4/ln(4) = 2/ln(2). So does x=4 or x=2?

[@ShivamBajpai219]

@@Notthatkindofdr Yup, you are right. I wrongly assumed it as a one on one function. My bad. X will still be equal to 2, since if the result is 4 then that's equal to 2^2. I have no idea how to explain this mathematically but the function will not be greater than or equal to 2^2.

[@pavelsolaris1901]

"...and we got this, and life is beautiful..." made my day :)

[@EIswazi]

That is so amazing! I audibly gasped at so many moments during this video and your attitude towards math is so beautiful! 9:40 and a beautiful checkmate indeed.

[@AngeloSpina-b8c]

Pay attention. W(-ln(sqrt2)) gives two results, not just one. Since -ln(sqrt2))=1/2 ln(1/2), it follows that W(-ln(sqrt2))=ln(1/2). On the other hand, since -ln(sqrt2))=1/2 ln(1/2))=1/4 ln(1/4), it also follows that W(-ln(sqrt2))=ln(1/4). Hence, actually the equation (sqrt2)^x=x has two solutions which are x=2 and x=4. Nevertheless, the infinite tetration of sqrt2 cannot be equal to 4 because the sequence a1=sqrt2, an=(sqrt2)^an-1 is increasing and an

[@samtux762]

> Tetration is useless Even more powerful operation is useful. Graham's number (really large number) was used for a practical proof of the upper bound in a problem in graph theory.

[@wrpios700]

Without showing that the limit converges, we could make the following logic true: 1+1+1+1+....=x x=1+(1+1+1+...)=x+1 x=x+1 0=1

[@sceneCatgirl]

"If you're still with me, give me a thumbs up" Me, who has not understood one thing since the very beginning: *thumbs up*

[@user-eu1zc1xm5k]

Knowledge is there for everyone to grab. But it takes a great teacher to guide you where to find it. Thank you for your outstanding style of teaching.

[@ozcanonal]

You brought me back to 40 years to my university ages. A big smile on my face. Thank you. You are very good at teaching btw. 👏👏👏

[@Cdictator]

The presentation is way more interesting than the problem itself

[@nazardovlatyan5126]

Good manipulation. Good articulation. Thank you for reasonable moving bit by bit (w/t the charcoal, but not ink). Thanks for your style & pace.

[@tzisorey]

Viewer: Asks "Cool, but what can we do with it" Viewer: Expects "It helps us plot trajectories of asteroids" or "It helps design Artificial Intelligences" or "It helps simulate quantum effects" Actual answer: It helps you turn root-2 into 2 - aint that cool!!!

[@slyfox743]

What kind of mathematical magnificence did I just witness?

[@jarskiboi]

That handwriting is so gorgeous

[@RuyVuusen]

I'm still early in my freshman year in a university, so I could be very wrong, but I wondered: could you simply after xln(sqrt(2)) = ln(x) divide both sides by x, so that you get ln(sqrt(2)) = (1/x)*ln(x), then write it ln(sqrt(2)) = ln(x^(1/x)) and by removing the natural log from both sides get sqrt(2) = x^(1/x), which is 2^(1/2) = x^(1/x)-where you get x = 2 trivially?

[@PrimeNewtons]

Yes!

[@RuyVuusen]

@@PrimeNewtons Ok, thank you!

[@cwldoc4958]

2^(1/2)=x^(1/x) does not imply x=2, because x=4 is also a solution.

[@RuyVuusen]

@@cwldoc4958 Oh yes, but since the solution can be showed to converge towards 2 and not even close to 4, x = 2 is the only valid result. The same explanation was omitted from the main video, too.

[@DmytroMiller]

Good video, thanks! The only thing, which somehow escaped from the discussion, is the fact that equation 2^(x/2)=x has two solutions: 2 and 4... which would be probably valid to address, at least, and also it somehow compromises the idea that sqrt(2) power infiniti has a fixed value, as it probably can't take 2 values the same time

[@cwldoc4958]

I enjoyed this video, which was very well done, and was fascinated by the unexpected result that the infinite tower converges to 2. However, there is a mistake at 8:22. The W(x)function as stated here is only well defined if x is restricted to nonnegative values. Otherwise, it is not defined because the map x -> xe^x is not injective. In particular, (-ln2)e^(-ln2)=(-2ln2)e(-2ln2). 2 and 4 are both solutions to the equation (sqrt(2))^x =x and this mistake incorrectly eliminates the solution x=4. As it turns out, the tetration does in fact converge to 2, as the sequence is increasing and is bounded above by 2.

[@kvanccapar8906]

sir, the thing you did there was just unbelievable at all meanings. WOW. that's mindblowing thank you so much

[@pbhowes]

You have beautiful handwriting

[@PrimeNewtons]

Thank you so much 😀

[@JossoJJossoJ]

Though it's irrelevant, I kind of want to hear him say "YEAH BOI"

[@manganeseheptoxide7825]

It is entirely possible to do without the use of lambert W. Just need to perform the same operations you did for converting -ln(√2) for converting into the W function, and then compare both sides. You end up with: ln(1/2)e^(ln(1/2)) = -ln(x)e^(-ln(x)) By comparing both sides you find: ln(1/2)=-ln(x) x=2

[@Andyg2g]

As a math teacher, the misspeak at 6:30 is so relatable. 😂

[@Galileosays]

Nice demonstration of the W-function, though @4:06 it is easy to see that x=2 since sqrt(2)^2=2.

[@user-ud1zv2yh3r]

it is relatively easy to prove that Y to the Yth power to the Yth power etc. = X has solutions for X only between 1/e and e using simple college mathematics. For anyone interested in experimenting with this, write an Excel program implementing the recursion A1 = Y,, A2 = Y ^ A1, A3 = Y ^ A2, A4 =Y ^A3, etc. This will converge for Y values between (1/e)^(e) and e^(1/e).

[@andrewyu2890]

You know what, I didn't listen your class. but I like your expression from your face so got a subscribe!

[@mertcankayamc]

How is he able to make me so interested in mathematics again. It’s almost 00:00 am and i was just watching this for fun

[@algirdasltu1389]

Damn when i realised where its going this blew my mind

[@davidcastillo4938]

Prime Newtons: flower BPRP: fish

[@sloosh2188]

your voice, the way you explain stuff, omfg. you are a perfect math teacher.

[@tomctutor]

@8:01 you can use online _Wolfram Alpha_ to solve (if you like to cheat like me). _Wα_ uses W(x) to represent the Lambert W-Function but you can't just put that into the text input field, use... Lambert function (-log(sqrt(2)) and it will return the output W(-log(√ 2) = - log(2) which must equal - log(x) from our hosts analysis so since -log(2) = -log(x) and log is bijective mapped function, x=2 😁 PS a solution to (√ 2)^x=x is not necessarily the solution to (√ 2)⥣∞ =x for every x that satisfies the former!

[@moonwatcher2001]

I love specially the use of a tradicional blackboard❤ +1 subscriptor!!!

[@PrimeNewtons]

Glad you like it!

[@chandranisahanone]

His Experience is just like papa Lambert💀💀💀. Love u from India❤❤❤❤❤❤

[@user-ue3eh1rh6r]

ok guys and sir hear me out at 3:50 x = (√2)^x x^(1/x) = √2 simply x = 2 as 2^(1/2) = √2 thats a very long procedure innit 😇

[@tomctutor]

Flawed logic, x=4 works as many have learnt here? 😤

[@novaz4768]

ok tetrations are useful for taking the limits of tetrations got it. also convenient I got recommended this after ur lamber function video

[@swarnajeetpaul7743]

After (√2)^x = x We can square both sides and get 2^x = x^2, and so by inspection we can conclude that 2 is the only answer for x

[@nigerianprinceajani]

There is an easy proof by induction that this converges (if you use the sequence a(1)=sqrt(2), a(n+1)=a(0)^a(n) it should be easy to see) and using tetr(sqrt(2),n)>tetr(sqrt(2),n-1). My problem is that if it converges, why is x=4 another valid solution to sqrt(2)^x=x? Why is the sequence tetr(sqrt(2),n) bounded by 2 when there is an algebraic way to show that this limit =4 when it should by all means exist? Are limits not unique when using tetration or is there something off about my math?

[@PrimeNewtons]

These are valid questions you raised. They are worth researching. Thank you 😊

[@ibcavid]

If you suppose sequence converges to x, then x satisfies sqrt(2) ^ x = x. But it doesn't mean that all roots of sqrt(2) ^ x = x are limits of our initial sequence. Also you showed that sequence is increasing, which isn't enough for it to have finite limit, it should also be bounded from above. It could be shown that the sequence is bounded by 2 by induction.

[@nigerianprinceajani]

@@ibcavid I was referring to an easy proof that it's increasing and bounded in my original comment. Since sqrt(2)

[@dhdusidjdjso]

I’ve seen the proof oppositely where you assume x^x^x… =2 and solve for x, which is very simple and you get sqrt2. But never seen the opposite of this. Very cool

[@user-rr1jk1ws2n]

I believe we first need to discuss whether the tetration converges rather than diverges before letting x = tetration, and that's not so trivial.

[@feierlord]

A more simple way to think of it, although it doesn't necessarily prove it, is that the sqrt(2)^(2-) < 2, where (2-) is a number slightly less than 2.

[@esmaa450]

I love thissss. Is there any chance you can explain why the infinites are acctually the equal despite the minute difference? I just feel like since this proof depends on this fact, itd be great to understanding it. Is it similar logic as why 0.9999999......... =1?

[@williamspostoronnim9845]

Lambert W is effective, but too complicated for unsophisticated. Luckily it's pretty clear from the get-go that the answer is 2.

[@francoislechampi2002]

guy, you are so charismatic !!

[@mohamedlekbir6086]

Bonjour professeur. Je suis fan des mathématiques, c est mon sport préféré. J aime beaucoup votre façon de nous enseigner cet art.

[@g.yohannes1848]

never stop learning...you bring life back

[@ericschehlmann4589]

How do you justify the step sqr(x) to the power of x equals x? The 2 x's do not have to be same number or am I missing something in the logic ? I would say it should be sqr(2) to the power of y equals x. I am not a mathmetician but I miss a logical explanation for that step.

[@ibn_klingschor]

"what does it cost to give your neighbor a dollar" none cause i can have it back after hes done using it

[@victormanuelpatosilva1637]

Great. You're the TEACHER. A huge hug, to u & thanks a lot.....

[@christopherward2748]

Wow! That was very clever!! I never really learned maths well, it is only now 1 appreciate it.

[@user-ud1zv2yh3r]

This result only works for x between 1/e and e. This is a math problem I first saw in 1965. Y to the Yth power to the Yth power etc. = 2 was the problem. Students were expected to solve this in less than a minute. I wondered if there was a solution for Y to the Yth power to the Yth power etc = 4. If you follow the trick to solve this, you get Y = sqrt(2) which converges to 2 (not 4). The trick only works when X is between 1/e and e.

[@oliviercomte7624]

1) the W function wasn’t necessary, 2) you forget to proove that the limit exists

[@KTC88]

This is a really surprising fact and an interesting proof; thank you for sharing!

[@tuliothx]

What a joy. He is to mathematics as Bob Marley was to music.

[@RexxSchneider]

What a lot of work. And what a pity you missed the second solution to (√2)^x = x, which is x=4. Lambert-W functions are multivalued. So what is the value of (√2)↑↑∞ ? Is it 2 or is it 4? If it's not 4, you might want to explain why not.

[@PrimeNewtons]

Sometimes, I leave out tangential ideas to avoid lengthening the video. I only wanted to show that it was 2. It doesn't converge to 4, but that's another long explanation.

[@robertkelly5025]

You've got to be a great father!

[@_im_also_here_]

I love this guy's energy

[@Negawisp]

4 is an answer too tho. The W function must be a multi-value function

[@tomctutor]

4 is not a solution to the infinite tower though?

[@edgar_eats_pi]

Wow! I was actually using the lambert function here, but i didnt see it that way. I just had an equation with a lambert function that you had to conpute. That's really smart!

[@surendrakverma555]

Very good lecture Sir. Thanks 🙏🙏🙏🙏🙏🙏

[@Brandon-sc3rz]

this was great im surprised that i understood every line. if this is how you teach classes your students are lucky to have you.

[@OlavRH]

i love your vids so much friend

[@kaderen8461]

you sound like some mad scientist endlessly scribbling incoherent thoughts on a chalkboard and then getting a revolutionary idea

[@ollieoniel]

He maths like a pro wrestler.

[@Universal-linguist]

Too much respect to the best ever maths teacher I ever seen❤🌹💌😁 But I also wanted to suggest something:- √2^x = x.... Until this equation, I am there with you, but after that, I think there is a simpler way to solve it, which is as below:- That (^x) will move to right hand side and will become [^(1/x)], ie √2 = x^(1/x) And √2 can be written as 2^(1/2) So,. x^(1/x)=2^(1/2), hence x=2... Please Correct me if I am wrong Yours Truly Piyush

[@olegskoropad5315]

Unbelievable! I once believed that any number greater than one in an infinite tetration should result in infinity. Your example inspired me to do a computer simulation and I saw that this series is convergent only if the result is

[@TheFrewah]

I have something for you! arxiv.org/pdf/1908.05559.pdf

[@isidorolorenzo802]

It's so nice an explanation! Really good job, my friend!

[@justinhillard62]

So.... Sqrt(2)to the power of 2 = 2, or x = 2., Does it follow that the cubed root of 3 infinitely tetrated is equal to 3? and the infinite sum of all positive natural numbers is -1/12. one infinity is not equal to another infinity.

[@takoau]

Your blackboard writing is very good

[@Phatphantuan]

This math is the inverse of 2=(√2)² , change ^2 by (√2)² , and we come to endless .

[@CampanerJRafael]

I followed every step, by two of them. 1. How do you prove that if you take out the base, the tower becomes the same? 2. I’m not familiar with the w function, so I would need another proof for that.

[@xgx899]

I would fail a student for this ``proof". This is a limit, so before computing it one should prove that it exists. Apart from this, Lambert W function is in no way needed for this simple problem .

[@antonvakhitov9477]

Infinite power tower of sqrt(2) indeed converges. However, it obviously does not converges for higher sqrt(4) or sqrt(3). It turns out that the max a such that tower of sqrt(a) still converges is like 2.087. And I have no clue what that number means :)

[@PrimeNewtons]

I am reading some materials on this. Hopefully, I get something concrete.

[@rainerzufall42]

That's very easy! 2,087065229 is the square of 1,444667861

[@rainerzufall42]

Okay, you may now ask, what's the fuzz about 1,444667861! First, I'll answer, why it has to be the square: Well, you've tried sqrt(a) for a = 4, 3, 2, 2.087. But you'd better searched for a-th root of a (like second root of 2), not the square root. Okay, now you've probably guessed, what 1,444667861 is! It's the e-th root of e. 1,444667861 = e^(1/e).

[@rainerzufall42]

Maybe one open question: Why e^(1/e)? Well, as I stated in my other comment above about W(y), for y between - 1/e and 0, it has two solutions (two branches W_0() and W_-1()) with W(-1/e) = -1. and W(0) = 0. For y < - 1/e, it has NO solution in |R. For the tower, that means, it diverges! W(1/e ln(1/e)) = W(-1 e^-1) = -1 is the last value, for which it will converge. But 1/e ln(1/e) = - ln(e^(1/e)), where the parameter to - ln() is the result of the tower (2 or 4, dependent on the branch, meaning if you take W_0() or W_-1()).

[@rainerzufall42]

en.wikipedia.org/wiki/Lambert_W_function#/media/File:Mplwp_lambert_W_branches.svg

[@thepoliticalschizophrenics3555]

If only I had found your channel earlier

[@handmiker988]

"n power n" is going to infinite for every n>1 -> inf. Tetrations are much harder of this rule and (inf tetration n^^^n) for n>1 is infinite. As 2^(1/2) > 1 so there is cannot be "inf tetr 2^(1/2) = 2". You use W-function out of its using limitations and get result formally correct but wrong.

[@RookFox]

I have no idea what he's doing but it looks cool as shit

[@lethalsub]

I'll need to learn about the Lambert W function, I just did this: limit of 2^(1/2) /\ n as n -> infinity = z 2^(1/2)^z=z log(2)/2=log(z)/z z=2 Not as rigorous as the video, I know. For the generic formula - limit of y^(1/y) /\ n as n -> infinity = y... but what if y=4? The maximum possible positive real value for y^(1/y) in the above equation is e^(1/e).

[@CoyMcBob]

Not sure why you need the Lambert function. You can easily manipulate x ln(sqrt(2)) = ln(x) to ln(2^x) = ln(x^2), or in other words 2^x = x^2. by inspection, x =2 trivially satisfies this

[@Notthatkindofdr]

So does x=4. 🙂

[@CoyMcBob]

@@Notthatkindofdr Good observation, looks like the original equation had two solutions.

[@SylvainB314]

It seems there are some manipulations yet to be proven correct. From your initial x, I can also write by applying the power of sqrt2, x=x^sqrt2, hence x=1. Seems to be a necessary condition but not sufficient!

[@SylvainB314]

If I apply the power of 2, I get, x=x^2, once again x=1 or 0 actually