Power Tower with

Grant Sanderson of 3Blue1Brown brings a power tower puzzle to Oxford for us to solve! Featuring self-similarity, convergence versus divergence and cobweb maps, the solution is not quite as straightforward as it first seems...
Find Grant's amazing channel 3Blue1Brown here: / @3blue1brown
Produced by Dr Tom Crawford at the University of Oxford.
For more maths content check out Tom's website tomrocksmaths.com/
You can also follow Tom on Facebook, Twitter and Instagram @tomrocksmaths.
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[@TomRocksMaths]

If you missed the first video in the series, you can watch me and grant take part in 'Maths Speed Dating' here: kzbin.info/www/bejne/jpmVfYmvbKdop5I

[@KramerEspinoza]

I love the way you guys presented this, showing that math is a human endeavor, that even the brightest among us can be momentary “fooled” and be humble and enthusiastic about it.

[@TomRocksMaths]

[@admiralhyperspace0015]

I just don't know. Whenever I see grant I am filled with this calm feeling. I really like his personality and vibe. I wish I was that gentle and soothing. I really do.

[@NP-zl7dz]

Damn, Grant's wearing a wedding ring, I'm too late.

[@farazriyaz9078]

Polygamy is a thing. Now go for it!

[@webgpu]

Grant's a Conservative. That's the proper way of constituting a family. (the freaks that like to behave out of Judeo-Christian ethics could all gather in their own country and disintegrate there by themselves)

[@devenhull3677]

Two of the most attractive men in math

[@hannahsrzich5110]

Bars down here bruv

[@ImaginaryMdA]

Tom and the camera man? jkjk

[@bryanfuentes1452]

love this joke

[@arlenestanton9955]

Tom is a freak

[@webgpu]

@@arlenestanton9955 you mean his piercings and tattoos?

[@garygallagher7341]

This is beautiful - the humility and collegiality on display here is inspiring.

[@varunraju1569]

This is so good! Love the collab!

[@TomRocksMaths]

Thanks Varun - glad you enjoyed it!

[@noonesperfect]

Mr. Grant also covered this topic in one of his lock-down series, good to see Mr. Tom collabed on same topic, it refreshes a lot things. Surely one of the rare topic we can get over internet so precisely :)

[@TomRocksMaths]

Glad you enjoyed it :)

[@MohitKumar-ge9hw]

I love how this puzzle links to iterated maps and dynamical systems! At first the connection isn't obvious but then Grant brings in the idea of seeing it as a function being repeated. This allows one to bring in the machinery of iterated maps and dynamical systems. As a novice of the said topics, it was nice to realise that one of the solutions is stable and the other is unstable and they coalesce in the form of a saddle node bifurcation at the value found in the video.

[@TomRocksMaths]

Grant picked a great puzzle :)

[@bonzaiii3]

Two such great persons in maths and charismatic, hope to see more vids of these guys together.

[@TomRocksMaths]

[@motivationnation999]

This is so above my level of knowledge and I thoroughly enjoyed watching you both talk through it.

[@TomRocksMaths]

as long as you had fun!

[@esb1874]

"To solve an equation, you have to know that the equation has a solution" would be fitting for both videos seeing as it is precisely one of open problems of Navier-Stokes equations!

[@jocantu]

Towers in two math channels at the same time? This is a conspiracy. (⌐■_■)

[@TomRocksMaths]

This I did not know, which other channel??

[@dekunut6416]

@@TomRocksMaths numberphile :)

[@TomRocksMaths]

Clearly we are both right on topic without even realising it...

[@shk439]

@@TomRocksMaths Also I think blackpenredpen

[@bachirblackers7299]

Blackpen. Redpen made it in a different way with the limits of convergence for both x n y values . More over still there is more to tell in this videos at least showing the case where a to the power of x crosses the y=x twice when we would be facing that bounded area between the curve n the y=x . Can we even think of integration !!!

[@AEmiliusLives]

Love these collabs with Grant :D

[@TomRocksMaths]

[@paradoxica424]

hi tom. the lower bound of the base is exp(-e), a proof can be found in a mock exam that i co-authored. BoS Trials, 2019 Mathematics Extension 1, Q13 (a)

[@TomRocksMaths]

Awesome thanks Jack!

[@LeonKayombo-yk4tc]

I like Dr Tom's glittering laughter.... Always...showing humility in mathematics! Ganbate kudasai !

[@emilianpopa1402]

13:45 "What would it mean actually?" 😅 The way he said it made me smile.

[@olivierbeaudry-ogden938]

I love the fact that 2 and 4 is possible if you look at the graoh depending where u start u can get 2 or 4

[@samayahone3497]

this is so far above my level and yet i thoroughly enjoyed the whole thing, hopefull one day i will be able to see this and understand what all your maths means lmao

[@ShromannMajumder]

I remember doing this question in high school, and the way I tackled it was. x^x^x… = 2 => log2(x^x…) = log2(2) => x^x^x… * log2(x) = 1 => 2 * log2(x) = 1 => log2(x) = 1/2 => x = 2^(1/2) That’s how I got sqrt(2) haha

[@ShromannMajumder]

so in general, k^(1/k), k > 0

[@sumchief]

this is awesome

[@em8136]

I dont know how i got here but I'm not disappointed

[@TomRocksMaths]

the YT algorithm works in mysterious ways

[@davidherrera8432]

This is beautiful, what a great video!

[@TomRocksMaths]

Glad you enjoyed it!

[@Holobrine]

9:46 That would appear to intersect at x=4, actually. But if you tried to approach it, you can’t because it’s like an unstable equilibrium. The fact remains that sqrt(2)^4 = 4.

[@TomRocksMaths]

It's true that it is an intersection point, but the iterative procedure that Grant defines will never end up there (unless you start at exactly that point).

[@Holobrine]

Tom Rocks Maths Yeah, that’s what I meant by unstable equilibrium.

[@tsvibenschar4135]

the domain max of x^y=y is the range max of y^x=x, which is point e,e^(1/e) by the first and second derivative tests (they get messy). So the domain max of the original x^y=y is e^(1/e)~1.44466, which maps to e.

[@christianorlandosilvaforer3451]

absolutly dope collabo

[@TomRocksMaths]

[@R3DD10N]

You are making maths cool

[@TomRocksMaths]

Thanks Eric!!

[@R3DD10N]

@@TomRocksMaths could I by any chance talk to you about the Navier-Stokes equations, I have some ideas about them and would like to bring them to an expert. My email is [e-mail]

[@extremestudy101]

This should be an Oxford Interview Question for students trying to apply for maths

[@ishaanjain4211]

Math men math men Math math math men men With Tom Crawford and/or Grant Sanderson

[@TomRocksMaths]

LOVE this. What's the tune?

[@ishaanjain4211]

Check out Jay Foreman's series 'Map Men'

[@asheep7797]

men

[@ishaanjain4211]

@@asheep7797 of course how could I have missed that

[@axelperezmachado3500]

Great vid! Another way to quickly find the range of convergence is to plot the y=x^y graph. That's using the fact that x^x^... = y. You will find that the only x's that satisfy that equation are in (0 ; e^(1/e)]

[@TomRocksMaths]

Nice!

[@aprendiendoC]

I tried plotting y=x^y in Wolfram Alpha but I don't know where you get the result that the only x's that satisfy that equation are in (0 ; e^(1/e)]

[@axelperezmachado3500]

@@aprendiendoC By graphing y=x^y you get a curve that is equivalent to x=y^x (which defines a real function f). We are looking for a subdomain in which this function is inyective beacuse x^x^...= y defined as a limit must be unique (a limit can`t have multiple values). It turns out that that subdomain is (0; argmax(f)] where argmax(f) turns out to be e^(1/e) by just using standard calculus tools.

[@wsw70]

Great video (as always from both of you). Would you know which program was used to plot the curves?

[@TomRocksMaths]

Desmos - free on their website

[@pinakichatterjee705]

@Tom @3B1B I found a brilliant idea to show the lower and upper bound for a for the function a^(x) there will be two bounds naturally, the lower bound is when the curve y=x is normal to the curve a^x and the upper bound is when the curve y=x is tangent to the curve a^x. Find out the rest :3

[@redpepper74]

Currently in calculus so it was cool to use it to solve this puzzle! I did f(x) = a^x *f(c)* = a^c = *c* a = c^(1/c) *f'(c)* = a^c • ln(a) = *1* c • ln(c^(1/c)) = 1 ln(c) = 1 c = e a = e^(1/e)

[@marshmelows]

"The whole property of infinity is that, you add 1 and you don't change size" Wow, so simple

[@letspass3465]

*add

[@TIO540S1]

I learned that an infinite set is defined as "any set for which there exists a one to one correspondence with a proper subset of that set."

[@marshmelows]

@@letspass3465 thanks!

[@melvinnjoroge5497]

@@TIO540S1 cool!

[@Someone-cr8cj]

i think i watched a video some time ago saying that if x^x^x^x.... converges, it converges to -W(-ln(x))/ln(x) (or something like that) where W is lambert's w function.

[@Eric4372]

Yes, that’s correct x^x^x^x^x... = LambertW(-ln(x))/-ln(x) Ther answer will only be finite for x=(0, e^(1/e)] or (0, ~1.44466786] A power tower of e^(1/e) will converge to e. Anything greater than that will diverge to infinity.

[@Someone-cr8cj]

@@Eric4372 so cool. I love this obscure part of mathematics

[@TomRocksMaths]

Agreed!

[@kaivalya931]

when he said I didnt think about this earlier and we would have to figure it on the way I thought it would some scripted moment but am genuinely happy that it took longer this has happened to me on many occasions but thats the interesting part, the brainstorming

[@joseville]

It wasn't explicitly stated, but the way in which function f was defined means x^x^x^x ^... means x^(x^(x^(x^(...)))) Does that mean the ^ operator is taken to be right associative? I think it does. By contrast, had they defined f(x) as return x**sqrt(2), then that would mean they're treating exponentiation as left associative.

[@ChronusZed]

I don't think the expression x^x^x^x^... really has a standard interpretation, but when you write it as a tower as in the video it's unambiguously right associative. Regardless, the left associative version is pretty boring. The n-th iterate just becomes x_0^(a^n), which has very easily analyzed behavior (you can end up with either 0, 1, x_0, infinity, or a simple period 2 oscillation).

[@karthikmanasali9343]

Name that website you guys used for graphs pls

[@Seba_World]

Tetration (towers) remainds me about how to imagine Graham's Number. Cheers from Poland. 🇵🇱🍻

[@TomRocksMaths]

Hello Poland :)

[@javieracruz7172]

Filooooo, los amo

[@alexanderying1558]

You can also ask yourself “To what power do you have to take sqrt2 to exceed 2 and how would you reach that value?” which would give you a contradiction, given you start at something smaller than 2.

[@TomRocksMaths]

Be careful - it doesn't quite work like that... for example, sqrt2 to the power 8 gives you 2^4.

[@alexanderying1558]

@@TomRocksMaths Yeah, but to get to 8 in the first place you would need a value bigger than 2 to start with.

[@vinujavihansith1123]

RIP bug.

[@TomRocksMaths]

I think he missed.

[@justitroyal7032]

What bug

[@UrielZyx]

So if you look at the graph at 9:37, you can see why the answer of sqrt(2) popped out for the second equation. If you start with c_0=4, you'll be at a stable state and will continue getting 4.

[@aprendiendoC]

Isn't that just a coincidence?

[@chrisclarkeML]

Does this hold true for all n-dimension numbers, that there is a number where the power tower does not explode to infinity or go to zero?

[@khajiit92]

x=4 seems to be the other intersection point of the graph, if you start your iteration at e.g. 2.5 instead of starting at 1, i think it would instead converge to 4 maybe? and then if you start iterating above 4 then it blows up. does it have to be just the first intersection? sqrt(2)^x = x has 2 solutions at x=2 and x=4

[@TomRocksMaths]

The intersection point at x=4 is unstable when analysing it using cobweb maps, which is why that 'solution' isn't valid.

[@ANunes06]

Are we just not going to talk about the fact that the graph of y= sqrt(2)^x and y= x also intersect at the value of 4? Because I really wanted you to bring that home.

[@TomRocksMaths]

It is a fixed point yes, BUT it is unstable so the iterative procedure will never end up there unless you start at exactly that point.

[@donaldhobson8873]

4 is a fixed point, of the iteration, just an unstable one.

[@TomRocksMaths]

Exactly.

[@Zoxesyr]

Grant seems a lot less uncomfortable around Tom this time.

[@justitroyal7032]

ARE YOU SURE ABOUT THAT

[@rosuav]

9:41 and 15:47 - 4 shows up as the other point where the graphs cross. Can you elaborate on whether it is, in some way, a valid solution?

[@TomRocksMaths]

As Grant mentions at the end of the video, it is an intersection point yes, BUT it is what we call 'unstable' which means that the 'power tower' process will never converge to it unless you start on exactly that point. Initiating the sequence from any other value will see you either go to 2 (as it is a stable point) or to infinity. If you look up 'cobweb maps' you should get more of an idea about what's happening.

[@rosuav]

@@TomRocksMaths Ah yep, thanks. So it's like why phi is 1.618 and not -0.618, cf kzbin.info/www/bejne/eZe6aWdrgbOCmK8

[@jasonbattermann9982]

Thanks for this question and explanation. I can see now how the region of a(x) between 2 and 4 will cobweb back to 2.

[@trojanlol]

Isn't there a mistake? (1) ln(a)*a^x = 1 (2) a^x = 1/(ln(a)) (3) x = ln(1/(ln(a))) If you apply ln to both sides on line (2), and since ln(a^x) = x*ln(a), shouldn't line (3) be: x = ln(1/(ln(a)))/ln(a) ???

[@TomRocksMaths]

There could well be, but fortunately we didn't actually use that part in the final calculation! As we show in the video the limiting value for convergence is e^(1/e)

[@justitroyal7032]

If we divide one tower into two we should still get 2 for both towers so 2^2 equals 4 right

[@TomRocksMaths]

Interesting idea, but I'm not sure it works quite like that... you have to be very careful when dealing with the concept of infinity, which is what Grant is hinting at when explaining the idea of defining a function that converges to 2.

[@justitroyal7032]

@@TomRocksMaths thanks for the reply

[@channelnamechannel]

i was wondering about this too, but the problem is that exponentiation isn’t associative, which means that you can’t move parentheses around without changing the result. in the video, they are talking about: x^(x^(x^(x^(x… which, for example, is not equal to: (x^x)^(x^x)^(x^x)… nor is it equal to: (x^x^x^x…)^(x^x^x^x…) which would be equal to 4 for x = √2. i thought i could be clever, so i tried rewriting the python code in the video starting with the function: >>> def f(x): return x**x but iterating this with √2 just puts larger and larger inputs into x**x. if you draw out the cobweb diagram for this, you see that it only converges for (real, non-negative) x < 1, with x = 1 being the only value for which x^x = x. all other negative/complex inputs seem to converge to 1??? (assuming you force the complex arguments to lie between -pi < Arg(z) < +pi)

[@justitroyal7032]

@@channelnamechannel cool

[@siddharthberera8480]

@@channelnamechannel yes exactly!!!

[@laviekolchinsky9441]

In terms of integer tetrations greater than 2, e.g. x^x^x, there is no general inverse like we have for the cases at 1,2, and infinity. It's transcendental. However, I've seen that it's possible to extend the inverse of the function (specifically the "super-root" inverse, instead of the logarithmic one) to these by using an extended form of the W-Lambert function. There's also no elementary integral, but I found the derivative of the integer case expressed in terms of the previous derivative. As far as I could tell and with my methods I couldn't find a neat universal form to express it, but the form I could express it in was quite simple and elegant. I'll paste it here if you're interested. As far as the integral is concerned, I currently have no idea. Do you have any knowledge on how to analyse these functions? Is there anything we can do with non-elementary/transcendental functions to generally work with them (maybe through series expressions?)? Thanks.

[@sergiokorochinsky49]

Nice surname.

[@laviekolchinsky9441]

@@sergiokorochinsky49 You too

[@arthurcpiazzi]

This video changed my mind. It makes me think about the nature of infinity and the quantum nature of reality. I hope to write a paper and have you both as co-authors

[@arthurcpiazzi]

Either way thanks for the inspiration. Maths is beautiful

[@shreyanshbhandari6215]

Was quite easy though

[@buIlmarket]

hello nice to meet you

[@TomRocksMaths]

You too :)

[@danielrhouck]

Okay, but at 9:37 you show that graph with a = √2, and it intersects at 2 **and at 4**. Why do we say that x^x^x^x^x^… is equal to 2 in that case instead of the other possible solution; why do we have to start at 1 instead of some other place where it converges to 4? (I assume this has something to do with 1 being the multiplicative identity and everything, but it still seems like there should be *some* way to express that the less-preferred solution is 4 or something)

[@OscarCunningham]

The Taylor expansion of sqrt(2)^(4+x) begins 4 + xlog(4) + ... . Since log(4) is greater than 1 this means that if you start with a point near 4 it will move *away* from 4 when you apply the function. Whereas if you start with a point near 2 it will move towards 2. So 2 is a stable equilibrium and 4 is unstable.

[@danielrhouck]

@@OscarCunningham Ah, that makes sesnse.

[@TomRocksMaths]

What he said ^^^^^ Thank you Oscar!

[@dr_rich_r]

We really just need to start at sqrt(2). To see what happens to x^x^x... when x=sqrt(2), we set f(x)=sqrt(2)^x. Then f(sqrt(2)) = sqrt(2)^sqrt(2). And f(answer) = sqrt(2)^sqrt(2)^sqr(2) {somewhat ambiguous, but this problem means sqrt(2)^(sqrt(2)^sqrt(2))}. Keep taking f(answer) to check convergence. But note that starting at x=1 gives the same tower, but just one iteration behind. (And it's easier to change bases but start at x=1 for a general principle.) So there is nowhere else to start if you want to check convergence of x^x^x... at sqrt(2) using iterations of the function f(x)=sqrt(2)^x.

[@talhabedir3812]

I did not know 3blue1brown was human

[@TomRocksMaths]

can confirm

[@ashwinkarthikeyan2355]

Aren't the first 2 proofs also correct? Because the statements you proved were "If [x^(x^(x^(... ] = 4, then x = sqrt(2)" and "If [x^(x^(x^(... ] = 2, then x = sqrt(2)". Both are true statements, it's just that only one of the 2 hypotheses is true.

[@TomRocksMaths]

The idea is that only one of them is possible within the setup of the problem... this is what we are alluding to when talking about cobweb diagrams and stable/unstable points.

[@williamrutherford553]

I don't think you ever answered why it works for 2 and not 4! The answer is, you assumed the seed was 1. Since the value wasn't e^(1/e) where it crossed at exactly one point, there's two solutions. When you come from below 2 (like 1 is) you converge upto 2. When you come from above 4, you converge down to 4.

[@marcospasa7535]

Actually you will diverge to infinity if you come from above 4 For example, starting with 5: Step 1: (2^(1/2))^5 ≈ 5,65 Step 2: (2^(1/2))^5,65 ≈ 7,10 Step 3: (2^(1/2))^7,10 ≈ 11,72 Step 4: (2^(1/2))^11,72 ≈ 58,17 Step 5: (2^(1/2))^58,17 ≈ 82,27

[@TomRocksMaths]

The 'solution' at 4 is an unstable point on the cobweb diagram which is why it will diverge.

[@BetaTestingUrGf]

so is it just a coincidence that the sqrt(2)^x graph intersect the y=x graph at 2 AND 4 ? 9:40

[@TomRocksMaths]

4 is a solution, but it is unstable which means the iterative process that grant talks about will never end up there unless you start at exactly that point. If you start at 4 + epsilon then you never get back to 4.

[@BetaTestingUrGf]

Thank you for taking the time to reply! Yeah i found the same question in other comments, and saw that you had replyed there :) im still not sure i understand it though 😅

[@Pedozzi]

6:39 here i understood that this had to do with the mandelbrot set and the population equation

[@TomRocksMaths]

Yes, that's another famous use of cobweb maps.

[@PritishMishra]

Video Length 16:15 Golden Ratio 🤨🤨

[@TomRocksMaths]

And I thought no-one would spot it...

[@frankiee1550]

5:43 Pov: You joined the Matrix

[@Frits34000]

They didn't explain what was wronng with the 4 answer of the inteo

[@adityakhanna113]

They did. You can't solve an equation before knowing if the solution exists or not.

[@gurkiratsingh7tha993]

if a = e^(1/e) then a^a^a^.... = e!!!

[@gabitheancient7664]

they are math mates

[@TomRocksMaths]

we sure are!

[@DestroManiak]

the felt tip pen noises kind of shredded inside my skull. great video otherwise.

[@joseville]

12:51 where does ln(a)x_0 = 1 come from??? Why gloss over the derivation?

[@danaily.2415]

We know ln(a)*a^x_0 = 1, but a^x_0 = x_0 since the graphs y = x and y = a^x intersect at that point i.e they have the same outputs. And since the output of y = x when we input x_0 is x_0, so is the one from a^x (a^x_0 = x_0). In other words, at that exact point a^x outputs its input(x_0) because it intersects y = x.

[@PritishMishra]

Best Colab present in KZbin !!

[@TomRocksMaths]

[@rtrtrtrtrt1909]

Actually it’s not a paradox. Cause you just prove that IF x^x......^x = 4 THEN x should be equal to sqrt(2). But you still have to verify that sqrt(2) is indeed a solution. It appears that is not so this equation has no solution.

[@tyronearnold6692]

There is some terrible electronic noise throughout this

[@DeadJDona]

💙💙💙 OTP 🤎

[@anaselabid1842]

At 2:32, x = Sqrt(sqrt2).

[@TomRocksMaths]

I think it's still sqrt(2) as we are taking 4 to the power of 1/4, which is the same as square rotting twice.

[@Abish_]

I understood nothing🤐