Power Tower with @3Blue1Brown

so tom yes i wanted to share with you uh i don't think you've heard this it's a lot of people will have heard this it's kind of a fun little puzzle about repeated exponentiation but you open this can of worms and it can get as deep as you want we'll just do things kind of quickly right now all right um but let's say i take x to the power of x to the power of x to the power of x and i go on and on and on and on right um if i just do this forever and i say i want to find some value of x so that this is equal to two okay right um how about how much you go about finding an answer to something like this um i feel like it wouldn't be helpful to move x's around yeah yeah you'd find yourself doing something like oh all we have to do is take roots we're just going to take the right side of the second side what's the route that we take oh it's the yeah it's the same that's not going to be helpful i'll just do that a bunch until you're like um so in that case i'm having an inkling it's something to do with the fact that it goes on and on and on yes is that it i feel like the fact that it's so we're assuming here that there's this infinite interest so that to me that's hinting at something that's going to play a role classic problem-solving uh strategy when you have this sort of infinite thing is to leverage self-similarity where inside the expression is a copy of itself which depends on the infinity which is kind of funny yeah yeah because the whole property of infinity is that if you add one you don't change size in some sense right you're an infinite set it means if you add one you can still have a one-to-one correspondence with yourself um so you leverage this idea and what that means is uh we have a copy of the original and say oh well we know what that has to be it's got to be 2. so x x squared equals 2. which means uh x is going to be the square root of 2 right yeah okay which seems lovely and beautiful right until i ask you a similar question which is we're gonna solve the same equation xxx we just keep going on and on i'm gonna kill that fly for you um equals four so how would you proceed here i'm going to guess that we're going to try and use the same kind of method great right so what's the next step so then it's going to be x sub 4 is equal to four and so then we're going to want to take fourth roots so that's going to give us back to root two again right so that doesn't seem to make sense because what he seems to say is if we take root two to its own power to its own power on and on and on and on that that equals two right yes but it also seems to say that that equals four so i've probably done something i'm not allowed to do right so when you're dealing with an infinite process like this uh it's important to be very clear on exactly what you mean yeah uh one way that you might think about this is let's say we've got the function f of x is equal to the square root of two to the x okay yeah this is like one point four one four to this to itself or something yeah and what you do is you start with a seed of one and then you run it through f and you'll get root two then you run it through f again and that's root two to the power root two and if you keep going on you're saying what we have is a certain infinite sequence and often when people see dot dot dot and then equal we're meant to assume that this sequence approaches something right so maybe let's just verify that experimentally it seems a good idea yeah so i can pull up some python just to all right so we're going to define a to be the square root of two yep all right so that's what it looks like and now i'm going to define um f of x to be a to the power x yep yep so if i take f of one uh oh gotta return that return a to the power x great so f of one should be the square root of two f of f of one should be something else yes so we could do is say we're going to start with um uh let's call it c is equal to one and then we're going to set c equal to f of c and then print it out right and presumably as i repeat this process over and over we'll get something kind of keep repeating maybe get tired and write a for loop i don't know 100. um let's just keep doing this c is equal to f c and then i want to print out what it is and it looks actually pretty good right it does yeah i'm seeing um a bunch of one point nine nine nines and it gets really close to two i think it's slightly above two yeah at this point yeah it might be numerical error right because computers can't know um irrational numbers perfectly yeah um so it's it seems like the the empirical answer is two but not four yeah so our logic got us the right answer but it can't be completely for the right reason because the same logic applies to that yeah yeah aside from finding the exact answer to this it's a little interesting that it converged at all yeah because i mean okay let's just uh let's just do this with a different value like um if i take z equal to two and then c is going to be two to the power of whatever c already is right and maybe we'll do that and then print c right it's like we get 16 and then we'll take two to the power of that which is six five five three six right so we take it very very big very very quickly right actually this is what i love about python right that it doesn't care about the size it's like yeah that can take a lot of space so the next step right which is just like the fifth step or whatever is two to the power of this wall which i don't i don't want to die so it explodes super quickly and yet square root of two it seemed to converge right so we could play this game actually um like what if it was 1.5 right what if we were taking we started c at one and then we take c equals 1.5 to the power of c and look at what it is right so we first get 1.5 then 1.8 2.1 we kind of keep doing this and we can do this so they're certainly getting bigger but yeah it's whether or not they're going to go to infinity i guess yeah um yeah result too large yeah because yeah at some point we're taking 1.5 to this uh so 1.5 blows up right and yet square root of 2 which is 1.41 didn't blow up right um so what you should ask is when does it blow up and when does it not mm-hmm absolutely um which is kind of this is very visual to think about where uh you know about cobweb diagrams yeah i do yes yes i teach them in math biology yeah awesome sorry because when you have like an iterative repeated process like a population who where the rule for going from one phase to another is the same you're just kind of yeah and you want to say yeah see if it blows up i see if it converges to a stable population number that's exactly what we do so let's say this is x i'm going to draw the line y equals x right and then let's say we have some kind of exponential curve that looks like a to the x right yes so when we want to play this process of we've got a zero notion called a0 c 0 is equal to 1 and then c 1 is equal to f c 0. yep and we sort of repeat that we're feeding it into itself sfc one on and on um the way we could think about this is okay uh our first value is an input of one sitting here so that's c so the output is wherever it is on that graph right but the output right now is represented with a y value so if we want to turn that into an input we say what is the point where we can see where the x value is the same as this y value so we hop over to the graph where y equals x yeah right and that gets us here so that kind of represents where c one is and then we apply the function again yes exactly we land on the graph and then again we say oh let's make that our input value so we have to make it x and i actually think cobweb diagrams are a slightly awkward way to think about iteration sometimes but can be very helpful yeah um now in this context yeah now it just it goes up it's going to go up and you're not going to get let's go to infinity isn't it yeah um but some exponential curves will look a little bit different okay so an exponential curve passes through one this will be here and let's say it we're going to have it go and like gently come in and then out so it it's something that just barely got into touching it so when we start at uh one we go up to the graph okay great great then we cobweb over here and then go up and over and up and over and we approach wherever it intersects great that makes sense um so this would be like two uh certainly in the case that we were looking at here oh yeah exactly evidently when we do root two so that's actually just let's straight up do this um i'll go back to my computer which i've got a little desmos situation going on here um where let's say a is equal to square root of two okay yeah awesome so in this context i have a graph yeah so we've got a graph that is uh uh square root of two to the power x and it actually looks like they intersect right at two which we know is true right because square root of two to the power two is that yes um so it's a little bit more fun is if we try different values of a right so i had a slider for a and in that context let me make this range from one up to three sometimes it doesn't hit yeah and that's where we know things are going to blow up and then sometimes it does right and if we take a look notice that remember where converge is gonna is gonna be where they intersect yeah right so here that intersection seems to happen a little above two but by the time it separates right it's separating here that intersection never got as high as four no yeah so there doesn't even exist an x such that this repeated process will land you on four instead it's like i converge to something a little above two yeah and then without passing through without passing go right you go straight to jail the jail of divergence to infinity yeah right um that seems to be what happens let's figure out what value of a yes this is right that's what i want to know because as i'm watching this like we i want to like you've just shown that that for those you know we know it works route 2 1.4 ish if we go to like 1.5 it's too big so like what's that value in between if it's because it seemed to be a little bit more than route 2. so there's nothing yeah so it looks like it's what 1.45 ish that seems to be the right neighborhood um i actually did not really think this through before coming here today but i think we can solve it together because we're not math folks yeah presumably we've got y equals x yeah exactly and over here we've got um a to the x yeah so uh if their tangents are going to be equal to each other uh then we know that the derivative of a to the x is equal to the natural log of a times a to the x and so we're looking for a point x naught so that this is equal to one because its slope has to be yes all right so ln of a times a to the x naught has to equal one we don't know what a is we also don't know what x naught is we need a second equation right yes clearly which is that we're sitting on the line y equals x right so it's got to be the case that x naught um they're intersecting as well as their gradients they have to be the same intersection point yeah exactly so their gradients are equal and so therefore x naught must equal a to the x naught yeah exactly okay must equal a to the x map great then you take that put that into here okay great great so now we know that x naught is x naught is equal to a no no not a e to the e to the x x naught to the negative one both the power of song it's not great um which is good so they're multiplying in there so that's x naught is oh so x naught is equal to e okay now we get e to the power of one over e yeah um which let's just go ahead and check that if we take um e to the power of one over e yeah okay so that's 1.44 which seems to be what we're seeing on the graph for down there awesome so that's the cutoff and we could uh this is one e is whatever there's going to be a lower bound too right which because there should be multiple solutions to this um that will give us like the other end for our range of convergence first of all just to give an indication of what what it would mean yeah um it's if what would it mean actually um so the idea being that we need this intersection yeah we need that as we make a sufficiently small except it's not just that we need them to intersect at some point it's got to be that the slope of uh the slope of our a to the x graph is less than one so that when we do our cop webbing we go towards the intersection and not away from it that's all situation i see but then if that slope was uh let's see so if that slope was too big um or if we're coming at it from the other direction then it's too small yeah we would cobweb away from it yes oh i see yeah yeah yeah so i think i'd be looking for the point yeah okay because let's say we've got um on some graph really zoomed in this is what like a to the x looks like as it passes through um y equals x and then when we're cobwebbing we'll like start here and then go to the right and then come down and then go until we hit it yeah and then go up and then it's a question yeah so as soon as a to the x as a graph is a little too steep then it'll diverge away right so i think i think we'd be looking for when it's intersecting with a curve and also has a slope of negative one okay that so in the same way that here we're solving for when the slope is one just substitute negative negative one yeah um and then you can see the lower bound on it which is fun it is fun i agree and the last thing i'd leave someone with is to google tetration fractal okay because you can ask okay we're doing this on the real number line which complex numbers for a will make this process converge versus diverge and very interesting but as a solution to um the original puzzle on why does it why does it seem to be two and four simultaneously the the mere assumption that there even exists an x such that this process could yield four is wrong yes to solve an equation you have to know that the equation has a solution right are you sure yeah that's what i got for you that was awesome that was great thank you so much you

Channel: Tom Rocks Maths

Views: 63,922

Keywords: 3blue1brown, grant sanderson, grant sanderson interview, 3blue1brown interview, tom rocks maths, tomrocksmaths, tom crawford, tom crawford numberphile, tom crawford maths, 3 blue 1 brown, 3 blue 1 brown interview, three blue one brown, mathematics, oxford mathematician, oxford math professor, math professor, 3blue1brown equation, 3blue1brown math, 3blue1brown math channels, 3blue1brown person, power tower, infinite power, oxbridge interview question, repeated exponentiation

Id: dnZ3xlif9VA

Length: 16min 15sec (975 seconds)

Published: Wed Jul 29 2020

Reddit Comments

[u/tomrocksmaths]

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: https://www.youtube.com/channel/UCYO_jab_esuFRV4b17AJtAw

Produced by Dr Tom Crawford at the University of Oxford. For more maths content check out Tom's website https://tomrocksmaths.com/

[u/iNinjaNic]

I think its interesting that when Grant pulls up the desmos graph and inserts a = sqrt(2) we find that the exponential crosses the line not only as 2, but also at 4. I wonder if that is somehow linked to why the wrong approach also gave you the answer of sqrt(2)