Derivative of infinite power tower x^x^x^... Can you solve x^x^x^...=2 vs. x^x^x^...=3 • Video Subscribe to @blackpenredpen for more fun math videos #calculus #blackpenredpen #math
Did a bit of research and now I know that the function x^x^x^x^..... (which can also be viewed as the solution for y in the equation x^y=y or x=y^(1/y)) is called "tetration of infinite height" and this function converges for e^(-e) < x < e^(1/e), roughly the interval from 0.066 to 1.44. Also this function is related to the Lambert W-function. In fact y = WC-In x)/ (In x) where Wis the Lambert W-function, which is defined as the functional inverse of xe^x with respect to x.
@@manuelferrer6501 no, x * ln(x^x^ . . .) = ln((x^x^ . . .)^x) (1), which is not ln(x^x^ . . .) (2) (e. g. if we substitute √2 for x in (1) we'll get ln(2^√2) ≈ 0.980, and if we substitute that in (2) we'll get ln(2) ≈ 0.693.
The infinite power tower converges on: { e^(-e) < x < e^(1/e) }, approx. { 0.066 ~< x ~< 1.445 } Also fun fact, the square root of 2 raised to its own power infinitely many times is actually equal to 2. Try it in a calculator!
Maybe I made an error, but my efforts to evaluate sqrt(2)^sqrt(2)^sqrt(2)^... aren’t giving me a clear answer. Here’s what I did: Let x = sqrt(2)^(sqrt(2)^(sqrt(2)^...)) x = sqrt(2)^x x = 2^(x/2) Squaring both sides gives: x^2 = 2^x Which has two positive solutions: x = 2 or x = 4 I know that there is another point where the functions x^2 and 2^x intersect, where x < 0, but since sqrt(2)^sqrt(2)... > 0, I disregarded it. I assume there is a similar process of elimination that rules out x = 4 as a solution, but what is it?
@@Riolupai the third root of 3 gets you two solutions: The true solution, 2.478 The second solution that you have to rule out, 3 The fourth root of 4 is sqrt(2) 😂 The fifth root of 5, is 1.7649 (true solution) Or 5 (other solution) I'm refering by the "true solution", to the result of a^a^a^... But if you take for example 5=x^x^x^... Then you will find x=5^1/5 So then 5 would be either "the true solution" or "the other solution" to 5^1/5 (And we found out it is the "other solution")
Interesting! I tried on Georgebra to add more and more Xs in the denominator, but I observed that the value of f(0) was oscillating between 1 and 0 when I add more Xs, so it is not converging. But if you plug 0 in your incredible derivative, we see that it gives a division by 0, which correlates to the impossible value of f(0)! Awesome mate!
this is 2 when x=sqrt(2): let Y= x^x^x^x^... -> Y=x^Y, take ln -> ln(Y)=Y*ln(x), now let Y be 2, then ln(2)=2*ln(x) ln(2)/2=ln(x) take the exponential and swap places x=e^(ln(2)/2) x=(e^ln(2))^(1/2) x=(2)^(1/2)=sqrt(2)
because that what is what he ask us in the video... and as this is a infinite exponentiation there is a value of x to which this no longer a constant, I don't know exactly where this thing explode, but is obvious that it does... maybe for anything greater than sqrt(2)? but it sure does for x>=2
I believe it blows up for x > e^(1/e) The function y = x^y Can be rewritten as x = y^(1/y) Of which the maximum is at x = e^(1/e), y = e He did found this maximum on his video about e^pi vs. pi^e.
Okay, I'll throw here my approach to the max. value where this function converges. First of all let's call y=x^x^x..... Therefore, y=x^y like he does in the video. After that, we take natural logarithms on both sides, so we get ln(y)=y*ln(x). Isolate ln(x) and we have ln(x)=ln(y)/y. Let's call f(y)=ln x = ln(y)/y and differentiate respect to y. f'(y)=(1-ln(y))/y^2 (after some simplification). The maximum of this function, is when f'(y)=0, (1-ln(y))/y^2=0 --> 1-ln(y)=0 --> y=e With the second derivative we can see it's a maximum and not a minimum. Coming back to ln(x)=ln(y)/y substitute y=e to obtain x=e^(1/e). Therefore x=e^(1/e) is the biggest value of x where the function y=x^x^x... converges, to y=e.
Link_Z - Hardcore Indie Game Achievements You have only shown that x^x^x^... does not converge when x>e^(1/e), but you have not shown when it does converge. It turns out, for example that it does not converge for 0,05. To be exact, x^x^x^... converges exactly if 1/e^e≤x≤e^(1/e) or if x=-1.
love your videos, theyre pretty much the main reason i love calculus. slight request if u havent already done so, integrate it (if its possible im not sure)
You can find a formula for when it does converge. x^...^x = y_n has y_n = x^y_(n-1). In order for each of these to converge, it must be the case that y_(n+1) = y_n, meaning x^y_n = y_n. So, if it does converge, it'll converge to a solution of that. Now, y is actually is the Lambert-W function, the inverse function of xe^x, of -ln(x) over -x. This is because -W(-ln(x)) / ln(x) = y, and -y * ln(x) = W(-ln(x)), so x^(-y) = e^(W(-ln(x))), and thus x^(-y) * W(-ln(x)) = W(-ln(x)) * e^(W(-ln(x))) = -ln(x), so you get -W(-ln(x)) / ln(x) * x^(-y) = 1. or y = x^(y). You find that there are no solutions to the Lambert-W function outside of the interval [-1/e, e], See here: math.stackexchange.com/questions/1693561/for-what-values-does-this-method-converge-on-the-lambert-w-function. So, we must have -1 / e
I think y = x^y is only a valid function for 0< x ≤ 1 and x = e^(1/e). For 1 < x < e^(1/e) there are two possible values of y for each x, and for x > e^(1/e) y doesn't exist. You can see this by considering the inverse function, which is simply y = x^(1/x).
It is, he didn’t prove the necessary assumptions to do it. I think this one requires uniform convergence, and this function series doesn’t even converge pointwise for most inputs in R. Edit: After further thought, maybe pointwise convergence is enough for this approach, I’m not sure though since I haven’t tried it on paper.
It's not even math. Firstly, he had do define what is x^x^x.... , because it's not a function from R to R. Then he had to define what is d/dx, and only after that he is allowed to do math
You can also rewrite xy ln(x) at the end as x ln(x^y), which also equals x ln(y) because y = x^y. That would make the derivative in terms of x be [x^(x^(x^...))]^2 / [x - x ln(x^(x^(x^...)))].
Totally expecting this! Haha, I really love the humor of your channel and how enthusiastic are you in your videos, greetings from Colombia, keep it up :P
Only converges when X = 1. When X=1 dy/dx = 1 If X is 0 the bottom will be undefined due to the ln function. If X = -1 the bottom becomes imaginary (also due to ln function)
Can’t it still be simplified to dy/dx = ( x^(2 * x^x^x....) )/( x - ln(x^x^x...) ) If you bring the 2 into the top tower and you bring the “xy” part into the ln and then use exponent laws to get the tower again
No idea what all that means but you sounded smart. But in all seriousness, I am trying to self-teach myself calculus in hopes of getting a degree in physics. Although, it is very confusing hopefully, one day I will be able to understand all of your videos. Until then keep up the awesome videos ^-^
Calculus isn't hard (espesially differentiation). I think the best way to self-teach calculus is watch whole calculus course somewhere like MIT OpenCourseWare or coursera.org Good luck!
Im teaching myself with the AP calculus BC course on khanacademy.org The calc 1 topics are really cool and not difficult, Im gonna start infinite series soon If you want a more college-like course watch lectures and use the college course on khan Good luck!
I would love some comments about which functionspace this function belongs to. It is not clear to me that it is differentiable. It looks to me like it is only defined between 0
I cam up with y^2/(y^(1/y)*(1-ln(y)) I did it by drawing a line tangent to y=x^y (since that is equivalent to the derivative) and then making a right hand triangle out of it with dx and dy as the legs. From their I just imagined dx/dy(x=y^(1/y)). It turned out to be equivalent to 1/tan(z) where is the angle formed by the dx leg going through the tangent line. From their I just worked out tan(z) so that I get dy/dx. That was equivalent to just taking the reciprocal of dx/dy(x=y^(1/y)). And that gave me dy/dx(y=x^y)=1/(1/y^2*y^(1/y)*(1-ln(y))) I feel skeptical of my answer but it checked out with the 2 tests I did. Can anyone confirm that my trigonometry method works?
I plotted the first 71 tetrations of x, really quite interesting. Also, the first time I tried I made the mistake of iterating (((x^x)^x)^x ..... Is there a term for this? The limit is equal to 1 for 0
Say you have an even number of infinite exponentials, if you plug in x = 0, then y = 1, however if there are an odd number of exponentials than if you plug in x = 0, then y = 0. It's like saying whats the answer to Grandi's series if you start with 0 as he first term.
The function y only exists on the interval (0,1]. And it appears to always be equal to 1. Therefore it's a constant and the derivative is zero. Note: pay attention to the domain of the function.
Weird. In the denominator of the final answer, if you bring the factors of ln(x) inside as its exponents, won't you just end up with ln(y) ? I feel like there's an infinite amount of answers
I have a question. I haven't studied that kind of derivative yet so I came up with this: f'(x)=((f^(-1))')^-1 (x) The inverse of f(x)=x^x^x^x^x^... -> y=x^y -> x=y^x; f^-1(x)=y=e^(lnx /x) y=f^-1 ' (x)=e^(lnx /x) * (1-lnx)/x^2 f'=((f^-1)' )^-1-> x=e^(lny /y) * (1-ln y)/y^2 How horrendous is that answer? (It actually works if you want to find the tangend line on y=a)
Why not factor out the common factor of x in the denominator and then make all those red x’s the power of the x that is the argument of the ln function. Then the denominator would be x(1 - ln [infinite tetration of x]). You could also move that 2 in your numerator inside the parentheses and subtract 1 by factoring the x out of the denominator. The numerator would be x to [(twice the infinite tetration of x) - 1)]
If x^x^x^x^...=2 Then x²=2 So x=√2 √2^√2^√2^...=2 If you plug that to the calculator, you have to first write √2, and then do √2^ANS a lot of times, and it's true, it converges :) Interestingly enough, tough... x^x^x^...=4 x⁴=4 x=⁴√4=4^¼=(2²)^¼=2^½ x=√2 Can someone explain this paradox? Could it have something to do with the ±?
There is no paradox. Remember that what you actually have is the expression y = x^y. Now notice that 2 = sqrt(2)^2 but also 4 = sqrt(2)^4. There is no paradox, that is just how numbers work.
Well I will add that there is a "paradox". The thing is that for some x (in your case sqrt(2)) the equation y = x^y has multiple solutions. You can see that y=4 and y=2 are both solutions when x= sqrt(2). But that is algebraic. When you see y as the limit of the sequence x,x^x,x^x^x,x^x^x^x... then that sequence either converges or not. And if it converges it can be proven that the limit must be unique. Therefore, if x^x^x^x... converges then it will only be able to reach one of the solutions to y = x^y, not all of them. You haven't really "proven" that sqrt(2)^sqrt(2)^sqrt(2) = 4. What you actually did was start with x^x^x^x... = 4 but then you replaced that to x^4 = 4, so now you are not doing a sequence, you are solving the algebraic equation y= x^y.
I found something curious. if you graph y=x^y you can see that y->0 as x->0, while if you graph x^x^x.... with a finite amount of Xs when x->0, y->0 if you have an odd number of total Xs and y->1 if you have an even number of total Xs. This seems to imply that infinity is odd since it behaves like an odd number. obviously infinity isnt odd or even so why does it aact like an odd number?
If you graph this, you will have a flat line between 0 and 1(with those points excluded), It will jump to 1 at 1 and then be gone, because how do you differentiate infinities? For 0 it won't exist. Between -1 and 0 the original function converges to 1, so you will have a flat line at 0 again. Left of -1, you have a diverging, oscillating mess. Good luck differentiating that.
When x is very large, (x^x^x^...)^2 will also be very large and positive, while x-x(x^x^x^...)ln x will be very large and negative, implying that dy/dx itself is negative (I guess for all x > 1?). That feels really counter-intuitive to me; the curve looks like it should be growing at a vast rate. Could you explain what's going on there? Also, is dy/dx defined in the reals for 0 < x < 1? Is there a clever way to find y(1/2) or y'(1/2) for example? Thankyou for the interesting video :)
I’m in Geometry. I am not sure why I am listening to Calculus stuff, but I can what is going on a little bit. Comment section: Should I stop? If I should not, please explain why this is a good idea to continue watching this.
Ken Garner I tried it and I think it does converge, but really slowly. I got x=0.067 to converge (reasonably) when the stack is about 2000 x's high. I think it converges for the half-open interval (0, e^(1/e)].
Angel Mendez-Rivera Hm that seems right. I couldn't get 0.65 to converge really, which is just outside the range. Do you mean the open or closed interval?
Michiel Horikx It is an open interval. Although that’s irrelevant, since regardless of the interval, no number outside of it should yield convergence. Not sure why would you test it with a number outside.
Angel Mendez-Rivera Open and closed refer to the endpoints, so in other words, are e^-e and e^(1/e) included or not? The second one works as far as I know.
