the most fun derivative of x^x^x^...

Can you solve x^x^x^...=2 vs. x^x^x^...=3 https://youtu.be/WAjgupg3hDk

Girl : So how many exes do you have? Me : Well, I have an infinite amount of Xs

Instructions unclear; impressed too many girls

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.

4:52 You was smiling and that all. But all of sudden you became serious and said "Blue." Lmao haha

BPRP does know the right kind of girls^^

3:37 now that is what I cal huge integration

In the rendez-vous : girl : "So, what can you do?" me : "I can differentiate x^x^x^x^... " she falls in your arms

Freshman boys, if you want to have the girls rolling in, show them the full solution. I cannot stress this enough. 👏

You can technically reduce that even more, as ylnx is equal to lny, so in the denominator you'd simply have x-xln(x^x^x^...)

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.

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!

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!

Ya boi finna go impress some girls

this feels like cheating...

blackpenredpenbluepen

"I have no idea what this is anymore" Your videos are awesome!!

This was very entertaining, thanks!

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: https://math.stackexchange.com/questions/1693561/for-what-values-does-this-method-converge-on-the-lambert-w-function. So, we must have -1 / e <= -ln(x) <= e, and thus e^(-e) <= x <= e^(1/e).

Your videos are fascinating. Brings back my Calculus memories.