Infinite Power Tower Equations Battle!

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This is the part 2 of the Infinite Power Tower!
start: 0:00
Solve x^x^x^...=2 0:18
vs. Solve x^x^x^...=3 2:17
Domain & Range of the infinite power tower: 4:19
Brilliant: 7:38

**The Satisfyingly Strange Journey to the Infinite Power Tower**

blackpenredpen,
math for fun

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Hi all, thanks for the 500k subs! I will make a post of the winners of the secret giveaway tonight. After that I will be taking a break from YT.

blackpenredpen

You got me here from the community post

ricsix.

Can't wait for part 3, I need that domain explanation!

peternelson

Will you explain why the domain is like that in another video?

SeeTv.

Interestingly, if you have any quantity of the form a^(1/a), where a is positive and real, the corresponding power tower will always converge, because if for all such a not equal to e, a^(1/a) < e^(1/e), since at x = e, x^(1/x) attains its global maximum. So, despite the fact z^^♾ = a only converges if a < e or a = e, the power tower with z = a^(1/a) does converge. It just does not converge to a.

angelmendez-rivera

When are you going to post .. isn't your break over ? 😔

monke

Currently it's 2 am here, and I don't know why I'm watching this at this solemn night, but still I'm enjoying Idk why.

spandanhalder

5:44 *technically speaking*

Ah yes, the engineering approximation.

Aldiyawak

For anyone who is curious, if you want to know what x^^♾ is equal to for any x in the domain, then notice that y = x^^♾, and x^^♾ = x^(x^^♾) = x^y. Therefore, y = x^y.

To solve for y, take the natural logarithm here. We can do this because are assuming x is positive in this initial exercise. Hence ln(y) = y·ln(x), implying ln(y)/y = ln(y)·exp[-ln(y)] = ln(x), hence -ln(x) = -ln(y)·exp[-ln(y)]. If -ln(x) > -1/e or -ln(x) = -1/e, then the above equation implies W[-ln(x)] = -ln(y), where W is the Lambert W function, in this case, the principal branch of the W map. Therefore, y = exp(-W[-ln(x)]) = 1/exp(W[-ln(x)]) = 1/(-ln(x)/W[-ln(x)]) = -W[-ln(x)]/ln(x).

The condition that -ln(x) > -1/e or -ln(x) = -1/e implies that ln(x) < 1/e or ln(x) = 1/e, which implies that x = e^(1/e) or x < e^(1/e), which agrees with what was stated during the video.

angelmendez-rivera

Managed to got half of bounds. Namely, x^x^x^...=y means x^y=y, this solves using Lambert W function: y=e^(-W(-ln x)) (I've seen your other videos where you explain what is W and how to solve such equations). But, W(x) domain is x≥-1/e (there are two real branches: W₀ domain is [-1/e; ∞), with values in the range [-1; ∞), W-₁ domain is [-1/e; 0), with values in the range (-∞; -1], that's a multivalued function). This means, there must be -ln(x)≥-1/e, solving that for x gives x≤e^(1/e). Then, if I put that value into equation, I've got y = e^(-W(-ln x)) = e^(-W(-1/e) = e¯¹ = 1/e.

Now I need to prove this is lowest possible value of y. Also still haven't figured out how to find out other bounds.

nikitakipriyanov

You're probably one of my fav youtubers of the last 2 years, bro! Your work is awesome.
I'm really passionate about maths and you just feed my hype! And in these days of quarantine and isolation, you and Dr. Peyam sometimes even feel like my "math friends"!
Hey listen; have you ever thought about a good video series about Stochastic Calculus? I'm quite into it right these days! (financial markets and stuff). Just learned how to derive Black-Scholes' PDE. But I get lost when certain subjects such as Ito's integral or similar come around. It would be excellent if you play some of that sort of things!
Long live BPRP!

gastonsolaril.

5:40
Engineering students don't know why, because π=e=3

balazscsillag

I really like your channel. It is way more entertaining than the other math channels.

dan_was_here

I'm a math hobbyist, I don't have advanced education in math, but I love them and I'm able to understand most of your videos (they're awesome). I have a question due to my ignorance: When you have a tower exponent of real numbers, it has to be solved up-to-down? I mean, the solution you give to x^x^3 only works if we solve the exponents up-to-down, and then the same answer fits to any x^x^x^...^x^3 form. Is it correct? Thank you very much in advance!

alexandrefrancoalcaraz