Limit of The Tower (Strike Back)

Calculus Level 4

a 0 = a and a n + 1 = a a n \large a_0=a \quad \text{and} \quad a_{n+1}=a^{a_n}

What is the smallest a > 0 a>0 such that a n a_n is convergent?

Note: This is a follow up question from Limit of The Tower


The answer is 0.065988.

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Albert Yiyi
Oct 24, 2018

e e \large e^{-e}

let f ( x ) = a x f(x) = a^x , if f n ( x ) f^n(x) converges to L L , then f ( L ) = a L = L , f ( L ) = a L ln a = ln L f(L) = a^L = L, f'(L) = a^L \ln a = \ln L .

when x L x \approx L we approximate f ( x ) f(x) using its tangent at x = L x=L , f ( x ) L + f ( L ) ( x L ) = L + ( ln L ) ( x L ) f ( L + ε ) L + ( ln L ) ε f 2 ( L + ε ) L + ( ln L ) 2 ε f 3 ( L + ε ) L + ( ln L ) 3 ε \begin{aligned} f(x) &\approx L + f'(L)(x-L) \\ &= L + (\ln L)(x-L) \\ f(L+\varepsilon) &\approx L + (\ln L) \ \varepsilon \\ f^2(L+\varepsilon) &\approx L + (\ln L)^2 \ \varepsilon \\ f^3(L+\varepsilon) &\approx L + (\ln L)^3 \ \varepsilon \\ \end{aligned} for f n ( L + ε ) f^n(L+\varepsilon) to converge, we must have 1 < ln L < 1 e 1 < L < e ( e 1 ) 1 e 1 < a < e 1 e ( a = L 1 L increases monotonically between e 1 < L < e ) e e < a < e 1 e \begin{aligned} -1&<\ln L<1 \\ e^{-1}&<L<e \\ (e^{-1})^\frac{1}{e^{-1}}&<a<e^\frac{1}{e} \ (\because a=L^\frac{1}{L} \text{ increases monotonically between } e^{-1}<L<e)\\ e^{-e}&<a<e^\frac{1}{e} \end{aligned}

p/s: im not sure if this solution is valid, so the result is double checked using computer program, which seems to support my answer.

Can you provide a bit more details?

At the very least, explain your previous misconception of why simply graphing x 1 x x^ { \frac{1}{x} } and seeing that it tends to 0 at x = 0 x = 0 , doesn't give the right answer.

Calvin Lin Staff - 2 years, 7 months ago

Log in to reply

well, i use non-rigorous method, its kinda weird, mixing some intuition, and use computer program to check my final answer is valid.

im waiting others to give a more rigorous solution, or prove my answer is wrong.

edit: solution added

albert yiyi - 2 years, 7 months ago

Log in to reply

For a rigorous solution, I refer you to read the award-winning paper by R. Arthur Knoebel.

Brian Lie - 2 years, 7 months ago

Log in to reply

@Brian Lie thanks for your link Brian, but it seems like a dead link to me.

"This site can’t be reached"

albert yiyi - 2 years, 7 months ago

Can we state that a > 0 a > 0 here.

a = 1 a = -1 is a perfectly valid solution as is.

Alex Burgess - 1 year, 12 months ago

Log in to reply

you're right. changed. thanks

albert yiyi - 1 year, 12 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...