Double Diagonal Power Tower

Algebra Level 3

Let $A=\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\cdot^{\cdot^{\cdot}}}}}$ and $B = ^{\sqrt[{}^\ddots\!\!\sqrt{2}]{2}}\!\!\!\!\!\sqrt{2}$

Then it is true that:

Note: $B$ is no t a nested tetration.

A < B

A = B

A>B

2 solutions

Daniel Liu
Apr 21, 2015

First, the value of $A$ can be computed via classical means to arrive at $A=2$ .

We see that $B$ satisfies $B=\sqrt[B]{2}$ .

Note that if $B=2$ , then $2 > \sqrt{2}\implies LHS > RHS$ .

Also, note that $f(B)=B$ is continually increasing while $f(B)=\sqrt[B]{2}$ is continually decreasing for $B>0$ ; thus, along with the fact that when $B=2\implies LHS > RHS$ , the (unique) root of $B=\sqrt[B]{2}$ satisfies $B<2$ .

Thus, $\boxed{A>B}$

How do you prove that $B$ is defined? (The limit might not exist, which means they are not comparable and thus there's no answer among the options.)

Likewise with $A$ , although you sidestepped the entire thing with "computed via classical means".

Ivan Koswara - 6 years, 1 month ago

$\sqrt[\sqrt{2}]{2} = 2^{\frac{1}{\sqrt 2}}$

Alex Zhong - 6 years, 1 month ago

Well for $A$ I sidestepped it because it is a classic problem (the 2 versus 4 dilemma).

For $B$ , well, I was too lazy to check that it was defined. But it had only one solution and it looked about right, so I assumed. I'm sure it can be proven to exist in some way or another. Please tell me if this is false though.

Daniel Liu - 6 years, 1 month ago

I think you should prove one of the four requirements here .

Pi Han Goh - 6 years, 1 month ago

@Pi Han Goh – I'm pretty sure by $\sqrt[\sqrt{2}]{2}$ he meant $2^\frac{1}{\sqrt{2}}$ ; the $\sqrt{2}$ at the left is the index of the root.

Ivan Koswara - 6 years, 1 month ago

@Ivan Koswara – I interpreted as a nested tetration function.

Pi Han Goh - 6 years, 1 month ago

@Pi Han Goh – This is what fooled me aswell! Overthought the problem.

Sualeh Asif - 6 years, 1 month ago

Could someone tell me how one can go about computing the value of A by classical means?

Shashank Rammoorthy - 5 years, 9 months ago

This is relevant.

Daniel Liu - 5 years, 9 months ago

A can also be equal to 4

Vighnesh Raut - 6 years, 1 month ago

-1Fps Hryc
Apr 15, 2019

as it "goes by" A is going to get higher values of the exponent and it'll get bigger while B is going to get smaller fractions as an exponent (so it'll be 'rooting' harder over time) So basically A = sqrt2 ^ x when x > 1 and B = sqrt ^ x' when x' < 1 which means A > B

Double Diagonal Power Tower

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