When You Have Too Many Ex's Part II

Calculus Level 5

If f(x) is defined on all x in the interval I, where

$f(x) = x^{x^{x^{\cdot^{\cdot^\cdot}}}},$

find $\lfloor 10000 \cdot \sup I \rfloor$ .

Note: $\lfloor L \rfloor$ is the greatest integer less than or equal to $L$ .

The answer is 14446.

1 solution

Trevor Arashiro
Jan 4, 2016

$y=x^y$

$y^{\frac{1}{y}}=x$

$y^{\frac{1}{y}}\frac{dx}{dy}=0$

$-y^{\frac{1}{y}-2}(\ln(y)-1)=0$

Since $-y^{\frac{1}{y}-2}$ has no roots, we need $\ln(y)-1=0\longrightarrow y=e$

$\large{\therefore x=y^{\frac{1}{y}}=e^{\frac{1}{e}}=1.4446...}$

I have a feeling I missed something here. Also, I can't explain why this function is undefined when $x<e^{-e}$

I'm not quite following your steps, but the answer is indeed $\large e^{\frac{1}{e}}$ . You might remember this question of mine; the solutions and subsequent discussions might prove edifying. As for the lower limit of $\large e^{-e}$ , this is apparently dealt with in Euler's (who else?) infinite tetration theorem, but I'm having a hard time finding a version of this theorem that is not in Latin. :P

Brian Charlesworth - 5 years, 5 months ago

Oh, sorry. I completely missed this.

I've had to make many changes to this problem and hopefully it is worded correctly now.

Trevor Arashiro - 5 years, 4 months ago

Hello @Trevor Arashiro , I found the problem very interesting. I think you should specify that $I$ is the largest interval of positive numbers (the x cannot be negative or zero) where the function is defined.

Arturo Presa - 10 months ago

When You Have Too Many Ex's Part II

The answer is 14446.

1 solution

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