Exploring the Equation: x^x^x^x^x^...=n

[MAGNIBORO]

hi, sorry for bad english,
watching videos on youtube I found this video:


and ask me if that equation works for any number and comes to the conclusion that

x^x^x^x^x^...=n
x=n^(1/n)

but this only works if the number is 1/e ≤ n ≤ e
and I wonder if this is a property of e or happens if for some other reason.
thak you

 

[axmls]

but this only works if the number is 1/e ≤ n ≤ e

Why? I don't see it.

 

[MAGNIBORO]

Why? I don't see it.

The video says that
x^x^x^x^...=2
x= 2^(1/2)

then I wanted to know if it works for any number.

example: x^x^x^x^...=1.5
x=1.5^(1/1.5)

x^x^x^x^...=n
x=n^(1/n)

but using my calculator , I realized that this only works if 1/e ≤ n ≤ e

and I would like to know why this happens

 

[pwsnafu]

$z^{z^{z..}} = \frac{W(-\log z)}{-\log(z)}$ for complex z, where we have used Lambert W-function and the principal branch of the logarithm. The W function is real valued only for $x \geq -1/e$. I'm assuming it has something to do with that.

 

[forcefield]

x^x^x^x^x^...=n
x=n^(1/n)

but this only works if the number is 1/e ≤ n ≤ e
and I wonder if this is a property of e or happens if for some other reason.

x^x^x^x = x^(x^(x^x))

At each step (when you add more x = n^(1/n)) the calculation approaches it's destination slower and slower.

The further away n is from 1 the slower the calculation approaches it's destination. If n is less than 1/e or more than e, the calculation approaches it's
destination so slow and still slowing that you can't say it's approaching n.

This is a property of n^(1/n). It has it's maximum at n = e.

 

[MAGNIBORO]

$z^{z^{z..}} = \frac{W(-\log z)}{-\log(z)}$ for complex z, where we have used Lambert W-function and the principal branch of the logarithm. The W function is real valued only for $x \geq -1/e$. I'm assuming it has something to do with that.

Thanks for your input , and sorry for late response

 

[MAGNIBORO]

x^x^x^x = x^(x^(x^x))

At each step (when you add more x = n^(1/n)) the calculation approaches it's destination slower and slower.

The further away n is from 1 the slower the calculation approaches it's destination. If n is less than 1/e or more than e, the calculation approaches it's
destination so slow and still slowing that you can't say it's approaching n.

This is a property of n^(1/n). It has it's maximum at n = e.

thanks and sorry for late answer

 

[Ssnow]

Now we can try to solve

$\log_{x}{(\log_{x}{(\log_{x}(...))})}=2$

so $x^{2}=2$, huao is the same ... :- D