# E is Everywhere

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

but this only works if the number is 1/e ≤ n ≤ e
Why? I don't see it.

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.

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.

##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

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
Gold Member
Now we can try to solve

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

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