# E is Everywhere

1. Dec 10, 2015

### 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

2. Dec 10, 2015

### axmls

Why? I don't see it.

3. Dec 11, 2015

### MAGNIBORO

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

4. Dec 11, 2015

### 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.

5. Dec 16, 2015

### forcefield

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.

6. Dec 30, 2015

### MAGNIBORO

Thanks for your input , and sorry for late response

7. Dec 30, 2015

### MAGNIBORO

thanks and sorry for late answer

8. Dec 30, 2015

### Ssnow

Now we can try to solve

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

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