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E is Everywhere

  1. Dec 10, 2015 #1
    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


    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. jcsd
  3. Dec 10, 2015 #2
    Why? I don't see it.
  4. Dec 11, 2015 #3
    The video says that
    x= 2^(1/2)

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

    example: x^x^x^x^.....=1.5


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

    and I would like to know why this happens
  5. Dec 11, 2015 #4


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    Science Advisor

    ##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.
  6. Dec 16, 2015 #5
    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.
  7. Dec 30, 2015 #6
    Thanks for your input , and sorry for late response
  8. Dec 30, 2015 #7
    thanks and sorry for late answer
  9. Dec 30, 2015 #8


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    Gold Member

    Now we can try to solve

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

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