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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

    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. jcsd
  3. Dec 10, 2015 #2
    Why? I don't see it.
     
  4. Dec 11, 2015 #3
    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
     
  5. Dec 11, 2015 #4

    pwsnafu

    User Avatar
    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

    Ssnow

    User Avatar
    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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