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[tex] f(x) = x^x [/tex]Given this function, defined, let's say for

  1. Jun 19, 2004 #1
    [tex] f(x) = x^x [/tex]
    Given this function, defined, let's say for all real numbers, is there any way to tell when x is rational versus irrational for integer values of f(x)?
    e.g.
    [tex] x^x = 4 [/tex]
    x = 2
    [tex] x^x = 27 [/tex]
    x = 3
    [tex] x^x = 3 [/tex]
    x = 1.825455054...

    Thanks!
     
    Last edited: Jun 19, 2004
  2. jcsd
  3. Jun 19, 2004 #2

    Tom Mattson

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    You mean other that simply taking the xth root of x?
     
  4. Jun 19, 2004 #3
    Do you mean the xth root of f(x)? Or am I misunderstanding? And yes, I meant other than simply looking at specific argument and function values.
     
  5. Jun 19, 2004 #4

    Tom Mattson

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    Whoops, I sure do. :redface:

    I'm not sure off the top of my head, but let me play with it.
     
  6. Jun 19, 2004 #5
    3^3 doesnt equal, so x=3 is irrational. sorry to nitpick.

    interesting problem though. i bet there isnt a way besides applying the interger set for x and assuming all others will be irrational. is there any value for f(x) that results in a non interger rational number? if so, that negates using the interger set to find the y.
     
  7. Jun 19, 2004 #6
    Thanks, I corrected the typo. And I guess that would put us on the path to a solution. Essentially, given that x is not an element of Z (the integer set), is [tex] x^x \in Z [/tex] possible?
     
  8. Jun 19, 2004 #7
    Okay, suppose x=a/b, where a,b are integers. Assume that the fraction is reduced, ie. gcd(a,b)=1. Then x^x= (a/b)^(a/b)=[tex](\frac{a^a}{b^a})^\frac{1}{b}[/tex]

    But [tex]gcd(a^a,b^a)=1[/tex], and therefore [tex](\frac{a^a}{b^a})^\frac{1}{b}[/tex] is irrational. It follows that if x^x is an integer, then either x is an integer or transcendental.
     
  9. Jun 19, 2004 #8
    so there are also no solutions that are non intergers but rational, like 1/3 and 1/4?
    i understand that x^x is an interger if x is an interger.

    that tells us that the set of intergers for x gives us the solution set of y? there are no fractional xs for interger ys? sorry im dumb. i want to learn though!
     
  10. Jun 20, 2004 #9
    I was wondering the same thing about taking the logarithm or ln of a function.
     
  11. Jun 20, 2004 #10

    Zurtex

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    I was interested in this quite some time ago. I finally found an article saying there is no known way of rearranging the equation:

    [tex]y=x^x[/tex]

    In to some function of y in terms of x. But I did find an iterative formulae so you could approximate to as much accuracy as you wanted if you had y and wanted to know x. I'll see if I can find it again.
     
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