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Graph of y = x^x

  1. May 21, 2006 #1
    Heyyhey...just wondering, is the graph of y = x^x significant in anyway?

    it looks kinda weird...?
     
  2. jcsd
  3. May 21, 2006 #2

    arildno

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    Chinese cooks usually put the spaghetti threads in that particular shape on your plate.

    Other than that, I don't know if that graph is "significant".
     
  4. May 21, 2006 #3

    benorin

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    it is particularly weird for x<0, being that it takes complex values there...
     
  5. May 21, 2006 #4

    arildno

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    The Chinese have never liked the negatives.
     
  6. May 22, 2006 #5
    I can't get mathematica to plot this function for negative values. Anyone know how I can do it?
     
  7. May 22, 2006 #6
    I think someone has already answered your question.
     
  8. May 22, 2006 #7
    I fail to see how I would be unable to plot it.
     
  9. May 22, 2006 #8
    Complex meaning they are imaginary.
    Try x=-1/2
     
  10. May 22, 2006 #9
    Yep, just do Plot[{y=x^2},{x,-10,10}] and you get all the values from -10 to 10 of this super significant function x^x.
     
  11. May 22, 2006 #10
    Mathematica can plot complex functions, and this function in particular because it's R->C.

    How does plotting x^2 give me all the values of x^x?
     
    Last edited: May 22, 2006
  12. May 23, 2006 #11
    thnx cool guys... whats the derivative of y=x^x ?
     
  13. May 23, 2006 #12

    LeonhardEuler

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    [tex]y=x^x=e^{\ln{x^x}}=e^{x\ln{x}}[/tex]
    [tex]\frac{dy}{dx}=(1+\ln{x})e^{x\ln{x}}=(1+\ln{x})x^x[/tex]
     
  14. May 23, 2006 #13

    Curious3141

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    It is much more interesting and informative to plot the hyperpower function [tex]f(x) = x^{x^{x^{x^{...}}}}[/tex]

    Find the upper bound of x for which that function is defined and see if you can spot the relationship of that bound to a famous constant.
     
    Last edited: May 23, 2006
  15. May 24, 2006 #14

    benorin

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    [tex]\frac{dy}{dx}=x^x(1+\ln{x})[/tex] is not real when x is a real negative number, yet if x is negative and of the form [tex]x=\frac{p}{2q+1}[/tex], where p,q are positive or negative integers, then y is real. Curious, no? It has to do with the complex branch-cut structure of [tex]y=x^x=e^{x\ln{x}+2k\pi ix},k=0,\pm 1, \pm 2,\ldots[/tex].

    Source: "A Course of Modern Analysis" by Whittaker & Watson, pg. 107.
     
  16. May 25, 2006 #15

    Curious3141

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    I'm not sure about this, certainly for x in that domain and of that form, a real value of y exists if y is defined to be multivalued.

    But if the principal value of ln(x) is used, which is [tex]Ln(x) + \pi i[/tex], then the value of y returned won't necessarily be real, right? :confused:

    Sorry, I don't know that much about complex analysis, just the basics. I do know the principal branch for ln x, but not the one for x^x. I would've assumed it would be based on the branch cut of the log function, giving [tex]x^x = e^{x Ln(x)} = e^{x ln(|x|) + i\pi x} = \frac{\cos{(\pi |x|)} - i\sin{(\pi |x|)}}{|x|^{|x|}}[/tex] (for negative real x) which would not necessarily return real values even for x of the form [tex]\frac{n}{2k+1}[/tex]
     
    Last edited: May 25, 2006
  17. Jun 6, 2006 #16
    f(x) = x^(x^(x^(x^x)))...
     
  18. Jun 7, 2006 #17
    I've always liked that the only critical point of [tex]f(x) = x^x[/tex] is
    [tex]x = \frac{1}{e}[/tex]
     
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