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Y^x = x^y

  1. Jun 14, 2005 #1

    cronxeh

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    An interesting function. If plotting you would probably omit the 0,0 point too.

    How do I get the general solution for this? like y = [function of x]? Is it just y=x ?

    What about the boundary conditions near origin? How does it look like there??
     
  2. jcsd
  3. Jun 14, 2005 #2

    HallsofIvy

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    If you mean y being defined implicitely as a function of x, then it certainly is not "y= x". I doubt that it could be solved in terms of elementary function but you might be able to use a version the the Lambert's W function.
     
  4. Jun 14, 2005 #3

    saltydog

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    Hello guys. I don't think it's just y=x. I've attached a rough plot of a curve I think all real solutions (in first quadrant) are on but can't prove it.

    Oh yea, I think it would be an interesting exercise to figure out the complex solutions.
     

    Attached Files:

    Last edited: Jun 14, 2005
  5. Jun 15, 2005 #4

    cronxeh

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    Some particularly interesting similar function was [tex]y=x^x[/tex] When I plotted it from 0 to 1 there was what seemed like an inflection point - although I dont think its really called that but from a graph it does look pretty neat.

    Could this be the plot for y^x = x^y from 0 to 1 ?
     
    Last edited: Oct 8, 2005
  6. Jun 15, 2005 #5
    It's not just y=x: here's a counterexample! 2^4 = 4^2
     
  7. Jun 15, 2005 #6
    there are an infinite number of counterexamples. slatydog graphed them in #3
     
  8. Jun 15, 2005 #7
  9. Jun 15, 2005 #8
    Yes, I know. I never said it was the only counterexample, now did I?
     
  10. Jun 15, 2005 #9

    Hurkyl

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    I think there should be a straightforward proof that there are exactly two (real) solutions for y, for each x, except for when x is... (reaches into my magic hat)... e. I just can't remember what the proof is. :biggrin:

    If you happen to recall the proof that 2^y = y^2 has exactly two solutions, that should generalize to this problem.
     
  11. Jun 15, 2005 #10

    saltydog

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    Hello Bicycle. I checked out your reference. Thanks. I found it an interesting exercise to convert:

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

    using the substitution y=kx yielding the parametric equations:

    [tex]x=k^{\frac{1}{k-1}}[/tex]

    [tex]y=k^{\frac{k}{k-1}}[/tex]

    However, when I plotted this, I did not obtain the line y=x. Attached is the parametric plot.
     

    Attached Files:

  12. Jun 16, 2005 #11

    Hurkyl

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    The reason you don't get the y=x solution is that you assumed k was not 1 when you took the (k-1)-th root of the equation.
     
  13. Jun 16, 2005 #12
    my simple attempt at proving nothing...

    y^x = x^y

    xlogy=ylogx

    (logy) = (ylogx)/x

    10^(ylogx) = y

    ylogxlog10 = y

    logxlog10 = 1

    log(x+10) = 1

    10^1 = x+10

    10 = x + 10

    x = 0

    y^0 = 0^y

    y = 0

    0^0 = 0^0

    1 = 1

    so, the only solutions I can find for y and x are = 0

    (don't flame, i'm a noob and i'm probably wrong)

    however...

    y^0 = 0^y

    0logy = ylog0 <-- that doesn't work...

    0 = undefined ?
     
  14. Jun 16, 2005 #13
    y=x=0 is the only point on y=x that's not a solution.
     
  15. Jun 16, 2005 #14

    cronxeh

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    z = f(x,y) = x^y - y^x = 0

    Can anyone plot this. I got some weird looking graph, maybe someone else will have a better one

    Oh man this is exciting. Check out this plot!
     
    Last edited: Oct 8, 2005
  16. Jun 16, 2005 #15

    Zurtex

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    Given some value fixed value of x then a solution of y will be:

    [tex]y = - \frac{x \text{ProductLog} \left( -\frac{\log x}{x} \right)}{\log x}[/tex]

    According to mathematica. This gives you a graph like so:
     

    Attached Files:

  17. Jun 16, 2005 #16

    cronxeh

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    Forget the positive side its too boring! Plot all stuff in QIII !

    Same interesting effect but without the positive side can be observed in [tex]x^y=-y^x[/tex]
     
    Last edited: Jun 16, 2005
  18. Jun 26, 2005 #17
    I think if x^y=y^x and x,y both integer then y=x

    If u need the proof please tell
     
  19. Jun 26, 2005 #18

    Hurkyl

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    A proof would be interesting to see, since your claim is false...
     
  20. Jun 26, 2005 #19

    cronxeh

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  21. Jun 26, 2005 #20
    [tex]2^4=4^2=16[/tex]

    edit: Just noticed Icebreaker said it before me in this same thread, giving him the credit.
     
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