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Unsolvable exponental?

  1. Oct 2, 2008 #1
    Is it possible to solve x = y^y for y?
     
  2. jcsd
  3. Oct 2, 2008 #2

    CRGreathouse

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    There's a closed form in Lambert's W, or you could use Newton's method. (Actually, depending on how fast you can evaluate logarithms and exponentials, the secant method is probably faster.)
     
  4. Oct 2, 2008 #3
    I'm sure I'm missing something here, but isn't the solution simply y equals the yth root of x?.
     
  5. Oct 2, 2008 #4

    mathman

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    Since y is unknown, the yth root of x is unknown.
     
  6. Oct 2, 2008 #5
    Just some clarification:

    is there a way to get an exact value, or just a numeric approximation?
    also, i do not really understand the Lambert W function. the link just says that it is the inverse function, but not how.

    also, how is the following true (or is it false?):
    Sqrt[x]^Sqrt[x]==1/x^Sqrt[x]
     
    Last edited: Oct 2, 2008
  7. Oct 3, 2008 #6
    False. 1/x = x^(-1), so you'd have [x^(1/2)]^[x^(1/2)] = [x^(-1)]^[x^(1/2)]

    Also, a quick numerical check can show you it's false. Take 4, for example. Square root 4 = 2, so you have 2^2 = (1/4)^2, so that's false.
     
  8. Oct 3, 2008 #7

    CRGreathouse

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    Is there a way to get an exact value for [itex]\sqrt 2[/itex], or just a numeric approximation? Because Newton's method is used for square roots as well.

    This is a philosophical question more than a mathematical one, I suspect.
     
  9. Oct 3, 2008 #8
    *erroneous post*
     
  10. Oct 3, 2008 #9
    to get an exact answer, i think the only real way is with a continuous fraction. could someone please explain to me exactly what newtons method is?
     
  11. Oct 3, 2008 #10

    CRGreathouse

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  12. Oct 4, 2008 #11
    I am not sure that i understand. if the root is negative, then it will never work, as the derivatives of the function are x^x (1 + \text {Log}[x]) for the first derivative, and x^{-1 + x} + x^x (1 + \text {Log}[x])^2 for the second derivative. i believe that both of these are non-continuous at any negative. is this right, or am is making a mistake?
     
  13. Oct 4, 2008 #12

    CRGreathouse

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    You didn't sy you were looking at x < 0. You'll need to use the secant method (as I recommended in my first post!) or something similar instead of Newton in that case. Otherwise you could use Newton's method or Halley's method to compute W and get the answer that way.

    I generally recommend Newton's method only because it is what is taught. The secant method is often better. Of course you can always fall back onto bisection if you have trouble... or don't want to use something complex like regula falsi.
     
  14. Oct 4, 2008 #13
    i am not sure that i understand. i thought from what i read that the secant method has the same kind of problem that newton's method does, namely that if there is no continuous second derivative, then there is no guarantee that it will converge.
    also, from what i understand of Halley's function, is that it is dependent on newtons method, so for -x it will also fail. in fact, i belive that all of these methods are for continuous functions only.
    is there a non-iterative way of doing this, or at least one that will work for negative x?
     
  15. Oct 5, 2008 #14

    CRGreathouse

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    I suggested Halley's method for solving W, not x^x.

    Your function is continuous, what's the problem?
     
  16. Oct 6, 2008 #15
    i thought that it is not continious at all negative x.
     
  17. Oct 6, 2008 #16

    CRGreathouse

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    Well, just take absolute values and find the magnitude, then see if a solution is possible with a sign change.
     
  18. Oct 6, 2008 #17
    how could that work if the magnitude of the absolute value is a different expresion? even with a sign change, the absolute value is changing more than that.
     
  19. Oct 7, 2008 #18
    also, how could someone compute W using halleys mehtod?
     
  20. Oct 7, 2008 #19

    CRGreathouse

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  21. Oct 7, 2008 #20
    thanks. but what about the previous question?
     
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