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Solve x^x=x

  1. Nov 3, 2006 #1
    How would you solve
    [tex]x^x = ax[/tex]
    for x (where a is any postive constant >1)?

    I am trying the Lambert W function but I just can't get it to the right form. I am starting to think that you can't solve this using the Lambert W function. Any help?

    edit: Sorry, I really wanted the expression to be x^x = a*x
     
    Last edited: Nov 4, 2006
  2. jcsd
  3. Nov 3, 2006 #2

    quasar987

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    You're looking for solutions to [itex]x^x=cx[/itex].

    Set x=a+ib=|x|exp(i[itex]\Phi[/itex]) and look to solutions to

    [tex]|x|\exp(i\Phi)^{a+ib}=|x|\exp(i\Phi)[/tex]

    knowing that two complex numbers are equal iff their modulus are equal and their phase differ by at most a factor of 2n[itex]\pi[/itex], [itex]n\in \mathbb{Z}[/itex]. For instance, I find that the condition of equality of modulus imposes the following relation btw a,b and c

    [tex]\frac{b-ab}{2}\ln(a^2+b^2)\tan^{-1}(b/a)=c[/tex]

    The condition on the phase will restrict the possible solutions some more.
     
    Last edited: Nov 3, 2006
  4. Nov 3, 2006 #3

    quasar987

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    I have a little doubt about what I wrote though because I don't remember how a complex number raised to a complex number "looks" like but I think

    [tex](e^{z})^w=e^{wz}[/tex]

    is correct.
     
  5. Nov 4, 2006 #4
    I don't know how is this going to help. Instead of having one equation and one unknown we would now have two equations and two unknowns. And it doesn't seem like this approach is going to be any easy looking how the a's and b's are "trapped" inside.
     
  6. Nov 4, 2006 #5

    quasar987

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    Wheter you find it pretty or not it's the solution nonetheless. If you give that to a computer, he will indiscriminately find all solutions in a fraction of second.
     
    Last edited: Nov 4, 2006
  7. Nov 4, 2006 #6
    That's the only clever way to go here.
    Closed form solution is impossible for this equation I think.Just numerically ,and it depends on constant [tex]a[/tex].
     
  8. Nov 4, 2006 #7
    So it is NOT possible to use the Lambert W function here. I mean, we can use the Lambert W function to find a closed-form solution to the equation x^x = a. But we can't, supposedly, solve for x^x=ax using the same function?

    Anyways, does anyone have clever way to solve this equation in CLOSED-FORM?
     
  9. Nov 4, 2006 #8
    Possible or not,this equation can't be solved explicitly.I haven't worked with "W" function much in past ,but I think you are right:
    X^(X)=a is one thing and X^(X-1)=a quite another one.
    There are numerical methods though.
    Some may be better suited for this type of equation (ie. faster convergence) than others. It depends what precision you want.
     
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