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Homework Help: Solving y=x^x for x

  1. Feb 13, 2007 #1
    Problem:
    I'm trying to figure out how to solve [tex]y=x^x[/tex] for x as a function of y.

    Related Equations:
    [tex]\log_n a^b=b \ast \log_n a[/tex]

    Attempt at Solution:
    I took the natural logarithm of both sides and got: [tex]\ln y=x \ast \ln x[/tex] I don't really have any idea where to go from here.
     
  2. jcsd
  3. Feb 13, 2007 #2

    Dr Transport

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    How about plotting both sides.....

    Remember that the [tex] \ln(x) [/tex] is nonsense for [tex] x \leq 0 [/tex].
     
  4. Feb 13, 2007 #3
    Uh... by plot, do you mean graph? I'm not trying to find where the graphs intersect or anything... I'm just trying to change the equation from [tex]y=f(x)[/tex] to [tex]x=f(y)[/tex]
     
    Last edited: Feb 13, 2007
  5. Feb 13, 2007 #4

    Dr Transport

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    You can't......
     
  6. Feb 13, 2007 #5
    You mean to say that [tex]x^x[/tex] has no inverse that can be defined in terms of elementary functions? Then what about using calculus? Can it be defined then?
     
  7. Feb 14, 2007 #6

    HallsofIvy

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    Depends upon what you mean by "using Calculus". The inverse of y= xx is the "Lambert W function".
     
  8. Feb 14, 2007 #7
    By "using calculus," I mean, for example, defining the function with sigma notation, the derivative of something, etc. Here's a good example: the indefinite integral [tex] \int x^x dx [/tex] can not be defined in terms of elementary functions, but can be defined as the indefinite integral of [tex] x^x [/tex] So, what I'm asking is, can the inverse of [tex] x^x [/tex] be defined in terms of elementary functions as well as indefinite integrals, series, sums (i.e. sigma notation), derivatives, and so on?
     
  9. Feb 14, 2007 #8
    I did some research on the "Lambert W function," and it's not the inverse of [tex]x^x[/tex]. It's the inverse of [tex]x \ast e^x[/tex].
     
  10. Jan 17, 2008 #9
    Nchimy

    Solving y=x^x for x does have a solution in terms of Lambert W Function. Although this reply may have come late, one could still make use of it later.

    To make the problem easy, lets assume that x and y are real. Then we can proceed as follows:

    If y = x^x, .........(1)

    then since we also have

    x = exp[ln[x]], .........(2)

    we may conviniently express (1) as

    y=x^exp[ln[x]]. .........(3)

    Now, taking logs on both sides of (3) (and noting that ln[d^c]=cln[d]) gives

    ln[y] = ln[x] * exp[ln[x]] ..... (4).

    Imediately, we see that (4) can be solved for ln[x], using Lambert W Function, as

    ln[x] = W[ln[y]] ..... (5)

    so that

    x = exp[W[ln[y]]] .... (6) .

    The solution in (6) is valid for y > 0 because ln[y] for real values of y makes sense only within this range. From (6) (and taking note that W[0]=1while W[e] =1) we see that

    (a) x = 0 when y = 1,
    (b) x = 1 when y = e
    (c) x is only real when ln[y] >= -(1/e) (or equivalently x is real for y >= exp[-(1/e)) but complex and multivalued otherwise.
    (d) x increases monotonically with increasing value of y.
    (e) The exists a taylor series expansion of x, about ln[y]=0, with a radius of converges equal to -(1/e).

    For more details on th Lambert W function, please refer to Corless R M et al, “On the Lambert W function”, Adv. Comput. Math, Vol. 5, pp.329-359.


    Cheers.....
     
    Last edited: Jan 17, 2008
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