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X to the xth power and other indices

  1. Aug 24, 2003 #1

    jcsd

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    Recently I've considering the two functions xy= yx and y = xx.

    1) For xy= yx can y be found in terms of x? I'm sure I've seen such a solution before.

    2) In y = xx when y (and therfore x)is postive what value of y gives the minium value for x? I know it's rougly 0.45 and that dy/dx = 0 at this point, but it's a long time since I've done more advanced differentiation and I don't even know if you can differentiate xx
     
  2. jcsd
  3. Aug 24, 2003 #2

    Hurkyl

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    2)

    y(x) = xx
    ln y(x) = x ln x

    (1 / y(x)) * y'(x) = 1 * ln x + x (1 / x)
    y'(x) = y(x) (1 + ln x)
    y'(x) = xx(1 + ln x)

    Since xx > 0, y'(x) can be 0 iff
    1 + ln x = 0
    ln x = -1
    x = 1/e

    So the minimum of xx occurs at x = 1/e = 0.368


    1)

    y = x

    Ok, ok, that's not the only solution. I highly doubt there is an elementary solution for this, but I can get you started:

    for x, y > 0:

    xy = yx
    y ln x = x ln y
    y / ln y = x / ln x
    ln y / y = ln x / x

    (note each step is reversible)

    so consider f(t) = ln t / t
    then f'(t) = (1 - ln t) / t2

    So we see that f(t) is strictly increasing for t < e and strictly decreasing for t > e. This means that for any z, f(t)=z has at most two solutions. More specifically one can fairly easily show that:

    the equation f(t) = z has:
    exactly 1 solution if z <= 0
    exactly 2 solutions if 0 < z < 1/e
    exactly 1 solution if z = 1/e
    exactly 0 solutions if z > 1/e


    We've seen that if xy = yx iff f(x)=f(y).

    y = x is clearly a solution... meaning that it is the only solution (for a given x) iff f(x) <= 0 or f(x) = 1/e... that is if x <= 1 or x = e.

    Also, we can see that if there are two solutions of xy = yx for a given x, then either
    1 < x < e < y
    or
    1 < y < e < x

    For example, the only nontrivial solution I know off the top of my head is 24 = 42... clearly 1 < 2 < e < 4


    But as I mentioned, I strongly suspect no elementary closed form solution is possible.
     
  4. Aug 26, 2003 #3
    I believe it is

    y(x) = (x productlog(-ln(x)/x))/ln(x);
    where prductlog(x) gives the principal solution for w in x=w e^w.
    Not all solutions are found.
     
  5. Aug 27, 2003 #4
    well, i am not quite sure about it so dont take it for granted...
     
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