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Fractional exponents of negative numbers?

  1. Jan 28, 2007 #1

    DaveC426913

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    I was just playing around in my head. I wanted to plot this graph:

    y=x^2.5; x=-2

    This is valid right? My calc says it's invalid input.
     
  2. jcsd
  3. Jan 28, 2007 #2

    DaveC426913

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    Oh I see.

    x^2.5 is the same as x^2 * x^.5

    So you're taking the square root of a negative number.

    Uh, I don't know quite enough about imaginary arithmetic to figure out the answer but I'll take a stab.

    -2^2 * -2^.5
    = 4 * 2i
    = 8i ?
     
  4. Jan 28, 2007 #3
    It's valid, but it's complex. So if you can't evaluate complex numbers on your calculator that would explain why the calculator says it's invalid.
     
  5. Jan 28, 2007 #4
    (-2)^.5 is actually either plus or minus 2.5i
     
  6. Jan 28, 2007 #5

    Hurkyl

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    Well, technically, it's multivalued:

    [tex]
    (-2)^{2.5} = \pm 4 i \sqrt{2}
    [/tex]

    I think the principal value is the one with the + sign.
     
  7. Jan 28, 2007 #6

    DaveC426913

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    OK, I think that actually sort of answers the original question I was going to ask.

    The graph of x^2 is a parabola, never crossing below the x-axis, yet the graph of x^3 does. Since the range from 2 to 3 is a continuum, you should be able to draw a sequence of graphs that shows where and how the "negative x" portion of one graph flips about the X-axis to the other graph.

    So, it seems that answer is that it doesn't discontinuously jump from one the other, it actually passes through imaginary space to get there.

    If this is true, then I have managed to, just through my own logic, discover the 3D space wherein real numbers and imaginary numbers exist together...

    I wish I'd gone on to post-secondary math...
     
  8. Jan 28, 2007 #7

    Hurkyl

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    In case you're curious:

    (-2)t := exp(t ln (-2))
    = exp(t (ln |-2| + i arg -2)) = exp(t ln 2) exp(t i (pi + 2 pi n))
    = 2t ( cos(t pi) + i sin(t pi) ) ( cos(2 pi n t) + i sin(2 pi n t))

    The principal value occurs when n = 0. (or maybe n=-1... but I think it's n=0) Note that usually there are infinitely many values to the exponential; that 2.5 is rational makes it special.
     
  9. Jan 29, 2007 #8

    DaveC426913

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    Ultimately what I want to do is graph the change from x^2 to x^3 with the real numbers as the exponent.
     
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