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Is it possible to evaluate x^(2/3) for x<0?

  1. Aug 10, 2009 #1
    I am very confused. Here I was thinking I know the basics of algebra, and then I get two questions in my homework today with exponents like this.

    [tex](-1)^{\frac{2}{3}} = undefined[/tex]

    when I type it into my calculator.

    But with manipulation,

    [tex](-1)^{2/3} = [(-1)^{2}]^{\frac{1}{3}} = 1[/tex]

    Am I missing something here? My textbook solutions manual says that you can graph such a function for negative values, but when I plug it into, for example, my Mac's graphing software, it only graphs it for positive values of x.

    Which one is correct?

    Thanks!
     
  2. jcsd
  3. Aug 10, 2009 #2

    mathman

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    I can't speak for the calculators. However the solution can be obtained using complex variables:
    -1=e[pi]i.
    We the have 3 possible answers for (-1)2/3:
    e2(n+1)[pi]i/3, where n=0,1,2.
     
  4. Aug 10, 2009 #3
    My TI-89 gives an answer of 1 no matter how I use the parenthesis.

    (-1)2/3 = 1
    ((-1)1/3)2 = 1
    ((-1)2)1/3 = 1

    Which should make sense since:
    - It is possible to square a negative and then take the cube root of the resulting number.
    - It is possible to find the cube root of a negative number and square the resulting number.
     
  5. Aug 10, 2009 #4
    Thank you, I guess I should buy a new calculator.
     
  6. Aug 11, 2009 #5

    mathman

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    (Details of previous comment)

    3 solurions: 1, (-1+(-3)1/2)/2, (-1-(-3)1/2)/2
     
  7. Aug 11, 2009 #6
    When I went to school it was considered conventional to consider, the square root of a square as its absolute value and hence positive. Of course, this is only a convention; but calculators may be constructed to honor this convention.
     
    Last edited: Aug 11, 2009
  8. Aug 12, 2009 #7
    I believe what is happening is a restriction of the computing device. Power functions with fractional powers are frequently restricted to nonnegative real numbers (Maple for example).

    But the fractional power identity

    [tex]x^{m/n}=\sqrt[n]{x^m}={(\sqrt[n]{x})}^m[/tex]

    holds provided that the radicals on the right are defined.

    However this assumes m and n are in lowest terms.

    For example:

    [tex](-1)^{2/3}={(\sqrt[3]{-1})^2=(-1)^2=1[/tex]

    but

    [tex](-1)^{4/6}={(\sqrt[6]{-1})^4=\text{ undefined}[/tex]

    This may be part of why calculating devices shy away from fractional powers of negative bases.

    The extension of roots to complex values is usually only done if the inputs are also complex. This subtely changes the meaning of some radicals.

    [tex]\sqrt[3]{-8}=-2[/tex] in the real number sense.

    [tex]\sqrt[3]{-8}=1+i\sqrt{3}[/tex] in the complex number sense.

    In general, provided m and n are in lowest terms and n is odd, the expression [itex]x^{m/n}[/itex] is defined for all real numbers and can be graphed.

    --Elucidus
     
  9. Aug 17, 2009 #8
    Use x^y = e^(y.log(x)). If x=r.e^(i.t) then log(x)=log(r)+i.(t+2.pi.k) where k is any integer. This will give you all possible values of x^y.
     
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