Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Fractional Exponent of Negative Numbers

  1. Aug 24, 2012 #1
    I have been thinking on this topic for a while.

    As I have seen on various sites:
    1) a^(b/c) means "c"th root of "a^b".
    2) Also an even root of a negative number does not exist in real numbers.

    Then I want to investigate this formula:

    a=(-4)^(1/2)

    Mustn't it be equal to b=(-4)^(2/4)

    However if we calculate the power first;

    a=square root of [(-4)^1]= Not a real number
    b=4th root of[(-4)^2]= 2

    What do I miss? I have some confussion about decimal exponents but I hope the answer of this question will solve it.
     
  2. jcsd
  3. Aug 24, 2012 #2
    That specific law of exponents does not work in negative numbers.
    A way to understand this is that your method entails [itex]\sqrt{-a}=(-a)^{2/4}=a^{1/4}[/itex], but clearly [itex]a^{1/4}\cdot a^{1/4}=\sqrt{a}\neq \sqrt{-a}[/itex].
    Your mistake is that you are applying the law [itex](ab)^{c}=a^c\cdot b^c[/itex], which isn't valid for negative a or negative b.
     
    Last edited: Aug 24, 2012
  4. Aug 24, 2012 #3
    Thanks for your answer.

    But then, how to handle fractional exponent problems:

    For example how to handle this one;

    (-2)^(4/5)
     
  5. Aug 24, 2012 #4
    The number -2 can be represented in polar form as [itex]2e^{i\pi}[/itex], and hence the problem of exponentiation reduces to evaluating [itex]2^{4/5}e^{4i\pi/5}=2^{4/5}(\cos(4\pi/5)+i\sin(4\pi/5))[/itex].

    Euler's identity is often employed to give the definition of complex exponentiation.
     
  6. Aug 24, 2012 #5
    @Millennial

    I really like your posts, they are succinct (reducing the chance of confusion) and very accurate.

    Thanks,
     
  7. Aug 24, 2012 #6
    Thanks for your answer.

    So, If I want to determine if an negative base has a real root or not, then should I approach it by expressing it as an complex number?

    Couldn't I say if, for example, -2^1.0954 has a real root or not easily?
     
  8. Aug 24, 2012 #7
    Well sure you could. But you are going to be able to say that based on Millennial's answer.

    Based on his formula, n^(x) where n < 0 will have a real root iff sin(x*pi) = 0. Otherwise it will be a complex number.
     
  9. Aug 24, 2012 #8

    uart

    User Avatar
    Science Advisor

    Before answering that, let me just add some more info to the previous question regarding (-2)^(4/5).

    Using the complex definitions of exp and log we can write it as,
    [tex]e^{0.8 \log(-2)} = e^{0.8 (\log(2) + i (1 + 2k) \pi))}[/tex]

    So we see that if take this expression over the complex numbers, then it's multi-valued. Basically we get a bunch of different answers, for k = 0, 1, 2, etc. It's easy to see however, that in this case they repeat after k=4, so there are 5 unique values. We also note that for k=2 we get a purely real solution (since 0.8 * i 5 pi = i 4 pi).

    In general, say we have (-x)^(p/q) where "x" is a positive real and fraction p/q is reduced to it's simplest form. Then we can write it as

    [tex]e^{\frac{p}{q} \log(-x)} = e^{\frac{p}{q} (\log(x) + i (1 + 2k) \pi))}[/tex]

    Here we can see that if "q" is odd then there must be some value "k" for which [itex]1+ 2k = q[/itex], and the corresponding value will be real. Conversely we can show that for "q" even there will be no real valued solutions.
     
  10. Aug 24, 2012 #9
    Complex exponentiation is not multi-valued, complex logarithm is. It is just that complex exponentiation gives the same answer for different numbers, to be precise, [itex]p^q=e^{p\log(q)}[/itex] is equal to [itex]\left(p+\frac{2\pi in}{\log(q)}\right)^{q}=e^{p\log(q)+2\pi in}[/itex] for [itex]n\in \mathbb{Z}[/itex] through the periodicity of trigonometric functions. However, since complex exponention is not a bijection, complex logarithm is multivalued, if [itex]\log(z)=w[/itex] holds, then for [itex]n\in \mathbb{Z}[/itex], [itex]\log(z)=w+2\pi in[/itex] also holds.

    Just wanted to fix that error.

    Also, if you don't know it, complex logarithm on the principal branch is defined by [itex]\log(z)=\log|z|+i\text{arg}(z)[/itex]. It is easily derivable from the polar form of a complex number. Of course, through my last argument above, a general form that covers not just the principal branch but all branches is given by [itex]\log(z)=\log|z|+i\text{arg}(z)+2\pi in[/itex] for [itex]n\in \mathbb{Z}[/itex].

    I'd like to thank you for the complement :)
     
    Last edited: Aug 24, 2012
  11. Aug 24, 2012 #10
    Firstly thanks all for your wise answers.

    It is clearer to me now.

    I was thinking about "the simplest form". I have got the logic now. But as I am not good at complex numbers it will take a while to understand why it is required to be simpliest form. I think I can find it after some study.

    Thank you all again for your great help.
     
  12. Aug 24, 2012 #11
    In complex numbers, the rule:
    [tex]
    (z_1 \, z_2)^{\alpha} \stackrel{?}{=} z^\alpha_1 \, z^\alpha_2
    [/tex]
    does not hold in general when α is not an integer. The reason behind this is because non-integer powers in complex numbers are defined through the principal branch of the complex logarithm (Log):
    [tex]
    z^\alpha \equiv \exp \left( \alpha \, \mathrm{Log} (z) \right)
    [/tex]
    and, for this branch, the product rule:
    [tex]
    \mathrm{Log}(z_1 \, z_2) \stackrel{?}{=} \mathrm{Log}(z_1) + \mathrm{Log}(z_2)
    [/tex]
    does not hold in general. This is because:
    [tex]
    \mathrm{Log} (z) \equiv \ln \vert z \vert + i \, \mathrm{Arg} (z)\,, \ -\pi < \mathrm{Arg} (z) \le \pi
    [/tex]
    and the principal arguments (Arg) do not add:
    [tex]
    \mathrm{Arg}(z_1 \, z_2) \neq \mathrm{Arg}(z_1) + \mathrm{Arg}(z_2)\,,
    [/tex]
    since they are restricted to the above stated interval.

    For example:
    [tex]
    \vert -4 \vert = 4\,, \ \mathrm{Arg}(-4) = \pi
    [/tex]
    [tex]
    (-4)^{\frac{1}{2}} = \exp \left(\frac{1}{2} \, ( \ln 4 + i \, \pi ) \right) = e^{\ln 2} \, e^{i \, \frac{\pi}{2}} = 2 \, i
    [/tex]
    On the other hand, [itex](-4)^2 = 16[/itex], and:
    [tex]
    \vert 16 \vert = 16\,, \ \mathrm{Arg}(16) = 0
    [/tex]
    [tex]
    (16)^{\frac{1}{4}} = \exp \left(\frac{1}{4} \, ( \ln 16 ) \right)= e^{\ln 2} = 2
    [/tex]
     
    Last edited: Aug 24, 2012
  13. Aug 24, 2012 #12

    uart

    User Avatar
    Science Advisor

    Looking at Millenial's and Dickfore's reply I realize I was wrong to say is was multi-valued, since the complex power is defined using only the principle branch of log. That is, using only k=0 in the equation I wrote.

    Think of this as being a bit like the distinction between the (multiple) solutions to the equation [itex]x^2 = 4[/itex] and and the (single) value of [itex]\sqrt{4}[/itex]. To keep sqrt as a single valued function it is usually defined as only the positive value and we use an explicit plus or minus ([itex]\pm[/itex]) to write all the solutions to the given equation.

    In the same manner we can see a close relation to the value of [itex](-2)^\frac{p}{q}[/itex] and the solutions to the equation [itex]x^q - (-2)^p = 0[/itex]. The equation does have "q" solutions, but [itex](-2)^\frac{p}{q}[/itex] is defined as just the one of these values corresponding to the logarithm principle branch.

    No problems, thanks for that. I hope that the above also helps to clarify the situation to the OP. :smile:
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Fractional Exponent of Negative Numbers
  1. Negative Exponent (Replies: 8)

  2. Fractional exponents (Replies: 4)

Loading...