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Clarification on Inverse of x^3

  1. Nov 21, 2004 #1
    Clarification on "Inverse" of x^3

    Hello. I was seeking some clarification on what should be a trivial problem.

    Consider [itex] f(x) = x^3 [/itex]. Does this function have an inverse?

    Well....many websites seem to say so...as do the solution sets for some abstract algebra courses. Certainly, my TI-89 has no trouble plotting the inverse. Without thinking about it too much, I would have to conclude that

    [tex] f^{-1}(x) = x^{1/3} [/tex]

    But is this actually correct? I have convinced myself that it is not. Let me make my case...and then hopefully someone can shed light on my eyes.

    First, I want to take a tangent and summarize some of the properties of exponents. Please comment on these---are they correct? too strong? not strong enough?


    Typically, we are told that [itex] (x^a)^b = x^{ab} [/itex] without any proper restictions on a and b. So here is my shot:

    Let [itex] x\in \mathbb{R} [/itex]. Unless stated otherwise, [itex] a [/itex] and [itex] b [/itex] are real numbers.

    \text{If } b \in \mathbb{Z}, &\qquad\text{then } (x^a)^b = x^{ab}.\\
    \text{If } b \notin \mathbb{Z}, &\qquad\text{then } (x^a)^b \neq x^{ab}.\\
    \text{If } a \in \{2,4,6,\ldots\}, &\qquad\text{then } (x^a)^b = |x|^{ab}.\\
    \text{If } a \in \{2,4,6,\ldots\}, &\qquad\text{then } (x^a)^{1/a} = |x|.\\&\\
    \text{In general, } &\qquad(x^a)^{1/a} \neq x.

    For example, [itex] (x^{1/2})^2 = |x| [/itex]. However, [itex] (x^2)^{1/2} \neq x [/itex].


    Now, back to the problem. Let me clairfy first: Let [itex] f:\mathbb{R} \rightarrow \mathbb{R} [/itex].

    Now, consider the follow example:
    f \left( f^{-1}(3) \right) &= \left(3^{1/3}\right)^3 = 3\\
    f \left( f^{-1}(-3) \right) &= \left((-3)^{1/3}\right)^3 = -3\\&\\
    f^{-1}\left( f(3) \right) &= \left( 3^3 \right)^{1/3} = 27^{1/3} = 3\\
    f^{-1}\left( f(-3) \right) &= \left( (-3)^3 \right)^{1/3} =
    (-27)^{1/3} = 3\,e^{i\pi/3}

    Essentially, this example shows that

    f(f^{-1}(x)) \neq f^{-1}(f(x))

    but this must be true for the inverse function! Thus, it would appear that all those websites which claimed an inverse for x^3 were wrong.

    What is going on here? When I look at f(x)=x^3, I see that it is onto and that it is 1-1. Could the problem be that f(x) = x^3 does not have an inverse when the range is the complex numbers---thus, it only has an inverse when [itex] f:\mathbb{R}\rightarrow \mathbb{R} [/itex] rather than [itex] f:\mathbb{R} \rightarrow \mathbb{C} [/itex]. Or is the problem that [itex] f^{-1}(x) [/itex] is only valid for x>0? If so, why does this restriction exist. Again, [itex] f(x) = x^3 [/itex] is 1-1 and onto.

    This problem came up while I was trying to show that the Legendre transformation of [itex] f(x) = x^\alpha / \alpha [/itex] was [itex] g(y) = y^\beta/\beta [/itex] where [itex] 1 = \alpha^{-1} + \beta^{-1} [/itex]. Since f must be convex, [itex] \alpha [/itex] must be even.

    To do this, I say that [itex] y = f'(x) = x^{\alpha - 1} [/itex]. This is x raised to an odd power. Basically, I have to solve for x in terms of y---which means I need to get rid of that exponent. I can get the result I seek, but it requires that I say that [itex] y^{1/(\alpha - 1)} = x [/itex]. I guess I don't feel comfortable with that statement...it seems like I am throwing away possible solutions.

    Last edited: Nov 21, 2004
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  3. Nov 21, 2004 #2

    matt grime

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    Before you start asking all these questions you ought to have written down the domain (and codomain).

    cubing is a bijection from R to R so it has an inverse, again from R to R.

    From R to C it cannot be invertible since it is not surjective, from C to C it cannot it be invertible since it is not injective.

    The domain (and codomain) is important, and you can't omit it if you are asking such questions.

    The inverse exists and is easy to write down as long as you do it sensibly.
  4. Nov 21, 2004 #3


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    This is true.

    You just said [itex] (x^a)^b = x^{ab} [/itex] for all real a and b, so why are you now saying [itex](x^a)^b \neq x^{ab}[/itex] if b is not an integer?

    So far so good and exactly what you would expect.
    Here you are moving into the complex numbers. In the complex numbers, any number has n n-th roots. There are n solutions to [itex]x^n=a[/itex]. (They say [itex]\sqrt[n]x[/itex] is a multi-valued or one-to-many 'function')
    The three values of [itex]\sqrt[3]{-27}[/itex] are [itex]3e^{i\pi/3}, 3e^{-i\pi/3}[/itex] and [itex]3e^{i\pi}[/itex] (which equals -3).
    You don't have this problem in the restriction to real numbers, where [itex](-27)^{1/3}[/itex] is unambiguously equal to -3 (the only real solution of the above 3).
    Last edited: Nov 21, 2004
  5. Nov 21, 2004 #4
    Ok...so it was a domain issue.

    Sorry for not being clear. I was actually trying to state that [itex] (x^a)^b \neq x^{ab} [/itex], in general for any real numbers [itex] a [/itex] and [itex] b [/itex]. This is why I then proceeded to show the conditions under which it was true. Concerning the case when [itex] b \notin \mathbb{Z} [/itex] consider this: [itex] (x^2)^{1/2} \neq x [/itex]. In fact, [itex] (x^2)^{1/2} = |x| [/itex] (I had a typo on my original post). Hence, we must restrict [itex] b [/itex] to an integer.

    However, as pointed out...I should state the domain and codomain. For all real numbers [itex] x [/itex], if we restrict our solutions to the real numbers, then we have the following specific cases:

    \text{If } b \in \mathbb{Z}, &\qquad\text{then } (x^a)^b = x^{ab}.\\
    \text{If } a \in \{2,4,6,\ldots\}, &\qquad\text{then } (x^a)^b = |x|^{ab}.\\
    \text{If } a \in \{2,4,6,\ldots\}, &\qquad\text{then } (x^a)^{1/a} = |x|.\\
    \text{If } a \in \{3,5,7,\ldots\}, &\qquad\text{then } (x^a)^{1/a} = x.\\

    And if when we include complex results, then the last statement is no longer true.

  6. Nov 22, 2004 #5
    How do you define [tex]x^a[/tex] when [itex]a[/itex] is irrational? [Hint: I would personally use the [itex]exp[/itex] and [itex]ln[/itex] functions.]

    Also, exponents and their properties are usually looked at for positive numbers. Your example:
    Your approach here is wrong. In this case, the function you're using is not an inverse. If a function is [tex]f : A \to B[/tex] and bijective, its inverse function [tex]f^{-1}[/tex] goes from [tex]B \to A[/tex] (by definition). Therefore, if you define a function to be [tex]f: \mathbb{R} \to \mathbb{R}^+[/tex] by [tex]f(x) = x^2[/tex], then that function has no inverse. That is because it is not bijective. Therefore you are taking two different functions that are not inverses of each other, and then taking their composition and expecting the identity. That is why your example doesn't really work out.

    In all above cases you should have used only positive real numbers as your domain and range for both functions. In this case, [tex]|x| = x[/tex].
    Last edited: Nov 22, 2004
  7. Nov 23, 2004 #6

    So [itex] 3^\pi = e^{3 \ln \pi} [/itex]...is that what you were suggesting?

    To evaluate, we could just expand [itex]e^x[/itex] in a Taylor series...though, we could also expand [itex] 3^x [/itex].
  8. Nov 23, 2004 #7
    Yes that is exactly what I am suggesting (except [tex]3^\pi = e^{\pi \ln 3}[/tex]). If you define [tex]\exp: \mathbb{R} \to \mathbb{R}^+[/tex] to be [tex]\exp(x) = \sum_{i=0}^\infty \frac{x^n}{n!}[/tex] then you can show it has an inverse (called [tex]\ln[/tex] usually). You can use these functions to show quite easilly all the properties you speak of. But the thing that is necessary is that you do it for positive numbers.

    We can then define [tex]a^b = e^{b \ln a}[/tex]. All properties of exponents can be proved pretty easilly. The one important fact is that [itex]a[/itex] must be greater than 0 because that is where [itex]\ln[/itex] is defined. In this fashion exponents can be defined for any real numbers but it does require that the numbers (not the exponents) must not be negative.
  9. Nov 24, 2004 #8


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    Just a minor quibble, cogito:
    Actually, I think you should define Exp(x) as:
    This is because it is unproblematic to define inductively the meaning of [tex]x^{n}[/tex] where n is a natural number, whereas the case [tex]x^{0}[/tex] ought to be a result:
    For [tex]x>0[/tex] we may define an inverse Ln, so that Exp(Ln(x))=x
    We may then define (for [tex]x>0[/tex] ) the a'th power of x as:
    It then trivially follows that [tex]x^{0}=1[/tex] for all [tex]x>0[/tex]

    (i've only considered the reals in this approach)
    Last edited: Nov 24, 2004
  10. Nov 24, 2004 #9
    Ahh that's pretty good. I've never seen this way but it really gets rid of that problematic first term. Thanks for the info.

    My only objection was that I thought at first that when you differentiate [itex]exp[/itex] you would then have to define [tex]x^0 = 1[/tex] so that the derivative would be equal but then I realized that [tex]x^0 = 1[/tex] is only dependent on the first [itex]exp[/itex] function so the derivatives are equal....or does this reasoning become circular somewhere?

    Either way, thanks for showing me this. It's sort of been a thing that's bothered me for quite a while.
  11. Nov 24, 2004 #10


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    I can't see that differentiation poses any trouble.
    A few ideas:
    1) At the start of our inductive definition of the natural powers of x, we DEFINE:
    [tex]x^{n}=x*x^{n-1}, n\geq{2}[/tex]
    2) For derivatives, we may derive the relations:
    3) Using these definitions when differentiating Exp(x) should eschew any troublesome nitpicks, if these exist.
  12. Nov 29, 2004 #11
    Yeah it didn't really seem like a problem to me. I just need to work out any issues before I'll accept something like this. There doesn't seem to be any issues (also less than my original definition) but I don't actually have any time to think about it. Either way thanks for showing me a new way of looking at it.
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