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Function whose cube is smooth

  1. Oct 24, 2011 #1

    I want to charectize the function whose cube is smooth from R to R. For example x^1/3 is smooth and olsa any polynomial but how can i charectrize it?

  2. jcsd
  3. Oct 25, 2011 #2


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    Do you have any special form in mind? You could just take the cube root of smooth functions.
  4. Oct 25, 2011 #3
    i just want to ask what are the functions whose cube is smooth?
  5. Oct 25, 2011 #4
    Last edited: Oct 25, 2011
  6. Oct 30, 2011 #5
    Wait, isn't the function f(x)=x^3 a diffeomorphism? So any function for which its cube is smooth must be smooth itself (just apply the inverse of f to it).
  7. Nov 1, 2011 #6


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    f(x)=x is smooth, but f(x)=x1/3 is not. And f(x)=x3 is not a diffeo. from ℝ to ℝ , since its inverse x1/3 is not differentiable at 0.
  8. Nov 2, 2011 #7
    Oops, oh yeah, the OP wrote that and I just blurted it out without checking.

    It is a local diffeomorphism elsewhere though, so if we're looking for an example of a function which isn't smooth but whose cube is, it'll have to be smooth everywhere except the origin where it must not be but is after we've cubed it. I'm not sure how to characterise such things- but they clearly exist e.g. the cube root of x is such a function- it is not smooth but its cube is (you can also have the cube root of any smooth function as such a function).
    Last edited: Nov 2, 2011
  9. Nov 2, 2011 #8
    Wait, clearly all such functions are of this form:

    Suppose we have a function f for which its cube is smooth. This means that f^3 is smooth. But then the cube root of f^3 is the original function and is in the form "cube root of a smooth function". Conversely, the cube root of a smooth function will clearly cube to a smooth function, so there's your answer. Funny how sometimes you can completely miss the obvious!
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