- #1

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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?

Thanks

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- #1

- 39

- 0

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?

Thanks

- #2

Science Advisor

- 1,089

- 10

Do you have any special form in mind? You could just take the cube root of smooth functions.

- #3

- 39

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i just want to ask what are the functions whose cube is smooth?

- #4

- 136

- 1

Its just the cube root of any smooth function, what else do you want to know? How to characterise a smooth function? http://en.wikipedia.org/wiki/Smooth_function

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- #5

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- #6

Science Advisor

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f(x)=x is smooth, but f(x)=x

- #7

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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).

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).

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- #8

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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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