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function whose cube is smooth

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Amit Kumar
Amit Kumar 2011-11-15
Hi,
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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采纳的回答

Jan
Jan 2011-11-15
The cube of smooth function is smooth.

更多回答（2 个）

Walter Roberson
Walter Roberson 2011-11-15
Note that the polynomial roots does not necessarily have to be restricted to reals in order to map R->R . For example,
x*(x-i)*(x+i)
is
x*(x^2+1)
which is
x^3 + x
which maps R -> R
One cannot simply say "polynomials" because not every polynomial with complex roots is going to map R -> R . One thus might need to characterize which polynomials with complex roots are suitable.
  2 个评论
Amit Kumar
Amit Kumar 2011-11-16
Is my concept right ???
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.
Walter Roberson
Walter Roberson 2011-11-16
I do not recognize the term "diffeo." ?

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Amit Kumar
Amit Kumar 2011-11-16
i just want to ask what are the functions whose cube is smooth?

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