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function whose cube is smooth

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Amit Kumar
Amit Kumar le 15 Nov 2011
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

Réponse acceptée

Jan
Jan le 15 Nov 2011
The cube of smooth function is smooth.

Plus de réponses (2)

Walter Roberson
Walter Roberson le 15 Nov 2011
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 commentaires
Amit Kumar
Amit Kumar le 16 Nov 2011
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 le 16 Nov 2011
I do not recognize the term "diffeo." ?

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

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