function whose cube is smooth

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Amit Kumar
Amit Kumar il 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

Risposta accettata

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

Più risposte (2)

Walter Roberson
Walter Roberson il 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 Commenti
Amit Kumar
Amit Kumar il 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 il 16 Nov 2011
I do not recognize the term "diffeo." ?

Accedi per commentare.


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

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