function whose cube is smooth

1 次查看（过去 30 天）
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

请先登录，再回答此问题。

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." ?

请先登录，再进行评论。

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

类别

在 Help Center 和 File Exchange 中查找有关 Polynomials 的更多信息

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by