function whose cube is smooth

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

 채택된 답변

Jan
Jan 2011년 11월 15일

0 개 추천

The cube of smooth function is smooth.

추가 답변 (2개)

Walter Roberson
Walter Roberson 2011년 11월 15일

0 개 추천

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.

댓글 수: 2

Amit Kumar
Amit Kumar 2011년 11월 16일
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 2011년 11월 16일
I do not recognize the term "diffeo." ?

댓글을 달려면 로그인하십시오.

Amit Kumar
Amit Kumar 2011년 11월 16일

0 개 추천

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

카테고리

도움말 센터File Exchange에서 Polynomials에 대해 자세히 알아보기

태그

질문:

2011년 11월 15일

Community Treasure Hunt

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

Start Hunting!

Translated by