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function whose cube is smooth

조회 수: 1 (최근 30일)
Amit Kumar
Amit Kumar 2011년 11월 15일
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
The cube of smooth function is smooth.

추가 답변 (2개)

Walter Roberson
Walter Roberson 2011년 11월 15일
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." ?

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

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