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Thread: smooth function

  1. #1
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    smooth function

    show that

    $\displaystyle f(x,y,z)=\frac{1}{z^2+1}(x^2-y^2,2xy,2z)$ defines a smooth function $\displaystyle f: S^{2}\to S^{2}$ where $\displaystyle S^{2}=\{p\in \mathbb{R}^3, |p|=1\}$ so I am guessing it is a unit sphere.

    How do we proceed with such problem?

    f is a smooth function if all partial derivatives of all possible orders are defined in all points of the domain of f right?

    I started with computing these partial derivatives

    $\displaystyle \frac{\partial f}{\partial x}=\frac{1}{z^2+1}(2x,2y,0)$


    $\displaystyle \frac{\partial f}{\partial y}=\frac{1}{z^2+1}(-2y,2x,0)$


    $\displaystyle \frac{\partial f}{\partial z}=\frac{-2(z^2-1)}{(z^2+1)^2}$

    and these partial derivatives are all defined at any point from the domain of f

    so I compute second order partial derivatives

    $\displaystyle \frac{\partial^2 f}{\partial x^2}=\frac{2}{z^2+1}$

    $\displaystyle \frac{\partial^2 f}{\partial y^2}=\frac{-2}{z^2+1}$

    $\displaystyle \frac{\partial^2 f}{\partial z^2}=\frac{4z(z^2-3)}{(z^2+1)^3}$

    and these 3 are also defined for all the points of the domain.



    if I continue differentiating

    $\displaystyle \frac{\partial^3 f}{\partial x^3}=\frac{\partial^3 f}{\partial y^3}=0$ and

    $\displaystyle \frac{\partial^3 f}{\partial z^3}=-\frac{12(z^4-6z^2+1)}{(z^2+1)^4}$ which is defined for all points

    What should I do next? can I conclude based on the calculations that f is certainly smooth?


    Thanks in advance
    Last edited by rayman; Feb 18th 2013 at 01:53 AM.
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  2. #2
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    Re: smooth function

    f is obviously smooth on R^3. So what you need to show is, when f is restricted to S^2, its image is in S^2.
    That is, to show that [(x^2-y^2)^2 + (2xy)^2 + (2z)^2]/(1+z^2)^2 = 1, given that x^2+y^2+z^2=1.
    Which is easy to show.
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  3. #3
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    Re: smooth function

    so showing all the computations above were not necessary if I am supposed to restrict myself to $\displaystyle S^{2}$ right?
    But I still don't get how to show $\displaystyle \Biggl[(x^2-y^2)^2+(2xy)^2+(2z)^2\Biggr]\frac{1}{(z^2+1)^2}=1$ how does this imply that f will be smooth on the sphere???
    Last edited by rayman; Feb 18th 2013 at 10:25 AM.
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  4. #4
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    Re: smooth function

    $\displaystyle (x^2-y^2)^2 + (2xy)^2 + (2z)^2=(x^2+y^2)^2 + (2z)^2$
    =$\displaystyle (1-z^2)^2+(2z)^2=(1+z^2)^2$

    This shows that $\displaystyle f|_{S^2}$ indeed maps to $\displaystyle S^2$.

    Then $\displaystyle f$ is smooth because "the restriction of a smooth function on a smooth sub-manifold is smooth".
    To prove the above statement, let $\displaystyle i: S^2 \rightarrow R^3$ is the smooth embedding of $\displaystyle S^2$ in $\displaystyle R^3$,
    $\displaystyle f : R^3 \rightarrow S^2 $ is a smooth map on $\displaystyle R^3$, then $\displaystyle f|_{S^2} = f \circ i$ is smooth on $\displaystyle S^2$.
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