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Math Help - Showing a piecewise function is smooth

  1. #1
    MHF Contributor Swlabr's Avatar
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    Showing a piecewise function is smooth

    I have a function,

    f:= \left\{ \begin{array}{lr}<br />
(g(x, y, z), h(x, y, z)) & x \neq 0 \text{ or } y \neq 0 \\<br />
(0, 0) & x=y=0<br />
\end{array}<br />
\right.<br />

    and I want to show that it is smooth. I know g and h are smooth. Does this imply that f is smooth? If so, why?
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by Swlabr View Post
    I have a function,

    f:= \left\{ \begin{array}{lr}<br />
(g(x, y, z), h(x, y, z)) & x \neq 0 \text{ or } y \neq 0 \\<br />
(0, 0) & x=y=0<br />
\end{array}<br />
\right.<br />

    and I want to show that it is smooth. I know g and h are smooth. Does this imply that f is smooth? If so, why?
    No, not without further conditions, for instance g(x,y,z)=1, h(x,y,z)=1 are smooth, but f(x,y,z) is not even continuous.

    CB
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  3. #3
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by CaptainBlack View Post
    No, not without further conditions, for instance g(x,y,z)=1, h(x,y,z)=1 are smooth, but f(x,y,z) is not even continuous.

    CB
    Good point. If I told you the function was

    \left(<br />
\frac{2xz-y+yx^2+y^3+yz^2}{2(x^2+y^2)},<br />
\frac{-2yz-x+x^3+xy^2+xz^2}{2(x^2+y^2)} \right)

    would that help?...
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by Swlabr View Post
    Good point. If I told you the function was

    \left(<br />
\frac{2xz-y+yx^2+y^3+yz^2}{2(x^2+y^2)},<br />
\frac{-2yz-x+x^3+xy^2+xz^2}{2(x^2+y^2)} \right)

    would that help?...
    That does not look continuous to me, put z=7, and x=y and consider the limit as x \to 0.

    CB
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  5. #5
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by CaptainBlack View Post
    That does not look continuous to me, put z=7, and x=y and consider the limit as x \to 0.

    CB
    Forgetting the z=7, we have \left(\frac{z_1 + 2x^2}{4x}, \frac{z_2+2x^2}{4x}\right). If we apply l'Hopital's rule we surely get that this tends to (0,0) as x \rightarrow 0. So that isn't a counter-example...
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  6. #6
    Grand Panjandrum
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    Quote Originally Posted by Swlabr View Post
    Forgetting the z=7, we have \left(\frac{z_1 + 2x^2}{4x}, \frac{z_2+2x^2}{4x}\right). If we apply l'Hopital's rule we surely get that this tends to (0,0) as x \rightarrow 0. So that isn't a counter-example...
    Coming in along a ray where x=y the limit is (\infty, \infty) for any value of z \ne 0. (and how come you now have two z's?)

    L'Hopital's rule is not applicable if z \ne 0 as the numerators are not zero.

    CB
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  7. #7
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by CaptainBlack View Post
    Coming in along a ray where x=y the limit is (\infty, \infty) for any value of z \ne 0. (and how come you now have two z's?)

    L'Hopital's rule is not applicable if z \ne 0 as the numerators are not zero.

    CB
    Hmm...I had forgotten about the conditions for l'Hopital's rule!

    I understand now. That is quite annoying!

    (my z_1 and z_2 are functions of z which I couldn't be bothered writing out).
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