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Thread: 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,

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

    and I want to show that it is smooth. I know $\displaystyle g$ and $\displaystyle h$ are smooth. Does this imply that $\displaystyle f$ 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,

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

    and I want to show that it is smooth. I know $\displaystyle g$ and $\displaystyle h$ are smooth. Does this imply that $\displaystyle f$ 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

    $\displaystyle \left(
    \frac{2xz-y+yx^2+y^3+yz^2}{2(x^2+y^2)},
    \frac{-2yz-x+x^3+xy^2+xz^2}{2(x^2+y^2)} \right)$

    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

    $\displaystyle \left(
    \frac{2xz-y+yx^2+y^3+yz^2}{2(x^2+y^2)},
    \frac{-2yz-x+x^3+xy^2+xz^2}{2(x^2+y^2)} \right)$

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

    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 $\displaystyle z=7$, and $\displaystyle x=y$ and consider the limit as $\displaystyle x \to 0$.

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

    L'Hopital's rule is not applicable if $\displaystyle z \ne 0$ 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 $\displaystyle x=y$ the limit is $\displaystyle (\infty, \infty)$ for any value of $\displaystyle z \ne 0$. (and how come you now have two $\displaystyle z$'s?)

    L'Hopital's rule is not applicable if $\displaystyle z \ne 0$ 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 $\displaystyle z_1$ and $\displaystyle z_2$ are functions of $\displaystyle z$ which I couldn't be bothered writing out).
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