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Math Help - [SOLVED] The Gradient Vector

  1. #1
    Member billa's Avatar
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    [SOLVED] The Gradient Vector

    I have to show that if one of the following statements is true, then the other statement is also true.
    Let f be a scalar field and let P be a point in the domain of f.

    There is some Cartesian system in which f is represented by a function which has a linear approximation at P.

    For every choice of Cartesian system, f is represented by a function which has linear approximation at P.


    I don't really know what either statement means by itself, so it is difficult to show how one implies the other. In particular, what does it mean to choose a different Cartesian system (that sounds like just moving the origin around). Further, I don't know what the book is saying in an earlier paragraph when they say "the gradient vector can be visualized without reference to a particular coordinate system"

    The only thing that I do know is that to have a linear approximation at a point means that the function is both continuous at the point and has partial derivatives at the point (there might be cases where this is not enough to say that a linear approximation exists however).

    Thanks for helpful responses
    Last edited by billa; June 9th 2009 at 08:35 AM. Reason: changed to better reflect my meaning
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  2. #2
    Member Ruun's Avatar
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    For some function f \in C^1, \nabla f = (\frac{\partial f}{dx},\frac{\partial f}{dy},\frac{\partial f}{dz}) in rectangular coordinates. From the definition of derivate can be shown that \nabla f it's a good aproximation of f. Further reading, Marsden & Tromba Vector Calculus, section 2.3 differentiation.
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  3. #3
    Member billa's Avatar
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    I don't know what to say
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  4. #4
    Member Ruun's Avatar
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    A)There is some Cartesian system in which f is represented by a function which has a linear approximation at P.

    B)For every choice of Cartesian system, f is represented by a function which has linear approximation at P.

    If we prove that B is true, then A is true.

    What book are you talking about? It will be helpful to know, for useful help lol. Sorry for the non-helping previous post.
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  5. #5
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    I don't like the way this question is worded- there is no statement of how good the approximation is. "Having a linear approximation" doesn't mean anything by itself. Any function can be "approximated", perhaps with large error, by a linear or even constant function.

    But if that linear approximation has the property that E(h), the error within distance h of the point, goes to 0 faster than h itself, then the function is differentiable at that point. And being differentiable is a property of the function, not the coordinate system. If a function is differentiable at a point in one coordinate system, it is differentiable there in any coordinate system.
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