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Directional derivative without geometry

  1. Oct 2, 2006 #1
    I have checked not few books in regard to directional derivatives. Practically all of them use vectors in the derivative definition. I would like to read a good explanation of this concept but keeping out the geometrical interpretations.

    Suppose we got a function of two variables. I understand that a directional derivative has to do with increments of the two variables. Would someone mind explain how to build this concept employing only analytical concepts, without turning to directions in a plane? Or maybe my wish is a nonsense ?
     
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  3. Oct 3, 2006 #2

    HallsofIvy

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    I'm not sure how one would define a directional derivative without mentioning direction. Exactly what is it you want defined?
     
  4. Oct 3, 2006 #3
    It is possible that what I asking for has no sense. I dont know.:confused:

    I would like to know an approach based on this: a partial derivative is merely a simple derivative because only one independent variable increases (or decreases) and the other one is fixed, but obviously there are infinity of cases when the two independent variables vary. (I am restricting to functions of two variables). I understand that the concept of directional derivative meets these cases, and I would like to know if it is possible to grasp the concept without resorting to geometry concepts.:uhh:
     
  5. Oct 12, 2006 #4
    Uh.. the question is can we "define" for an scalar function in n-dimensions

    [tex] f(x_1 ,x_2 , x_3,..............,x_n) [/tex] and study its "directional

    integral" [tex] D^{-1}_{v} [/tex] in the direction of vector "v" so it satisfies that:

    [tex] (D^{-1}_{v}o D_{v})f=f [/tex]
     
  6. Oct 12, 2006 #5

    Hurkyl

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    Vectors are algebraic objects. If you don't want to deal with a geometric interpretation, then don't interpret your vectors geometrically!
     
  7. Oct 12, 2006 #6

    HallsofIvy

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    All I can say is that since "direction" is itself a geometric concept, I don't know what you mean by "directional derivative" without using geometry!
     
  8. Oct 12, 2006 #7

    Hurkyl

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    A name is just a name! Algebra borrows from geometry all the time, and vice versa. :smile: The algebraic definition of the directional derivative is, after all, algebraic, and doesn't require one to invoke a notion of geometry.

    That said, I don't think it's fruitful to discard the geometric interpretation, even when working with an abstract vector space -- IMHO the notion of "direction" is intuitively useful, even though it would have to be abstracted.
     
  9. Oct 13, 2006 #8
    also, IMHO, i seem to see the trend that we always argue on terms and vocabularies... lol
     
  10. Oct 14, 2006 #9
    Please, can you post what you call the algebraic definition of the directional derivative?

    Thanks.
     
  11. Oct 14, 2006 #10

    Hurkyl

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    [tex]
    \nabla_{\vec{v}} f(\vec{x}) =
    \lim_{t \rightarrow 0} \frac{f(\vec{x} + t \vec{v}) - f(\vec{x})}{t}
    [/tex]

    Or, less rigorously, it is the function that satisfies

    [tex]
    f(\vec{x} + t\vec{v}) \approx f(\vec{x}) + t\nabla_{\vec{v}} f(\vec{x})
    [/tex]

    Equivalently,

    [tex]
    \nabla_{\vec{v}} f(\vec{x}) = (\nabla f(\vec{x})) \vec{v}
    [/tex]
     
    Last edited: Oct 14, 2006
  12. Oct 15, 2006 #11
    can't we define its "inverse" to get the "Directional integral"?.

    I'm amazed with this, for almost every space (finite infinite even functional ones) you can define a "derivative" however you find lots of problems in finding a meaning for "integration" in most of cases.
     
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