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Definition2 of derivative

  1. Sep 11, 2006 #1
    "Definition2 of derivative...

    We have the defintion (taking the limit) for hte derivative:

    [tex] \frac{f(x)-f(a)}{x-a} [/tex] for an Euclidean Space...

    But what keeps us from defining another metric (on an Euclidean or other abstract space) so the derivative takes de form:

    [tex] \frac{(df(x),f(a))}{d(x,a} [/tex] so "d" is a distance in the form that you can define "infinitesimal2 elements on an space and that for x=a ,d=0

    Then the "abstract" definition of integral is:

    [tex] \sum_{i} f(X_i ) d(X_{i+1},X_{i}) [/tex]

    of course in the limit that the distance [tex] || X_{i+1}-X_í}||\rightarrow 0 [/tex] and ¿what happens if we had an "Infinite" dimensional space...so

    - it is "numerable" (ie: R^{n})
    - it's not "numerable" (function space)
  2. jcsd
  3. Sep 11, 2006 #2


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    Do you ever proofread your posts? :grumpy:

    I don't see any technical problem with defining

    f'(a) := \lim_{d(x, a) \rightarrow 0} \frac{d(f(x), f(a))}{d(x, a)}

    on an arbitrary metric space... but it won't really behave algebraically like a derivative. I don't know if it will exist very often for something that doesn't look like a linear space. It also doesn't agree with the ordinary derivative on Euclidean space: by your definition, the derivative of -x is 1. (and not -1) (because, for the real numbers, d(x, y) = |x - y|)
    Last edited: Sep 11, 2006
  4. Sep 11, 2006 #3
    But the metric space has to be smooth right?
  5. Sep 11, 2006 #4


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    The definition makes literal sense for any metric space. (Though I suspect such limits might not exist very often)

    I'm not sure what a "smooth metric space" is -- if you've heard that, then that's probably a good criterion that ensures lots of these "derivatives" will exist.
  6. Sep 11, 2006 #5


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    What is (or is not) a "smooth" metric space?

    A standard definition, given in most "Calculus III" courses, for the derivative is

    A function, f:X-> Y, where X and Y are metric spaces, is said to be "differentiable at a" if and only if there exist a linear function, L:X->Y, and a function [itex]\epsilon[/itex] such that
    [tex]f(x)= f(a)+ L(x-a)+ \epsilon(x-a)[/tex]
    [tex]\lim_{x\rightarrow a}\epsilon(x-a)||x-a||= 0[/tex]

    What more do you want?
  7. Sep 11, 2006 #6


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    Hrm. I had assumed that he wanted to talk about derivatives for something that isn't a linear space, but now that I read again, that wasn't a good assumption!
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