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Evaluation of the derivative

  1. Nov 3, 2006 #1
    Let be a function so the derivative of any order n exist, my question is if there is a way to evaluate:

    [tex] \frac{d^{n}f(x)}{dx^{n}} [/tex] as [tex] n\rightarrow \infty [/tex]
  2. jcsd
  3. Nov 3, 2006 #2


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    Depends on the function. If it's ex, it's trivial. If it's a polynomial, it's trivial. If it's cos(x), it can't be done (cyclic, so there's no limit). If it's something that's a complete mess of a rational function, you're probably righteously too lazy to figure it out
  4. Nov 3, 2006 #3


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    Why should it exist some general way apart from differentiating the function?
    The derivatives of the functions f(x)=cos(x) g(x)=1 "evolve" in totally different manners.
    Last edited: Nov 3, 2006
  5. Nov 4, 2006 #4
    well i was thinking about "Cauchy's theorem" so you have that the derivative of any order should satisfy:

    [tex] 2\pi i f^{(n)}(a)= n! \oint_{C}dzf(z)(z-a)^{-n-1} [/tex]

    and from this integral, if C is a circle of unit radius and centered at z=a then, the contour integral just becomes:

    [tex] 2\pi f^{(n)}(a)= n! \int_{-\pi}^{\pi} dxf(e^{ix}+a)e^{-inx} [/tex]

    EDIT: another good idea would be perhaps to use the "generalized difference" operator so..

    [tex] f^{(n)}(x)=\frac{\nabla ^{n}}{h^{n}} [/tex] as h-->0 (small h)
    Last edited: Nov 4, 2006
  6. Nov 4, 2006 #5


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    Why do you assume that Cauchy's theorem in general gives you a neat way to actually evaluate the derivative??

    You are not Eljose, are you?
  7. Nov 4, 2006 #6
    I'm supposing that f(z) is analytic with NO poles or at least that the point z=a is not a pole of f, then if f has no poles on the unit circle centered at z=a then "Cauchy Theorem" for derivatives holds.

    I don't know what's this stuff about someone called eljose ..and what has to do with me or the forums....
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