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Discrete derivatives with finite-differences

  1. May 4, 2010 #1
    Hello,
    I have a function in discrete domain [tex]f:\mathbb{Z}\rightarrow \mathbb{R}[/tex], and I assume that f is an approximation of another differentiable function [tex]g:\mathbb{R}\rightarrow \mathbb{R}[/tex].
    In other words [tex]f(n)=g(n)[/tex], [tex]n\in \mathbb{Z}[/tex].

    When one wants to approximate the first derivative of g, it is possible to use the forward difference or backward difference operators, which are respecively:

    [tex]\Delta f(n)=f(n+1)-f(n)[/tex]
    [tex]\nabla f(n)=f(n)-f(n-1)[/tex]

    My question is: is it common or allowed to use a mixture of these two operators in the following way:

    [tex]g'(x) \approx \Delta f(x)[/tex] for [tex]x\geq 0[/tex]
    [tex]g'(x) \approx \nabla f(x)[/tex] for [tex]x<0[/tex]
     
  2. jcsd
  3. May 4, 2010 #2

    HallsofIvy

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    No, it is not "common" and I can see no reason to do that. Why would you use one for positive x and the other for negative x? It is the change that is positive or negative in the two differences and whether x is positive or negative has nothing to do with that.
     
  4. May 4, 2010 #3
    I would like to use different operators for positive/negative x's in order to obtain a symmetry I need. In fact, in my case the continous function g has the property:

    [tex]g'(x)=-g'(-x)[/tex], for [tex]x\in \mathbb{R}[/tex]

    If f(n)=g(n), [itex]n\in \mathbb{Z}[/itex], you get [tex]\Delta f(n) \neq -\Delta f(-n)[/tex].
    Instead, given g I was able to obtain:

    [tex]\Delta f(n) = -\nabla f(-n)[/tex], [itex]n\in \mathbb{Z}^+ - \{ 0 \}[/itex]

    EDIT: in simpler words g is an even function and its first derivative is an odd function. The discretized f is even too, but [itex]\Delta f[/itex] is not odd.
     
    Last edited: May 4, 2010
  5. May 4, 2010 #4
    You could presumably use [itex]\frac{1}{2}(\Delta f+\nabla f)=(f(n+1)-f(n-1))/2[/itex], but that may be too much arithmetic (the shift operation is usually quite quick if you're doing it by computer).
     
    Last edited: May 4, 2010
  6. May 5, 2010 #5
    Thanks Martin,
    yours is probably a better idea. In fact, the central-difference operator keeps the property "f is even [itex]\Rightarrow[/itex] f' is odd".
     
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