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Taylor series in terms of discrete derivative

  1. Feb 25, 2014 #1
    All analitic function can be express how: [tex]f(x) = \frac{1}{0!} \frac{d^0f}{dx^0}(x_0) (x - x_0)^0 + \frac{1}{1!} \frac{d^1 f}{dx^1}(x_0) (x - x_0)^1 + \frac{1}{2!} \frac{d^2f}{dx^2}(x_0) (x - x_0)^2 + \frac{1}{3!} \frac{d^3f}{dx^3}(x_0) (x - x_0)^3 + ...[/tex] that is the taylor series of the function f(x).

    Analogously, should exist a taylor series that use the discrete derivative ##\frac{\Delta f}{\Delta x} = f[x+1] - f[x]##. I found this NumberedEquation4.gif in wolframpage http://mathworld.wolfram.com/ForwardDifference.html but appears strange. I didn't undertood what such series means. Someone can explain me?
     
    Last edited by a moderator: May 8, 2017
  2. jcsd
  3. Feb 26, 2014 #2

    pwsnafu

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    I'm going to use notation from time scale calculus, which is a unification of continuous and discrete calculus. It makes my message more obvious.

    A time scale is a non-empty closed subset ##\mathbb{T}\subset\mathbb{R}##.
    Given a function ##f : \mathbb{T}\to\mathbb{R}##, we define the (Hilger) derivative ##f^{\Delta}## in such a way such that
    • if ##\mathbb{T} = \mathbb{R}## then Hilger derivative is equal to ##f^{\Delta}(t) = f'(t)## and
    • if ##\mathbb{T} = \mathbb{Z}## then Hilger derivative is equal to ##f^{\Delta}(t) = \Delta f(t)##.

    To study the Taylor theorem, we also have "h-polynomials" which are defined
    ##h_{0}(t,s) = 1## and
    ##h_{k+1}(t,s) = \int_s^t h_k(\tau, s) \, \Delta \tau##.
    In words "start with the constant and keep integrating over and over".
    • if ##\mathbb{T} = \mathbb{R}## then ##h_k(t,s) = \frac{(t-s)^k}{k!}## and
    • if ##\mathbb{T} = \mathbb{Z}## then ##h_k(t,s) = \frac{(t-s)(t-s-1)(t-s-2)\ldots(t-s-k+1)}{k!}##

    The Taylor theorem itself is
    ##f(t) = \sum_{k=0}^{n-1}f^{\Delta^k}(a)\, h_{k}(t, a) \, + R##
    where R is the remainder (I'm not going to write down the remainder term).
    • if ##\mathbb{T} = \mathbb{R}## then
      ##f(t) = f(a) + f'(a) \frac{t-s}{1!} + f''(a) \frac{(t-s)^2}{2!}+\ldots+f^{(k)}(a) \frac{(t-s)^k}{k!} + \ldots## and
    • if ##\mathbb{T} = \mathbb{Z}## then
      ##f(t) = f(a) + \Delta f(a) \frac{t-s}{1!} + \Delta^2 f(a) \frac{(t-s)(t-s-1)}{2!}+\ldots+\Delta^{k}f(a) \frac{(t-s)(t-s-1)\ldots(t-s-k+1)}{k!} + \ldots##.


    The important thing is that in discrete calculus these ##\frac{(t-s)(t-s-1)(t-s-2)\ldots(t-s-k+1)}{k!}## serve the same role as ##\frac{(t-s)^k}{k!}## does in the continuous calculus.
     
  4. Feb 26, 2014 #3
    good answer!
     
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