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Finite differences on scalar? Matrix?

  1. Sep 10, 2010 #1

    In a paper I have

    [tex]v_{n,k} = \Delta^K ( (-1)^n n^k y_n )[/tex]

    with [tex]n = K, \dots , N-1[/tex], [tex]k = 0, \dots, K[/tex] and [tex]N = 2K[/tex]

    where [tex]\Delta^K[/tex] is the Kth finite difference operator.

    As you can see, all [tex]v_{n,k}[/tex] consistute an [tex](N-K) \times (K+1)[/tex] matrix.

    So without the [tex]\Delta[/tex]'s, each [tex]v_{n,k}[/tex] would be a scalar. I do not see how to calculate the finite difference of a scalar?!

    Well, probably it is not a finite difference. But can anybody tell me what could be meant with that?

  2. jcsd
  3. Sep 11, 2010 #2


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    I don't understand. You say that [itex]\Delta^K[/itex] is the "Kth finite difference operator" and ask what would [itex]v_{n, k}[/itex] be without the [itex]\Delta[/itex]? It would no longer be a finite difference!

    (The finite difference of a scalar function, at some n, would be f(n+1)- f(n).)
    Last edited by a moderator: Sep 12, 2010
  4. Sep 11, 2010 #3

    Yes, this is exactly my question! I do not understand how it is meant!

    Maybe you can take a look at http://biblion.epfl.ch/EPFL/theses/2001/2369/EPFL_TH2369.pdf, page 55-56.

    I again try to explain: Suppose you have a sequence [tex]y_n[/tex] consisting of N values. Now consider the expression

    [tex](-1)^n P(n) y_n[/tex] where [tex]P(n)[/tex] is a polynomial of order K in the variable n. The authors argue that this term vanishes if K finite differences are applied. This yields equation 3.14-3.16 in the link above and can be written as matrix equation:

    \Delta^K ((-1)^n P(n) y_n) = \sum_{k=0}^K p_k \underbrace{\Delta^K ((-1)^n n^k y_n)}_{v_{n,k}} = \mathbf{V} \cdot \mathbf{p} = 0

    My question is how to calculate the matrix V. The algorithm 3.1 in the link above tells me exactly what I have asked in the original post but I do not know how to actually calculate it.

    It would be easier to understand if there would be just one dimension.

    But this brings me to an idea: The finite difference is the difference dependent on n, isn't it?

    So for the first column I set k=0 and have the sequence [tex](-1)^n n^0 y_n[/tex] and the column are just the discrete differences of the sequence?

    E.g. if [tex](-1)^n n^0 y_n = \left\{1, 4, 9, 8, 10\right\}[/tex] then the first column would be [tex]\left[3 , 5 , -1 , 2\right]^T[/tex]?

    And the same for columns 1,...K?

    Maybe this is the solution?

    Regards, divB
    Last edited by a moderator: Apr 25, 2017
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