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A Action of an anti-difference operator

  1. Aug 27, 2016 #1
    hi, i want to ask you that how to perform an anti-difference operator on the product of two functions? i.e.
    D^{-1}(f_{n} g_{n})
     
  2. jcsd
  3. Aug 27, 2016 #2

    Math_QED

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    What do you mean, find the antiderivative of a function ## (fg)(x) ##? There is no standard formula for this, but you might want to google: integration by parts.
     
  4. Aug 27, 2016 #3
    no this is not a usual calculas, this is discrete calculus, i.e. Df_{n} = f_{n+1}-f_{n}, where D is the forward difference operator, then how an anti-difference act on the product of two discrete functions?
     
  5. Aug 28, 2016 #4

    Svein

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    An anti-difference operator is a sum (operator).
     
  6. Aug 29, 2016 #5
    ok, if it is a sum operator then how it is act?
     
  7. Aug 29, 2016 #6

    Stephen Tashi

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    As math_QED said, there is no general formula.

    In the special case where ##g(n) = \triangle h(n) ## we can derive a formula. I suspect it is just a special case of the formula for "summation by parts".

    Integration by parts is derived from the formula for differentiating a product.

    By analogy, try:

    ##\triangle (f(k) h(k)) = f(k+1) h(k+1) - f(k) h(k) ##
    ## = f(k+1) h(k+1) - f(k)h(k+1) + f(k)h(k+1) - f(k) h(k) ##
    ##= h(k+1)\triangle f(k) + f(k)\triangle h(k)##

    ##f(k) \triangle h(k) = \triangle (f(k) h(k)) - h(k+1) \triangle f(k) ##

    ## \triangle^{-1} ( f(k) \triangle h(k) ) = \triangle^{-1} \triangle ( f(k) h(k)) - \triangle^{-1} ( h(k+1)\triangle f(k) ) ##
    ## \triangle^{-1} ( f(k) \triangle h(k)) = f(k) h(k) - \triangle^{-1} ( h(k+1) \triangle f(k) ) ##


    So if ##g(k) = \triangle h(k) ##

    ##\triangle^{-1} ( f(k) g(k) ) = f(k) h(k) - \triangle^{-1} ( h(k+1) \triangle f(k) ) ##
     
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