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Math Help - method of differnce

  1. #1
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    method of differnce

    the 2 equations above are from my books, can i modify the equation to the fourth one, and use it?method of differnce-dsc_0001-5-1-.jpg
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  2. #2
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    Re: method of differnce

    If you are asking "Can I go from \sum [f(r)- f(r+1)] to \sum [f(r- 1)- f(r)?" the answer is "that depends".

    It depends upon exactly what that sum is- what values of r is it over?

    If you have \sum_{r= a}^b [f(r)- f(r+ 1), then letting n= r+1, so that r= n- 1, f(r)- f(r+ 1) becomes f(n- 1)- f(n). But when r= a, n= a+ 1 and when r= b, n= b+ 1. So we have \sum_{r= a}^b [f(r)- f(r+ 1)]= \sum_{n= a+1}^{b+1} [f(n)- f(n-1)]. Of course we could change the "dummy index" at will so we can write \sum_{r= a}^b [f(r)- f(r+ 1)]= \sum_{r= a+1}^{b+1} [f(r)- f(r-1)].
    But notice that the limits on the sum have changed.

    Of course, if the sum were from -\infty to \infty that wouldn't matter.
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