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Thread: method of differnce

  1. #1
    Apr 2014

    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
    MHF Contributor

    Apr 2005

    Re: method of differnce

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

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

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

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