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I find it difficult to remember the different forms of summation by parts: where the indices begin, end, whether to take forward/backward differences, etc. For example, Wikipedia has one form $$\sum_{k=m}^n f_k(g_{k+1}-g_k) = \left[f_{n+1}g_{n+1} - f_m g_m\right] - \sum_{k=m}^n g_{k+1}(f_{k+1}- f_k)$$ and I've seen another $$\sum_{k=m}^n f_k(g_k-g_{k-1}) = [f_n g_n - f_mg_{m-1}] - \sum_{k=m}^{n-1} g_k (f_{k+1} - f_k)$$ Even though I can derive one from the other, it takes some time. So I'm wondering if all this can be easily seen in a general setting with counting measure. Or perhaps if there are some nice heuristics which can help me remember the different forms quickly.

user879123
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1 Answers1

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The same proof idea should work. For example, starting with

$$\Delta(f_n g_n) = f_{n+1} g_{n+1} - f_n g_n $$

and using any of the usual tricks to convert this into, e.g.,

$$\ldots = f_{n+1} \Delta g_n + g_n \Delta f_n $$

and then you can just add up both sides.

With a little practice, you can quickly go from some $f_n \Delta g_{n+7}$ and then figure out what you need to add in to "complete the delta"; e.g. what you need to add to make it $\Delta(f_n g_{n+7})$. (Why is a good delta to complete it to? Because both have $f_n g_{n+7}$ with a negative coefficient)

I'm used to $\Delta$ always meaning a forward difference; if you pick one meaning for $\Delta$ and stick with it, it's easier to work with these things.

  • Aha. So I have the following "rule": $f_k \Delta g_k + \Delta f_k g_{k+1} = \Delta(f_k g_k)$, and $f_k \Delta g_k + \Delta f_{k-1} g_k = \Delta(f_{k-1} g_k)$. So in finding "completing the delta" we can proceed as follows:

    1. Move the $\Delta$ to the other factor, then

    2. Increment the index of the original $\Delta$ term, or decrement the new $\Delta$ term.

    This will give $\Delta(f_m g_n)$, where $m, n$ are the minimums of the indices seen in the sum.

     – user879123 Mar 22 '14 at 09:34
  • Moreover, using this rule gives similar forms for both the first and second equation in the question. The second equation just has an additional cancellation of terms, so they seem more different than they actually are. – user879123 Mar 22 '14 at 09:41
  • Right, and one advantage of using the forward difference is that you get a nicer expression for the Newton series. – user21820 Jul 08 '15 at 09:20