Talk:Summation by parts

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First, there is a typo in the formula for the partial sums, in the last formula of the "Methods" section, the first b_N should be a B_N. Next, I believe that the hypotheses in the "Applications" section is not sufficient for the result stated. In the current case, you can take the sequence a_n to be a constant sequence, say identically 1, and the sequence b_n to be the following: 1, -1/2, -1/2, 1/3, 1/3, 1/3, ... (there are n terms each of 1/n, with alternating signs between each group). The sequence a_nb_n approaches 0, the series (a_(n+1)-a_n) is absolutely convergent, and the partial sums B_n are uniformly bounded (they are all between 0 and 1). However, the sum a_nb_n does not converge, since it is just the sum of b_n which oscillates infinitely often between 0 and 1. The correct hypothesis for convergence should be to replace "a_nb_n approches 0", by the condition that "a_n approaches 0". This can be seen by writing the Cauchy sequence of the partial sums in the summation by parts formulation and then estimating each part.

I think there's some typo in the second equation, introducing the \Delta operator. The first term on the right hand side, f_{k+1} g_{k+1} should be f_{n+1} g_{n+1} - f_m g_m. The summation variable k is not bound... / nisse@lysator.liu.se

[edit] formula for shifted index

The formula is listed for this sum:

\sum_{k=m}^n f_k(g_{k+1}-g_k)

but I think it'd be nice to also list the formula for this sum:

\sum_{k=m}^n f_{k+1}(g_{k+1}-g_k)

which is slightly different, but in the same general form. Lavaka (talk) 18:11, 15 January 2008 (UTC)

which results from iterated application of the initial formula.



"The auxiliary quantities are Newton series:" Here, the link to the newton series is actually a link to the difference operator. I think it should be fixed. I don't know how though..