Talk:Riemann sum

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[edit] Merge?

I think the Riemann sum and the Riemann integral have too much in common. I suggest merging information from them into Riemann integral, and make Riemann sum a redirect. Please see discussion at talk:Riemann integral.(Igny 21:51, 5 December 2005 (UTC))

[edit] Clarification

I think it would be worth clarifying that the distance between the points (x1, x2, xi-1, xi) have to be a uniform distance, and it is done for simplicities sake. The graphs would give that idea as well, which isn't really true of Riemann's original "Riemann sums." --AstoVidatu 04:47, 7 December 2006 (UTC)

[edit] Error Estimation

Are the error estimation formulas correct for the "middle sum" and "trapezoidal sum" methods? The "middle sum" error estimate is currently quoted as :\left \vert \int_{a}^{b} f(x) - A_\mathrm{mid} \right \vert \le \frac{M_2(b-a)^3}{(24n^2)},

...and the "trapezoidal sum" error estimate is :\left \vert \int_{a}^{b} f(x) - A_\mathrm{trap} \right \vert \le \frac{M_2(b-a)^3}{(12n^2)},

I don't have a calculus book handy, but it doesn't make intuitive sense that the "trapezoidal sum" error could be twice the size of the "middle sum" error. Is this right? Is there a handy reference online where these formulas are derived?

--Imperpay 22:30, 26 March 2007 (UTC)


The formulae are correct. http://people.hofstra.edu/stefan_Waner/realworld/integral/numint.html Accuracy of Trapezoid and Simpson Approximations lists the formula for the trapezoidal error as \left \vert \int_{a}^{b} f(x) - A_\mathrm{trap} \right \vert \le \frac{M_2(b-a)^3}{(12n^2)},


It may not be intuitive, but oddly enough, the consideration of multiple derivatives and the fact that the middle sum method overlaps in both directions makes the error bound for the middle sum method smaller.

-- Icedemon —Preceding unsigned comment added by 98.226.21.88 (talk) 09:35, 25 December 2007 (UTC)