The discussion revolves around the approximation of a discrete sum by an integral, specifically the expression ##\sum_{k=0}^m\left(\frac{k}{m}\right)^n\approx\int_0^m\left(\frac{x}{m}\right)^ndx##. Participants explore the conceptual understanding of this approximation through graphical representations and the concept of Riemann sums.
Participants generally agree on the validity of the Riemann sum interpretation, but there is no explicit consensus on the details of the approximation method or its implications.
Participants do not delve into specific mathematical limitations or assumptions regarding the approximation, nor do they clarify the conditions under which the approximation holds.
We first think about the sum as the area under the rectangles having unit width.andrewkirk said:Choose a moderate value of m, not too small, say m=20.
Then set n=1, and draw the graph of ##f(x) = (\frac xm)^n## between 0 and ##m##. On the same set of axes draw the series of m rectangles such that the k-th rectangle (k = 0, ..., m-1) is the set of points (x,y) with ##\frac km\le x\le \frac{k+1}m## and ##(\frac xm)^n\le y\le (\frac{x+1}m)^n##.
Then do the same thing on a new set of axes, for ##n=2##.
That should give you a strong sense for why the approximation works.
thank you so muchJFerreira said:Geogebra can help you see the graphs: https://www.geogebra.org/graphing/sfgvm5zu
Yes, that's correct. It is called a Riemann Sum. You can read more about them here, including some nice pictures.Kashmir said:We first think about the sum as the area under the rectangles having unit width.
This area is roughly the integral of the corresponding continuous function.
Am i right?
Thank you :)andrewkirk said:Yes, that's correct. It is called a Riemann Sum. You can read more about them here, including some nice pictures.