The discussion focuses on approximating the discrete sum ##\sum_{k=0}^m\left(\frac{k}{m}\right)^n## using the integral ##\int_0^m\left(\frac{x}{m}\right)^n dx##. Participants suggest using Geogebra to visualize the approximation by drawing graphs for moderate values of m, specifically m=20, and for n values of 1 and 2. The concept of Riemann Sums is introduced as a method to understand the relationship between the sum and the integral, emphasizing the area under rectangles representing the discrete sum.
PREREQUISITESStudents and educators in mathematics, particularly those studying calculus, as well as anyone interested in visualizing mathematical concepts through graphing tools like Geogebra.
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.