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Kashmir

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In summary, the conversation discusses the approximation of the Riemann Sum and how it works by using a moderate value of m and drawing the graph of a continuous function. The sum is represented by the area under the rectangles and is roughly equal to the integral of the function. Geogebra can be used to visualize this concept.

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Kashmir

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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.

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JFerreira

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Geogebra can help you see the graphs: https://www.geogebra.org/graphing/sfgvm5zu

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Kashmir

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We first think about the sum as the area under the rectangles having unit width.andrewkirk said:

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.

This area is roughly the integral of the corresponding continuous function.

Am i right?

- #5

Kashmir

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thank you so muchJFerreira said:Geogebra can help you see the graphs: https://www.geogebra.org/graphing/sfgvm5zu

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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?

- #7

Kashmir

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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.

The purpose of approximating a discrete sum by an integral is to find an easier way to calculate the sum of a large number of discrete data points. By using the integral, we can represent the sum as a continuous function, making it easier to manipulate and analyze.

To approximate a discrete sum by an integral, we first divide the interval of the sum into smaller intervals. Then, we use the midpoint or trapezoidal rule to estimate the area under the curve of each interval. Finally, we add up all of these estimated areas to get an approximation of the original discrete sum.

A discrete sum is a finite sum of individual data points, while an integral is a continuous function that represents the sum of an infinite number of data points. In other words, a discrete sum is a collection of individual values, while an integral is a representation of the sum as a whole.

It is appropriate to use an approximation of a discrete sum by an integral when the number of data points is large and the function representing the data is smooth and continuous. This method is also useful when the exact value of the sum is difficult to calculate or unknown.

One limitation of approximating a discrete sum by an integral is that it can only provide an estimation of the original sum, not the exact value. Additionally, this method may not be accurate if the function representing the data is not smooth or if the number of data points is too small. It is also important to note that this method is only applicable for one-dimensional data and cannot be used for multi-dimensional data.

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