lightweightx
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- Dec 5, 2015
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I am currently working on a question:
Consider the function y= e^x on the interval [0,1] divided into n intervals of length [1/n]. Compute the following Riemann sum;
. . . . .\(\displaystyle \displaystyle \sum_{j\, =\, 0}^{\frac{n}{2}\, -\, 1}\, \left[\,\left(\dfrac{1}{3n}\right)\, \cdot\, \left( \,e^{\frac{2j}{n}}\, +\, 4e^{\frac{(2j\, +\, 1)}{n}}\, +\, e^{\left(2j\, +\, \frac{2}{n}\right)}\, \right)\,\right]\)
Which is supposedly a summation of the Simpson's rule.
As i continue computing when n tends to infinity, I still cannot get rid of the (1/n), which would give me an infinity value to my area.
Would appreciate any advice on this!!
Consider the function y= e^x on the interval [0,1] divided into n intervals of length [1/n]. Compute the following Riemann sum;
. . . . .\(\displaystyle \displaystyle \sum_{j\, =\, 0}^{\frac{n}{2}\, -\, 1}\, \left[\,\left(\dfrac{1}{3n}\right)\, \cdot\, \left( \,e^{\frac{2j}{n}}\, +\, 4e^{\frac{(2j\, +\, 1)}{n}}\, +\, e^{\left(2j\, +\, \frac{2}{n}\right)}\, \right)\,\right]\)
Which is supposedly a summation of the Simpson's rule.
As i continue computing when n tends to infinity, I still cannot get rid of the (1/n), which would give me an infinity value to my area.
Would appreciate any advice on this!!
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