Simpson Rule summation help

lightweightx

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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!!
 

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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!!
So you think that the limit as n--> infinity of (1/n) is infinity.
So if you have one glass of beer and you share it with lots and lots of people then every one gets a LOT of beer?

1/( a very very large number) is close to 0! The ability to see this is a major pre-requiste to calculus.

Please show us your work and someone will find your algebra mistake (another pre-req to Calcul
 
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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!!
First I understand that you probably must answer this question. So my answer will not help.

Advice: Ask whoever set this question why are we still wasting time with such outdated questions?
Given today's computing power, what is the point? In either 1979 or 80, I heard Leonard Gillman then pres. of the MAA ask why we are still using ideas from the 18th and 19th centuries? Think about that, back then the most powerful calculator or computer algebra system had only one hundredth the calculation power as does any smart phone today. As Keith Devlin has written, if we want people to use mathematics let us teach in such a way they understand that what the machines give us. Do not turn them off with pointless calculations. In his essay on back to the basics, he says to look forward. That was written in 1998.
 
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First I understand that you probably must answer this question. So my answer will not help.

Advice: Ask whoever set this question why are we still wasting time with such outdated questions?
Given today's computing power, what is the point? In either 1979 or 80, I heard Leonard Gillman then pres. of the MAA ask why we are still using ideas from the 18th and 19th centuries? Think about that, back then the most powerful calculator or computer algebra system had only one hundredth the calculation power as does any smart phone today. As Keith Devlin has written, if we want people to use mathematics let us teach in such a way they understand that what the machines give us. Do not turn them off with pointless calculations. In his essay on back to the basics, he says to look forward. That was written in 1998.
I 100% agree with you about this problem being outdated. My concern is that this student (and many others) can't do the algebra necessary to finish this problem and needs to practice their algebra so they can solve such problems. Possibly there are other, more current type, problems that will also strengthen their algebra skills. Either way (whether they formally do problems like these or not) students need to have the ability to do these type of problems.
 
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!!
First let n = 2 k, to get rid of that n/2 and break your summation into three parts. The first part is
S(k) = \(\displaystyle \frac{1}{6\, k}\, \underset{j=0}{\overset{j=k-1}{\Sigma}}\, t^j\)
where
t = \(\displaystyle e^{\frac{1}{k}}\)
The summation is the sum of a geometric series which can be solved in closed form. What you will end up with is a denominator
d(k) = \(\displaystyle {6\, k\, [ e^{\frac{1}{k}}\, -\, 1] }\)
which, as k goes to infinity is of the form 0*\(\displaystyle \infty\). What tools have you been given to evaluate a limit of that form?

Hint: You could let x=1/k and see where that leads. You could also write ex in the standard power series form
ex = 1 + x + (1/2) x2 + (1/6) x3 + (1/24) x4 + ...


EDIT: I strongly suspect that the third term is incorrect, especially if that sum is supposed to represent the integral of ex on [0,1]
 
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First I understand that you probably must answer this question. So my answer will not help.

Advice: Ask whoever set this question why are we still wasting time with such outdated questions?
Given today's computing power, what is the point? In either 1979 or 80, I heard Leonard Gillman then pres. of the MAA ask why we are still using ideas from the 18th and 19th centuries? Think about that, back then the most powerful calculator or computer algebra system had only one hundredth the calculation power as does any smart phone today. As Keith Devlin has written, if we want people to use mathematics let us teach in such a way they understand that what the machines give us. Do not turn them off with pointless calculations. In his essay on back to the basics, he says to look forward. That was written in 1998.

If you will look at my reply above, you might change your mind about this being an outdated question. It seems to me that the point is not 'summing the series' but rather to recognize that sometimes things like L'Hospital's rule might be buried in a problem.
 
So you think that the limit as n--> infinity of (1/n) is infinity.
So if you have one glass of beer and you share it with lots and lots of people then every one gets a LOT of beer?

1/( a very very large number) is close to 0! The ability to see this is a major pre-requiste to calculus.

Please show us your work and someone will find your algebra mistake (another pre-req to Calcul

Sorry for the mistake, I seemed to have expressed myself wrongly regarding the limit.

My current work is like this: (I was given a summation formula to use)

attachment.php



I was unable to arrive at a number for the area.




I will now follow Ishuda's comments and try to work it out using k = n/2 through out. Thank you very much.
 

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First let n = 2 k, to get rid of that n/2 and break your summation into three parts. The first part is
S(k) = \(\displaystyle \frac{1}{6\, k}\, \underset{j=0}{\overset{j=k-1}{\Sigma}}\, t^j\)
where
t = \(\displaystyle e^{\frac{1}{k}}\)
The summation is the sum of a geometric series which can be solved in closed form. What you will end up with is a denominator
d(k) = \(\displaystyle {6\, k\, [ e^{\frac{1}{k}}\, -\, 1] }\)
which, as k goes to infinity is of the form 0*\(\displaystyle \infty\). What tools have you been given to evaluate a limit of that form?

Hint: You could let x=1/k and see where that leads. You could also write ex in the standard power series form
ex = 1 + x + (1/2) x2 + (1/6) x3 + (1/24) x4 + ...


EDIT: I strongly suspect that the third term is incorrect, especially if that sum is supposed to represent the integral of ex on [0,1]


\
attachment.php


I tried working it out, is it something along these lines?

Thank you so much for your reply!
 

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...I tried working it out, is it something along these lines?

Thank you so much for your reply!
Yes. However, you did make one a minor mistake. Minor only because it did not affect the final answer.
\(\displaystyle \underset{j=0}{\overset{j=k-1}{\Sigma}}\, t^j\)
is
\(\displaystyle \frac{t^k\, -\, 1}{t\, -\, 1}\)
not
\(\displaystyle \frac{t^{k-1}\, -\, 1}{t\, -\, 1}\)

Now do the other two pieces and sum them. I think the sum remains the same, just the constant out front changes.
 
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