Construct a table of Riemann sums to show that sums with right-endpoint, midpoint, and left-endpoint evaluation all converge to the same value as n approaches infinity.

f(x) = sin x, [0, π/2]

Thanks in advance!

Kind of lost here.. (Headbang)

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- Dec 7th 2012, 09:45 PMJDSConstruct a table of Riemann sums to show that...
Construct a table of Riemann sums to show that sums with right-endpoint, midpoint, and left-endpoint evaluation all converge to the same value as n approaches infinity.

f(x) = sin x, [0, π/2]

Thanks in advance!

Kind of lost here.. (Headbang) - Dec 7th 2012, 10:34 PMMarkFLRe: Construct a table of Riemann sums to show that...
Let's look at the left sum first:

$\displaystyle \Delta=\frac{\frac{\pi}{2}-0}{n}=\frac{\pi}{2n}$

and so we have:

$\displaystyle A_n=\frac{\pi}{2n}\sum_{k=0}^{n-1}\sin\left(k\cdot\frac{\pi}{2n} \right)=\frac{\pi}{2n}\cdot\frac{1}{2}\left(\cot \left(\frac{\pi}{4n} \right)-1 \right)=$

$\displaystyle \frac{\pi\left(\cot\left(\frac{\pi}{4n} \right)-1 \right)}{4n}$

To compute the sum, I used an identity of Lagrange:

$\displaystyle \sum_{n=1}^N \sin n\theta & = \frac{1}{2}\cot\frac{\theta}{2}-\frac{\cos(N+\frac{1}{2})\theta}{2\sin\frac{1}{2} \theta}$

Now, to compute the limit:

$\displaystyle \lim_{n\to\infty}\left( \frac{\pi\left(\cot\left( \frac{\pi}{4n} \right)-1 \right)}{4n} \right)=$

$\displaystyle \frac{\pi}{4}\lim_{n\to\infty} \left( \frac{ \cot \left( \frac{\pi}{4n} \right)-1}{n} \right)=L$

We have the indeterminate form $\displaystyle \frac{\infty}{\infty}$ so applying L'Hôpital's rule, we find:

$\displaystyle L=\frac{\pi}{4}\lim_{n\to\infty} \left( \frac{-\csc^2\left( \frac{\pi}{4n} \right)\left(-\frac{\pi}{4n^2} \right)}{1} \right)=\left( \frac{\pi}{4} \right)^2\lim_{n\to\infty} \left( \frac{\frac{1}{n^2}}{\sin^2\left( \frac{\pi}{4n} \right)} \right)$

Now we have the indeterminate form $\displaystyle \frac{0}{0}$, so applying L'Hôpital's rule again:

$\displaystyle L=\frac{\pi}{4}\lim_{n\to\infty} \left( \frac{\frac{2}{n}}{\sin\left( \frac{\pi}{2n} \right)} \right)=\lim_{n\to\infty} \left( \frac{\frac{\pi}{2n}}{\sin\left( \frac{\pi}{2n} \right)} \right)=1$

Now, see if you can do the same for the remaining two sums. - Dec 8th 2012, 12:49 PMJDSRe: Construct a table of Riemann sums to show that...
Thanks for the reply, I got a bit busy so late getting back to this one. In the book, they give an example where it shows a table with values of n (that I am supposed to select) and the corresponding left endpoint, midpoint and right end point evaluation all converge to the same value as n approaches infinity. So, i know that they supposedly converge to 1 (based on the info you have so graciously shown), however, how do I show this in the table. I mean, I obviously know how to make a table, and of course I pick my own (ever increasing) n values, but what method am I supposed to use to then substitute "my" N values in and then solve for left-mid-right end points. (which then should show that they all converge to 1) I hope that makes sense! If not I can try and explain further

- Dec 8th 2012, 06:59 PMMarkFLRe: Construct a table of Riemann sums to show that...
Before constructing the table, can you compute the other two sums?

- Dec 9th 2012, 10:07 AMJDSRe: Construct a table of Riemann sums to show that...
oh, yeah, I think so....I will give it a shot, but I might not get to it today, Kind of a busy schedule but as soon as I get it done I will post here. Thanks in advance!

- Dec 12th 2012, 11:19 AMJDSRe: Construct a table of Riemann sums to show that...
Been crazy busy, but this is all I could manage to figure out....

(see attached picture) - Dec 16th 2012, 01:44 PMJDSRe: Construct a table of Riemann sums to show that...
well, I am not sure what I am supposed to do here....any Ideas/Suggestions?