I couldn't get LaTEX to display this one, it kept erroring out after about 3/4 of the equation was entered. So I pasted a jpg instead. Thanks for looking.

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- Aug 26th 2006, 06:32 PMYogi_Bear_79Help with Riemann sums for a continuous function
I couldn't get LaTEX to display this one, it kept erroring out after about 3/4 of the equation was entered. So I pasted a jpg instead. Thanks for looking.

- Aug 26th 2006, 06:40 PMThePerfectHacker
I think there is a mistake. That is not expressable as a Riemann sum.

- Aug 26th 2006, 06:50 PMYogi_Bear_79
That's not good, I double-checked it and it is as typed in the course guide!

- Aug 26th 2006, 07:02 PMThePerfectHacker
Sorry forgive me. I saw something else in the problem and made a mistake.

The function is,

$\displaystyle \sin \pi x$

Then, the convergent riemann sum is, (by definition)

$\displaystyle \int_0^1 \sin \pi x dx$

Now,

$\displaystyle f(x)=-\frac{1}{\pi} \cos \pi x$

Has property that,

$\displaystyle f'(x)=\sin \pi x$

Then by the Fundamental Theorem of Calculus (eventhough I am angry :mad: when it is referred to as like this. It certainly is not a fundamental theorem at all!)

$\displaystyle -\frac{1}{\pi} \cos \pi +\frac{1}{\pi} \cos 0=2\pi$ - Aug 27th 2006, 05:40 AMJameson
Rewrite the sum like so.

$\displaystyle \lim_{n \rightarrow \infty}\sum_{i=1}^{n} \frac{\sin(\frac{\pi{i}}{n})}{n}$

Now let $\displaystyle x=\frac{i}{n}$ and $\displaystyle dx=\frac{1}{n}$. Now all of PH's work follows. - Aug 27th 2006, 09:33 AMYogi_Bear_79
The only example from my book that I have to go by is like the first image below. I tried to rewrite your work in the second image to anwser the question. Can you look at it and let me know if I messed up? Thanks

- Aug 27th 2006, 04:48 PMJameson
You must first write the sum in a general way as I did for you. You are setting up rectangles partitioned $\displaystyle \frac{1}{n},\frac{2}{n}$...etc and letting n approach infinity for the rectangles height, and the width of each rectangle is $\displaystyle \frac{1}{n}$. Now when you apply the limit, the infinite sum tranforms into the integral $\displaystyle \int_{0}^{1}\sin(\pi{x})dx$.