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Math Help - Interesting (reverse?) Riemann Sum Problem

  1. #1
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    Interesting (reverse?) Riemann Sum Problem

    Hi,

    I'm working on this question and having trouble coming up with a solution, mainly due to the cos. I have no idea how to type sums in here so I am just going to post two screenshots of the original problem.




    Basically all I have to do is solve the Reimann sum that is already provided, but it's way too complicated to work out so I have to convert it back to the integral then solve the integral. I'm having trouble finding what f should be in the original integral. If I set it as cos(t) I almost get the right Riemann sum but there is still a (4pi)/(21) leftover from the delta x part of the Riemann sum, and I don't know how to change this because it's all doomed to be "trapped" inside the cos.

    Help!

    Thank you!

    Interesting (reverse?) Riemann Sum Problem-screen-shot-2011-01-21-7.07.20-pm.png
    Interesting (reverse?) Riemann Sum Problem-screen-shot-2011-01-21-7.07.17-pm.png
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  2. #2
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    Quote Originally Posted by metaname90 View Post
    Hi,

    I'm working on this question and having trouble coming up with a solution, mainly due to the cos. I have no idea how to type sums in here so I am just going to post two screenshots of the original problem.




    Basically all I have to do is solve the Reimann sum that is already provided, but it's way too complicated to work out so I have to convert it back to the integral then solve the integral. I'm having trouble finding what f should be in the original integral. If I set it as cos(t) I almost get the right Riemann sum but there is still a (4pi)/(21) leftover from the delta x part of the Riemann sum, and I don't know how to change this because it's all doomed to be "trapped" inside the cos.

    Help!

    Thank you!

    Click image for larger version. 

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    Click image for larger version. 

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    Well, put \displaystyle{f(x):=\cos x\Longrightarrow \frac{1}{n}\sum\limits^n_{k=1}\cos\left(\frac{4k\p  i}{21n}\right)} is just the Riemann

    sum of the function in the interval \displaystyle{[0,\frac{4\pi}{21}] partitioned in n equal subintervals and taking the rightmost

    end of each subinterval to evaluate the function on it (why can we do this?)

    Tonio
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  3. #3
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    Alright, so that's what I originally tried, and you get sin(4pi/21) as the solution (integral on that interval of cos(x)), but that is the wrong solution. And when I actually work it out I dont' see how cos(x) can be f(x), or entirely, there must be something else. Here is my working below. The part I was talking about struggling with was that leftover 4pi/21:



    Interesting (reverse?) Riemann Sum Problem-screen-shot-2011-01-21-8.05.39-pm.png
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  4. #4
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    Quote Originally Posted by metaname90 View Post
    Alright, so that's what I originally tried, and you get sin(4pi/21) as the solution (integral on that interval of cos(x)), but that is the wrong solution. And when I actually work it out I dont' see how cos(x) can be f(x), or entirely, there must be something else. Here is my working below. The part I was talking about struggling with was that leftover 4pi/21:



    Click image for larger version. 

Name:	Screen shot 2011-01-21 at 8.05.39 PM.png 
Views:	130 
Size:	22.0 KB 
ID:	20541

    Ok, that looks right...now just take the constant factor \displaystyle{\frac{4\pi}{21}} out of the sum

    and of the limit and what's left is the integral of cosine over the given interval.

    Tonio
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