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Math Help - Evaulating Riemann Summations

  1. #1
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    Evaulating Riemann Summations

    <br /> <br />
\sum _{k=7}^{77} (k)sin\frac{(\pi)}{2} + \frac{cos(\pi)}{2}<br />


    Here is my logic, not sure if its 100% concrete. Let z =  \frac{(k)sin(\pi)}{2} + \frac{cos(\pi)}{2}

    then the first term of k will be 7.. 8 ... 9 .... 77. Obviously summing this up one at a time would be a daunting calculating task which is prone to error. So, im looking for a quicker way to do it. Since we can assume that i is incrementing by one each time and i use the formula n(n+1)/2 and then multiply it by the equation given? If thats the case, i got 3003 when i did that, but this seems like too large of a number to multiply this fraction by. Any suggestions?
    Last edited by tokio; January 21st 2010 at 07:34 PM. Reason: mistake writing the promblem
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by tokio View Post
    <br /> <br />
\sum _{k=7}^{77} = \frac{(k)sin(\pi)}{2} + \frac{cos(\pi)}{2}<br />


    Here is my logic, not sure if its 100% concrete. Let z =  \frac{(k)sin(\pi)}{2} + \frac{cos(\pi)}{2}

    then the first term of k will be 7.. 8 ... 9 .... 77. Obviously summing this up one at a time would be a daunting calculating task which is prone to error. So, im looking for a quicker way to do it. Since we can assume that i is incrementing by one each time and i use the formula n(n+1)/2 and then multiply it by the equation given? If thats the case, i got 3003 when i did that, but this seems like too large of a number to multiply this fraction by. Any suggestions?
    Check your post. what you wrote makes no sense. (By the way, \sin \pi = 0 \text{ and } \cos \pi = -1)
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    Quote Originally Posted by Jhevon View Post
    Check your post. what you wrote makes no sense. (By the way, \sin \pi = 0 \text{ and } \cos \pi = -1)
    Ok, i admit i have no idea what im doing! Im trying to think of a way to figure this out, any suggestions?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by tokio View Post
    Ok, i admit i have no idea what im doing! Im trying to think of a way to figure this out, any suggestions?
    i mean your question, the problem you posted itself does not make sense.

    \sum_{k = 7}^{77} makes no sense by itself. You need something after the summation sign.
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  5. #5
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    Quote Originally Posted by Jhevon View Post
    i mean your question, the problem you posted itself does not make sense.

    \sum_{k = 7}^{77} makes no sense by itself. You need something after the summation sign.
    Sorry, there is no equal sign.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by tokio View Post
    <br /> <br />
\sum _{k=7}^{77} \frac{(k)sin(\pi)}{2} + \frac{cos(\pi)}{2}<br />


    Here is my logic, not sure if its 100% concrete. Let z =  \frac{(k)sin(\pi)}{2} + \frac{cos(\pi)}{2}

    then the first term of k will be 7.. 8 ... 9 .... 77. Obviously summing this up one at a time would be a daunting calculating task which is prone to error. So, im looking for a quicker way to do it. Since we can assume that i is incrementing by one each time and i use the formula n(n+1)/2 and then multiply it by the equation given? If thats the case, i got 3003 when i did that, but this seems like too large of a number to multiply this fraction by. Any suggestions?
    As in \sum_{k = 7}^{77} \frac {k \sin \pi}2 + \frac {\cos \pi}2 or \sum_{k = 7}^{77} \left( \frac {k \sin \pi}2 + \frac {\cos \pi}2 \right) ?


    Since \sin \pi = 0 and \cos \pi = -1, in the first case, the answer is just -1/2.

    in the second case, what you have is \sum_{k = 7}^{77} (-1/2) = - \frac 12 \sum_{k = 7}^{77}1, and i suppose you know how to deal with that.
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  7. #7
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    Quote Originally Posted by Jhevon View Post
    As in \sum_{k = 7}^{77} \frac {k \sin \pi}2 + \frac {\cos \pi}2 or \sum_{k = 7}^{77} \left( \frac {k \sin \pi}2 + \frac {\cos \pi}2 \right) ?


    Since \sin \pi = 0 and \cos \pi = -1, in the first case, the answer is just -1/2.

    in the second case, what you have is \sum_{k = 7}^{77} (-1/2) = - \frac 12 \sum_{k = 7}^{77}1, and i suppose you know how to deal with that.

    Tried to edit the original promblem once again with latex, but cant figure it out . The original problem is ksin (pi/2) + cos(pi/2). In other words it looks like its sin(function) as opposed to \frac{(sin * \pi)}{2}. . I end up getting 3003 again once i was done.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by tokio View Post
    Tried to edit the original promblem once again with latex, but cant figure it out . The original problem is ksin (pi/2) + cos(pi/2). In other words it looks like its sin(function) as opposed to \frac{(sin * \pi)}{2}. . I end up getting 3003 again once i was done.
    *sigh* ok, one more time. (by the way, \sin * \pi makes no sense. you mean \sin \pi or \sin (\pi)--not multiplication).

    Anyway, \sin \frac {\pi}2 = 1,~\cos \frac {\pi}2 = 0, so your series is:

    \sum_{k = 7}^{77} k = \sum_{k = 1}^{77}k - \sum_{k = 1}^6k

    and you know the formula to calculate each of those
    Last edited by Jhevon; January 21st 2010 at 08:36 PM.
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