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Math Help - Evaulating Riemann Sums

  1. #1
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    Evaulating Riemann Sums

    I have another questions about thes,e no very good with them as you may seen from my previous posts . Anyhow, here is the problem below.

    <br /> <br />
\sum _{i=1}^n(ni + \sum _{j=i}^nnj)<br />

    Ok, so what i do know is that i should expand these summations and then look for a pattern? Then, i assume that im suppose to derive sometype of formula? Im kind of unsure whats the whole purpose of riemann sums ecept for just summing a whole bunch of numbers.

    1. Start with inner summation

    Kind of like nested integrals i was told that you should evaluate the inner summation, and then work your way to the outer.

    So when i expand the inner loop i get  n+2n+3n+4n +...+ jn, and\ this\ is\ when\ j=n

    2. Expand outer summation
     n+2n+3n+4n+ ... +ni, and\ this\ is\ when\ i = n

    Im confused on what to do next.


    PS: I will have one Riemann sum problem to ask tonight, but i will make sure to try and get far on it as much as possible before i post to the forums. Thanks!
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  2. #2
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    Quote Originally Posted by tokio View Post
    I have another questions about thes,e no very good with them as you may seen from my previous posts . Anyhow, here is the problem below.

    <br /> <br />
\sum _{i=1}^n(ni + \sum _{j=i}^nnj)<br />
    n is fixed and can be take outside the sums:
    n\sum_{i=1}^n(i+ \sum_{j=1}^n j)

    Now, a very useful formula, I am surprised you didn't already know it:
    \sum_{j=1}^n j= \frac{n(n+1)}{2}

    You can prove that easily by writing the sum as 1+ 2+ 3+ ...+ (n-2)+ n-1+ n, reversing it as n+ (n-1)+ (n-2) +...+ 3+ 2+ 1 and adding each term to the one above: n+ 1, n-1+ 2= n+ 1, n-2+ 3= n+1, etc. Each pair adds to n+1 and there are n terms- that sum is n(n+1). Since we added the sum twice, divide by 2 to get the formula above.

    So now you have \sum_{i=1}(i+ \frac{n(n+1)}{2}).
    Break that into two sums: \sum_{i=1}^n i is, of course, \frac{n(n+1)}{2} again. And \sum_{i= 1}^n \frac{n(n+1)}{n} is just a constant added to itself n times- that is n(\frac{n(n+1)}{2})= \frac{n^2(n+1)}{2}

    Ok, so what i do know is that i should expand these summations and then look for a pattern? Then, i assume that im suppose to derive sometype of formula? Im kind of unsure whats the whole purpose of riemann sums ecept for just summing a whole bunch of numbers.

    1. Start with inner summation

    Kind of like nested integrals i was told that you should evaluate the inner summation, and then work your way to the outer.

    So when i expand the inner loop i get  n+2n+3n+4n +...+ jn, and\ this\ is\ when\ j=n

    2. Expand outer summation
     n+2n+3n+4n+ ... +ni, and\ this\ is\ when\ i = n

    Im confused on what to do next.


    PS: I will have one Riemann sum problem to ask tonight, but i will make sure to try and get far on it as much as possible before i post to the forums. Thanks!
    Last edited by Jhevon; January 25th 2010 at 09:10 AM. Reason: inserted LaTeX tags where they were missing
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