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Trouble with Riemann sums

  1. Sep 8, 2008 #1
    Alright, I started doing Riemann sums and I am ripping my hair out in frustration. I just can't wrap my head around how I'm supposed to do it, and my woefully vague textbook isn't helping either. I'm wondering how I'm supposed to solve a Riemann sum question with sigma notation (no limits), and with only the width and height of the rectangles provided (no curve function). I used the image as an example. I have absolutely no clue how/why they get the 30 and 27. If anyone could provide a step by step explanation, it would be great. Thanks

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  2. jcsd
  3. Sep 9, 2008 #2


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    Welcome to PF!

    Hi guynoone! Welcome to PF! :smile:

    ok … for some reason, the 0th Riemann sum, from 1 to n, is defined as being 1/n times the ∑, and the 1st Riemann sum as 1/n² times ∑.

    In this case, you're expecting 10/3 times the 0th sum, and 1/3 times the 1st, but this particular definition means that you multiply the 0th one by 9, = 10/3 * 9 = 30, and the 1st one by 81, = 81 * 1/3 = 27. :smile:

    (it isn't the only way of working out this problem … it's only really useful if you happen to have tables of Riemann sums :wink:

    personally, i prefer to remember ∑i = n(n+1)/2)
  4. Sep 9, 2008 #3
    Thanks for the answer. To be honest I still don't really get what is happening for the solution I posted, but I just plan on sticking to the ∑i = n(n+1)/2) method which is way better (they must be trying to confuse us on purpose with these identities!)
  5. Sep 20, 2008 #4


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    Re: Welcome to PF!

    ''ok … for some reason, the 0th Riemann sum, from 1 to n, is defined as being 1/n times the ∑, and the 1st Riemann sum as 1/n² times ∑.''

    Little abuse of notation here, even though I'm sure tiny-tim understands the entire problem.
    The 0th order term refers to a sum which involves [tex] i^0[/tex], or [tex] 1[/tex], a first order term involves [tex] i^1 = i [/tex], and so on. Writing [tex] \sum [/tex] without any explicit is common, but writing it without any arguments (summands) is not.
    Actually, all that is done in this work a (very poor, in my opinion) condensation of the basic steps.

    \sum_{i=1}^9 {\frac{10-i} 3} & = \frac 1 3 \sum_{i=1}^9 {(10-i)}\\
    & = \frac 1 3 \left(\sum_{i=1}^9 10 - \sum_{i=1}^9 i \right) \\
    & = \frac 1 3 \left( 10 \sum_{i=1}^9 1 - \sum_{i=1}^9 i \right)\\
    & = \frac 1 3 \left(10 \cdot 9 - \frac{10\cdot 9}{2} \right) \\
    & = \frac 1 3 \left(90 - 45\right) = \frac{45}{3} = 15

    Note that the formula

    \sum_{i=1}^n =\frac{n(n+1)} 2

    was used, with [tex] n = 9 [/tex], was used.

    As I indicated at the top of my post, I believe the solution for this problem (the one posted by the OP) is very poorly presented and typeset
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