Trouble with Riemann sums

  • Thread starter guynoone
  • Start date
  • #1
5
0
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
 

Attachments

  • Untitled-1.jpg
    Untitled-1.jpg
    48.7 KB · Views: 364

Answers and Replies

  • #2
tiny-tim
Science Advisor
Homework Helper
25,836
251
Welcome to PF!

… I have absolutely no clue how/why they get the 30 and 27.

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)
 
  • #3
5
0
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!)
 
  • #4
statdad
Homework Helper
1,495
35


''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.

[tex]
\begin{align*}
\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
\end{align*}
[/tex]


Note that the formula

[tex]
\sum_{i=1}^n =\frac{n(n+1)} 2
[/tex]

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
 

Related Threads on Trouble with Riemann sums

  • Last Post
Replies
5
Views
1K
  • Last Post
Replies
3
Views
7K
  • Last Post
Replies
4
Views
3K
  • Last Post
Replies
10
Views
5K
  • Last Post
Replies
2
Views
830
  • Last Post
Replies
1
Views
3K
Replies
3
Views
2K
  • Last Post
Replies
2
Views
3K
Replies
2
Views
4K
Top