March 16th, 2018, 05:48 PM  #1 
Newbie Joined: Mar 2018 From: Canada Posts: 5 Thanks: 0  A Reimann Sum (I think)
I am struggling to solve this. I know it converges. Calculus.gif But, I don't even know whether I have described the equation properly. Any help appreciated, Thanks, Colin. Last edited by skipjack; March 22nd, 2018 at 02:04 AM. 
March 21st, 2018, 05:02 PM  #2 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 894 
Yes, what you have is a "Riemann sum" and its limit is an integral. A "Riemann sum" is always of the form $\displaystyle \sum_{i= 0}^n f(i(b a)/n)$ which, in the limit as n goes to infinity, becomes the integral $\displaystyle \int_a^b f(x)dx$. Here, the sum is $\displaystyle \sum_{i= 0}^n \frac{d}{(d r^2(i/n)^2)^{3/2}}$. Taking b= 1 and a= 0, so that "i(b a)/n" is "i/n", that limit is $\displaystyle \int_0^1 f(x)dx$ where $\displaystyle f(x)= \frac{d}{(d r^2x^2)^{3/2}}$. Integrate that. 
March 24th, 2018, 07:25 PM  #3 
Newbie Joined: Mar 2018 From: Canada Posts: 5 Thanks: 0 
You completely lost me. Why is b=1? Or, is there a good tutorial somewhere that covers solving Riemann Sums? Thanks, Colin. 
March 24th, 2018, 08:10 PM  #4 
Newbie Joined: Mar 2018 From: Canada Posts: 5 Thanks: 0 
I ran that integral against a few online integral calculators without much success. https://www.integralcalculator.com/ has the solution as  ((dr^2)^0.5) / (r^2d) https://www.symbolab.com has the solution as needing a constant https://www.emathhelp.net has the solution as (dr^2)^(0.5) https://www.mathportal.org dies trying. I know by using a spreadsheet or a few lines of code that for 5=5 and r=2 this evaluates to 0.0436496. Calling it quits for the day. Colin. 
March 25th, 2018, 03:30 AM  #5 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 894 
The denominator in the sum is $\displaystyle (d r^2(i/n)^2)^{3/2}$, with i going from 0 to n, and I replaced that by $\displaystyle (d r^2x^2)^{3/2}$. The only change is that I replaced $\displaystyle i/n$ by $\displaystyle x$. When i= 0, x= i/n= 0 and when i= n x= i/n= 1. That's why the limits of integration are 0 and 1 You might try "M.I.T OpenCourseware". https://ocw.mit.edu/courses/mathemat...riemannsums/ 
August 6th, 2018, 02:49 PM  #6 
Newbie Joined: Mar 2018 From: Canada Posts: 5 Thanks: 0  I think I got it
$\displaystyle d\sqrt{d^{2}r^{2}}$

August 6th, 2018, 03:29 PM  #7 
Global Moderator Joined: Dec 2006 Posts: 19,704 Thanks: 1804  
August 7th, 2018, 04:25 AM  #8 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 894  

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