My Math Forum A Reimann Sum (I think)

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 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: 3,264 Thanks: 902 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.integral-calculator.com/ has the solution as - ((d-r^2)^0.5) / (r^2-d) https://www.symbolab.com has the solution as needing a constant https://www.emathhelp.net has the solution as (d-r^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: 3,264 Thanks: 902 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...-riemann-sums/
 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
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Quote:
 Originally Posted by Country Boy The denominator in the sum is $\displaystyle (d- r^2(i/n)^2)^{3/2}$
No, it's $\displaystyle \left(d^2\! - r^2(i/n)^2\right)^{3/2}\!$.

August 7th, 2018, 04:25 AM   #8
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 Originally Posted by skipjack No, it's $\displaystyle \left(d^2\! - r^2(i/n)^2\right)^{3/2}\!$. Check your integration, Colin Connelly.
Yes, thanks.

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