Calculus Calculus Math Forum

 March 16th, 2018, 06:48 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 03:04 AM. March 21st, 2018, 06:02 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, 08:25 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, 09:10 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, 04:30 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, 03:49 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, 04:29 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, 05:25 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. Tags reimann, sum Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post cummings123 Real Analysis 0 October 22nd, 2012 04:09 PM Rejjy Calculus 15 October 2nd, 2012 12:52 PM lovetolearn Calculus 4 April 14th, 2012 04:44 PM davy89 Real Analysis 1 November 27th, 2008 03:02 PM

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