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March 16th, 2018, 06:48 PM   #1
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Question 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.
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March 21st, 2018, 06:02 PM   #2
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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.
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March 24th, 2018, 08:25 PM   #3
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You completely lost me. Why is b=1? Or, is there a good tutorial somewhere that covers solving Riemann Sums?

Thanks,

Colin.
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March 24th, 2018, 09:10 PM   #4
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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.
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March 25th, 2018, 04:30 AM   #5
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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/
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August 6th, 2018, 03:49 PM   #6
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I think I got it

$\displaystyle d\sqrt{d^{2}-r^{2}}$
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August 6th, 2018, 04:29 PM   #7
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Quote:
Originally Posted by Country Boy View Post
The denominator in the sum is $\displaystyle (d- r^2(i/n)^2)^{3/2}$
No, it's $\displaystyle \left(d^2\! - r^2(i/n)^2\right)^{3/2}\!$.

Check your integration, Colin Connelly.
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August 7th, 2018, 05:25 AM   #8
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Quote:
Originally Posted by skipjack View Post
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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