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 December 27th, 2016, 12:41 AM #1 Newbie   Joined: Dec 2016 From: turkey Posts: 1 Thanks: 0 Explaining the curious phenomenon that occurs Could you please help me on explaining the curious phenomenon that occurs for this question ?
 December 27th, 2016, 03:28 PM #2 Member   Joined: Dec 2016 From: USA Posts: 46 Thanks: 11 If done in floating point, you will quickly lose accuracy, to the point where the output is not only wrong, but blatantly wrong. I'm pretty sure that's the "curious phenomenon" you were advised about. There's nothing mathematically curious about the true values of $x_n$ -- they simply approach zero, slowly and boringly. What programming language are you planning to use? What data type will you use for the values of $x_n$?
 December 27th, 2016, 04:04 PM #3 Senior Member     Joined: Sep 2007 From: USA Posts: 349 Thanks: 67 Math Focus: Calculus This simple algorithm can be programmed on a graphing calculator easily. Using a CAS on the computer, however, it can be done with much greater precision.
 December 27th, 2016, 07:41 PM #4 Member   Joined: Aug 2016 From: Used to be Earth Posts: 64 Thanks: 14 @Compendium Where can I find a proof of your profile picture? Thanks
December 27th, 2016, 07:50 PM   #5
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 Originally Posted by sKebess @Compendium Where can I find a proof of your profile picture? Thanks
just massage the standard normal probability distribution a bit and integrate

December 27th, 2016, 08:19 PM   #6
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 Originally Posted by romsek just massage the standard normal probability distribution a bit and integrate
Massage. Excellent

December 28th, 2016, 02:50 AM   #7
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 Originally Posted by sKebess @Compendium Where can I find a proof of your profile picture? Thanks
Equate the volume of revolution around the y-axis to the double integral over the entire x-y plane,

$\displaystyle 2\pi\int_{0}^{\infty}xe^{-x^2}dx\,=\,\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-(x^2+y^2)}dx\,dy$.

The left hand side should be easy to integrate. On the right hand side, the double integral is separable. Then the solution should become obvious.

Last edited by Compendium; December 28th, 2016 at 03:19 AM.

December 28th, 2016, 02:26 PM   #8
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 Originally Posted by Compendium Equate the volume of revolution around the y-axis to the double integral over the entire x-y plane, $\displaystyle 2\pi\int_{0}^{\infty}xe^{-x^2}dx\,=\,\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-(x^2+y^2)}dx\,dy$. The left hand side should be easy to integrate. On the right hand side, the double integral is separable. Then the solution should become obvious.
Awesome, cheers.

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