|By Redpepper (Redpepper) on Tuesday, March 23, 2004 - 11:58 am: Edit|
|By Angrod (Angrod) on Tuesday, March 23, 2004 - 12:44 pm: Edit|
OK, how about this:
(1/-2x)(e^-(x^2)) + C
(e^-(x^2))/(-2x) + C
whichever way you like it.
What do y'all think? I just divided the problem by the derivative of the exponent of e. Looks good to me! lol
|By Punjabiashiq (Punjabiashiq) on Tuesday, March 23, 2004 - 12:54 pm: Edit|
The function e^(-x^2) does not have a straight forward antiderivative. This is a bell shape curve used widely in statistics, the is an apporximation for it it involves a squaroot of 2 pi in it somewhere but i don't know it of the top of my head, you can search for it online.
|By Angrod (Angrod) on Tuesday, March 23, 2004 - 01:12 pm: Edit|
Oops! Oh well, shows how much I know.
|By Hankinator3000 (Hankinator3000) on Tuesday, March 23, 2004 - 01:41 pm: Edit|
|By Angrod (Angrod) on Tuesday, March 23, 2004 - 01:59 pm: Edit|
Oops, sorry guys, I just realized what I did wrong! lol, pretty obvious now that I look at it. Well, if y'all don't see me for the next three months it's because I am hiding in shame.
|By Dinotopia (Dinotopia) on Tuesday, March 23, 2004 - 02:58 pm: Edit|
That antiderivative is the same as the antiderivative of the Maclaurin Series from 0 to infinity of [x^(n-2)]/n!, so basically, the antiderivative is C + the Maclaurin Series from 0 to infinity of [x^(n-1)]/(n-1)n!, where |x|<1
It's not an elementary function, and using Taylor/Maclaurin series is one way of finding an antiderivative... of course I may be wrong ;-)
|By Nhlgoalie (Nhlgoalie) on Tuesday, March 23, 2004 - 04:57 pm: Edit|
it's only really useful to integrate that function (previously identified as the bell curve) if you have definite endpoints. For statistics these endpoints are usually negative infinity to infinity, in which case the answer is root pi. the integration is pretty intense (it requires a substitution that is truly brilliant), but if you'd like i can scan it and send it to you.
|By Punjabiashiq (Punjabiashiq) on Tuesday, March 23, 2004 - 05:23 pm: Edit|
I am not sure how familiar you guys are with calculus, but with the help of a double integral and a simple conversion to polar coordinates the integral reduces down nicely.
|By Vinny919 (Vinny919) on Wednesday, March 24, 2004 - 04:21 pm: Edit|
ooo scan and email it to me please! email firstname.lastname@example.org too lazy to put it in profile unless its already there.
|By Creon (Creon) on Wednesday, March 24, 2004 - 10:54 pm: Edit|
Punjabiashiq is right, using that method, the integral from negative infinity to positive infinity of e^(-x^2) comes out to the square root of pi.
Although the MacLaurin series is a clever solution as well.
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