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Topic: intergal of e^-x^2 (Read 2038 times) |
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EzisEz
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intergal of e^-x^2
« on: Jul 20th, 2005, 6:49am » |
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ok, i am still working on this one i think i am close but the problem is as follows :"using only basic calculus, no complex numbers or series expansion, find the exact value of intergal of e^-x^2,in respect to x, from - infinity to infinity"
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EzisEz
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Re: intergal of e^-x^2
« Reply #2 on: Jul 20th, 2005, 7:48am » |
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Yea, my best guess would be polar coordinates
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River Phoenix
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Re: intergal of e^-x^2
« Reply #3 on: Jul 20th, 2005, 7:50am » |
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The answer should be sqrt(pi). The integrand is just sqrt(pi) times the density of the standard normal probability distribution. And of course the integral of Z from -inf to inf is 1.
The integral of the normal distribution can actually be evaluated and shown to equal 1 though, although I forget how.
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EzisEz
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Re: intergal of e^-x^2
« Reply #4 on: Jul 20th, 2005, 8:36am » |
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I got it , don't wanna ruin it , plus i don't know how to write the intergal symbols on here, if anybody knows how to attach an image i could attach it hidden...but let me know how
Ez
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| « Last Edit: Jul 20th, 2005, 8:43am by EzisEz » |
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Michael Dagg
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Re: intergal of e^-x^2
« Reply #5 on: Jul 20th, 2005, 9:53am » |
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Square the integral and use, say, y as the dummy variable in the second integral. Using the rule for the product of integrals gives a double integral whose integrand is exp(-x^2-y^2). Convert the integrand to polar form ( looks like exp(-r^2) r dr d[Angle] ) and replace the upper and low limits of integration with their polar equivalent and then integrate. The result follows.
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| « Last Edit: Jul 21st, 2005, 8:36am by Michael Dagg » |
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Michael Dagg
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EzisEz
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yep thats pretty much what i did, just couldn explain it as well....here it is in detail
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| « Last Edit: Jul 20th, 2005, 12:03pm by EzisEz » |
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Michael Dagg
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Re: intergal of e^-x^2
« Reply #7 on: Jul 20th, 2005, 4:03pm » |
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Indeed. The square term in the integrand gives us the liberty to use symmetry about the vertical axis.
Just for fun make note that exp(-x^2) = cosh(x^2) - sinh(x^2).
Let me add: now, using complex analysis show that
int[exp(-x^2) dx, -inf, inf] = sqrt(pi).
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| « Last Edit: Jul 20th, 2005, 8:52pm by Michael Dagg » |
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Michael Dagg
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ThudnBlunder
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Re: intergal of e^-x^2
« Reply #8 on: Jul 20th, 2005, 6:33pm » |
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This belongs in Putnam, mod.
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towr
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Re: intergal of e^-x^2
« Reply #9 on: Jul 21st, 2005, 8:08am » |
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on Jul 20th, 2005, 6:33pm, THUDandBLUNDER wrote:| This belongs in Putnam, mod. |
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Your wish is my command 
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Wikipedia, Google, Mathworld, Integer sequence DB
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uMRod
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on Jul 20th, 2005, 4:03pm, Michael_Dagg wrote:Indeed.
Let me add: now, using complex analysis show that
int[exp(-x^2) dx, -inf, inf] = sqrt(pi).
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I have been working with Euler formula exp(ix) = cos(x) + i sin(x) to try get a complex integrand for exp(-x^2) but have not been successful.
Can you up a hint?
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