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        riddles >> putnam exam (pure math) >> intergal of e^-x^2
        
(Message started by: EzisEz on Jul 20^th, 2005, 6:49am)

Title: intergal of e^-x^2
Post by EzisEz on Jul 20^th, 2005, 6:49am

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"

Title: Re: intergal of e^-x^2
Post by towr on Jul 20^th, 2005, 7:40am

I think you can do it by using a transformation.

Title: Re: intergal of e^-x^2
Post by EzisEz on Jul 20^th, 2005, 7:48am

Yea, my best guess would be [hide]polar coordinates[/hide]

Title: Re: intergal of e^-x^2
Post by River_Phoenix on Jul 20^th, 2005, 7:50am

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.

Title: Re: intergal of e^-x^2
Post by EzisEz on Jul 20^th, 2005, 8:36am

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

Title: Re: intergal of e^-x^2
Post by Michael_Dagg on Jul 20^th, 2005, 9:53am

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.

Title: Re: intergal of e^-x^2
Post by EzisEz on Jul 20^th, 2005, 10:05am

yep thats pretty much what i did, just couldn explain it as well....here it is in detail

Title: Re: intergal of e^-x^2
Post by Michael_Dagg on Jul 20^th, 2005, 4:03pm

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).

Title: Re: intergal of e^-x^2
Post by THUDandBLUNDER on Jul 20^th, 2005, 6:33pm

This belongs in Putnam, mod.

Title: Re: intergal of e^-x^2
Post by towr on Jul 21^st, 2005, 8:08am

on 07/20/05 at 18:33:51, THUDandBLUNDER wrote:

This belongs in Putnam, mod.

Your wish is my command ;)

Title: Re: intergal of e^-x^2
Post by uMRod on Jul 22^nd, 2005, 9:33am

on 07/20/05 at 16:03:17, Michael_Dagg wrote:

Indeed.
Let me add: now, using complex analysis show that
int[exp(-x^2) dx, -inf, inf] = sqrt(pi).

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?