{{
`${userDetail.user.displayName}${userDetail.user.cachedBungieGlobalDisplayNameCode ? `#${standardizeBungieGlobalCode(userDetail.user.cachedBungieGlobalDisplayNameCode)}` : ""}`}}
{{getUserClans()}}
2/13/2012 5:27:40 AM
107
A rather striking and errant mathematical fact
Whence one might presume that the integral, from the negative of infinity towards the infinite, of the simple function exp(-x^2) might equate to the root of pi? I shall set forth to prove to you this assertion, that you should be struck by its audacity as much as I on this Sunday's eve.
We begin by considering the square, such that the integral under examination would be the product of an integral of exp(-x^2) and of an integral of exp(-y^2), for {x:-r,r} and {y:-r,r}.
As we should desire r to tend toward an infinite quantity, it necessarily must be that we have, indeed, a product of two limits. As one must recall from elementary calculus,
Lim(r->oo) int[-r,r]exp(-x^2)dx * Lim(r->oo) int[-r,r]e^(-y^2)dy
is equivalent to
Lim(r->oo) { int[-r,r]exp(-x^2) int[-r,r]exp(-y^2)dy }
One may then be struck with an occurrence: a theorem of Fubini may be applied to conjoin the integration. The result follows as
Lim[r->oo] int[(-r <= x <= r), (-r <= y <= r)]exp(-(x^2+y^2))dxdy
This is, of course, the same as int[R^2]exp(-(x^2+y^2))dxdy; we are therefore not constrained by a square modicum of integration, and indeed we should choose a circular region to evaluate this integral. Thus, we have the result
Lim(q->oo) int[x^2+y^2<=q^2]exp(-(x^2+y^2))dxdy
When considering circular regions it is most convenient to use a polar co-ordinate system, thus we apply the angular transformation to result in
Lim(q->oo) int[0,q]int[0,2pi]q*exp(-q^2)dtdq
Which can be evaluated by elementary calculus techniques to yield Lim(q->oo) (2pi)*(1 - exp(-q^2))/2 = pi.
We recall that we have calculated the square of the integral of interest. Thus, it follows by taking the principal root that the infinite-integral of exp(-x^2) is the root of pi.
[url=http://www.wolframalpha.com/input/?i=integrate+e%5E(-x%5E2)+from+x+%3D+-infinity+to+infinity]tl; dr[/url]
[Edited on 02.12.2012 11:49 PM PST]
-
[quote][b]Posted by:[/b] Alt account1 I read the OP, I don't understand. Someone explain the fact in short? I don't need to know why, just tell me the damn fact. [/quote] Basically, what is happening is that a surface is being made. The surface is a square with infinitely long sides (think of it more philosophically than literally). The height of the surface above the ground of zero at any point (x,y) is determined by e^-(x^2 + y^2). [url=http://www.wolframalpha.com/input/?i=e^-%28x^2+%2B+y^2%29]Here's[/url] a small portion of this surface. What integrals do is they find the area below a curve, from point a to b. In this instance, the bounds are from negative infinity to infinity, which means you are finding the area beneath the [i]entirety[/i] of the curve. By having a double integral (that's what dx and dy imply), you are finding the volume under the infinitely large square, or the entirety of the volume. X and Y (and Z) coordinates (what is referred to as the "Cartesian coordinate system" isn't the only way to represent space. It could also be done in polar coordinates, where a position is identified by a radius (usually from the 0,0 origin) and angle (from the x axis counter-clockwise). The same space of an infinitely wide and tall square to a circle of infinite radius (again, "philosophically" the same thing) spun about the 0,0 point a full 360 degrees (though we use radians. It's simpler mathematically. 360 degrees = 2pi radians) When you switch coordinates, the variables need to be adjusted. See in the OP where e^1(x^2+y^2) switches to e^-r^2? Think of the Pythagorean Theorem. x^2 + y^2 = z^2. z is the equivalent of a radius (Are you familiar with the Unit Circle), so this Theorem is tasked in this scenario as x^2 + y^2 = r^2, which you see in the OP. Solving integrals is a bit complicated for those unfamiliar with it. Once a certain amount of familiarity is acquired with them, it's mostly pattern recognition. At this point you're just going to have to trust me/us on the issue. So the solution here solves out to be pi. It is then assumed that since the two integrals at the beginning are so similar, you could treat them as if they are equivalent to one of them squared. This means that The square root of one of them squared, lim(a->oo)int(-a,a)e^(-x^2)dx, is equal to the square root of the answer, equating to sqrt(pi). I have some issues with this proof, even though it mathematically is sound. They can be seen in the document linked in [url=http://www.bungie.net/Forums/posts.aspx?postID=70094942&postRepeater1-p=4#70136471]this post[/url].
