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  • Subject: A rather striking and errant mathematical fact
Subject: A rather striking and errant mathematical fact

"The individual has always had to struggle to keep from being overwhelmed by the tribe. If you try it, you will be lonely often, and sometimes frightened. But no price is too high to pay for the privilege of owning yourself."
-Nietzsche

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.


tl; dr

[Edited on 02.12.2012 11:49 PM PST]

  • 02.12.2012 9:27 PM PDT

797,700 threads in the Flood and counting

Posted by: Disambiguation
the simple function exp(-x^2)


Posted by: Disambiguation
simple

Posted by: Disambiguation
(-x^2)

lol

  • 02.12.2012 9:32 PM PDT

The beta is super awesome. RogueAssassin27 is now the person who got my Totodile's nickname. Hoo-hah! Relevant.inb4lock

Dude, Disam..... I'm sorry, but you and I both know this is something the Flood can't even begin to comprehend.

  • 02.12.2012 9:32 PM PDT

And he entered the service of the Dark Tower when it first rose again, and because of his cunning he grew ever higher in the Lord's favour; and he learned great sorcery, and knew much of the mind of Sauron; and he was more cruel than any orc.

By (-x^2) do you mean -x^2 or (-x)^2?

  • 02.12.2012 9:32 PM PDT

Posted by: mamamia
Posted by: Disambiguation
the simple function exp(-x^2)

Posted by: Disambiguation
simple

Posted by: Disambiguation
(-x^2)

lol

Yes. You take a number, you square it, and then make it negative. What's there not to understand?

  • 02.12.2012 9:33 PM PDT

Stop trying to validate yourself through knowing more math than people on the Flood.

  • 02.12.2012 9:33 PM PDT

797,700 threads in the Flood and counting


Posted by: Hylebos
Posted by: mamamia
Posted by: Disambiguation
the simple function exp(-x^2)

Posted by: Disambiguation
simple

Posted by: Disambiguation
(-x^2)

lol

Yes. You take a number, you square it, and then make it negative. What's there not to understand?

In sentence form, that made much more sense. But I can't read math that is on a calculus level, for I am only in eighth grade.

  • 02.12.2012 9:34 PM PDT

"The individual has always had to struggle to keep from being overwhelmed by the tribe. If you try it, you will be lonely often, and sometimes frightened. But no price is too high to pay for the privilege of owning yourself."
-Nietzsche


Posted by: quickdog93
Stop trying to validate yourself through knowing more math than people on the Flood.
'Tis but elementary calculus with multiple variables, my good chap! I would expect any lad aged at least fourteen to understand such mathematics.

  • 02.12.2012 9:35 PM PDT

hmmm very interesting indeed -- While he was busy figuring this out, I stole his girlfriend

  • 02.12.2012 9:36 PM PDT


Posted by: Disambiguation

Posted by: quickdog93
Stop trying to validate yourself through knowing more math than people on the Flood.
'Tis but elementary calculus with multiple variables, my good chap! I would expect any lad aged at least fourteen to understand such mathematics.

*sigh*

  • 02.12.2012 9:36 PM PDT

The beta is super awesome. RogueAssassin27 is now the person who got my Totodile's nickname. Hoo-hah! Relevant.inb4lock


Posted by: Disambiguation

Posted by: quickdog93
Stop trying to validate yourself through knowing more math than people on the Flood.
'Tis but elementary calculus with multiple variables, my good chap! I would expect any lad aged at least fourteen to understand such mathematics.
>14
>calculus
>wat

  • 02.12.2012 9:37 PM PDT

lol

These are the threads that keep me coming back to Bnet.

  • 02.12.2012 9:38 PM PDT

If I don't believe in him...



why should he believe in me?

This is why I hate calculus.

  • 02.12.2012 9:39 PM PDT


Posted by: Disambiguation
Whence one might presume that the integral, from the negative of infinity towards the infinite
Lost me there. Still in pre calc here.

[Edited on 02.12.2012 9:43 PM PST]

  • 02.12.2012 9:39 PM PDT

Conquer the musical world Hatsune Miku!

PS: loyalty forever-The Crimson Empire

co-leader

i wouldnr be surprised with that though... although you lost me in that-


the forumal of 2pi is equal to 360 degrees or a full circle. Circles are also infinite.

exponents multiplied can form semi circles given other equations to make it a circle. Given that, there can be two poles in a circle at any point- and thus infinite can be both reach in a negative and positive direction, they form a circle.





[Edited on 02.12.2012 9:41 PM PST]

  • 02.12.2012 9:40 PM PDT

Generalizations.
Helping idiots hate other idiots since people have existed.

Well, that's all fine and dandy...

...but will it blend?

  • 02.12.2012 9:40 PM PDT

Man, math is not my subject.

  • 02.12.2012 9:42 PM PDT

"The individual has always had to struggle to keep from being overwhelmed by the tribe. If you try it, you will be lonely often, and sometimes frightened. But no price is too high to pay for the privilege of owning yourself."
-Nietzsche

Posted by: forcereconFR13
i wouldnr be surprised with that though... although you lost me in that-

the forumal of 2pi is equal to 360 degrees or a full circle. Circles are also infinite.

exponents multiplied can form semi circles given other equations to make it a circle. Given that, there can be two poles in a circle at any point- and thus infinite can be both reach in a negative and positive direction, they form a circle.
You want to think of it as an integral in two dimensions, rather than one. The purpose of considering exp(-x^2)*exp(-y^2) is so that we can apply this transformation. By default, the limits of integration form a rectangular region in the xy-plane; however, because we are tending toward infinity we can transform this into a circular region and allow the radius to tend toward infinity instead. The polar coordinate transformation into radial and angular variables then lets us evaluate the integral, and we take the square root to get the final result.

[Edited on 02.12.2012 9:47 PM PST]

  • 02.12.2012 9:46 PM PDT

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Disambiguation: 1
Everyone who isn't a math major: 0

  • 02.12.2012 9:46 PM PDT

-/
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[[[[[[[[]]]]]]]

I think I should understand this, but I can't be bothered.

  • 02.12.2012 9:49 PM PDT

TL:DR

  • 02.12.2012 9:51 PM PDT
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Hey, that's hip, man.

  • 02.12.2012 9:52 PM PDT

"The individual has always had to struggle to keep from being overwhelmed by the tribe. If you try it, you will be lonely often, and sometimes frightened. But no price is too high to pay for the privilege of owning yourself."
-Nietzsche

Posted by: The Tempun
TL:DR
Integral from minus infinity to infinity of e(-x^2)dx = sqrt(pi)

  • 02.12.2012 9:52 PM PDT

"All science is either physics or stamp collecting."

- Ernest Rutherford

Posted by: Dustin 6047

Posted by: Chalupa King117
Dude, Disam..... I'm sorry, but you and I both know this is something the Flood can't even begin to comprehend.
He uses his superior knowledge of mathematics and physics to troll us into confusion. Basically he's saying that negative infinity to infinity equals pi, but that's just silly. I'm not smart enough to read his evidence but it probably doesn't support his claim, which is fine because no one will understand regardless.


Uhhh no.

The integral from -infinity to infinity of e^-x^2 certainly does equal pi.

  • 02.12.2012 9:58 PM PDT