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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

Posted by: Dustin 6047
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.
I resent that, sir. I can assure you my mathematics are performed at the highest standards of rigour!

  • 02.12.2012 9:59 PM PDT

"All science is either physics or stamp collecting."

- Ernest Rutherford

Posted by: Dustin 6047

Posted by: Eternal Way
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.

I have a math level of basic algebra and geometry, give me a break.


Then I implore you to not pass judgement in those areas which you, my good and noble sire, have no knowledge of.

*curtsies*

[Edited on 02.12.2012 10:23 PM PST]

  • 02.12.2012 10:03 PM PDT

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Stole many a man's soul and faith

I've got a snake mang! I fed it some beer one time. It was slithering this way and that. It was all -blam!- up!

  • 02.12.2012 10:06 PM PDT

Flours2012
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Math: the method by which the Universe is explained.

*confusing the majority of the human population since 35,000 BCE [Citation Needed]

  • 02.12.2012 10:10 PM PDT
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Posted by: Disambiguation
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]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.


We have a badass over here.

  • 02.12.2012 10:13 PM PDT

All the world's for living when love is what you find
Despair and loneliness you've got to leave them far behind
Tell the clouds to go let the feeling show
It's great when life is full of living

I didn't know you speak Japanese.

  • 02.12.2012 10:15 PM PDT


Posted by: TheMouthOfSauron
By (-x^2) do you mean -x^2 or (-x)^2?

if he meant (-x)^2 he would have put that

  • 02.12.2012 10:15 PM PDT

I think that disambiguation is trying to compensate for something...

  • 02.12.2012 10:15 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: Ratheld
I think that disambiguation is trying to compensate for something...
I can assure you that all the uni lasses at the pub are quite taken by my cranial capacity and dashing good looks.

  • 02.12.2012 10:19 PM PDT

Old school Bungie, born and raised,
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Got in one little argument, and the mods got scared,
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You forgot that, when switching from Cartesian integrals to polar integrals, dx*dy becomes r*dr*dtheta (or in your case, q*dt*dq)

Anyway, I just worked it out on paper. Showing in my own terms here, but is the same.

lim(a->oo)int[-a,a]int[0,2pi]r*e^(-r^2)dthetadr
=2pi*lim(a->oo)int[-a,a]r*e^(-r^2)dr, int(r*e^(-r^2)) = -(e^(-r^2))/2
=-pi*lim(a->oo)(e^(-r^2))](-a,a), lim(a->oo)e^(-a^2)=0
=-pi(0-0)
=0

  • 02.12.2012 10:35 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: prometheus25
You forgot that, when switching from Cartesian integrals to polar integrals, dx*dy becomes r*dr*dtheta (or in your case, q*dt*dq)
Whoops. I had that when I worked it out on paper; seems to have slipped out when I typed this up. Thanks for catching that.
Posted by: prometheus25
lim(a->oo)int[-a,a]int[0,2pi]r*e^(-r^2)dthetadr
=2pi*lim(a->oo)int[-a,a]r*e^(-r^2)dr, int(r*e^(-r^2)) = -(e^(-r^2))/2
=-pi*lim(a->oo)(e^(-r^2))](-a,a), lim(a->oo)e^(-a^2)=0
=-pi(0-0)
=0
Note that by transforming into radial terms we lose symmetry of the integral bounds. The outer integral is from zero to a, not -a to a; thus it doesn't equate to zero.

[Edited on 02.12.2012 10:43 PM PST]

  • 02.12.2012 10:37 PM PDT

Old school Bungie, born and raised,
In the Septagon is where I spend most of my days.
Relaxin', maxin', posting all cool,
Talking about Halo, life and some school.
Got in one little argument, and the mods got scared,
they said "You're gonna get banned and your member title'll be bare!"

You're correct. But there's an issue. You attempt to prove that int(e^-x^2) is the root of pi by creating a "mirror" function of int(e^-y^2) and multiplying the two. With the solution being pi, you're making the assumption that:

int (e^-x^2) = [int (e^-x^2)*int(e^-y^2)]/int(e^-y^2). This is, fundamentally, not incorrect.

When you switch to polar coordinates, you are taking the assumption that the variables x and y are orthogonal, meaning that they exist 90 degrees out of phase in a two dimensional surface.

You then to make the assumption that x and y are interchangeable because they exist with the same parameter space (-oo to oo). You attempt to equate that x*y/y = x^2/x, which it is not in the space of the function we are working in, due to the assumption of orthogonality that is forced when the integral is taken into polar coordinates.

Thus, (e^-x^2) does not equal sqrt(int (r^-r^2)), and so does not equal sqrt(pi)

[Edited on 02.12.2012 10:58 PM PST]

  • 02.12.2012 10:57 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

I can assure you my proof is quite sound (though devoid of significant detail -- this is The Flood), and you may want to brush up on your analysis.

Here.

[Edited on 02.12.2012 11:01 PM PST]

  • 02.12.2012 11:00 PM PDT

Old school Bungie, born and raised,
In the Septagon is where I spend most of my days.
Relaxin', maxin', posting all cool,
Talking about Halo, life and some school.
Got in one little argument, and the mods got scared,
they said "You're gonna get banned and your member title'll be bare!"

I don't like it.

But it is math, so I can't argue...

  • 02.12.2012 11:01 PM PDT
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>Implying everyone in the Flood is a 1337 math major

  • 02.12.2012 11:01 PM PDT

...yeah I'll take your word for it.

  • 02.12.2012 11:02 PM PDT

You're either first or you're last.

You'll never amount to anything, go back under your bridge.

Edit: I can't really follow your proof as I am tired and looking at integrals on a forum drives me nuts, but I do believe you. Wolfram agrees, that's good enough for me.

[Edited on 02.12.2012 11:15 PM PST]

  • 02.12.2012 11:11 PM PDT

This thread makes me feel stupid. Thanks OP.

  • 02.12.2012 11:16 PM PDT


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


TL;DR

Why aren't you curing cancer or something?

  • 02.12.2012 11:17 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
Why aren't you curing cancer or something?
Takes too long to get an M.D.

  • 02.12.2012 11:19 PM PDT

"No one thought up being;
he who thinks he has
Step forward"

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Uh.... Uh.....

24!

  • 02.12.2012 11:19 PM PDT

Posted by: bobcast
i'm drunk and habe the hiccups.

I love triguy and he like all the cool stuff.

respect him or die.

Woah, bro, I'm not an aero major anymore. This is now russian to me.

Scratch that, I'm a russian minor. It's greek to me.

[Edited on 02.12.2012 11:20 PM PST]

  • 02.12.2012 11:20 PM PDT

Wake me, when you need me.

>Is bad at math
>Opens thread anyways
>Leaves depressed

  • 02.12.2012 11:22 PM PDT