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Old 09-22-2007, 11:46 PM   #1
ljw5021
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Default Elegant Mathematics

Post your favorite mathematical proof or theorem.

Mine has to be Euler's.

g(x) is defined as e^ix

therefore g'(x) = ie^ix, and g''(x) = -e^ix

g''(x) = -g(x)
g''(x) + g(x) = 0

g1(x) = cosx
g2(x) = sinx

g(0) = A = 1
g'(0) = B = i

g(x) = e^ix = cosx + isinx

And for x=pi

g(pi) = e^(i*pi) + 1 = 0

The Taylor series proof and Calc proof are equally elegant.
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Old 09-23-2007, 12:08 AM   #2
Kilroy_x
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Default Re: Elegant Mathematics

Heh, I'm varying degrees of ignorant and idiotic when it comes to math, so I'll just say

1. Q -> R (premise)
2. P -> Q (premise)
3. P -> R (1,2, ->I)

In other words, the classical syllogism. Yes, lame I know. I'll get back to you after I've actually studied math. at all.
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Old 09-23-2007, 12:27 AM   #3
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Default Re: Elegant Mathematics

rofl
(vagina+penis)-penis=vagina+load

Last edited by perkeyone; 09-23-2007 at 12:27 AM.. Reason: dont ban me please
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Old 09-23-2007, 12:29 AM   #4
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Default Re: Elegant Mathematics

Not funny.
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Old 09-23-2007, 12:31 AM   #5
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Default Re: Elegant Mathematics

M i s s i s s i p p i
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Old 09-23-2007, 12:37 AM   #6
perkeyone
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Default Re: Elegant Mathematics

Quote:
Originally Posted by FRO View Post
Not funny.
wow im gona cry now because your opinion matters to me

squeeze theorem
Let I be an interval containing the point a. Let f, g, and h be functions defined on I, except possibly at a itself. Suppose that for every x in I not equal to a, we have:

g(x) \leq f(x) \leq h(x)

and also suppose that:

\lim_{x \to a} g(x) = \lim_{x \to a} h(x) = L.

Then \lim_{x \to a} f(x) = L.
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Old 09-23-2007, 12:48 AM   #7
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Default Re: Elegant Mathematics

Unfortunately, this is definately a Chit-Chat sort of topic, as are all "List your X" threads.
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Old 09-23-2007, 01:09 AM   #8
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Default Re: Elegant Mathematics

I'm gonna have to go with the ultimate inequality:

i > u

(Thanks ThinkGeek)

Oh, and yeah, CC material.
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Old 09-23-2007, 03:55 AM   #9
smartdude1212
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Default Re: Elegant Mathematics

a^2 + b^2 = c^2
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