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#1 |
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FFR Player
Join Date: Jun 2007
Age: 32
Posts: 40
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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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#2 |
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Little Chief Hare
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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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#3 |
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FFR Player
Join Date: Dec 2005
Age: 30
Posts: 240
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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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#4 |
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FFR Player
Join Date: Oct 2006
Age: 30
Posts: 332
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Not funny.
__________________
FRO __________________________________________________ Tournament Victories: Doug | Colt | u84 | rshadow8888 | Sweet_Feet |
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#5 |
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FFR Player
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M i s s i s s i p p i
__________________
![]() ![]() wewt10k aim: IMB3AU![]() http://video.google.com/videoplay?do...&q=vertex+beta I play Vertex BETA :O |
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#6 |
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FFR Player
Join Date: Dec 2005
Age: 30
Posts: 240
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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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#7 |
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Very Grave Indeed
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Unfortunately, this is definately a Chit-Chat sort of topic, as are all "List your X" threads.
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#8 |
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FFR Player
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I'm gonna have to go with the ultimate inequality:
i > u (Thanks ThinkGeek) Oh, and yeah, CC material. |
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#9 |
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2 is poo
Join Date: Sep 2005
Age: 28
Posts: 6,620
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a^2 + b^2 = c^2
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