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MeBeatYou
Member
Joined: 29 Sep 2005
Posts: 178
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 Posted: 18 Nov 2005 10:12:32 pm Post subject: |
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I've got no idea who it is in this forum, but somebody has an avatar which says that e^(pi*i)=-1
I've been over this about 20 times and it's been driving me crazy. Can anybody explain this to me? I saw it again on a stupid math movie today and my teacher offered a good deal of extra credit to the person who can explain it :biggrin: |
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Weregoose
Authentic INTJ
Super Elite (Last Title)
Joined: 25 Nov 2004
Posts: 3976
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| Posted: 18 Nov 2005 11:04:42 pm Post subject: |
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The following identities exist:
[font="courier new"]sin([font="times new roman"]π[font="courier new"]) = 0
cos([font="times new roman"]π[font="courier new"]) = -1
e^(xi) = cos(x) + sin(x) × i
Thus,
[font="courier new"]e^([font="times new roman"]π[font="courier new"]i) = cos([font="times new roman"]π[font="courier new"]) + sin([font="times new roman"]π[font="courier new"]) × i
e^([font="times new roman"]π[font="courier new"]i) = -1 + 0 × i
e^([font="times new roman"]π[font="courier new"]i) = -1 |
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thornahawk
μολών λαβέ
Active Member
Joined: 27 Mar 2005
Posts: 569
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| Posted: 19 Nov 2005 10:37:25 am Post subject: |
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FYI, the third relation Goose gave is the so-called "Euler identity". The easiest way to prove it is with power series, but IIRC there is a "more elementary proof" which I'll get back to you with if I can find my notes...
thornahawk
P.S. From here: "Gauss is reported to have commented that if this formula was not immediately obvious, the reader would never be a first-class mathematician."  |
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Liazon
title goes here
Bandwidth Hog
Joined: 01 Nov 2005
Posts: 2007
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| Posted: 19 Nov 2005 05:49:30 pm Post subject: |
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wow, too bad I'm not smart enough to explain it myself.
The kid who showed me the proof used log and exponent laws though. It was much simpler and did not require trig.
And i'm very sorry if you don't like my avatar.
Last edited by Guest on 19 Nov 2005 06:36:01 pm; edited 1 time in total |
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