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| Subject:
whats the story behind: (e^(-i * pi)) + 1 = 0
 Category: Science > Math Asked by: drkay4414-ga List Price: $3.50  | 
Posted:
11 Jan 2005 13:40 PST
 Expires: 18 Jan 2005 11:49 PST Question ID: 455688  | 
a friend told me that the equation (e^(-i * pi)) + 1 = 0 is evidence that the universe has intelligent design. i understand the equation but would like a few scientific articles and summaries throughly explaining and discussing it, and possibly how it was derived. the equation is correct: ://www.google.com/search?hl=en&lr=&safe=off&q=e%5E%28-i*pi%29%2B1&btnG=Search  | |
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| Subject:
Re: whats the story behind: (e^(-i * pi)) + 1 = 0
 From: ticbol-ga on 12 Jan 2005 03:35 PST  | 
(e^(-i * pi)) + 1 = 0 Let us play with that and convert it into a more familiar equation. (e^(-i * pi)) + 1 = 0 1/[e^(i*pi) +1 = 0 Multiply both sides by e^(i*pi), 1 +e^(i*pi) = 0 Or, e^(i*pi) +1 = 0 ----*** That is the special form of the Euler's Formula. It shows the relationship of 0, 1, pi, e, and i in one equation. The general form of the euler's formula is e^(i*x) = cos(x) +i*sin(x) So if x = pi, e^(i*pi) = cos(pi) +i*sin(pi) e^(i*pi) = -1 +0 e^(i*pi) +1 = 0 ----*** For the story, intelligent design, whatever, behind that formula, go search Google for euler's formula.  | 
| Subject:
Re: whats the story behind: (e^(-i * pi)) + 1 = 0
 From: touf-ga on 13 Jan 2005 13:53 PST  | 
If we look at the Taylor Series Expansion for the exp function, we see it is:
exp (x) = 1 + x/1! + x^2/2! + x^3/3! + x^4/4! + ...
If we look at the expansion for the sin and cos functions, we see it is:
sin (x) = x - x^3/3! + x^5/5! - x^7/7! + ...
cos (x) = 1 - x^2/2! + x^4/4! - x^6/6! + ...
Now, add the cos and sin terms
You get...
cos (x) + sin(x) = 1 + x - x^2/2! - x^3/3! ...
Substituting x = (i*pi), you get...
cos (i*pi) + sin(i*pi) = 1 + i*pi - (i*pi)^2/2! - (i*pi)^3/3! ...
But wait!  i^1 = i
           i^2 = -1
           i^3 = -i
           i^4 = +1 and so on...
So, cos (i*pi) + sin (i*pi) = 1 + i (pi) + pi^2/2 + i (pi^3)/3! + ...
Note all the sin terms from the Taylor expansion have a factor of i in them.
Now, note that this is almost exactly identical to the above expansion
of the exp term.  The only difference is those pesky i terms in the
odd powers, which are in the sin terms.
Well, we can just factor those out, and get...
exp (i*x) = cos (x) + i * sin(x), which was already pointed out by ticobal.
Doing the actual math, we find that
cos (pi) = -1
i* sin (pi) = 0
cos (pi) + i* sin(pi) = -1
exp (i*pi) = -1
Note that cos (-pi) = cos (pi) = -1, so it doesn't matter if you have
exp (i*pi) or exp (-i*pi) -- you get the same answer.
Hope that helped.
-touf | 
| Subject:
Re: whats the story behind: (e^(-i * pi)) + 1 = 0
 From: touf-ga on 13 Jan 2005 13:56 PST  | 
In response to whether this is Divine or defined (haha), in my opinion, probably a bit of both. Most of it has to do with definitions and the like, but I do believe that math is Divinely inspired, as its perfection cannot of human design.  | 
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