The Math Thread Options

[Enzd]

I don't use it, but some do. SO PUT IT HERE.

This post has been edited by Enzd: Apr 13 2011, 09:35 PM

[Raijinili]

e ^ (pi * i) + 1 = 0

Where z is a complex number, e^z is defined as
\sum_{n=0}^{\infinity} z^n / n!

Thus, where x is a real number, e^(ix) is thus defined as
\sum_{n=0}^{\infinity} i^n x^n / n!

Let's also define sine and cosine:
cos(x) := \sum_{n=0}^{\infinity} i^(2n) x^(2n) / (2n)!
sin(x) := \sum_{n=0}^{\infinity} i^(2n+1) x^(2n+1) / (2n+1)!

Or, simplified:
cos(x) := \sum_{n=0}^{\infinity} (-1)^n x^(2n) / (2n)!
sin(x) := \sum_{n=0}^{\infinity} i (-1)^n x^(2n+1) / (2n+1)!

So really, e^(ix) = cos(x) + i sin(x). Let x = pi. Then cos(pi) is -1, and sin(pi) = 0. So e^(i pi) = -1, so e^(i pi) + 1 = 0, which is kind of neat because we have addition, multiplication, exponentiation, and =, as well as 0, 1, pi, e, and i.