The Math Thread Options

[Kinsoku]

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

[Suzu Fujibayashi]

e^iπ + 1 = 0

I know not math, but this is apparently a special thing.

[Neon_the_Chao]

I've seen it before, not far enough in to know its relevance. Is that the formula though?

http://www.wolframalpha.com/input/?i=e^%28i*n%29%2B1

This post has been edited by Elnendil: Apr 17 2011, 12:37 AM

[whoyodaddy]

:/ Haven't you people ever heard of complex variables before?

[Brook]

I've seen it before, not far enough in to know its relevance. Is that the formula though?

http://www.wolframalpha.com/input/?i=e^%28i*n%29%2B1

I think so.

:/ Haven't you people ever heard of complex variables before?

nope.

[Astarte Mellia]

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.

[Takeru-kun]

:/ Haven't you people ever heard of complex variables before?

Uh, yes. I just haven't had any courses based on them. You think that they taught deeper meaning concepts of complex variables in high school? Lolno, they didn't.

[Another Necromancer]

so defensive

[fensti]

Maybe I just had a bad day, I dunno.

Looking over some geometric proofs, one of them is discussing some constructable number, sqrt(4+sqrt(7+sqrt(2))). The proof that such a number is irrational I thought was clever. First you prove that the square root of 2 is irrational. Then you have r equal that number and simply solve for the square root of 2, assuming that r is a rational number. You find a contradiction there.

The geometric one is to show that a collapsing compass is mathematically equivalent to the modern compass by proving that you can copy circles using only a collapsing compass, something it cannot easily do unlike the modern compass. You use radii of circles as triangle side lengths and show that many triangles are congruent, and thus their angles are congruent. A bit time consuming but clever.

This post has been edited by Elnendil: Apr 23 2011, 02:24 PM