e^(i*pi)

[sleeplessdragn]

In my math class a discussion arose concerning this interesting problem. The solution of e^(i*pi) is equal to -1, yet not even my math teacher could explain exactly why this was. All that he could say was that in order to prove this, a derivative of e and the fractional value of pi was involved. I know that it involves a certain trig identity but not much more than that.

After searching through google I found not much more information, although I admit that i did not really google too deeply. I found, though, that a correlation between e and pi has to exist.

Does anyone know of any correlation between e and pi? Are there any practical uses for this equation? Discuss.

[GuidoHunter]

It's been said that that equation is the most magical equation in all of math because it incorporates several different types of math and relates them all. That is, the transcendental, the algebraic, and the imaginary.

I don't know much about it myself, and if you already googled it, I can't really furnish much more.

--Guido

http://andy.mikee385.com

[Kefit]

This equation is incredibly important to physics and differential equations. Lemme see if I can remember the proof:

First off, this proof uses infinite series. Put basically, any differentiable function can be represented by the sum of an infinite series. No, I don't expect you to know what that means. Just accept that:

cos(x) = 1 - (x^2)/(2!) + (x^4)/(4!) - (x^6)/(6!) + . . .

sin(x) = x - (x^3)/(3!) + (x^5)/(5!) - (x^7)/(7!) + . . .

e^x = 1 + x + (x^2)/(2!) + (x^3)/(3!) + . . .

It follows that:

e^(i*x) = 1 + (i*x) - (x^2)/(2!) - (i*x^3)/(3!) + (x^4)/(4!) + (i*x^5)/(5!) - (x^6)/(6!) + . . .

i*sin(x) = (i*x) - (i*x^3)/(3!) + (i*x^5)/(5!) - (i*x^7)/(7!) + . . .

Now something interesting happens when you add cos(x) and i*sin(x) together. Lemme see if I can illustrate this clearly:

.....cos(x) = 1..........- (x^2)/(2!) ......................+ (x^4)/(4!) .................- (x^6)/(6!) + . . .
+ i*sin(x) = ..(i*x)..................... - (i*x^3)/(3!) .................+ (i*x^5)/(5!) - . . .
---------------------------------------------------------------------------------------------------------------
....e^(i*x) = 1 + (i*x) - (x^2)/(2!) - (i*x^3)/(3!) + (x^4)/(4!) + (i*x^5)/(5!) - (x^6)/(6!) + . . .

Don't be discouraged if the idea of infinite series is unfamiliar to you - just take the series I gave for sin(x), cos(x), and e^x for granted, and everything else I show above follows from simple arithmetic on the series.

Anyway, this creates the general equation:

e^(i*x) = cos(x) + i*sin(x)

if x = PI, then:

e^(i*PI) = cos(PI) + i*sin(PI) = -1 + i*0 = -1

Oh, and shame on your math teacher for not knowing this proof

[Cenright]

(It seems as you are just finding a way to rid yourself of the 'i', thus attempting to find a way to legally multiply by zero.)
I know I am looking at this from outside the equation, but still...

[GuidoHunter]

Uh, what? Where does zero come into this and when is it illegal to multiply by zero?

--Guido

http://andy.mikee385.com

[Kefit]

Well, some people use zero to set up this identity as:

e^(i*PI) + 1 = 0

And then they sometimes use that to try and prove the existence of God due to the existence of such a streamlined relation between what are more or less the five most important constants in all of math.

Of course, that is not what Cenright was talking about; rather this is the real way that zero gets involved with this equation.

[Anticrombie0909]

*head explodes*

[Omeganitros]

Actually, I think I got a grasp on that.

However, I can see that me and Infinity will be having a frustrating relationship in the future.

[sleeplessdragn]

I heard about the mathematical proof of the existence of God, but i had NO clue that this equation was the root of that proof. That simply astounds me. I actually did a one eye brow raise when i read that.

[Kilgamayan]

Jesus, I couldn't have produced that and I'm a math major.

I'll have to remember that.

[Tasselfoot]

All I remember about that proof is that my fat ass math god buddy used to geek around talking about e^(i*Pi) instead of -1 any time -1 came up, or some such multiplier of that.

Needless to say, yes... he's a math god. and yes... he's a virgin. and yes... he weighs 300lbs.

And, with that, I go back to hating math math and back to loving finance math.

[blahblah18]

OK ITS REAL SIMPLE...
kefit has it right...
they're called taylor series... anyone? anyone? Bueller? calculus?

[GuidoHunter]

Pfft, they're nothing but MacLaruin serieseses centered about zero.

--Guido

http://andy.mikee385.com

[igotrhythm]

Here's the proof I saw once for proving the existence of God. At least, I think this is how it goes.

0=0+0+0+0... and this stretches on to infinity. This can be rearranged to

0=(1-1)+(1-1)+(1-1)...

0=1-1+1-1+1-1...

=1+0+0+0+0...=1

Thus, something has been created out of nothing. However, this is not legitimate because the infinite series is divergent. That is, the series of partial sums (1, 0, 1, 0...) do not approach any specific number. If it did, this limit is defined as the sum of the infinite series.

[banditcom]

Here's the proof I saw once for proving the existence of God. At least, I think this is how it goes.

0=0+0+0+0... and this stretches on to infinity. This can be rearranged to

0=(1-1)+(1-1)+(1-1)...

0=1-1+1-1+1-1...

=1+0+0+0+0...=1

Thus, something has been created out of nothing. However, this is not legitimate because the infinite series is divergent. That is, the series of partial sums (1, 0, 1, 0...) do not approach any specific number. If it did, this limit is defined as the sum of the infinite series.

lol... That's for stupid people who fail to realize that the -1 is still in there to counter the 1.

Anyways, nice proof Kefit. I totally forgot about the different series, even though they're really cool. Currently, I'm so sick of proofs. Most of them I feel like "this doesn't seem to prove anything" but they do.