e^(i*pi) =

[bdude]

need some math help with this one

 

[HAL9000]

-1

 

[Matthiasa]

No one needs help with that one.

 

[Newbian]

The cake is a lie, as everyone knows i * pie = cake.

 

[bdude]

The cake is a lie, as everyone knows i * pie = cake.

cake is yummy but pie is better

 

[bdude]

 

[Minjin]

 

[shortylickens]

42

Everyone knows this, it cannot be denied.

 

[NuroMancer]

http://xkcd.com/179/

 

[bdude]

 

[Newbian]

 

[ElFenix]

only thing that gets you off anymore amirite?

 

[Newbian]

only thing that gets you off anymore amirite?

Sadly it's true, I blame the internet.

 

[guyver01]

 

[TuxDave]

How about i^i?

 

[SSSnail]

peach fuzz?

 

[Locut0s]

-1

This

 

[Matthiasa]

How about i^i?

i^i=x
i*ln(i)=ln(x)
i*i*π/2=ln(x)
-1*π/2=ln(x)
e^(-π/2)=x

 

[QuantumPion]

also e^(i*pi) = cos pi + i sin pi

 

[paulney]

I effing hate polar coordinates and all this i bullshit.
Live Descartes or die.

 

[Locut0s]

I effing hate polar coordinates and all this i bullshit.
Live Descartes or die.

Who said anything about polar coordinates?

 

[paulney]

Who said anything about polar coordinates?

'cause that's where e^<Phi>*i is used to transform from Cartesian to polar and back.

 

[Powermoloch]

silly people

it's -1

 

[Locut0s]

'cause that's where e^<Phi>*i is used to transform from Cartesian to polar and back.

Isn't it:

r = &#8730; ( x2 + y2 )

&#952; = tan-1 ( y / x )

and

x = r × cos( &#952; )

y = r × sin( &#952; )

 

[paulney]

Isn't it:

r = &#8730; ( x2 + y2 )

&#952; = tan-1 ( y / x )

and

x = r × cos( &#952; )

y = r × sin( &#952; )

That's for the entry-level calculus students.