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#3 - tittylovin
Reply +1
(09/13/2013) [-]
Someone explain this to me.

Amazing, even the most obscure and brainy SMBC gets thumbed up, these can't be the same people that thumb up the same 40 tumblr screenshots to the front page every day. Either there's a brainy crowd I'm not aware of, or people are just pretending.
#68 to #3 - anon
Reply 0
(09/14/2013) [-]
e^i*pi = cos(pi) + i * sin(pi) = -1 + i * 0= -1
#50 to #3 - hawaiianhappysauce
Reply +3
(09/14/2013) [-]
I will derive Eulers formula.

To do this, you need the following identities in the picture:
The one with sine and the one with cosine, you also need the power series of

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

Now we plug in e^(i*y) = 1 + i*y/(1!) +((iy)^2)/(2!)+ ((iy)^3)/(3!)+ ...+ ((iy)^n)/(n!) + ...
using the fact that i^2 = -1 ( i is the sqrt of -1), i^3 = - i and i^4 = (-1)^2 = 1
e^(i*y) = 1 + i*y/(1!) - ((y)^2)/(2!) - i (y)^3)/(3!)+ ...+ i^n((y)^n)/(n!) + ...

When n is even we get - 1 or 1, when n is odd we get -i or i

Splitting them apart gives us:
(1 - ((y)^2)/(2!) + (y)^4)/(4!)+ ...+ (-1)^(n)((y)^(2n))/(2n!) + ... )
+ ( iy/(1!) - i((y)^3)/(3!) + i(y)^5)/(5!)+ ...+ (-1)*(n-1) * i* ((y)^(2n+1)/((2n+1)!) + ...)
Using the identity in the picture, and factoring out i from the second series we get:
e^(iy) = Cos(y) + i*Sin(y)
plug y= pi we get
e^(i*pi) = Cos(pi) + i sin(pi) = -1 + 0 = -1
#33 to #3 - loluzer
Reply +1
(09/14/2013) [-]
It has to do with complex numbers. Euler discovered that a+bi can be rewritten as re^(i*theta). That's called the Euler form of a number. The Euler form of -1 is e^(i*pi). This is really cool because it can be written as e^(i*pi)+1=0. If you had to list the most important numbers in math, they would be e, pi, i, 1, and 0. This is really cool because they have all just been linked together with one formula, showing that all important numbers in math are related.
#24 to #3 - anon
Reply 0
(09/14/2013) [-]
Here's an easier explanation: First you need to know the index law that states (x^a)/(x^b) = x^a-b . Now imagine the exponents were equal, i.e. (x^2)/(x^2), this would mean that the answer is x^2-2, which is x^0... but as we all know, anything divided by itself is 1. Therefore, anything to the power of 0 is 1
#18 to #3 - harleycurnow
Reply +1
(09/13/2013) [-]
3 numbers that completely unrelated and cannot even be written down (one of which doesn't even exist) go together in an extremely simple way and create a really round number.
#8 to #3 - articulate
Reply +3
(09/13/2013) [-]
There's nothing to really "explain." It's just interesting how this works.
note: "i" is the square root of negative one.
#96 to #8 - revisandbutthead
Reply +2
(09/14/2013) [-]
There is something to "explain", it just involves pretty complex calculus.

To get an idea look at comment #50. Or you can watch this video from khan academy
Euler's Formula and Euler's Identity

But it requires some knowledge of power series
#49 to #8 - srapture
Reply 0
(09/14/2013) [-]
Ah, thanks. The i is the one bit there I knew and it's quite late so I couldn't be bothered to google it.
#7 to #3 - dazzl
Reply 0
(09/13/2013) [-]
What the woman says is actually wrong. big ****** suprise init?
e^2*pi*i (where i is the complex number sqrt(-1)) is actually 1.
This notation is commenly known as Euler's equation. en.wikipedia.org/wiki/Euler%27s_formula if you want to know more.
#10 to #7 - dazzl
Reply +2
(09/13/2013) [-]
oh read it wrong, thought it said e^2*pi in the picture, but it's actually e^i*pi. Same principle though, Euler still be tha boss
oh read it wrong, thought it said e^2*pi in the picture, but it's actually e^i*pi. Same principle though, Euler still be tha boss
#4 to #3 - seriff
Reply +23
(09/13/2013) [-]
math be crazy
#5 to #4 - tittylovin
Reply +1
(09/13/2013) [-]
SHEEEEEEIIIIIIIIITTTTTTT
#6 to #5 - seriff
Reply +4
(09/13/2013) [-]
i like you :o
#12 to #6 - tittylovin
Reply 0
(09/13/2013) [-]
and i like you too