Comments on Profile Post by Emerly_Nickel

I SEE THAT E^Iπ THERE

The -e^(i*pi) and only!

So this is what engineering math is like...

God, that's confusing! And I'm having problems with ORDER OF OPERATIONS WITH RATIONAL NUMBERS! Jeez...

just wait until complex numbers. THEN you'll be having problems.

gg

n^2 (aka n squared) divided by n is equal to n
n^3 (n cubed) divided by n^2 is equal to n
-e^(pi*i) is equal to 1
i is an imaginary number which is equal to the square root of -1
pi is the circumference of a circle divided by it's diameter

sooo it factors out to be:
(n)Spoopy(n+2)Me

ex. n=2 would produce: 2Spoopy4Me

I can explain the e^pi*i thing if you want, too :3

it's from Euler's formula which states for any real number x:
e^(ix)= cos(x) + isin(x).
Since cos(π) = -1 and sin(π) = 0, e^(iπ) = -1 + 0i, which gives:
e^(iπ) + 1 = 0 and e^(iπ) = -1, therefore:
-(e^(iπ)) = 1

awwww...I wanted to do it :[

;3

Fun fact: in electrical engineering, sqrt(-1) is commonly called j and not i because i stands for current in circuit theory.

god emerly, such basic engineering facts, do you think I'm 7 or something!????

no ;-;
I have no idea how old you are

14.

7 ......*2
:3

I think you mean
7*(-(e^(iπ))*2)

I think you mean 7 * (the derivative of(2*x^3/(x*((x^2)/x)))

*derivative with respect to x

OMG DA MATH

ITS KILLEN MEH

*heads explodes*