Prove you're white and solve for e

e is not a variable retard.
That's just a true expression

e^(i.π) is equal to cos(π)+sin(π) which is equal to -1 . So this statement is true. Don't be a nigger. Know some basic math.

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i could never understand math, am i a shitskin?

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-1

remove kebab

Whats the idea of "i" there... series of all integers?

it's nigger

it means its imaginary

>solve for e

you just proved you're not white, nigger.

-1^(1/(i*Pi))
>solving for a constant

You idiot

You're suppose to ask people to prove the equation is true not solve for e

I'm studying pharmacy, not math, this is irrelevant to me.

it's fuckin' gay as shit mang

i is the only variable in that equation.

i = ln(-1)/pi.

ln(-1) = pi

i = pi/pi

i = 1

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

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what is E^i because it confuses me, I'm starting Uni, this shit confuses me, in a few years I might get it, but now I'm confused

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e^j*pi=sin(pi)=-1

I'm an EE on smoke break at work

i isn't a variable

That's a Euler identity. You combine some Taylor expansions of geometric functions and add the imaginary unit i to get the signs right, and voilá you've connected three branches of mathematics that aren't supposed to have anything to do with each other. Some retards think this means God exists. I just chalk it up to reality being mathematical.

Oh, i is imaginary number i.

It's just a true expression.

e^(i*pi) = -1

>isn't e equal to (1 + 1/n)^n as n tends to infinity?
>>solve for e.
>doesn't make any sense
e^i*pi = -1
e^(i*pi-(i*pi-1)) = e = -1^(-(i*pi-1))

j=i in EE because we use i for current in other equations

e is not a number

well you already proved that you're nigger OP
kill yourself

sorry i had this shit like 1 year ago
i isn't a variable

retard

wait a sec solve for e?
wtf e is an actual number faggo

It's cos(π)+i sin(π).

So e is a number? Who is the retard?

>solving for e
>not knowing it's a mathematical constant
sage

square root of (-1)

>solve for a constant
American educamation at its finest.

yes

that's the entire idea of a constant

Take 1. Solved.

Well, technically it is impossible just from this identity. Because exponential of a complex number must be defined as an analytic continuation of series expansion for e^x into complex plane, but by knowing this series you automatically know e^1.

e is a natural logarithm you fucks
e=2.718

>implying e^(pi) = -1

everyone in this thread should go back to striking blows for the white race because you're dogshit at basic mathematics.

That is an expresion. That is only true when E^Pi*i=-1 and that is only true with complex numbers. Is known and the Euler's formula.

he's not wrong tho
e is a mathematical constant

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High school?

Base of a natural logarithm

a constant is just a specific number

A good number of you pooinloos are good at math

lim (1 + 1/n)^n
for n -> infinity

Is "e" the mathematical constant e, or is it just a variable?

>implying e is a variable
there is nothing to solve for, e is a defined number

e = the limit as n goes to infinity of [1 + 1/n]^n

>solve for a constant
murrican education everyone

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>ln(-1) = pi
Nope.
ln(-1)= Pi i, with i being the symbol for "imaginary numbers".

It's already solved. There's no variable. It's an identity.

e is always e 2.7.......

√-1

e is a constant. Expressed as a limit since it's a transcendental number like π. It's about 2.78 iirc. Stupid question.

t. Math major

Now to discredit the PISA meme.

e^( i*[pi] ) + 1 = 0 | e = x, c = i*pi
x^c + 1 = 0
x^c = -1
ln(x^c) = ln(-1)
c * ln(x) = ln(-1)
ln(x) = (1/c)*ln(-1)
ln(x) = ln(-1^(1/c))
x = -1^(1/c)
e = -1^(1/[i*pi])

It's multivalued. On one Riemann sheet it's iPi, on another it's 3iPi, 5iPi etc

Legit posts

z = cos(x) + i*sin(x)
dz/dx = -sin(x) + i*cos(x)
dz/dx = i*z
dz/z = i*dx
INT[dz/z] = INT[i*dx]
ln(z) = i*x + const
z = e^(i*x + const)
z = cos(x) + i.sin(x) when x=0, z=1 so the constant = 0
z = e^(i*x)
cos(x) + i*sin(x) = e^(i*x)
When x = pi : -1 = e^(i*pi)