/sci/ - Science & Math

[Anonymous No.4702325]

<div class="math"> e^{i\pi}+1 = 0 </div>

[Anonymous No.4702331]

>>4702325
do you know at least how to prove it?

[Anonymous No.4702330]

Oh hello, welcome to mathematics. We all hope you enjoy your stay.

[Anonymous No.4702333]

<span class="math">e^{-1/\pi}=1/2[/spoiler]

[Anonymous No.4702339]

>>4702331
do you?

[Anonymous No.4702352]

>>4702331
>>4702331
quit being such a faggot. Even an engineer could prove it with the maclaurin series.

[Anonymous No.4702367]

>>4702331
e^(i*pi) = -1. Then you add 1 to each side of the equation.

[Anonymous No.4702380]

The most accepted proof is the Taylor expansion of e^ix

[Anonymous No.4702386]

>>4702380
>The most common and easiest proof
fix'd

[Anonymous No.4702409]

>>4702325

e^(theta)i = CiS. Proof complete.

e^(pi)i = cos (pi) +isin(pi)
= -1 + i0
=-1
Thus, e^(pi)i +1 = 0.

[Anonymous No.4702410]

can someone explain to me why e^(i*x)=cos(x)+i sin(x)

[Anonymous No.4702414]

>>4702367
>romney
>got this

I guess if you mean making america a third world country then, YES. Romney got it!

[Anonymous No.4702427]

>>4702410
I bet Euler can. Ask him. it has to do with polar polar form of complex numbers. I dont know exactly how to derive it.

[Anonymous No.4702435]

>>4702410
see>>4702352

[Anonymous No.4702437]

>>4702410
Taylor e^ix boy, then compare it to the sin and cos functions.

[Anonymous No.4702443]

>>4702437
it's called Maclaurin series you stupid fucking engineer

[Anonymous No.4702454]

Does this constitute a proof?

Let <span class="math">y = \cos(x) + i\sin(x)[/spoiler]
<span class="math">y' = -\sin(x) + i\cos(x) = iy[/spoiler]
<span class="math">y = A e^{ix}[/spoiler]
Plugging in <span class="math">x = 0[/spoiler],
<span class="math">1 = A e^0 = A[/spoiler]
so <span class="math">A = 1[/spoiler] and
<span class="math">\cos(x) + i\sin(x) = e^{ix}[/spoiler]

[Anonymous No.4702461]

babby's first "deep" result

[Anonymous No.4702471]

>>4702443

A Maclaurin series IS a Taylor series by definition. It is just a special case of it.

durrrr

[Anonymous No.4702487]

e to the ix = cos x + isin x cause

e^x = 1+x+(x^2/2)+(x^3/6)+(x^n/n)
so e^ix = 1 + ix -(x^2/2) -i(x^3/6) + (x^4/4) etc.
all reals are cos x, all complexs are isin x

Are there other ways to prove this, though?