/sci/ - Science & Math

[Anonymous No.14824896]

What does it mean?

[Anonymous No.14824906]

2D, or not 2D, that is the question

[Anonymous No.14824909]

It means if you multiply e with itself ipi times you'll get -1.

brainlet

[Anonymous No.14824910]

r(cos(x) + isin(x))=-1
1(cos(pi) + isin(pi))
-1+0=-1

[Anonymous No.14824914]

>>14824896
First part left of the plus sign looks like Eulers' Identity to thepower of i of pi.
Adding 1 to it looks like compensationalism while trying to make it all equal zero.
This probably won't work in commutative mathematics because it defies the Hamilton Principle of ijk in clockwork 12th order degrees of rotationalism.
...or something like that. I really don't know what you call these equations or who discovered them.
It either makes sense to me or it doesn't.

[Anonymous No.14824916]

>>14824910
Nice. Where did e show up in the solution?

[Anonymous No.14824921]

>>14824916
[/math] e^{ix}=\cos x+i\sin x[/math]

[Anonymous No.14824924]

>>14824921
Derp. [math] e^{ix}=\cos x+i\sin x[/math]

[Anonymous No.14824926]

>>14824921
[math] e^{ix}=\cos x+i\sin x[/math]

[Anonymous No.14824928]

>>14824896
https://en.wikipedia.org/wiki/Euler%27s_identity

Deriving Euler's law is not hard, try it, it's fun.

[Anonymous No.14824991]

>>14824916
position around unit circle in complex plane can be described by >>14824926
If you go with argument pi, you just go to the other side of the real axis and land at -1

[Anonymous No.14825340]

>>14824896
the ((())) are trying to bambozzle you with symbols and arcane voodoo
don't fall for it

[math]\
\frac{ \pi ^0\begin{bmatrix}0 & -1 \\1 & 0 \end{bmatrix}^0 }{!0}+
\frac{\pi^1\begin{bmatrix}0 & -1 \\1 & 0 \end{bmatrix}^1 }{!1}+
\frac{ \pi ^2\begin{bmatrix}0 & -1 \\1 & 0 \end{bmatrix}^2 }{!2} + ... = -1 [/math]

[Anonymous No.14825400]

>>14824896
You can either consult the taylor theorem or the unique second derivatives but either way you confront analytic continuation, either in a unitary or functional form.

[Anonymous No.14825452]

>>14824896
counterclockwise rotation by pi in the complex plane moves 1 to -1

[Anonymous No.14825562]

>>14824896
IT'S OVER

[Anonymous No.14825577]

>>14824909
Then it would be e^pi=-1 not e^ipi=-1
tardlet

[Anonymous No.14825813]

>>14825577
read it again

[Anonymous No.14825877]

>>14824896
halfway around the circle takes you to the diametrically opposite point

[Anonymous No.14827146]

>>14824896
This may help you:
https://youtu.be/mvmuCPvRoWQ

[Anonymous No.14827293]

>>14824991
Ah. Thanks. I am terrible at mathematics.