Explain to a brainlet like me why e^iπ=-1

Explain to a brainlet like me why e^iπ=-1

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e^(ix)= cos(x) + isin(x)

you can prove it with series expansion

exp(ti) is a rotation around plane by a unit arc of t.

>a unit arc of t
...except you don't mean "unit".

Any function defined on the real line has a unique extension to an analytic function on the complex plane, if one exists. On the real line, [math]e^x, \sin(x), \cos(x)[/math] can all defined by power series, which make perfect sense if you plug in complex numbers, giving the analytic continuations of each. When you write the series for [math]e^{ix}[/math] down, [math]i\sin(x)[/math] has all the odd terms, and [math]\cos(x)[/math] has all the even terms, so [math]e^{ix} = \cos(x) + i\sin(x)[/math]. Plugging in [math]x = \pi[/math] gives the identity.

[math] \displaystyle
e=\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\cdots
\\ \\
e^x=\frac{x^0}{0!}+\frac{x^1}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\cdots
\\ \\
sin(x)=\frac{x^1}{1!}-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\frac{x^9}{9!}-\cdots
\\ \\
cos(x)=\frac{x^0}{0!}-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\frac{x^8}{8!}-\cdots
\\ \\
cos(x)+sin(x)=1+x-\frac{x^2}{2!}-\frac{x^3}{3!}+\frac{x^4}{4!}+\frac{x^5}{5!}-\frac{x^6}{6!}-\frac{x^7}{7!}+\frac{x^8}{8!}+\frac{x^9}{9!}-\cdots
\\ \\
e^{ix}=\frac{(ix)^0}{0!}+\frac{(ix)^1}{1!}+\frac{(ix)^2}{2!}+\frac{(ix)^3}{3!}+\frac{(ix)^4}{4!}+\cdots
\\ \\
e^{ix}=1+ix-\frac{x^2}{2!}-\frac{ix^3}{3!}+\frac{x^4}{4!}+\frac{ix^5}{5!}-\frac{x^6}{6!}-\frac{ix^7}{7!}+\frac{x^8}{8!}+\frac{ix^9}{9!}-\cdots
\\ \\
e^{ix}=\left ( 1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\frac{x^8}{8!}-\cdots \right )
+i \left ( x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\frac{x^9}{9!}-\cdots \right )
\\ \\
e^{ix}=cos(x)+i \, sin(x)
[/math]

So why is the "i" necessary? When we expand e^x into cos and sin, why isn't that enough to define e^pi as -1?
Obviously, if you put e^pi into a calculator, it isn't negative one but cos(pi) + sin(pi) = -1 so what makes the "i" important?

because we want to draw triangles on a complex plane

>When we expand e^x into cos and sin
That never happened. e^x expands into cosh and sinh.

posting the superior identity
[math]e^{2\pi i}-1=0[/math]
it's got e, pi, i, 0, 1 AND 2

think that is the definition of i
duh

Pythagorean theorem

Here
youtube.com/watch?v=-dhHrg-KbJ0

Not any function on the real line. It needs to be analytic.
en.wikipedia.org/wiki/Non-analytic_smooth_function

The series expansion of e^(i*pi) is equal to the sum of seres expansions of cos x and i*sin x

what is cosh(ix) and sinh(ix)

cosh(x) = (e^x + e^-x)/2
sinh(x) = (e^x - e^-x)/2

e^ix = cosh(ix)+sinh(ix) = e^ix with some arithmetic but this is trivial unless you know how we get these equations for cosh and sinh

cosh and sinh are functions of the inversive distance between two circles, which is d=ln(a/b) for circles inversed into concentric circles with radii b and a. we know that for a figure where n circles can be drawn tangent to an exterior and interior circle which are nonconcentric, these n circles invert into n congruent circles of diameter b-a. now sin(pi/n) = ((a-b)/2)/((a+b)/2) = (a-b)/(a+b) = (a/b - 1)/(a/b + 1) = (e^d - 1)/(e^d+1) = (e^.5d - e^-.5d)/(e^.5d+e^-.5d) = tanh(d/2).

So we can also obtain tan(pi/n) = sinh(d/2) and sec(pi/n) = cosh(d/2) using similar methods.

Because Oily Faggots Got Too Rough.

When will tau (2pi) replace pi
Fuck pi. Circles should be defined by radius not diameter. Everything clicks into place when you use tau instead of pi

>Everything clicks into place
you're a dork (2idiot)

wow, it really works!
thanks dork

Pi is wrong. youtu.be/jG7vhMMXagQ

>cosh and sinh are functions of the inversive distance between two circles
I prefer to think of it this way.

because sin^e = -1

e^(i(tau)) = 1

pi is gay

>hihi le nerdy girl xD
cant stand her vocal fry

I like her videos