Explain to a brainlet like me why e^iπ=-1
Explain to a brainlet like me why e^iπ=-1
Matthew Clark
Brody Mitchell
Juan Hernandez
Anthony Cruz
e^(ix)= cos(x) + isin(x)
you can prove it with series expansion
Cameron Rivera
exp(ti) is a rotation around plane by a unit arc of t.
Juan Baker
>a unit arc of t
...except you don't mean "unit".
Ryder Barnes
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.
Brody Hughes
[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]
James Evans
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?
Hunter Butler
because we want to draw triangles on a complex plane