From b7a72c9ad64a8f8f1c9495a66942d05369e6b49d Mon Sep 17 00:00:00 2001 From: bnewbold Date: Wed, 30 Jun 2010 06:23:15 +0000 Subject: tex --- ClassJune26.page | 7 ++++--- 1 file changed, 4 insertions(+), 3 deletions(-) diff --git a/ClassJune26.page b/ClassJune26.page index 3724f4e..18f2292 100644 --- a/ClassJune26.page +++ b/ClassJune26.page @@ -161,11 +161,12 @@ We can raise complex numbers to powers, divide by the real denominators, and add them up just fine, so we can exponentiate complex values of $z$. We know what happens to real values, what happens to pure imaginary ones? Let $y\in\mathbb{R}$. Then -$\begin{array}{}ee^{iy} & = & 1+iy+\frac{(iy)^{2}}{2!}+\frac{(iy)^{3}}{3!}+\frac{(iy)^{4}}{4!}+\frac{(iy)^{5}}{5!}+\cdots\\ + +$$\begin{array}{ccc} +ee^{iy} & = & 1+iy+\frac{(iy)^{2}}{2!}+\frac{(iy)^{3}}{3!}+\frac{(iy)^{4}}{4!}+\frac{(iy)^{5}}{5!}+\cdots\\ & = & 1+iy-\frac{y^{2}}{2!}-i\frac{y^{3}}{3!}+\frac{y^{4}}{4!}+i\frac{y^{5}}{5!}+\cdots\\ & = & (1-\frac{y^{2}}{2!}+\frac{y^{4}}{4!}+\cdots)+i(y-\frac{y^{3}}{3!}+\frac{y^{5}}{5!}-\cdots)\\ - & = & \cos y+i\sin y\end{array}$ - + & = & \cos y+i\sin y\end{array}$$ Substituting $y=\pi$, we recover Euler's famous identity, $e^{i\pi}=-1.$ -- cgit v1.2.3