Sunday 13 October 2013

calculus - Simplifying expression using Euler's formula



I'm trying to simplify this expression.



$$
a + be^{-iw} + be^{-2iw} + ae^{-3iw} =
$$



$$
e^{-iw3/2} \cdot [2 a \cdot \cos(3w/2) + 2b\cdot \cos(w/2) ]

$$



How do I go from the left hand side of the equation to the right hand side? I realize that I'm supposed to use Euler's formula, what I don't really get is how.


Answer



The exponents go from $-0iw$ to $-3iw$, so take the middle point $-3iw/2$, getting
$$
e^{-3iw/2}\cdot[ae^{+3iw/2}+be^{iw/2}+be^{-iw/2}+ae^{-3iw/2}]=\\
e^{-3iw/2}\cdot\left[2a\frac{e^{+3iw/2}+e^{-3iw/2}}{2}+2b\frac{e^{iw/2}+e^{-iw/2}}{2}\right]
$$


No comments:

Post a Comment

real analysis - How to find $lim_{hrightarrow 0}frac{sin(ha)}{h}$

How to find $\lim_{h\rightarrow 0}\frac{\sin(ha)}{h}$ without lhopital rule? I know when I use lhopital I easy get $$ \lim_{h\rightarrow 0}...