e^(j theta)
We've now defined for any positive real number
and any
complex number
. Setting
and
gives us the
special case we need for Euler's identity. Since
is its own
derivative, the Taylor series expansion for
is one of
the simplest imaginable infinite series:
![$\displaystyle e^x = \sum_{n=0}^\infty \frac{x^n}{n!}
= 1 + x + \frac{x^2}{2} + \frac{x^3}{3!} + \cdots
$](http://www.dsprelated.com/josimages_new/mdft/img315.png)
![$ f^{(n)}(0)=1$](http://www.dsprelated.com/josimages_new/mdft/img316.png)
![$ n$](http://www.dsprelated.com/josimages_new/mdft/img80.png)
![$ x=0$](http://www.dsprelated.com/josimages_new/mdft/img144.png)
![$\displaystyle e^{j\theta} \isdef \sum_{n=0}^\infty \frac{(j\theta)^n}{n!}
= 1 + j\theta - \frac{\theta^2}{2} - j\frac{\theta^3}{3!} + \cdots
\,.
$](http://www.dsprelated.com/josimages_new/mdft/img317.png)
![\begin{eqnarray*}
\mbox{re}\left\{e^{j\theta}\right\} &=& 1 - \theta^2/2 + \thet...
...heta}\right\} &=& \theta - \theta^3/3! + \theta^5/5! - \cdots\,.
\end{eqnarray*}](http://www.dsprelated.com/josimages_new/mdft/img318.png)
Comparing the Maclaurin expansion for
with that of
and
proves Euler's identity. Recall
from introductory calculus that
![\begin{eqnarray*}
\frac{d}{d\theta}\cos(\theta) &=& -\sin(\theta) \\ [5pt]
\frac{d}{d\theta}\sin(\theta) &=& \cos(\theta)
\end{eqnarray*}](http://www.dsprelated.com/josimages_new/mdft/img321.png)
so that
![\begin{eqnarray*}
\left.\frac{d^n}{d\theta^n}\cos(\theta)\right\vert _{\theta=0}...
...} \\ [5pt]
0, & n\;\mbox{\small even}. \\
\end{array} \right.
\end{eqnarray*}](http://www.dsprelated.com/josimages_new/mdft/img322.png)
Plugging into the general Maclaurin series gives
![\begin{eqnarray*}
\cos(\theta) &=& \sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!}\theta...
...mbox{\tiny$n$\ odd}}}^\infty \frac{(-1)^{(n-1)/2}}{n!} \theta^n.
\end{eqnarray*}](http://www.dsprelated.com/josimages_new/mdft/img323.png)
Separating the Maclaurin expansion for
into its even and odd
terms (real and imaginary parts) gives
![\begin{eqnarray*}
e^{j\theta} \isdef \sum_{n=0}^\infty \frac{(j\theta)^n}{n!}
&...
...-1)^{(n-1)/2}}{n!} \theta^n\\
&=& \cos(\theta) + j\sin(\theta)
\end{eqnarray*}](http://www.dsprelated.com/josimages_new/mdft/img324.png)
thus proving Euler's identity.
Next Section:
Back to Mth Roots
Previous Section:
Back to e