Line 1: Line 1:
 
Euler's identity
 
Euler's identity
  
<math>  e^{i \pi} + 1 = 0, \,\! </math>
+
<math>  e^{j \pi} + 1 = 0, \,\! </math>
  
 
Euler's formula
 
Euler's formula
  
<math>    e^{ix} = \cos x + i \sin x \,\! </math>
+
<math>    e^{jx} = \cos x + i \sin x \,\! </math>
  
<math>   e^{i \pi} = \cos \pi + i \sin \pi.\,\! </math>
+
<math>       \cos x = \mathrm{Re}\{e^{ix}\} ={e^{ix} + e^{-ix} \over 2}</math>
 +
 
 +
<math>  \sin x = \mathrm{Im}\{e^{ix}\} ={e^{ix} - e^{-ix} \over 2i}. </math>
 +
 
 +
<math>    \cos(y) = {e^{-jy} + e^{jy} \over 2}</math>

Revision as of 20:38, 22 July 2009

Euler's identity

$ e^{j \pi} + 1 = 0, \,\! $

Euler's formula

$ e^{jx} = \cos x + i \sin x \,\! $

$ \cos x = \mathrm{Re}\{e^{ix}\} ={e^{ix} + e^{-ix} \over 2} $

$ \sin x = \mathrm{Im}\{e^{ix}\} ={e^{ix} - e^{-ix} \over 2i}. $

$ \cos(y) = {e^{-jy} + e^{jy} \over 2} $

Alumni Liaison

Followed her dream after having raised her family.

Ruth Enoch, PhD Mathematics