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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 + j \sin x \,\! </math>
  
<math>   e^{i \pi} = \cos \pi + i \sin \pi.\,\! </math>
+
<math>       \cos x = \mathrm{Re}\{e^{jx}\} ={e^{jx} + e^{-jx} \over 2}</math>
 +
 
 +
<math>  \sin x = \mathrm{Im}\{e^{jx}\} ={e^{jx} - e^{-jx} \over 2i}. </math>
 +
 
 +
<math>    \cos(x) = {e^{-jx} + e^{jx} \over 2}</math>
 +
 
 +
<math>    \sin(x) = {e^{-jx} - e^{jx} \over 2j} </math>

Latest revision as of 19:40, 22 July 2009

Euler's identity

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

Euler's formula

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

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

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

$ \cos(x) = {e^{-jx} + e^{jx} \over 2} $

$ \sin(x) = {e^{-jx} - e^{jx} \over 2j} $

Alumni Liaison

Ph.D. 2007, working on developing cool imaging technologies for digital cameras, camera phones, and video surveillance cameras.

Buyue Zhang