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[[Image:7b1_Old Kiwi.jpg]]
 
[[Image:7b1_Old Kiwi.jpg]]
 +
[[Image:7b2_Old Kiwi.jpg]]
  
 
Therefore, <math> m_k = \left ( \frac {1}{k\pi} \sin ( \frac {k\pi}{2} ) \right) , n_k = \left( \frac {-1}{k\pi} \sin ( \frac {k\pi}{2} ) e^\frac{-j2k\pi2}{4} \right)</math>
 
Therefore, <math> m_k = \left ( \frac {1}{k\pi} \sin ( \frac {k\pi}{2} ) \right) , n_k = \left( \frac {-1}{k\pi} \sin ( \frac {k\pi}{2} ) e^\frac{-j2k\pi2}{4} \right)</math>

Revision as of 12:16, 1 July 2008

7b Old Kiwi.jpg

Let $ g(t) = \left ( \frac{dz}{dt} \right ) $

7b1 Old Kiwi.jpg 7b2 Old Kiwi.jpg

Therefore, $ m_k = \left ( \frac {1}{k\pi} \sin ( \frac {k\pi}{2} ) \right) , n_k = \left( \frac {-1}{k\pi} \sin ( \frac {k\pi}{2} ) e^\frac{-j2k\pi2}{4} \right) $

Alumni Liaison

Basic linear algebra uncovers and clarifies very important geometry and algebra.

Dr. Paul Garrett