Revision as of 16:29, 26 September 2008 by Jwise (Talk)

$ x[n]=-0.5cos(3\pin)+sin(3\pin)\! $.


$ x[n]=-\frac{1}{2}[\frac{e^{j3\pin}+e^{-j3\pin}}{2}]+\frac{e^{j3\pin}-e^{-3\pin}}{j2} $


$ x[n]=-\frac{1}{4}e^{j3\pin}-\frac{1}{4}e^{-j3\pin}+\frac{1}{j2}e^{3\pin}-\frac{1}{j2}e^{-j3\pin} $

Alumni Liaison

Abstract algebra continues the conceptual developments of linear algebra, on an even grander scale.

Dr. Paul Garrett