Revision as of 09:24, 22 July 2008 by Dvtran (Talk)

Notice that $ V $ is a positive and increasing function, therefore, $ f $ is decreasing. Hence $ f(x)-f(0)=-V_{0}^{x}) $.

We then have $ V_{0}^{x}=(V_{0}^{x})^{1/2} $

Alumni Liaison

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

Dr. Paul Garrett