Revision as of 09:31, 22 July 2008 by Dvtran (Talk)

From the identity $ f(0)-(V_{0}^{x})^{1/2} = f(x) $ $ \forall x\in[0,1] $ we 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})^{2} $

It means that there is a point $ a $ in $ [0,1] $ such that $ V $ jumps from $ 0 $ to $ 1 $ right after the point. (It has to occur like that in order to fulfill the identity.)

So $ f(x)= f(0) $ if $ x\in[0,a] $ and $ f(x)= f(0)-1 $ if $ x\in(a,1] $

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

Dr. Paul Garrett