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| − | From the identity <math>f(0)-(V_{0}^{x})^{1/2} = f(x) \forall x\in[0,1]</math> we notice that <math>V</math> is a positive and increasing function, therefore, <math>f</math> is decreasing. Hence <math>f(x)-f(0)=-V_{0}^{x})</math>. | + | From the identity <math>f(0)-(V_{0}^{x})^{1/2} = f(x)</math> <math>\forall x\in[0,1]</math> we notice that <math>V</math> is a positive and increasing function, therefore, <math>f</math> is decreasing. Hence <math>f(x)-f(0)=-V_{0}^{x})</math>. |
We then have <math>V_{0}^{x}=(V_{0}^{x})^{2}</math> | We then have <math>V_{0}^{x}=(V_{0}^{x})^{2}</math> | ||
It means that there is a point <math>a</math> in <math>[0,1]</math> such that <math>V</math> jumps from <math>0</math> to <math>1</math> right after the point. (It has to occur like that in order to fulfill the identity.) | It means that there is a point <math>a</math> in <math>[0,1]</math> such that <math>V</math> jumps from <math>0</math> to <math>1</math> right after the point. (It has to occur like that in order to fulfill the identity.) | ||
Revision as of 10:29, 22 July 2008
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.)
