(New page: 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>) |
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| − | Notice that <math>V</math> is a positive and increasing function, therefore, <math>f</math> is decreasing. Hence <math>f(x)-f(0)=- | + | 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>. |
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| + | We then have <math>V_{0}^{x}=(V_{0}^{x})^{1/2}</math> | ||
Revision as of 10:24, 22 July 2008
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} $
