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| − | Since all the <math>f_{n}</math> are AC, there exists <math>f_{n}^{'}</math> | + | Since all the <math>f_{n}</math> are AC, there exists <math>f_{n}^{'}</math> such that <math>f_{n}(x)=f_{n}(x)-f_{n}(0)=\int{0}{x}S</math> |
Revision as of 10:10, 10 July 2008
Since all the $ f_{n} $ are AC, there exists $ f_{n}^{'} $ such that $ f_{n}(x)=f_{n}(x)-f_{n}(0)=\int{0}{x}S $
