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Unformatted text preview: taking the magnitude, and equality holds iF and only iF u and v are pointing in the same direction. So in the case, let u = ( x 1 , x 2 , . . . , x n ), v = (1 , 1 , . . ., 1). Then we have  u v  =  n s i =1 x i  r R R n s i =1 x i r R R n s i =1 1 = r R R n s i =1 x 2 i n Squaring both sides, we get ( n i =1 x i ) 2 n n i =1 x 2 i . We can assume the x i are not all equal (as the problem would be trivial then), so we do indeed get that (1) holds. Thus, according to theorem C, this is indeed a minimum. 1...
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 Summer '08
 Rothamer
 Heat Transfer

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