# 15828 - taking the magnitude and equality holds iF and only...

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Solution to problem 28c in section 15.8 This is assuming you got part a) and b) correct. Now, we need to use theorem C to show that D = f mm f bb - f 2 mb > 0 along with f mm > 0 We fnd that f mm = 2 n i =1 x 2 i (in today’s class, i Forgot the two) and f bb = 2 n (again, i Forgot the two), while f mb = 2 n i =1 x i . Assuming that not all oF the x i are equal to 0, we get that f mm > 0. Also, we get that D = 4( n n s i =1 x 2 i - ( n s i =1 x i ) 2 ) Now, For D to be bigger than 0, we clearly just need n n s i =1 x 2 i - ( n s i =1 x i ) 2 > 0 (1) Now, this is a pretty tricky step. There’s a Famous theorem called the Cauchy-Schwarz Inequality, which states that For all vectors u and v , | u · v | ≤ | u || v | where on the leFt we are taking the absolute value, and on the right we are
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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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