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Midterm2practice_sol

# Midterm2practice_sol - Math 415 Midterm 2 practice...

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Math 415 — Midterm 2 practice solutions Problem 1: The given definition does not satisfy the positivity condition that h f, f i > 0 if f 6 = 0. In fact consider the continuous function defined as f ( x ) = ( x for - 1 < x < 0 0 for 0 < x < 1 Calculating h f, f i for this function we get h f, f i = Z 1 - 1 f 2 ( x ) x d x = Z 0 - 1 x 2 x d x = - 1 / 4 < 0 Problem 2: First notice that the limits of integration in the definition of the inner product in this problem are different from the ones in the previous problem. This one is a valid inner product. Using this inner product we start by calculating the norm of w 1 , as p h w 1 , w 1 i = 1 / 2. Therefore u 1 = w 1 || w 1 || = 2 Then we define v 2 = w 2 - h w 2 , u 1 i u 1 and divide it by its norm to get u 2 = v 2 || v 2 || = 6( x - 2 3 ) Finally v 3 = w 3 - h w 3 , u 1 i u 1 - h w 3 , u 2 i u 2 and u 3 = v 3 || v 3 || = 10 6 x 2 - 6 5 x + 3 10 Problem 3: a) The matrix is not symmetric, so it can’t be positive definite (We consider positive definite matrices to be symmetric).

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