A basic identity for quadratics shows y = A 1b as minimizing: P(y) = 1 2 y TAyy T b = 1
Chapter 6, Problem 6.5.6(choose chapter or problem)
A basic identity for quadratics shows y = A 1b as minimizing: P(y) = 1 2 y TAyy T b = 1 2 (yA 1 b) TA(yA 1 b) 1 2 b TA 1 b. The minimum over a subspace of trial functions is at the y nearest to A 1b. (That makes the first term on the right as small as possible; it is the key to convergence of U to u.) If A = I and b = (1,0,0), which multiple of V = (1,1,1) gives the smallest value of P(y) = 1 2 y T yy1?
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