solsmidterm - ECE 580 Math 587 Correspondence 12 FALL 2006...

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ECE 580 / Math 587 FALL 2006 Correspondence # 12 March 26, 2011 SOLUTION TO THE MIDTERM EXAM Problem 1 (i) Let x ( t ) = y ( t ) 1 t . The given constraint can then be written in terms of x as follows: 1 0 t ( 1 + t + x ( t ) ) dt = 1 1 0 tx ( t ) dt = 1 6 Then, the problem is equivalently minimization of the L 2 [0 , 1] norm of x subject to the equality constraint ( t, x ) = 1 / 6. By Theorems 1 and 2 on pp. 63-64 of the text, there exists a unique minimizing solution, x o , and it is in the span of t : x o ( t ) = αt for some α . Then, y o ( t ) = ( α + 1) t + 1 is the unique solution to the original problem. (ii) ( y o , t ) = 1 α = 1 / 2. Hence, y o ( t ) = 3 2 t + 1 , t [0 , 1] and F ( y o ) = 1 12 = 0 . 08333 (iii) Since t and t 2 are linearly independent in L 2 [0 , 1], again by the same theorems there exists a unique solution to the minimization problem, with the solution (in terms of x ) in the form: x o ( t ) = λ 1 t + λ 2 t 2 , for some λ 1 and λ 2 . Hence, y o ( t ) = 1 + (1 + λ 1 ) t + λ 2 t 2 . The coefficients can be computed uniquely from the equality constraints: ( t, y o ) = 1 and ( t 2 , y o ) = 1 1 2 + 1 3 (1 + λ 1 ) + 1 4 λ 2 = 1 and 1 3 + 1 4 (1 + λ 1 ) + 1 5 λ 2 = 1 λ 1 = 17 , λ 2 = 70 3 = 23 . 3333 . Hence,
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This note was uploaded on 10/11/2011 for the course ECE 580 taught by Professor Staff during the Spring '08 term at University of Illinois, Urbana Champaign.

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solsmidterm - ECE 580 Math 587 Correspondence 12 FALL 2006...

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