quiz7 and ans

quiz7 and ans - + x 2 y + xy-( y-a 2 ) . Then the Jacobian...

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AMSC/CMSC 660 Quiz 7 , Fall 2006 1. (10) Write Matlab code to apply 5 steps of Newton’s method to the problem x 2 y 3 + xy = 2 , 2 xy 2 + x 2 y + xy = 0 , starting at the point x = 5, y = 4. Answer: F ( x ) = x 2 y 3 + xy - 2 2 xy 2 + x 2 y + xy and J ( x ) = 2 xy 3 + y 3 x 2 y 2 + x 2 y 2 + 2 xy + y 4 xy + x 2 + x . x = [5;4]; for i=1:5, F = [x(1)^2*x(2)^3 + x(1)*x(2) - 2 2*x(1)*x(2)^2 + x(1)^2*x(2) + x(1)*x(2)]; J = [2*x(1)*x(2)^3 + x(2), 3*x(1)^2*x(2)^2 + x(1) 2*x(2)^2 + 2*x(1)*x(2) + x(2), 4*x(1)*x(2) + x(1)^2 + x(1)]; x = x - J\F; end 1
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2. Consider using a homotopy method to solve the problem F ( x ) = x 2 y 3 + xy - 2 2 xy 2 + x 2 y + xy = 0 . Our homotopy function is ρ a ( λ, x ) = λ F ( x ) + (1 - λ )( x - a ) , where x = [ x, y ] T . (a) (4) Compute the Jacobian matrix for ρ a ( λ, x ). (b) (6) What needs to hold in order that the function ρ a is transversal to zero on its domain? Why is this likely to be true? Answer: (a) We compute the partial of ρ a ( λ, x ) with respect to λ : s = x 2 y 3 + xy - 2 - ( x - a 1 ) 2 xy 2
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Unformatted text preview: + x 2 y + xy-( y-a 2 ) . Then the Jacobian of a is the 2 3 matrix J ( , x ) = s , (1- ) I + J ( x ) / where J ( x ) is the matrix from Problem 1. (b) In order for the function to be transversal to zero, the matrix J ( , x ) must be full rank (i.e., rank-2) at every point [0 , 1), x, y (- , ). The matrix J ( x ) has two eigenvalues call them 1 and 2 . The matrix K = (1- ) I + J ( x ) has eigenvalues (1- ) + i , so it is singular only if = 1 / (1- 1 ) or = 1 / (1- 2 ). Even if that happens, it is likely that the vector s will point in a diFerent direction, making the rank of J ( , x ) equal to 2. 2...
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quiz7 and ans - + x 2 y + xy-( y-a 2 ) . Then the Jacobian...

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