a3_sol.pdf

# 2 10 marks 5 assume you are given the linear system a

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2. ( 10 marks ) 5

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Assume you are given the linear system A x = b , with A = 1 - 2 2 4 and || b || 0 . 8. If ˆx = 0 . 5 - 0 . 1 is an approximate solution for x , then find an upper bound, U , for the error vector, e = x - ˆx , such that || e || U . We have A - 1 = 1 8 4 2 - 2 1 = 0 . 5 0 . 25 - 0 . 25 0 . 125 || A - 1 || = 0 . 75 Since e = A - 1 b - ˆx || e || = || A - 1 b - ˆx || ≤ || A - 1 b || + || ˆx || ≤ || A - 1 || || b || + || ˆx || 0 . 75 · 0 . 8 + 0 . 5 = 1 . 1 Also acceptable: Since A e = r we have e = A - 1 r = A - 1 ( b - A ˆx ), hence || e || = || A - 1 ( b - A ˆx ) || ≤ || A - 1 || || b - A ˆx || ≤ || A - 1 || ( || b || + || A ˆx || ) 0 . 75(0 . 8 + 0 . 7) = 1 . 125 since || A ˆx || = 0 . 7 0 . 6 = 0 . 7 3. ( 10 marks ) Let y be an approximate solution to A x = b , with residual r = b - A y , and define E = ry T || y || 2 2 . (a) Then prove y is the exact solution to ( A + E ) y = b . (b) Using definition of the matrix norm and | y T z | ≤ || y || 2 || z || 2 (Cauchy-Schwartz inequal- ity), show that || E || 2 || r || 2 || y || 2 6
(a) ( A + E ) y = A y + ( b - A y ) y T y || y || 2 2 = A y + ( b - A y ) || y || 2 2 || y || 2 2 = A y + ( b - A y ) = b (b) For any z R n , || E z || 2 || z || 2 = 1 || y || 2 2 || ry T z || 2 || z || 2 = | y T z | || y || 2 2 || r || 2 || z || 2 || y || 2 || z || 2 || y || 2 2 || r || 2 || z || 2 = || r || 2 || y || 2 || E z || 2 || z || 2 || r || 2 || y || 2 z = || E || 2 || r || 2 || y || 2 4. ( 10 marks ) Let λ 1 , ..., λ n be the eigenvalues of a matrix C R n × n . The spectral radius of C is defined as ρ ( C ) = max {| λ 1 | , ..., | λ n |} . Consider solving the linear system A x = b (2) where A R n × n is non-singular and x , b R n , using the iterative method x ( k ) = x ( k - 1) + B - 1 ( b - A x ( k - 1) ) (3) for some non-singular matrix B R n × n . Let C = I - B - 1 A . If ρ ( C ) < 1, show that ( 3 ) converges to the solution of ( 2 ) for any starting vector x (0) R n . What will the speed of convergence depend on? Hint: Show that the error vector, e ( k ) = x - x ( k ) 0 as k → ∞ by using simplifying assumption that C has n linearly independent eigenvectors. If x is the exact solution to ( 2 ) then b - A x = 0 and we have x = x + B - 1 ( b - A x ) (4) 7

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( 3 ) - ( 4 ) gives e ( k ) = e ( k - 1) - B - 1 A e ( k - 1) = ( I - B - 1 A ) e ( k - 1) = C e ( k - 1) = C k e (0) Let u l be an eigenvector of C with eigenvalue λ l . Assume that C has n linearly independent eigenvectors. Then we can find constants c l , such that e (0) = X l c l u l . Since C u l = λ l u = C k u l = λ k l u l , thus e ( k ) = C k e (0) = n X l =1 c l C k u l = n X l =1 c l λ k l u l From assumption, | λ l | < 1 , l = 1 , ..., n , so that λ k l 0 as k → ∞ . Hence, lim k →∞ e ( k ) = 0 for any e (0) = for any x (0) . The speed of convergence will depend on the largest magnitude eigenvalue, i.e. ρ ( C ), with smaller ρ ( C ) = faster convergence. 5. ( 10 marks ) Find (if possible) a, b, c, d so that S ( x ) = 1 + x x [ - 2 , - 1] a + bx + cx 2 + dx 3 x [ - 1 , 0] b + ax x [0 , 1] is a natural cubic spline? Let S 1 ( x ) = 1 + x = S 0 1 ( x ) = 1 S 00 1 ( x ) = 0 S 2 ( x ) = a + bx + cx 2 + dx 3 = S 0 2 ( x ) = b + 2 cx + 3 dx 2 S 00 2 ( x ) = 2 c + 6 dx S 3 ( x ) = b + ax = S 0 3 ( x ) = a S 3 ( x ) = 0 Since S 00 1 ( - 2) = 0 and S 00 3 (1) = 0, both endpoints are natural. 2nd derivative smoothness conditions: S 00 2 (0) = S 00 3 (0) = 2 c = 0 = c = 0 S 00 1 ( - 1) = S 00 2 ( - 1) = 0 = 2 c - 6 d = d = 0 8
1st derivative smoothness conditions: S 0 1 ( - 1) = S 0 2 ( - 1) = 1 = b - 2 c + 3 d = b = 1 S 0 2 (0) = S 0 3 (0) = a = b = a = 1 Check continuity conditions: S 2 (0) = S 3 (0) = a = b which is true S 1 ( - 1) = S 2 ( - 1) = 0 = a - b + c - d = c = d which is true So a = 1 , b = 1 , c = 0 , d = 0 satisfy all conditions for a natural cubic spline.

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