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Unformatted text preview: Math 471, Fall 2006 Homework 7 Solutions Assigned: Friday, October 27, 2005. Due: Friday, November 3, 2005 . 1. (Positive Definite Matrices) . Do page 221, #9. Some hints: Establish the 2 × 2 case. Then reduce the general problem to this special case using the definition of positive definiteness. Proof. Define a vector x whose entries are zero except for a one in the i th position. By the definition of a positive definite matrix, < x T Ax = a ii . The second part is harder. First, we prove the assertion in the 2 × 2 case. Suppose that A = a 11 a 12 a 12 a 22 is positive definite. Since the eigenvalues of a positivedefinite matrix are positive, the determinant of A is positive, and < det A = a 11 a 22 a 2 12 . Rearrange this equation to establish the 2 × 2 case. Next, observe that we can reduce the general problem to this special case. Suppose that A is an n × n positivedefinite matrix. Without loss of generality, let i = 1 and j = 2. By considering vectors x of the form x = [ x 1 x 2 ... 0] T , we see that x 1 x 2 T a 11 a 12 a 12 a 22 x 1 x 2 > . So this 2 × 2 submatrix is positive definite. Applying the result for this special case gives a 2 12 < a 11 a 22 . Since we can make the same argument for any i 6 = j , the result follows. 2. (Poisson BVP) Do page 741, #7. Mesh size h RMS Error RMS Error / h 2 1/4 0.00136076 0.02177211 1/6 0.00056606 0.02037815 1/8 0.00030714 0.01965715 1/10 0.00019220 0.01921973 1/12 0.00013144 0.01892701 1/14 0.00009550 0.01871766 1/16 0.00007250 0.01856061 RMS error is the ‘ 2 error over all the points in the mesh divided by the squareroot of the number of points. The RMS error divided by h 2 tends to a constant ≈ . 019, so we have secondorder convergence. Is this also true for the ‘ ∞ error?...
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This note was uploaded on 04/02/2008 for the course MATH 471 taught by Professor Linzhi during the Winter '08 term at University of Michigan.
 Winter '08
 LinZhi
 Numerical Analysis, Matrices

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