lecture05

lecture05 - INTRODUCTION TO NUMERICAL SIMULATION LECTURE 5....

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I NTRODUCTION TO N UMERICAL S IMULATION L ECTURE 5. Linear Systems – Conditioning 1 T ODAY S O UTLINE : Solution of Dense Linear Systems Hard to Solve Problems Perturbation Analysis and Conditioning Solution of Sparse Linear Systems LU Factorization Reminder. Example of Problems with Sparse Matrices Struts and joints, resistor grids, 3-D heat flow Tridiagonal Matrix Factorization General Sparse Factorization Fill-in and Reordering Graph Based Approach Sparse Matrix Data Structures Scattering S OLUTION OF D ENSE L INEAR S YSTEMS Hard to Solve Problems ± Fitting example Polynomial Interpolation Table of Data () N N t f t t f t t f t M M 1 1 0 0 Problem: fit data with an N th order polynomial N N t t t t t f α + + α + α + α + α = ... 3 3 2 2 1 0 Matrix Form ( ) = α α α N N M N N N N N N t f t f t f t t t t t t t t t 1 0 1 0 2 1 2 1 1 0 2 0 0 interp 1 1 1 4 4 4 43 4 4 4 42 1 L L L f t t 0 t 1 t 2 t N 2 2 || || M M r r
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I NTRODUCTION TO N UMERICAL S IMULATION L ECTURE 5. Linear Systems – Conditioning 2 ( ) () N N N N N N N N N N t f t t t t f t t t t f t t t = α + + α + α + α = α + + α + α + α = α + + α + α + α L M L L 2 2 1 0 1 1 2 1 2 1 1 0 0 0 2 0 2 0 1 0 t f ( t )= t t 0 = 3.5 f ( t 0 ) = 3.5 t 2 = 5.2 f ( t 2 ) = 5.2 t 1 = 7.8 f ( t 1 ) = 7.8 ( ) t t f t t = = + α + α + α L 2 2 1 0 0 0 1 0 2 1 0 = α = α = α = α N M α 0 α 1 α 2 α 3 α 4 α 5
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I NTRODUCTION TO N UMERICAL S IMULATION L ECTURE 5. Linear Systems – Conditioning 3 Coefficient number Fitting f(t) = t Coefficient Value Since we are looking at the coefficients, α values, for the linear plot f ( t ) = t , we expect that α 1 = 1 and all other values of α will be zero – this is accurately represented in this graph for 10 α values. Coefficient number Fitting f(t) = t Coefficient Value For 20 α values one begins to see some peculiar nonzero values for α 11, they are very close to zero.
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I NTRODUCTION TO N UMERICAL S IMULATION L ECTURE 5. Linear Systems – Conditioning 4 Coefficient number Fitting f(t) = t Coefficient Value For 100 α values it only gets worse. Coefficient number Fitting f(t) = t Coefficient Value
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I NTRODUCTION TO N UMERICAL S IMULATION L ECTURE 5. Linear Systems – Conditioning 5 Perturbation Analysis and Conditioning ± Geometric Approach is clearer f t For 100 α values, even though there are many higher order coefficient values that are non-zero, the graph still produces a linear result. 2 2 || || M M r r 1 1 || || M M r r 2 2 || || M M r r 1 1 || || M M r r Columns orthogonal Columns nearly aligned 1 x 2 x 1 x 2 x [ ] 2 1 , M M M r r = b M x M x r r r = + 2 2 1 1 Solving, M x = b is finding When vectors are nearly aligned, difficult to determine how much of versus how much of 1 M r 2 M r orthogonal columns = 6 10 0 0 1 M = 1 10 1 10 1 1 6 6 M columns nearly aligned
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I NTRODUCTION TO N UMERICAL S IMULATION L ECTURE 5.
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This note was uploaded on 09/26/2010 for the course AERO 16.910 taught by Professor Daniel during the Spring '10 term at MIT.

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lecture05 - INTRODUCTION TO NUMERICAL SIMULATION LECTURE 5....

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