mecn 3500 inter - bayamon lecture 8 numerical methods for engineering mecn 3500 professor: dr. omar...
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LectureLecture
88Numerical Methods for EngineeringNumerical Methods for Engineering
MECN 3500 MECN 3500
Professor: Dr. Omar E. Meza CastilloProfessor: Dr. Omar E. Meza [email protected]
http://www.bc.inter.edu/facultad/omeza
Department of Mechanical EngineeringDepartment of Mechanical Engineering
Inter American University of Puerto RicoInter American University of Puerto Rico
Bayamon CampusBayamon Campus
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Tentative Lectures ScheduleTentative Lectures Schedule
TopicTopic LectureLecture
Mathematical Modeling and Engineering Problem SolvingMathematical Modeling and Engineering Problem Solving 11
Introduction to MatlabIntroduction to Matlab 22
Numerical ErrorNumerical Error 33
Root FindingRoot Finding 4-5-64-5-6
System of Linear EquationsSystem of Linear Equations 77
Least Square Curve FittingLeast Square Curve Fitting
Polynomial Interpolation Polynomial Interpolation
Numerical IntegrationNumerical Integration
Ordinary Differential Equations Ordinary Differential Equations
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Elimination MethodElimination Method
LU Decomposition and Matrix LU Decomposition and Matrix InversionInversion
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To solve linear algebraic equations using To solve linear algebraic equations using the technique the technique LU DecompositionLU Decomposition in order in order to provides an efficient means to compute to provides an efficient means to compute the matrix inverse. the matrix inverse.
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Course ObjectivesCourse Objectives
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Provides an efficient way to compute Provides an efficient way to compute matrix inverse by separating the time matrix inverse by separating the time consuming elimination of the consuming elimination of the Matrix [A]Matrix [A] from manipulations of the from manipulations of the right-hand right-hand side {B}side {B}..
Gauss eliminationGauss elimination, in which the forward , in which the forward elimination comprises the bulk of the elimination comprises the bulk of the computational effort, can be computational effort, can be implemented as an LU decompositionimplemented as an LU decomposition
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IntroductionIntroduction
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LU DecompositionLU Decomposition
IfL- lower triangular matrixU- upper triangular matrixThen, [A]{X} = {B} or [A]{X} - {B} = 0
[L]{[U]{X}-{D}} = [A]{X} - {B}
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[L][U]=[L][U]=[A] [A] andand [L]{D}= [L]{D}={B}{B}
A two-step strategyA two-step strategy1.-LU Decomposition step: 1.-LU Decomposition step: [A] [A] can be can be
decomposed into two matrices decomposed into two matrices [L][L] and and [U][U] such that such that [L][U]=[A][L][U]=[A]
2.-Substitution step: 2.-Substitution step: Similar to first phase of Similar to first phase of Gauss eliminationGauss elimination, consider, consider[U]{X}={D}[U]{X}={D}[L]{D}={B}[L]{D}={B}[L]{D}={B}[L]{D}={B} is used to generate an is used to generate an intermediate vector intermediate vector {D}{D} by forward by forward substitutionsubstitutionThen, Then, [U]{X}={D}[U]{X}={D} is used to get is used to get {X}{X} by by back substitution.back substitution.
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Requires the same total FLOPS as for Requires the same total FLOPS as for Gauss elimination.Gauss elimination.
Saves computing time by separating time-Saves computing time by separating time-consuming elimination step from the consuming elimination step from the manipulations of the right hand side.manipulations of the right hand side.
Provides efficient means to compute the Provides efficient means to compute the matrix inversematrix inverse
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>>Enter Matrix A > [3 -0.1 -0.2; 0.1 7 -0.3; 0.3 -0.2 10]A = 3.0000 -0.1000 -0.2000 0.1000 7.0000 -0.3000 0.3000 -0.2000 10.0000>>Enter Solution Vector B > [7.85 -19.3 71.4]B = 7.8500 -19.3000 71.4000S = 3.0000 -2.5000 7.0000Bc = 7.8500 -19.3000 71.4000
>>Enter Matrix A > [3 -0.1 -0.2; 0.1 7 -0.3; 0.3 -0.2 10]A = 3.0000 -0.1000 -0.2000 0.1000 7.0000 -0.3000 0.3000 -0.2000 10.0000>>Enter Solution Vector B > [7.85 -19.3 71.4]B = 7.8500 -19.3000 71.4000S = 3.0000 -2.5000 7.0000Bc = 7.8500 -19.3000 71.4000
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