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****INSTANT DOWNLOAD****PDF****Solutions Manual for Numerical and Analytical Methods with MATLAB 1st Edition by Bober

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****INSTANT DOWNLOAD****PDF****Solutions Manual for Numerical and Analytical Methods with MATLAB 1st Edition by BoberSolutions Manual for Numerical and Analytical Methods with MATLAB 1st Edition by BoberSolutions Manual for Numerical and Analytical Methods with MATLAB 1st Edition by BoberSolutions Manual for Numerical and Analytical Methods with MATLAB 1st Edition by BoberSolutions Manual for Numerical and Analytical Methods with MATLAB 1st Edition by BoberSolutions Manual for Numerical and Analytical Methods with MATLAB 1st Edition by BoberSolutions Manual for Numerical and Analytical Methods with MATLAB 1st Edition by Bober

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Chapters 2 – 14 Covered
c c c c




SOLUTIONS

, SOLUTIONMANUAL c




NUMERICALANDANALYTICALMETHODSWITH
c c c c




MATLAB

Table of Contents
c c




Page

Chapter 2 c 1

Chapter 3 c 46

Chapter 4 c 58

Chapter 5 c 98

Chapter 6 c 107

Chapter 7 c 176

Chapter 8 c 180

Chapter 9 c 188

Chapter 10 c 214

Chapter 11 c 271

Chapter 12 c 303

Chapter 13 c 309

Chapter 14 c 339




@@SSee
isis
mmiciicsis
oolala
titoionn

, CHAPTER 2 c




P2.1. Taylor series expansion of f (x) about x = 0 is:
c c c c c c c c c c c c




f''(0) 2 f'''(0) 3 f 1V 4
f (x) = f (0)+ f '(0) x+ x + x + x +...
c c c c c c c c


c c c c c c c c c c c c c



4!
c c



2! 3! c c




For f (x)=cos(x), c c c c c c c f (0) =1,
c c c




f (x) =−sin(x), f '(0)=0,
c c c c c c c c c c c




f ''(x) = −cos(x), f ''(0)=−1,
c c c c c c c c c c c c c c




f '''(x) = +sin(x), f '''(0)=0,
c c c c c c c c c c c c c c c c




f1V(x)=+cos(x), f 1V(0)=1
c c c c c c c c c c c c




We can see that c c c




x x 4 x6 8 2
cos(x) =1− + − + x −+−+...
c c c


c c c c c




2! 4! 6! 8! c c c



c




and that c




x2
term(k)=−term(k −1)
c




2k (2k −1)
c c c c c c c c



c c c c




The following program evaluates cos(x)by both an arithmetic statement and bythe above
c c c c c c c c c c c c c c




c series for -π ≤ x ≤ π in step of 0.1  .
c c c c c c c c c c c c




% cosf.m
c




% This program evaluates cos(x) by both arithmetic statement and by
c c c c c c c c c c




% series for -π ≤ x ≤ π in steps of 0.1 π
c c c c c c c c c c c c




clear; clc; c




xi=-pi;dx=0.1*pi; c




c for j=1:21
c




x(j)=xi+(j-1)*dx;

cos_arith(j)= cos(x(j)); c




1iciicsisoolalatitoionn
@@SSeeisismm

, sum=1.0;term=1.0; c




for k=1:50
c c




den=2*k*(2*k-1);

term=-term*x(j)^2/den;

sum=sum+term;
c




test=abs(sum*1.0e-6);
c




if abs(term) <= test;
c c c c




break;

end

end

cos_ser(j)=sum;
c




nterms(j)=k;
c




end

fo=fopen('output.dat','w');
c




fprintf(fo,'x cos(x) cos (x) c terms inc c \n');

fprintf(fo,'
c by arith stm
c c by series
c the series
c c \n');

fprintf(fo,'=====================================================\n');
c




for j=1:21
c




fprintf(fo,'%10.5f %10.5f %10.5f %3i\n',...
c




x(j),cos_arith(j),cos_ser(j),nterms(j));
c




fprintf(fo,' \n');

end

fclose(fo);
c




plot(x,cos_arith),xlabel('x'),ylabel('cos(x)'),

title('cos(x) vs. x'),grid;
c c c




2iciicsisoolalatitoionn
@@SSeeisismm

Connected book
 image
William Bober, Chi-Tay Tsai, Oren Masory Numerical and Analytical Methods with MATLAB
Edition: 2009 ISBN: 9781420093568 Edition: Unknown

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