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Exam (elaborations)

Solutions Manual for Numerical and Analytical Methods with MATLAB 1st Edition by Bober

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Solutions Manual for Numerical and Analytical Methods with MATLAB 1st Edition by Bober Solutions Manual for Numerical and Analytical Methods with MATLAB 1st Edition by Bober Solutions Manual for Numerical and Analytical Methods with MATLAB 1st Edition by Bober Solutions Manual for Numerical and Analytical Methods with MATLAB 1st Edition by Bober Solutions Manual for Numerical and Analytical Methods with MATLAB 1st Edition by Bober

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Institution
Numerical And Analytical Meth
Course
Numerical and Analytical Meth











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Institution
Numerical and Analytical Meth
Course
Numerical and Analytical Meth

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Uploaded on
October 18, 2025
Number of pages
404
Written in
2025/2026
Type
Exam (elaborations)
Contains
Questions & answers

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Chapters 2 – 14 Covered
ks ks ks ks




SOLUTIONS

, SOLUTION MANUAL ks




NUMERICALANDANALYTICALMETHODS WITH
ks ks ks ks




MATLAB

Table of Contents
ks ks




Page

Chapter 2
ks 1

Chapter 3
ks 46

Chapter 4
ks 58

Chapter 5
ks 98

Chapter 6
ks 107

Chapter 7
ks 176

Chapter 8
ks 180

Chapter 9
ks 188

Chapter 10
ks 214

Chapter 11
ks 271

Chapter 12
ks 303

Chapter 13
ks 309

Chapter 14
ks 339




@
@SSeeisismm
ici icsisoolalati toi on n

, CHAPTER 2 ks




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




f 1V 4
f (x) = f (0) + f '(0) x+ f' '(0) x2 + f' ''(0) x3 + x +... ks ks ks ks ks ks ks ks


ks ks ks k s ks ks k s ks ks ks ks k s ks ks ks



2! 3! 4! ks ks




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




f ( x) = − sin(x),
ks ks ks ks ks ks k s f '(0) =0,
ks ks ks ks




f ''( x) = −cos(x),
ks ks ks ks ks k s ks k s f ''(0)=−1,
ks ks ks ks ks ks




f '''(x) = +sin(x),
ks ks ks ks ks ks k s ks ks k s f '''(0) =0,
ks ks ks ks ks ks




f1V(x) = + cos(x),
ks ks ks ks ks ks ks k s f1V(0) =1
ks ks ks ks




We can see that ks ks ks




x2 x4 x6 8
cos( x) =1− − +x
k s ks




6! 8! − + − +...
ks ks ks k s k s




4! ks ks ks




+
k s

ks




2!

and that ks




x2
term (k) = −term (k −1) 
ks




2k(2 k −1)
ks ks ks ks ks ks ks ks



ks ks ks ks




The following program evaluates cos(x) by both an arithmetic statement and by t he above
ks ks ks k s ks ks ks ks ks ks ks ks ks ks




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




% cosf.m
ks




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




% series for
ks ks ks -π ≤ x ≤ π ks ks ks ks k s in steps of 0.1 ks ks ks ks π

clear; clc; ks




xi=-pi;dx=0.1*pi; s
k




ks for j=1:21ks




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

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




@
@ 1i ci icsiso olalati toionn
SSeeisismm

, sum=1.0;term=1.0; s
k




ks for k=1:50
ks




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

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

ks sum=sum+term;

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

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




break;

end

end

ks cos_ser(j)=sum;

ks nterms(j)=k;

end

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

fprintf(fo,'x cos(x) cos (x) ks terms in ks k s \n');

ks fprintf(fo,' by arith stm
ks ks by series
ks the series
ks k s \n');

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

for j=1:21
ks




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




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

fprintf(fo,' \n');

end

ks fclose(fo);

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

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




@
@ 2i ci icsiso olalati toionn
SSeeisismm
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