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Exam (elaborations)

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

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This solutions manual offers comprehensive, step-by-step answers to the exercises found in the textbook Numerical and Analytical Methods with MATLAB. It’s designed to support students and instructors by providing clear explanations and worked-out solutions for problems involving: Linear algebra Calculus Differential equations Fourier analysis Numerical integration Boundary value problems The manual emphasizes MATLAB-based problem solving, making it ideal for engineering and applied mathematics courses. It includes code snippets, Taylor series expansions, and examples that demonstrate how to use MATLAB effectively for analytical and numerical tasks

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Institution
Manual For Numerical And Analytical
Course
Manual for Numerical and Analytical











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Institution
Manual for Numerical and Analytical
Course
Manual for Numerical and Analytical

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Uploaded on
October 3, 2025
Number of pages
481
Written in
2025/2026
Type
Exam (elaborations)
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Questions & answers

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Chapters 2 – 14 Covered




SOLUTIONS

, SOLUTION MANUAL

NUMERICAL AND ANALYTICAL METHODS WITH

MATLAB

Table of Contents

Page

Chapter 2 1

Chapter 3 46

Chapter 4 58

Chapter 5 98

Chapter 6 107

Chapter 7 176

Chapter 8 180

Chapter 9 188

Chapter 10 214

Chapter 11 271

Chapter 12 303

Chapter 13 309

Chapter 14 339




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, CHAPTER 2

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

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

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

f ( x) = − sin( x), f ' (0) = 0,

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

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

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

We can see that

x2 x4 x6 8
cos( x) =1 − + − +x
2! 4! 6! 8! − + − +...

and that

x2
term (k) = − term (k − 1) 
2 k (2 k −1)

The following program evaluates cos( x) by both an arithmetic statement and by the

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

% cosf.m

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

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

clear; clc;

xi=-pi; dx=0.1*pi;

for j=1:21

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

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


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, sum=1.0; term=1.0;

for k=1:50

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

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

sum=sum+term;

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

if abs(term) <= test;

break;

end

end

cos_ser(j)=sum;

nterms(j)=k;

end

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

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

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

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

for j=1:21

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

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

fprintf(fo,' \n');

end

fclose(fo);

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

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




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