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Solutions Manual for For Numerical and Analytical Methods with Matlab (1st Edition) by Bober - Graded A+ Compreh...

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### Complete Solutions Manual: For Numerical and Analytical Methods with Matlab (1st Edition) by Bober | **Format:** Instant PDF Download | **Pages:** 393 Pages Master complex textbook exercises and exam problems with the complete, official **Solutions Manual** for **For Numerical and Analytical Methods with Matlab** (1st Edition) by Bober. #### What is Included: - **100% Complete Worked Solutions:** Step-by-step mathematical derivations, conceptual reasoning, and formulas for all textbook exercises. - **All Chapter Coverage:** Detailed answers for all end-of-chapter problems, questions, and review sets. - **Homework & Exam Advantage:** Check your work, practice challenging problem sets, and prepare thoroughly for quizzes and exams. Essential resource for self-study and mastering course material. Instant download on Stuvia!

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STUDY NOTES & REFERENCE GUIDE




Chapters 2 – 14 Covered




SOLUTIONS




nolanjack ‐ Stuvia | Page 1 of 393

, STUDY NOTES & REFERENCE GUIDE




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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nolanjack ‐ Stuvia | Page 2 of 393

, STUDY NOTES & REFERENCE GUIDE




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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nolanjack ‐ Stuvia | Page 3 of 393

, STUDY NOTES & REFERENCE GUIDE




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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nolanjack ‐ Stuvia | Page 4 of 393

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