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

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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