Solutions Manual for Numerical and Analytical Methods with MATLAB 1st Edition by William Bober PDF | Step-by-Step Solutions and MATLAB Implementations for Engineering and Applied Mathematics Problems | Covers Differential Equations, Linear Algebra, Curve

Chapters 2 – 14 Covered




SOLUTIONS

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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

, CHAPTER 2

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

f ' ' (0) f ' ' ' (0) f
f ( x) f (0) f ' (0) x x2 x3 1V x 4 ...
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
... 8!
2! 4! 6!

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

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

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

, 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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Program result:


x cos(x) cos (x) by terms in the
by arith stm series series
=====================================================
-3.14159 -1.00000 -1.00000 9

-2.82743 -0.95106 -0.95106 8

-2.51327 -0.80902 -0.80902 8

-2.19911 -0.58779 -0.58779 8

-1.88496 -0.30902 -0.30902 7

-1.57080 0.00000 0.00000 14

-1.25664 0.30902 0.30902 6

-0.94248 0.58779 0.58779 5

-0.62832 0.80902 0.80902 4

-0.31416 0.95106 0.95106 4

0.00000 1.00000 1.00000 1

0.31416 0.95106 0.95106 4

0.62832 0.80902 0.80902 4

0.94248 0.58779 0.58779 5

1.25664 0.30902 0.30902 6

1.57080 0.00000 0.00000 14

1.88496 -0.30902 -0.30902 7

2.19911 -0.58779 -0.58779 8

2.51327 -0.80902 -0.80902 8

2.82743 -0.95106 -0.95106 8

3.14159 -1.00000 -1.00000 9

, cos(x) vs. x
1

0.8

0.6

0.4

0.2
cos(x)




0

-0.2

-0.4

-0.6

-0.8

-1
-4 -3 -2 -1 0 1 2 3 4
x




P2.2. Taylor series expansion of f ( x) about x = 0 is:

f ' ' (0) f ' ' ' (0) f
f ( x) f (0) f ' (0) x x2 x3 1V x4 ...
2! 3!
4!

For f ( x) sin ( x) f (0) 0
,

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

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

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

f 1V ( x) sin ( x), f 1V (0) 0

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

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We can see that

, x x5
3 x7 9
sin ( x) x x
... 9!
3! 5! 7!

and that

x2
term (k) term (k 1)
2 k (2 k
1)




The following program evaluates sin ( x) by both an arithmetic statement

and by the above series for -π ≤ x ≤ π in step of 0.1 .



Program

% sinf.m

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

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

clear; clc;

xi=-pi; dx=0.1*pi; for j=1:21

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

sin_arith(j)= sin(x(j));

sum=x(j); term=x(j); 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;

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end