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
@
@SS
eeisimiciicsiosolalatiotionn
sm
, CHAPTER 2
P2.1. Taylor series expansion of f ( x) about x = 0 is:
f ( x) f (0) f ' (0) x f ' ' (0) 2 f ' ' ' (0) 3 f 1V x 4 ...
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
x x
2 4 6 8
cos( x) 1 x 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 - x in step of 0.1 .
% cosf.m
% This program evaluates cos(x) by both arithmetic statement and by
% series for - x in steps of 0.1
clear; clc;
xi=-pi; dx=0.1*pi; for j=1:21
x(j)=xi+(j-1)*dx;
cos_arith(j)= cos(x(j));
@ @SS
eeisim1
sm
icisolation
, 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;
@ @SS
eeisim2
sm
icisolation
,Program result:
-----------------------------------------------------------------------------------------
x cos(x) cos (x) terms in
by arith stm by series the 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
-------------------------------------------------
@ @SS
eeisim3
sm
icisolation
, 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 ( x) f (0) f ' (0) x f ' ' (0) 2 f ' ' ' (0) 3 f 1V x 4 ...
x x
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
We can see that
@ @SS
eeisim4
sm
icisolation
, x x3 x5 x 7 9
sin ( x) x ...
3! 5! 7! 9!
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 - x in step of 0.1 .
Program
% sinf.m
% This program evaluates sin(x) by both arithmetic statement and by
% series for - x in steps of 0.1
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;
end
@ @SS
eeisim5
sm
icisolation
, end sin_ser(j)=sum;
nterms(j)= k;
end fo=fopen('output.dat','w');
fprintf(fo,'x sin(x) sin (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),sin_arith(j),sin_ser(j),nterms(j));
fprintf(fo,'-------------------------------------------------\n');
end fclose(fo);
plot(x,sin_arith),xlabel('x'),ylabel('sin(x)'),title('sin(x)vs. x'),grid;
Program Results
---------------------------------------------------------------------------------------
x sin(x) sin (x) terms in
by arith stm by series the series
=====================================================
-3.14159 -0.00000 -0.00000 17
-------------------------------------------------
-2.82743 -0.30902 -0.30902 8
-------------------------------------------------
-2.51327 -0.58779 -0.58779 8
-------------------------------------------------
-2.19911 -0.80902 -0.80902 7
-------------------------------------------------
-1.88496 -0.95106 -0.95106 6
-------------------------------------------------
-1.57080 -1.00000 -1.00000 6
-------------------------------------------------
-1.25664 -0.95106 -0.95106 5
-------------------------------------------------
-0.94248 -0.80902 -0.80902 5
-------------------------------------------------
@ @SS
eeisim6
sm
icisolation
, -0.62832 -0.58779 -0.58779 4
-------------------------------------------------
-0.31416 -0.30902 -0.30902 3
------------------------------------------------- 0.00000 0.00000
0.00000 1
------------------------------------------------- 0.31416 0.30902
0.30902 3
------------------------------------------------- 0.62832 0.58779
0.58779 4
------------------------------------------------- 0.94248 0.80902
0.80902 5
------------------------------------------------- 1.25664 0.95106
0.95106 5
------------------------------------------------- 1.57080 1.00000
1.00000 6
------------------------------------------------- 1.88496 0.95106
0.95106 6
------------------------------------------------- 2.19911 0.80902
0.80902 7
------------------------------------------------- 2.51327 0.58779
0.58779 8
------------------------------------------------- 2.82743 0.30902
0.30902 8
------------------------------------------------- 3.14159 0.00000
0.00000 17
-------------------------------------------------
P2.3. To show that eix = cos (x) + i sin (x) and e-ix = cos (x) – i sin (x) by a Taylor series
expansions about x = 0.
f ( x) f (0) f ' (0) x f ' ' (0) 2 f ' ' ' (0) 3 f 1V x 4 ...
x x
2! 3! 4!
For f ( x) eix , f (0) 1
f ' ( x) i eix , f ' (0) i
f ' ' ( x) eix , f ' ' (0) 1
f ' ' ' ( x) i eix , f ' ' ' (0) i
f 1V ( x) eix , f 1V (0) 1
f V ( x) i eix , f V (0) i
@ @SS
eeisim7
sm
icisolation
, THOSE WERE PREVIEW PAGES
TO DOWNLOAD THE FULL PDF
CLICK ON THE L.I.N.K
ON THE NEXT PAGE
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
@
@SS
eeisimiciicsiosolalatiotionn
sm
, CHAPTER 2
P2.1. Taylor series expansion of f ( x) about x = 0 is:
f ( x) f (0) f ' (0) x f ' ' (0) 2 f ' ' ' (0) 3 f 1V x 4 ...
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
x x
2 4 6 8
cos( x) 1 x 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 - x in step of 0.1 .
% cosf.m
% This program evaluates cos(x) by both arithmetic statement and by
% series for - x in steps of 0.1
clear; clc;
xi=-pi; dx=0.1*pi; for j=1:21
x(j)=xi+(j-1)*dx;
cos_arith(j)= cos(x(j));
@ @SS
eeisim1
sm
icisolation
, 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;
@ @SS
eeisim2
sm
icisolation
,Program result:
-----------------------------------------------------------------------------------------
x cos(x) cos (x) terms in
by arith stm by series the 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
-------------------------------------------------
@ @SS
eeisim3
sm
icisolation
, 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 ( x) f (0) f ' (0) x f ' ' (0) 2 f ' ' ' (0) 3 f 1V x 4 ...
x x
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
We can see that
@ @SS
eeisim4
sm
icisolation
, x x3 x5 x 7 9
sin ( x) x ...
3! 5! 7! 9!
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 - x in step of 0.1 .
Program
% sinf.m
% This program evaluates sin(x) by both arithmetic statement and by
% series for - x in steps of 0.1
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;
end
@ @SS
eeisim5
sm
icisolation
, end sin_ser(j)=sum;
nterms(j)= k;
end fo=fopen('output.dat','w');
fprintf(fo,'x sin(x) sin (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),sin_arith(j),sin_ser(j),nterms(j));
fprintf(fo,'-------------------------------------------------\n');
end fclose(fo);
plot(x,sin_arith),xlabel('x'),ylabel('sin(x)'),title('sin(x)vs. x'),grid;
Program Results
---------------------------------------------------------------------------------------
x sin(x) sin (x) terms in
by arith stm by series the series
=====================================================
-3.14159 -0.00000 -0.00000 17
-------------------------------------------------
-2.82743 -0.30902 -0.30902 8
-------------------------------------------------
-2.51327 -0.58779 -0.58779 8
-------------------------------------------------
-2.19911 -0.80902 -0.80902 7
-------------------------------------------------
-1.88496 -0.95106 -0.95106 6
-------------------------------------------------
-1.57080 -1.00000 -1.00000 6
-------------------------------------------------
-1.25664 -0.95106 -0.95106 5
-------------------------------------------------
-0.94248 -0.80902 -0.80902 5
-------------------------------------------------
@ @SS
eeisim6
sm
icisolation
, -0.62832 -0.58779 -0.58779 4
-------------------------------------------------
-0.31416 -0.30902 -0.30902 3
------------------------------------------------- 0.00000 0.00000
0.00000 1
------------------------------------------------- 0.31416 0.30902
0.30902 3
------------------------------------------------- 0.62832 0.58779
0.58779 4
------------------------------------------------- 0.94248 0.80902
0.80902 5
------------------------------------------------- 1.25664 0.95106
0.95106 5
------------------------------------------------- 1.57080 1.00000
1.00000 6
------------------------------------------------- 1.88496 0.95106
0.95106 6
------------------------------------------------- 2.19911 0.80902
0.80902 7
------------------------------------------------- 2.51327 0.58779
0.58779 8
------------------------------------------------- 2.82743 0.30902
0.30902 8
------------------------------------------------- 3.14159 0.00000
0.00000 17
-------------------------------------------------
P2.3. To show that eix = cos (x) + i sin (x) and e-ix = cos (x) – i sin (x) by a Taylor series
expansions about x = 0.
f ( x) f (0) f ' (0) x f ' ' (0) 2 f ' ' ' (0) 3 f 1V x 4 ...
x x
2! 3! 4!
For f ( x) eix , f (0) 1
f ' ( x) i eix , f ' (0) i
f ' ' ( x) eix , f ' ' (0) 1
f ' ' ' ( x) i eix , f ' ' ' (0) i
f 1V ( x) eix , f 1V (0) 1
f V ( x) i eix , f V (0) i
@ @SS
eeisim7
sm
icisolation
, THOSE WERE PREVIEW PAGES
TO DOWNLOAD THE FULL PDF
CLICK ON THE L.I.N.K
ON THE NEXT PAGE
