echo on; clc;
%---------------------------------------------------------------------------
%A1_17 MATLAB script file for investigating Theorem 1.17
%
% NUMERICAL METHODS: MATLAB Programs, (c) John H. Mathews 1995
% To accompany the text:
% NUMERICAL METHODS for Mathematics, Science and Engineering, 2nd Ed, 1992
% Prentice Hall, Englewood Cliffs, New Jersey, 07632, U.S.A.
% Prentice Hall, Inc.; USA, Canada, Mexico ISBN 0-13-624990-6
% Prentice Hall, International Editions: ISBN 0-13-625047-5
% This free software is compliments of the author.
% E-mail address: in%"mathews@fullerton.edu"
%
% Theorem 1.17 (Remainder term for Taylor's Theorem).
% Section 1.2, Error Analysis, Page 34
%---------------------------------------------------------------------------
clc; clear all; format long;
% - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
%
% Theorem 1.17 (Remainder term for Taylor's Theorem). Assume
% (n)
% that f(x) and its derivatives f'(x),f''(x),...,f (x) are
%
% are all continuous on [a,b]. If both x and x = x + h
% 0 0
% lie in the interval [a,b], then
%
% n
% f(x) = P (x) + O(h ), where
% n
%
% P (x) is the n-th degree Taylor polynomial expanded about x .
% n 0
%
% - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
pause % Press any key to continue.
clc;
% - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
%
% Example for page 34. Consider f(x) = sin(x) and the Taylor
%
% polynomials of degree n = 3, 5 and 7 expanded about x = 0:
% 0
% n-th degree Taylor polynomial Order of the approximation
%
% 3
% P (x) = x - x /6 x^5/120 = O(x^5)
% 3
% 3 5
% P (x) = x - x /6 + x /120 x^7/5040 = O(x^7)
% 5
% 3 5 7
% P (x) = x - x /6 + x /120 - x /5040 x^9/362880 = O(x^9)
% 7
pause % Press any key to continue.
clc;
% (i) Show that |sin(x) - P (x)| <= |x^5/120| over [-1,1].
% 3
% REMARK. The factor 1.04 was added to separate the curves.
clc;
% ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
% Prepare graphics arrays to plot y = f(x).
% ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
a = -1;
b = 1;
F = 'sin(X)-(X-1/6*X.^3)';
G = '1.04*X.^5/120';
h = (b-a)/100;
X = a:h:b;
Y = eval(F);
Z = eval(G);
clc; figure(1); clg;
%~~~~~~~~~~~~~~~~~~~~~~~
% Begin graphics section
%~~~~~~~~~~~~~~~~~~~~~~~
a = -1;
b = 1;
c = -1/120;
d = 1/120;
whitebg('w');
plot([a b],[0 0],'b',[0 0],[c d],'b');
axis([a b c d]);
axis(axis);
hold on;
title('The error bound |sin(x) - P3(x)| <= |x^5/120| over [-1,1]'); figure(gcf);
% fplot(' 1.04*x^5/120',[-1 1],'-g'); figure(gcf);
% fplot('-1.04*x^5/120',[-1 1],'-g'); figure(gcf);
plot(X,Z,'-g',X,-Z,'-g'); figure(gcf);
% fplot('sin(x)-(x-1/6*x^3)',[-1 1],'-r'); figure(gcf);
plot(X,Y,'-r');
grid
hold off
figure(gcf);
clc;
%....................................
% Begin section to print the results.
%....................................
clc;
% The function is f(x) = sin(x)
%
% 3-rd degree Taylor polynomial Order of the approximation
%
% 3
% P (x) = x - x /6 x^5/120 = O(x^5)
% 3
%
% The error bound is
%
% |sin(x) - P (x)| <= |x^5/120| over [-1,1].
% 3
pause % Press any key to continue.
clc;
% (ii) Show that |sin(x) - P (x)| <= |x^7/5040| over [-1,1].
% 5
% REMARK. The factor 1.06 was added to separate the curves.
clc;
% ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
% Prepare graphics arrays to plot y = f(x).
% ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
a = -1;
b = 1;
F = 'sin(X)-(X-1/6*X.^3+1/120*X.^5)';
G = '1.06*X.^7/5040';
h = (b-a)/100;
X = a:h:b;
Y = eval(F);
Z = eval(G);
clc; figure(2); clg;
%~~~~~~~~~~~~~~~~~~~~~~~
% Begin graphics section
%~~~~~~~~~~~~~~~~~~~~~~~
a = -1;
b = 1;
c = -1/5040;
d = 1/5040;
whitebg('w');
plot([a b],[0 0],'b',[0 0],[c d],'b');
axis([a b c d]);
axis(axis);
hold on;
title('The error bound |sin(x) - P5(x)| <= |x^7/5040| over [-1,1]'); figure(gcf);
% fplot(' 1.06*x^7/5040',[-1 1],'-g'); figure(gcf);
% fplot('-1.06*x^7/5040',[-1 1],'-g'); figure(gcf);
plot(X,Z,'-g',X,-Z,'-g'); figure(gcf);
% fplot('sin(x)-(x-1/6*x^3+1/120*x^5)',[-1 1],'-r'); figure(gcf);
plot(X,Y,'-r');
grid
hold off
figure(gcf);
clc;
%....................................
% Begin section to print the results.
%....................................
clc;
% The function is f(x) = sin(x)
%
% 5-th degree Taylor polynomial Order of the approximation
%
% 3 5
% P (x) = x - x /6 + x /120 x^7/5040 = O(x^7)
% 5
%
% The error bound is
%
% |sin(x) - P (x)| <= |x^7/5040| over [-1,1].
% 5
pause % Press any key to continue.
clc;
% (iii) Show that |sin(x) - P (x)| <= |x^9/362880| over [-1,1].
% 7
% REMARK. The factor 1.08 was added to separate the curves.
clc;
% ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
% Prepare graphics arrays to plot y = f(x).
% ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
F = 'sin(X)-(X-1/6*X.^3+1/120*X.^5-1/5040*X.^7)';
G = '1.08*X.^9/362880';
h = (b-a)/100;
X = a:h:b;
Y = eval(F);
Z = eval(G);
clc; figure(3); clg;
%~~~~~~~~~~~~~~~~~~~~~~~
% Begin graphics section
%~~~~~~~~~~~~~~~~~~~~~~~
a = -1;
b = 1;
c = -1/362880;
d = 1/362880;
whitebg('w');
plot([a b],[0 0],'b',[0 0],[c d],'b');
axis([a b c d]);
axis(axis);
hold on;
title('The error bound |sin(x) - P7(x)| <= |x^9/362880| over [-1,1]'); figure(gcf);
% fplot(' 1.08*x^9/362880',[-1 1],'-g'); figure(gcf);
% fplot('-1.08*x^9/362880',[-1 1],'-g'); figure(gcf);
plot(X,Z,'-g',X,-Z,'-g'); figure(gcf);
% fplot('sin(x)-(x-1/6*x^3+1/120*x^5-1/5040*x^7)',[-1 1],'-r'); figure(gcf);
plot(X,Y,'-r');
grid
hold off
figure(gcf);
clc;
%....................................
% Begin section to print the results.
%....................................
clc;
% The function is f(x) = sin(x)
%
% 7-th degree Taylor polynomial Order of the approximation
%
% 3 5 7
% P (x) = x - x /6 + x /120 - x /5040 x^9/362880 = O(x^9)
% 7
%
% The error bound is
%
% |sin(x) - P (x)| <= |x^9/362880| over [-1,1].
% 7
