# why my iteration prog doesn't work ?

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khalid boureeny on 23 Jan 2017
Commented: Walter Roberson on 25 Jan 2017
% Use NR method f(x)= x^3-5x^2+7x-3
clc
TV=3;
x=(4);
tol=0.0007;
format long
for i=1:5;
fx=(x(i))^3-5*(x(i))^2+7*(x(i))-3;
fxx=3*(x(i))^2-10*(x(i))+7;
x(i+1)=(x(i))-(fx/fxx);
E_T(i)=(abs((TV-x(i+1))/TV))*100;
end
for i=1:5;
e(i)=(x(6))-(x(i));
fx=(x(i))^3-5*(x(i))^2+7*(x(i))-3;
fxx=3*(x(i))^2-10*(x(i))+7;
fxxx=6*(x(i))-10;
e(i+1)=(-fxxx/2*fxx)*(e(i))^2;
end
if abs(TV-(x(i+1)))<tol
disp(' enough to here')
disp(' -----------')
disp(' x(i+1) ')
disp(' -----------')
x;
disp(' -----------')
disp(' T.V ')
disp(' -----------')
E_T;
disp(' -----------')
disp(' E(i+1) ')
disp(' -----------')
e;
disp(' ------------- ')
end
%-----------------------------------------------------------
Walter Roberson on 25 Jan 2017
If you need 50 or 80 digits you will need to switch to the Symbolic Toolbox, and display using vpa()
NumDig = 50;
x = sym(-0.1);
y = sym(0);
fprintf(' i x \n')
fprintf(' --- ------------- \n')
for i=1:5;
fx=(x(i))^3-(x(i))^2+2;
fxx=3*(x(i))^2-2*(x(i));
y(i)=x(i)-(fx/fxx);
fy=(y(i))^3-(y(i))^2+2;
fyy=3*(y(i))^2-2*(y(i));
x(i+1)=x(i)+([fy-fx]/fyy);
end
for i=1:length(x)
fprintf('%2i %s\n', i, char(vpa(x(i),NumDig)))
end

Walter Roberson on 23 Jan 2017
if abs(TV-(x(i+1)))<tol
disp(' enough to here')
disp(' -----------')
disp(' x(i+1) ')
disp(' -----------')
x;
disp(' -----------')
disp(' T.V ')
disp(' -----------')
E_T;
disp(' -----------')
disp(' E(i+1) ')
disp(' -----------')
e;
disp(' ------------- ')
end
remove the ';' from the 'x;' and 'E_T;' and 'e;' -- so
if abs(TV-(x(i+1)))<tol
disp(' enough to here')
disp(' -----------')
disp(' x(i+1) ')
disp(' -----------')
x
disp(' -----------')
disp(' T.V ')
disp(' -----------')
E_T
disp(' -----------')
disp(' E(i+1) ')
disp(' -----------')
e
disp(' ------------- ')
end
##### 2 CommentsShowHide 1 older comment
khalid boureeny on 23 Jan 2017
Hi , Walter Roberson .... YOU're a genius ... thanks again ...thanks thanks .

Lateef Adewale Kareem on 23 Jan 2017
increase your number of iteration, you will meet the tolerance target
khalid boureeny on 23 Jan 2017
Thanks , I will try ...