why my iteration prog doesn't work ?

2 visualizzazioni (ultimi 30 giorni)
% 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
%-----------------------------------------------------------
  7 Commenti
khalid boureeny
khalid boureeny il 25 Gen 2017
Modificato: Walter Roberson il 25 Gen 2017
Hi , Walter Roberson .... I'm deeply grateful for your help ... I still have a problem with decimal digits ...
What if I need more decimal digits in my answer than the default ? for example 50 or 80 digits .
PLEASE , HELP in this one ...
clear all
clc
x=(-0.1);
y=(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 %18.15f\n',i,x(i))
end
THE RESULTS ARE
i x
--- -------------
1 -0.100000000000000
2 -3.119140528362807
3 -1.474069695697443
4 -1.017351956185236
5 -1.000001047082565
6 -1.000000000000000
Walter Roberson
Walter Roberson il 25 Gen 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

Accedi per commentare.

Risposta accettata

Walter Roberson
Walter Roberson il 23 Gen 2017
In your section
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 Commenti
khalid boureeny
khalid boureeny il 23 Gen 2017
Hi, Walter Roberson .... thanks a lot , I will try ... thanks again ...
khalid boureeny
khalid boureeny il 23 Gen 2017
Hi , Walter Roberson .... YOU're a genius ... thanks again ...thanks thanks .

Accedi per commentare.

Più risposte (1)

Lateef Adewale Kareem
Lateef Adewale Kareem il 23 Gen 2017
increase your number of iteration, you will meet the tolerance target

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by