matlab code for iterative equation

Please I need matlab code to solve this iterative equation X (k+1)= c+ Tx(k) For k=0,1,2,3… with the input value c, T and x and stops when the iteration converges .

[Star Strider]

Use a while loop. Decide on what ‘converges’ means in this context.

Write your code. If you have problems, post it here, along with any error it throws (copy and paste all the red text in the Command Window to a Comment here).

Experiment! Unless your code somehow manages to connect to the nuclear missile command codes, the world will not come to an end if it throws an error.

Answer 1

[Roger Stafford]

If you like the lazy approach to problems, note that if abs(T) < 1, you can rewrite your equation as:

   x(k+1)-b = T*(x(k)-b)

where b = c/(1-T), and therefore

   x(k+1)-b = T^k*(x(1)-b).

In this form it is obvious what x(k) will converge to, namely b, since x(k)-b must converge to zero. Accordingly, carrying out all those tedious iterations becomes unnecessary. As I say, that is the lazy method.

[Alper Olca]

x1=5;
x2=5;
x3=5;
x4=5;
td=10^-2;
a=0;

for i= 1:10
    a=0+i;
    if( (abs(x1-x1)<td && abs(x2-x2)<td) && (abs(x3-x3)<td)&& abs(x4-x4)<td)
        x1=(-23+x2-x3+2*x4)/4;
        x2=(-21-2*x1+x3-3*x4)/6;
        x3=(-11+x1+2*x2-x4)/5;
        x4=(22+x1-2*x2+3*x3)/6;
    end
    
     k=(4*x1-x2+x3-2*x4);
     l=(2*x1+6*x2-x3+3*x4);
     m=(-x1-2*x2+5*x3+x4);
     n=(-x1+2*x2-3*x3+6*x4);
    end
rslt=[k l m n ; x1 x2 x3 x4]

Answer 2

[segun egbekunle]

x0=zeros(1,n)';
for k=1 :itmax+1

               x= c + T*x;

          disp(x)

need improvement for the iteration to stop after obtaining the same value twice

[Walter Roberson]

num_previous = 2;
xold = nan(1, num_previous);
for k = 1 : itmax + 1;
   x= c + T*x;
   if all(x == xold)
     break;
   end
   xold(1:end-1) = xold(2:end);
   xold(end) = x;
 end

This is, of course, not certain to stop any time before itmax + 1, as it is common for exact numeric function values to oscillate around the infinitely-precise point of theoretical convergence.

[segun egbekunle]

it gives error message at if all(x == xold) and xold(end) = x; i have modified it to num_previous = 2; xold = nan(1, num_previous); for k = 1 : itmax + 1; x= c + G*x;

   end
   xold(1:end-1) = xold(2:end);
   disp(x)
now it converges but i need little modification to display the number of iterations before convergence and to display no convergence when there is no convergence

[Alper Olca]

x1=5;
x2=5;
x3=5;
x4=5;
td=10^-2;
a=0;

for i= 1:10
    a=0+i;
    if( (abs(x1-x1)<td && abs(x2-x2)<td) && (abs(x3-x3)<td)&& abs(x4-x4)<td)
        x1=(-23+x2-x3+2*x4)/4;
        x2=(-21-2*x1+x3-3*x4)/6;
        x3=(-11+x1+2*x2-x4)/5;
        x4=(22+x1-2*x2+3*x3)/6;
    end
    
     k=(4*x1-x2+x3-2*x4);
     l=(2*x1+6*x2-x3+3*x4);
     m=(-x1-2*x2+5*x3+x4);
     n=(-x1+2*x2-3*x3+6*x4);
    end
rslt=[k l m n ; x1 x2 x3 x4]

Answer 3

[segun egbekunle]

 i have modified my previuos answer to
 num_previous = 2; xold = nan(1, num_previous); for k = 1 : itmax + 1; x= c + T*x;

   end
   xold(1:end-1) = xold(2:end);
   disp(x)
now it converges but i need little modification to display the number of iterations before convergence and to display no convergence when there is no convergence