matlab code for iterative equation

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segun egbekunle
segun egbekunle 2016-6-18
评论： Alper Olca 2020-3-27
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 .
  1 个评论
Star Strider
Star Strider 2016-6-18
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.

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回答（3 个）

Roger Stafford
Roger Stafford 2016-6-18
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.
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Alper Olca
Alper Olca 2020-3-27
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]

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segun egbekunle
segun egbekunle 2016-6-26
编辑：Walter Roberson 2016-6-26
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
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显示 1更早的评论
segun egbekunle
segun egbekunle 2016-6-29
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
Alper Olca 2020-3-27
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]

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segun egbekunle
segun egbekunle 2016-6-29
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

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