Equations which depend on each other with unknown initial value

Hi all,

I want to solve equations which depend on each other such as,

5*x_1 = 3*x_2 + 4

4*x_2 = 2*x_1 + 4*x_3 + 4

4*x_3 = 2*x_2 + 4*x_4 + 4

3*x_4 = 6*x_3 +4

I usually put the coefficients of the variables into matrix

A = [5 -3 0 0; -2 4 -4 0; 0 -2 4 4; 0 0 -6 3]

C = [4;4;4;4]

B = inv(A) * C

and get the result, since the number of variables can change and go up to 100, I wrote a 'for loop'

n=4;
for i = 2:n-1
    x(1) = (3*x(2) + 4)/5
    x(i) = (2 * x(i-1) + 4 *x(i+1) +4)/4
    x(n) = (3 * x(i-1) + 4) / 3
end

but the results are not same, i think in the for loop matlab assumes x(2) as zero and continues, anyway if someone can solve this problem I appreciate.

Thanks.

Answer 1

[Walter Roberson]

n=4;
x = sym('x', [1 n]);

x(1) = (3*x(2) + 4)/5;
for i = 2:n-1
    x(i) = (2 * x(i-1) + 4 *x(i+1) +4)/4;
end
x(n) = (3 * x(i-1) + 4) / 3;

[Erkin Karatas]

Thanks, but this does not solve for x values, just creates equations, I want to get

x = [2.2500

2.4167

0.2917

1.9167]

How can I modify it?

Thanks in advance.

[Walter Roberson]

n=4;
x = sym('x', [1 n]);
eq(1) = (3*x(2) + 4)/5;
for i = 2:n-1
    eq(i) = (2 * x(i-1) + 4 * x(i+1) +4)/4;
end
eq(n) = (3 * x(n-2) + 4) / 3;

xval = vpasolve(eq);

Now xval.x1, xval.x2, xval.x3, and so on.

[Erkin Karatas]

The results are not the same with the matrix calculation

xval.x1=0.6666

xval.x2=-1.33

xval.x3=-1.33

xval.x4=-0.33

instead of

x = [2.2500

2.4167

0.2917

1.9167]

[Walter Roberson]

You should go through the equations and make sure that they are correct.

Note: when you construct equations in symbolic form, it is not necessary to do the division by 4 and whatever. You can code things like 5*x(1) == 4*x(2) + 4 inside the eq vector.