Equations which depend on each other with unknown initial value

Question

Erkin Karatas 2019-11-5

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/489412-equations-which-depend-on-each-other-with-unknown-initial-value

评论： Walter Roberson 2019-11-6

在 MATLAB Online 中打开

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.

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

请先登录，再进行评论。

请先登录，再回答此问题。

Answer 1

Walter Roberson 2019-11-5

1
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/489412-equations-which-depend-on-each-other-with-unknown-initial-value#answer_400028

在 MATLAB Online 中打开

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;

4 个评论
显示 2更早的评论隐藏 2更早的评论

Erkin Karatas 2019-11-6

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 2019-11-6

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.

请先登录，再进行评论。