Math-Q: Solving linear equations complex variables

1 次查看（过去 30 天）

Wazy sky 2018-11-16

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/430188-math-q-solving-linear-equations-complex-variables

编辑： Wazy sky 2018-11-28

G'day all,

How to solve the following equation please, (A, B, C,.. real constants, while X, Y are complex variables: knowing that the equations are determinate (applied for several inputs of data so that the number of unknown=knowns)

A+imaginary(B)= Real[(C.Real(X)+D.Real(Y)+ E.imaginary(X)+F.imaginary(Y))]+ Imaginary[ (G.Real(X)+H.Real(Y)- I.imaginary(X)+J.imaginary(Y))] ----Eq(1)

I was thinking of Two possibilities :

sol-1

the real part for RS = the Real part in the LS:

the imaginary part for RS = the Real part in the LS

noting that each real part has both real&imaginary variables

After first attempt : it seems no solution exist for this case

A=[(C.Real(X)+D.Real(Y)+ E.imaginary(X)+F.imaginary(Y))]----1

B=[(G.Real(X)+H.Real(Y)+ I.imaginary(X)+J.imaginary(Y))]--2

Thanks

12 个评论
显示 10更早的评论隐藏 10更早的评论

Walter Roberson 2018-11-24

在 MATLAB Online 中打开

1) Like I posted before,

couldItBe = @(proposition) ~isAlways(~proposition)

2) xr_at_min is "candidate xreal" that can be substituted into the residue

3) "I have a vector of xreal, and y_real ..etc"

That is inconsistent with your "The equation has 2 separable parts (Real+Imaginary) with four real variabels."

4) Each cycle of differentiation, solve for a zero, differentiate a second time, substitute the roots into the second derivative, accept the solutions that are potentially positive, substitute those into the equation: each round of that reduces the number of free variables by 1 (but does potentially introduce multiple solution branches based upon the possibility of multiple solutions for the zero.) You may get to a point where you can demonstrate that all of the substitutions into the second derivative come out negative or zero: if so then you reject that and move backwards to the last point where you had a choice of solutions and take the next possible solution and move forward from there. Eventually you may arrive at a full list of substitutions that in chain result give you a minima: if so run back substitutions until you get a full list of values for the variables, and add that full list of values to your solution list, after which you go back to the last point you had a choice of solutions and move forwards with the next candidate. If you reach the end of all possible branches without having found at least one full list of solutions, then the equations do not have a minima, in which case the minima will be found at some mix of +/- infinity for all of the variables.

Wazy sky 2018-11-24

编辑：Wazy sky 2018-11-24

Walter Roberson

Thank you for your answer. Actually, It seems on higher level than my experience with Matlab. I am not sure I am capable to simulate the procedure correctly. I will give it try.

The equation that I want to be solve seems very complicated. It takes me while to have more accurate understanding to the real structure of the equation.

You are correct, my the original post was for kinda of prlimineary and might be unaccurate interpretation to the problem.

I am not sure if the suggested solution is still valid for the new definition of the problem (equaiton with a vectors of variable). I will the question in in new post.

Thank

请先登录，再进行评论。

请先登录，再回答此问题。