I have two equations:
y1 = 2sinx1;
y2 = 2cos^2(x1) + 3sin(2x2+3);
here y1 = 0 while y2 = 1.
Can anyone please tell me which approach would be the best to find out the values of x1 and x2.
Thank you.

 采纳的回答

Torsten
Torsten 2016-3-31
编辑:Torsten 2016-3-31
Try this:
function driver
y=zeros(1,2);
y(1)=0;
y(2)=1;
x0=zeros(1,2);
x=fsolve(@(x)HW1(x,y),x0)
function res = HW1(x,y)
res(1)=y(1)-2*sin(x(1));
res(2)=y(2)-(2*cos(x(1))^2 + 3*sin(2*x(2) + 3));
Best wishes
Torsten.

2 个评论

Thanks a lot. It works perfectly well. I am really thankful to you.
I am new to Matlab. Can you tell me which optimization will let me find the unique values of x1 and x2 for the given values of y1 and y2. If I change the interval from [0 0] to [2 2] it finds the zeros between this interval.
Your equations don't have a unique solution. So - depending on the starting guess x0 - you'll get different solutions for x. Do you have any condition on x that could make the solution unique ?
Best wishes
Torsten.

请先登录,再进行评论。

更多回答(1 个)

if you still need this
A=solve('0 = 2*sin(x1)','1 = 2*cos(x1)^2 + 3*sin(2*x2+3)','x1','x2');
% first ans
A.x1(1)
A.x2(1)
% second ans
A.x1(2)
A.x2(2)

5 个评论

Thank you for your answer.
Yes, It solves the equation but I want to do this using optimization. Maybe fzero or something. I made a function but when I use fzero or some other optimization method it gives the error of not enough input arguments.
Thank you.
You have two equations - thus you will need MATLAB's "fsolve".
Best wishes
Torsten.
Yes, Thank you.
I have tried to use fsolve when I input the function, I get an error of not enough input arguments and the optimization stops.
Please show your code.
Best wishes
Torsten.
function [y1,y2] = HW1(x1,x2)
y1 = 2*sin(x1);
y2 = 2*cos(x1)^2 + 3*sin(2*x2 + 3);
end
I call this function HW1 from optimization tool box and it stops after giving error of not enough input arguments.
Thank you for your help. I am stuck for two days with this error.

请先登录,再进行评论。

类别

帮助中心File Exchange 中查找有关 Surrogate Optimization 的更多信息

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by