I continue to receive an error message when I try to use fsolve. Can anyone tell me what I'm doing wrong? (I'm using MATLAB R2012a)

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Jarrett White
Jarrett White 2014-2-27
编辑： Matt J 2014-2-28
First I run the following script.
x1 = 15600; y1 = 7540; z1 = 20140;
x2 = 18760; y2 = 2750; z2 = 18610;
x3 = 17610; y3 = 14630; z3 = 13480;
x4 = 19170; y4 = 610; z4 = 18390;
t1 = 0.07074; t2 = 0.07220; t3 = 0.07690; t4 = 0.07242;
c = 299792458;
x0 = [1 1 1 1];
Now my function to find x,y,z, and d.
function F = myfun325_1(x,y,z,d)
F = [sqrt(((x - x(1)).^2) + ((y - y(1)).^2) + ((z - z(1)).^2)) - c*(t(1) - d); sqrt(((x - x(2)).^2) + ((y - y(2)).^2) + ((z - z(2)).^2)) - c*(t(2) - d); sqrt(((x - x(3)).^2) + ((y - y(3)).^2) + ((z - z(3)).^2)) - c*(t(3) - d); sqrt(((x - x(4)).^2) + ((y - y(4)).^2) + ((z - z(4)).^2)) - c*(t(4) - d);];
end
I run the code: fsolve(@myfun325_1,x0) and am returned the following error:
Error using myfun325_1 (line 3) Not enough input arguments.
Error in fsolve (line 241) fuser = feval(funfcn{3},x,varargin{:});
Caused by: Failure in initial user-supplied objective function evaluation. FSOLVE cannot continue.

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采纳的回答

Star Strider
Star Strider 2014-2-27
You made some coding errors. Note that x1 is not the same as x(1). Also, if you look closely at the documentation for fsolve, all your unknown variables have to be members of the same vector. I reformatted your code:
x(1) = 15600; y(1) = 7540; z(1) = 20140;
x(2) = 18760; y(2) = 2750; z(2) = 18610;
x(3) = 17610; y(3) = 14630; z(3) = 13480;
x(4) = 19170; y(4) = 610; z(4) = 18390;
t(1) = 0.07074; t(2) = 0.07220; t(3) = 0.07690; t(4) = 0.07242;
c = 299792458;
x0 = [1 1 1 1];
% myfun325_1 = @(x,y,z,d) [sqrt(((x - x(1)).^2) + ((y - y(1)).^2) + ((z - z(1)).^2)) - c*(t(1) - d); sqrt(((x - x(2)).^2) + ((y - y(2)).^2) + ((z - z(2)).^2)) - c*(t(2) - d); sqrt(((x - x(3)).^2) + ((y - y(3)).^2) + ((z - z(3)).^2)) - c*(t(3) - d); sqrt(((x - x(4)).^2) + ((y - y(4)).^2) + ((z - z(4)).^2)) - c*(t(4) - d);];
% Redefine x, y, z, d as p(1) ... p(4)
myfun325_1 = @(p) [sqrt(((p(1) - x(1)).^2) + ((p(2) - y(1)).^2) + ((p(3) - z(1)).^2)) - c*(t(1) - p(4)); sqrt(((p(1) - x(2)).^2) + ((p(2) - y(2)).^2) + ((p(3) - z(2)).^2)) - c*(t(2) - p(4)); sqrt(((p(1) - x(3)).^2) + ((p(2) - y(3)).^2) + ((p(3) - z(3)).^2)) - c*(t(3) - p(4)); sqrt(((p(1) - x(4)).^2) + ((p(2) - y(4)).^2) + ((p(3) - z(4)).^2)) - c*(t(4) - p(4));];
estp = fsolve(myfun325_1, x0);
x = estp(1)
y = estp(2)
z = estp(3)
d = estp(4)
It ran without problems. (I did not change anything else in myfun325_1.)
  2 个评论
Jarrett White
Jarrett White 2014-2-27
I ran the code, without all the ():
function F = myfun325_1(x,y,z,d)
x1 = 15600; y1 = 7540; z1 = 20140;
x2 = 18760; y2 = 2750; z2 = 18610;
x3 = 17610; y3 = 14630; z3 = 13480;
x4 = 19170; y4 = 610; z4 = 18390;
t1 = 0.07074; t2 = 0.07220; t3 = 0.07690; t4 = 0.07242;
c = 299792458;
x0 = [1 1 1 1];
%myfun325_1 = @(x,y,z,d) [sqrt(((x - x1).^2) + ((y - y1).^2) + ((z - z1).^2)) - c*(t1 - d); sqrt(((x - x2).^2) + ((y - y2).^2) + ((z - z2).^2)) - c*(t2 - d); sqrt(((x - x3).^2) + ((y - y3).^2) + ((z - z3).^2)) - c*(t3 - d); sqrt(((x - x4).^2) + ((y - y4).^2) + ((z - z4).^2)) - c*(t4 - d);];
%Redefine x, y, z, d as p(1) ... p(4) myfun325_1 = @(p) [sqrt(((p(1) - x1).^2) + ((p(2) - y1).^2) + ((p(3) - z1).^2)) - c*(t1 - p(4)); sqrt(((p(1) - x2).^2) + ((p(2) - y2).^2) + ((p(3) - z2).^2)) - c*(t2 - p(4)); sqrt(((p(1) - x3).^2) + ((p(2) - y3).^2) + ((p(3) - z3).^2)) - c*(t3 - p(4)); sqrt(((p(1) - x4).^2) + ((p(2) - y4).^2) + ((p(3) - z4).^2)) - c*(t4 - p(4));];
estp = fsolve(myfun325_1, x0);
x = estp(1) y = estp(2) z = estp(3) d = estp(4)
end
I didn't get an error message but MATLAB returned:
>>myfun325_1
No solution found.
fsolve stopped because the relative size of the current step is less than the default value of the step size tolerance squared, but the vector of function values is not near zero as measured by the default value of the function tolerance.
criteria details
x =
3.6317e+03
y =
400.3288
z =
3.5781e+04
d =
0.0730
Star Strider
Star Strider 2014-2-27
I get close to those same values (and a similar notification when the solver finishes) even when I add:
opts = optimoptions(@fsolve, 'FinDiffType','central', 'MaxFunEvals',50000, 'MaxIter',10000, 'TolFun',1E-10, 'TolX',1E-10);
estp = fsolve(myfun325_1, x0, opts);
The problem may be that your function has at least one complex zero. The optimization functions only return real values.
From the documentation: fsolve only handles real variables. When x has complex variables, the variables must be split into real and imaginary parts.
I haven’t done that in a while, so one way I suggest you might go about coding it would be to refer to the imaginary parts of your variables as p(5) ... p(8), so x becomes p(1)+j*p(5) and so on for the other variables. Beyond that, I have no suggestions.

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更多回答（2 个）

Paul
Paul 2014-2-27
change all x(1), y(1) etc to x1,y1 etc. Also put:
x1 = 15600; y1 = 7540; z1 = 20140;
x2 = 18760; y2 = 2750; z2 = 18610;
x3 = 17610; y3 = 14630; z3 = 13480;
x4 = 19170; y4 = 610; z4 = 18390;
t1 = 0.07074; t2 = 0.07220; t3 = 0.07690; t4 = 0.07242;
c = 299792458;
inside the function file myfun325_1.
  7 个评论
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Jarrett White
Jarrett White 2014-2-27
Yes I saved it but when I run it now it returns:
No solution found.
fsolve stopped because the relative size of the current step is less than the default value of the step size tolerance squared, but the vector of function values is not near zero as measured by the default value of the function tolerance.
criteria details
ans =
1.0e+04 *
0.3632 0.0400 3.5781 0.0000
Paul
Paul 2014-2-27
编辑：Paul 2014-2-27
Yes, either your equations dont have a solution (check if they are correct) or you should try other initial guesses (although I don't think this will help in this case).

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Matt J
Matt J 2014-2-27
编辑：Matt J 2014-2-27
Below is my suggestion for how to rewrite the problem. It basically converts things to more manageable units and gets rid of the sqrt() operations. Without it, you will have points of non-differentiability and also have to worry about the algorithm knowing that c*(t(1) - d) is supposed to be positive.
The rewrite also makes it pretty clear, I think, why the problem has no solution with the data given. The distance of (x,y,z) from (x1,y1,z1) is supposed to be D=c*(t(1) - d). Then the triangle inequality says that the distance from (x2,y2,z2) is at most
D+norm([x1,y1,z1]-[x2,y2,z2]) = D+5.9389
However, your second inequality insists that it be D+q2 = D+437.6970
function F = myfun325_1(p)
x1 = 15.600; y1 = 7.540; z1 = 20.140;
x2 = 18.760; y2 = 2.750; z2 = 18.610;
x3 = 17.610; y3 = 14.630; z3 = 13.480;
x4 = 19.170; y4 = .610; z4 = 18.390;
t1 = 0.07074; t2 = 0.07220; t3 = 0.07690; t4 = 0.07242;
c = 299792.458;
q2=c*(t2-t1);
q3=c*(t3-t1);
q4=c*(t4-t1);
x=p(1); y=p(2); z=p(3); D=p(4);
F = [((x - x1).^2) + ((y - y1).^2) + ((z - z1).^2) - D^2;
((x - x2).^2) + ((y - y2).^2) + ((z - z2).^2) - (D+q2)^2;
((x - x3).^2) + ((y - y3).^2) + ((z - z3).^2) - (D+q3)^2;
((x - x4).^2) + ((y - y4).^2) + ((z - z4).^2) - (D+q4)^2];
  3 个评论
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Jarrett White
Jarrett White 2014-2-27
Since there are four equations in four unknowns, subtracting the first equation from the last three would lead to three linear equations in (x,y,z). A solution could be found using Gaussian elimination. Then I could obtain a quadratic equation in d upon substitution. Should I modify the code to go this route?
Matt J
Matt J 2014-2-28
编辑：Matt J 2014-2-28
We've established that the equations have no solution. Re-organizing them into whatever equivalent form won't change that.

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