fsolve inputs

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Alex
Alex 2012-2-28
Hey all,
I'm trying to pass some equations to fsolve, but I'm getting nothing but errors. I have 3 complex equations, and to simplify input, I broke down the equations into component parts like so:
y_1 = 309;
n0_1 = 1;
n1_1 = x(1);
k1_1 = x(2);
n2_1 = 5.07;
k2_1 = 3.62;
d1_1 = x(3);
R_1 = .4335;
g1_1 = (n0_1.^2 - n1_1.^2 - k1_1.^2)./((n1_1 + n2_1).^2 + k1_1.^2);
g2_1 = (n1_1.^2 - n2_1.^2 + k1_1.^2 - k2_1.^2)./((n1_1 + n2_1).^2 + (k1_1 + k2_1).^2);
h1_1 = (2.*n0_1.*k1_1)./((n0_1 + n1_1).^2 + k1_1.^2);
h2_1 = (2.*(n1_1.*k2_1 - n2_1.*k1_1))./((n1_1 + n2_1).^2 + (k1_1 + k2_1).^2);
a_1 = (2.*pi().*k1_1.*d1_1)./y_1;
b_1 = (2.*pi().*n1_1.*d1_1)./y_1;
A_1 = 2.*(g1_1.*g2_1 + h1_1.*h2_1);
B_1 = 2.*(g1_1.*h2_1 - g2_1.*h1_1);
C_1 = 2.*(g1_1.*g2_1 - h1_1.*h2_1);
D_1 = 2.*(g1_1.*h2_1 + g2_1.*h1_1);
g1_1.^2 + h1_1.^2).*(exp(2.*a_1)) + (g2_1.^2 + h2_1.^2).*(exp(-2.*a_1)) + A_1.*(cos(2.*b_1)) + B_1.*(sin(2.*b_1)))./((exp(2.*a_1)) + (g1_1.^2 + h1_1.^2).*(g2_1.^2 + h2_1.^2).*(exp(-2.*a_1)) + C_1.*(cos(2.*b_1)) + D_1.*(sin(2.*b_1))) - R_1
This equations repeats itself two more times, with different values as constants. The problem I seem to be having is that Matlab doesn't like my definitions of the variables. I don't know how to express this information in a way that Matlab is able to understand, and despite all of the help I've had from the Mathworks community (thanks so much), I can't seem to get this to work. I've tried syms, I've troubleshot syms, I've tried setting up a seperate function file with no results, and I'm stuck at this point. How would you pass this through? I'm sure it would be easy enough to just create the entire system of equations without all of the substitution, and then pass that on to fsolve, but this is much user friendly, and I would like to learn how to make this work.
Thanks!
  2 个评论
Walter Roberson
Walter Roberson 2012-2-28
"Matlab doesn't like my definitions of the variables" is not very specific, makes it difficult for us to help you.
Walter Roberson
Walter Roberson 2012-2-28
Duplicate is at http://www.mathworks.com/matlabcentral/answers/30540-passing-data-to-fsolve

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采纳的回答

Walter Roberson
Walter Roberson 2012-2-28
I cannot test this as I do not have the toolbox with fsolve in it.
This is the structure I told you to adapt several threads ago.
function nkd = g3driver
Guess = [1.58, 0, 25];
nkd = fsolve(@g3, Guess);
end
function F = g3(X)
% This program aims to back-solve for reflectance (R) using given values
% input by hand.
x = X(1); y = X(2); z = X(3);
y_1 = 309;
n0_1 = 1;
%n1_1 = 1.580086;
n1_1 = x;
%k1_1 = 0;
k1_1 = y;
n2_1 = 5.07;
k2_1 = 3.62;
%d1_1 = 25;
d1_1 = z;
R_1 = .4335;
y_2 = 310;
n0_2 = 1;
%n1_2 = 1.579925;
n1_2 = x;
%k1_2 = 0;
k1_2 = y;
n2_2 = 5.07;
k2_2 = 3.56;
%d1_2 = 25;
d1_2 = z;
R_2 = .4294;
y_3 = 311;
n0_3 = 1;
%n1_3 = 1.579764;
n1_3 = x;
%k1_3 = 0;
k1_3 = y;
n2_3 = 5.08;
k2_3 = 3.53;
%d1_3 = 25;
d1_3 = z;
R_3 = .4277;
g1_1 = (n0_1.^2 - n1_1.^2 - k1_1.^2)./((n1_1 + n2_1).^2 + k1_1.^2);
g2_1 = (n1_1.^2 - n2_1.^2 + k1_1.^2 - k2_1.^2)./((n1_1 + n2_1).^2 + (k1_1 + k2_1).^2);
h1_1 = (2.*n0_1.*k1_1)./((n0_1 + n1_1).^2 + k1_1.^2);
h2_1 = (2.*(n1_1.*k2_1 - n2_1.*k1_1))./((n1_1 + n2_1).^2 + (k1_1 + k2_1).^2);
a_1 = (2.*pi().*k1_1.*d1_1)./y_1;
b_1 = (2.*pi().*n1_1.*d1_1)./y_1;
A_1 = 2.*(g1_1.*g2_1 + h1_1.*h2_1);
B_1 = 2.*(g1_1.*h2_1 - g2_1.*h1_1);
C_1 = 2.*(g1_1.*g2_1 - h1_1.*h2_1);
D_1 = 2.*(g1_1.*h2_1 + g2_1.*h1_1);
g1_2 = (n0_2.^2 - n1_2.^2 - k1_2.^2)./((n1_2 + n2_2).^2 + k1_2.^2);
g2_2 = (n1_2.^2 - n2_2.^2 + k1_2.^2 - k2_2.^2)./((n1_2 + n2_2).^2 + (k1_2 + k2_2).^2);
h1_2 = (2.*n0_2.*k1_2)./((n0_2 + n1_2).^2 + k1_2.^2);
h2_2 = (2.*(n1_2.*k2_2 - n2_2.*k1_2))./((n1_2 + n2_2).^2 + (k1_2 + k2_2).^2);
a_2 = (2.*pi().*k1_2.*d1_2)./y_2;
b_2 = (2.*pi().*n1_2.*d1_2)./y_2;
A_2 = 2.*(g1_2.*g2_2 + h1_2.*h2_2);
B_2 = 2.*(g1_2.*h2_2 - g2_2.*h1_2);
C_2 = 2.*(g1_2.*g2_2 - h1_2.*h2_2);
D_2 = 2.*(g1_2.*h2_2 + g2_2.*h1_2);
g1_3 = (n0_3.^2 - n1_3.^2 - k1_3.^2)./((n1_3 + n2_3).^2 + k1_3.^2);
g2_3 = (n1_3.^2 - n2_3.^2 + k1_3.^2 - k2_3.^2)./((n1_3 + n2_3).^2 + (k1_3 + k2_3).^2);
h1_3 = (2.*n0_3.*k1_3)./((n0_3 + n1_3).^2 + k1_3.^2);
h2_3 = (2.*(n1_3.*k2_3 - n2_3.*k1_3))./((n1_3 + n2_3).^2 + (k1_3 + k2_3).^2);
a_3 = (2.*pi().*k1_3.*d1_3)./y_3;
b_3 = (2.*pi().*n1_3.*d1_3)./y_3;
A_3 = 2.*(g1_3.*g2_3 + h1_3.*h2_3);
B_3 = 2.*(g1_3.*h2_3 - g2_3.*h1_3);
C_3 = 2.*(g1_3.*g2_3 - h1_3.*h2_3);
D_3 = 2.*(g1_3.*h2_3 + g2_3.*h1_3);
F = [((g1_1.^2 + h1_1.^2).*(exp(2.*a_1)) + (g2_1.^2 + h2_1.^2).*(exp(-2.*a_1)) + A_1.*(cos(2.*b_1)) + B_1.*(sin(2.*b_1)))./((exp(2.*a_1)) + (g1_1.^2 + h1_1.^2).*(g2_1.^2 + h2_1.^2).*(exp(-2.*a_1)) + C_1.*(cos(2.*b_1)) + D_1.*(sin(2.*b_1))) - R_1; ...
((g1_2.^2 + h1_2.^2).*(exp(2.*a_2)) + (g2_2.^2 + h2_2.^2).*(exp(-2.*a_2)) + A_2.*(cos(2.*b_2)) + B_2.*(sin(2.*b_2)))./((exp(2.*a_2)) + (g1_2.^2 + h1_2.^2).*(g2_2.^2 + h2_2.^2).*(exp(-2.*a_2)) + C_2.*(cos(2.*b_2)) + D_2.*(sin(2.*b_2))) - R_2; ...
((g1_3.^2 + h1_3.^2).*(exp(2.*a_3)) + (g2_3.^2 + h2_3.^2).*(exp(-2.*a_3)) + A_3.*(cos(2.*b_3)) + B_3.*(sin(2.*b_3)))./((exp(2.*a_3)) + (g1_3.^2 + h1_3.^2).*(g2_3.^2 + h2_3.^2).*(exp(-2.*a_3)) + C_3.*(cos(2.*b_3)) + D_3.*(sin(2.*b_3))) - R_3];
end

更多回答（1 个）

Walter Roberson
Walter Roberson 2012-2-28
Your line that starts g1_1.^2 is missing an opening '(', and has an extra ')' before the './'
  4 个评论
显示 2更早的评论
Alex
Alex 2012-2-28
Just add a parenthesis in front of the g1_1.^2 and it should be fine. I don't think you need to run the code to answer my question, this isn't even the full program. I just want to know how to use fsolve with substitution as seen above.
Walter Roberson
Walter Roberson 2012-2-28
That still leaves the extra ')' before the './' .
When I put in the needed '(' and remove the unneeded ')', I have no problems with the code when I initialize x = rand(3,1)
Your code shown is not doing any substitution.

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