# solve nonlinear equations + numerical integration

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john birt on 25 Apr 2012
Hi, I just have no idea how to do this.. I have code
x = [1.4334 1.46 0.1; 1.435 1.46 0.1];
t = 0.1:0.1:0.6;
for i=1:length(x)
z(i) = trapz(t,normcdf(((log(x(i,1)./x(i,2))+t.*x(i,3).^2)),0,1));
end
But I do not know what x(:,3) is, above I have just used x(1,3)=x(2,3)=0.1 but that is just a guess! I do now what the above function z(i) should be equal to for each 'i', z(1)=0.0213 and z(2)=0.0222 and I want to solve to find a x(1,3)=x(2,3). I do not want to find a single numeric value x(1,3)=x(2,3) that will not solve the equations exactly but just the best x(1,3)=x(2,3) that fits both equations in terms of mean squared error. I do not want two values x(1,3) and x(2,3) that are different, I am not trying to solve each equation individual but a x(.,3) that solves them all best in terms of mse.
Heres my code
x = [1.4334 1.46 0.1; 1.435 1.46 0.1];
y = [0.0213, 0.0222]
t = 0.1:0.1:0.6;
xdata = x(:,1); ydata = y;
for i=1:length(x)
z(i) = trapz(t,normcdf(((log(x(i,1)./x(i,2))+t.*c.^2)),0,1));;
end
c = 0.1; %initial guess for x(:,3)
cfit = nlinfit(xdata,ydata,z,c)
which does not work. I just dont understand how to simultaneously solve a set of equations when there is numerical integration involved.
Any ideas?
john birt on 26 Apr 2012
I think you answered my question, but I think I asked the wrong question :-)

Richard Brown on 25 Apr 2012
You could pose it as an optimization problem. First set up a function that will be zero when z1,z2 are equal to their correct values. I'm going to define the vector y to contain your two decision variables (x(1,3) and x(2,3))
t = 0.1:0.1:0.6;
z = [0.0213; 0.0222]
x = [1.4334 1.46; 1.435 1.46]; %unknowns ommited
f = @(y) sum(([trapz(t,normcdf(((log(x(1,1)./x(1,2))+t.*y(1).^2)),0,1));
trapz(t,normcdf(((log(x(2,1)./x(2,2))+t.*y(2).^2)),0,1))] - z).^2);
And then use fminsearch to find it:
[y, fval] = fminsearch(f, [0.1; 0.1])
If fval == 0, then y contains a solution to your equations. Disclaimer: I don't have the stats toolbox on my computer, so I can't run this, but with a bit of luck (and a good enough initial guess), it will hopefully work.
john birt on 26 Apr 2012
I'm going to have to delete this question a re-post it in a better way to get people to understand what my question is.

Walter Roberson on 25 Apr 2012
Looks to me like this code would be appropriate:
lnx12 = log(x(:,1) ./ x(:,2));
cguess = 0.1;
C = zeros(1,length(x));
for i=1:length(x)
C(i) = fzero( @(c) z(i) - trapz(t, normcdf(lnx12(i)+t.*c.^2), 0, 1), cguess);
end
cfit = nlinfit(xdata,ydata,z,C);
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john birt on 26 Apr 2012
thanks, It was 50:50 whose answer I accepted, I would have liked to be able to click accepted yours as well :-) Anyhow, I asked the wrong question. My code is huge and I have tried to ask a simple question but asked the wrong thing, so I've put up a new question trying to make it really simple.

Richard Brown on 25 Apr 2012
t = 0.1:0.1:0.6;
z = [0.0213; 0.0222];
x = [1.4334 1.46; 1.435 1.46];
f1 = @(y) trapz(t, normcdf(log(x(1,1)/x(1,2) + t*y^2),0,1))-z(1)
f2 = @(y) trapz(t, normcdf(log(x(2,1)/x(2,2) + t*y^2),0,1))-z(2)
Now plot both functions - neither has a zero
fplot(f1, [-10, 10])
hold on
fplot(f2, [-10, 10])
john birt on 26 Apr 2012
thank you again. Back at my desk now so will try shortly, the issue that I did not make clear is that I am not trying to find a 'c' that solves each equation exactly, but just a single 'c' that fits both equations the best it can, they will not solve these equations but give the best 'c' in terms of mean squared error.
Anyhow, than kyou for your advice and I will have another try today to get some results.