# How can I solve a set of equations using a function handle?

17 views (last 30 days)
Janna Hinchliff on 11 Dec 2019
Commented: Star Strider on 11 Dec 2019
I have a set of data made up of values between 0 and 1:
biglambda = [ 0.0455 0.0476 0.0500 0.0526 0.0556 0.0588 0.0625 0.0667 0.0714 0.0769 0.0833 0.0909 0.1000
0.1111 0.1250 0.1429 0.1667 0.2000 0.2500 0.3333 0.5000 1.0000]
I want to fit this to a functional form, given by
where a and b are free parameters, t is a time parameter:
t = 1*10^3*[0.5156 0.5693 0.6444 0.6698 0.6730 0.6850 0.8073 0.8615 0.9927 1.0587 1.0791 1.1747 1.1827 1.2286
1.2917 1.2967 1.3735 1.3735 1.3735 1.3924 1.7004 1.9965]
such that each value of biglambda will have its own value of t (in the order shown). T is the time bin, i.e. the time between different values of t. I want to use these parameter sets to find the correct values for a and b that give a suitable fit to the data. My first thought for how to solve this would be to do something of the form
where i is the ith value of biglambda or t, and then minimise this for a and b. However, I'm not sure which is the best approach to take in Matlab for calculating this, as I would normally write a function for , but as I have a set of values I need to take the sum of this seems to be more complicated. I have started with
for j = 1:length(t)
BigLambdaFunction = @(x)biglambda(j)-(1-x(1)*(x(2)-(t(j)+(t(j+1)-t(j))))/(x(2)-t(j))*exp(x(1)*(t(j+1)-t(j))/x(2)));
end
for each j, but I'm not sure how I can convert this to calculate the whole sum and minimise for x(1) and x(2). What would be a good place to start?

Star Strider on 11 Dec 2019
Try this:
BigLambdaFunction = @(p,t,T) 1 - p(1).*((p(2)-(t-T))./(p(2)-t)).*exp((p(1).*T)./p(2))
biglambda = [ 0.0455 0.0476 0.0500 0.0526 0.0556 0.0588 0.0625 0.0667 0.0714 0.0769 0.0833 0.0909 0.1000 0.1111 0.1250 0.1429 0.1667 0.2000 0.2500 0.3333 0.5000 1.0000];
t = 1;
T = 42;
p0 = rand(2,1);
Parms = fminsearch(@(p) norm(biglambda - BigLambdaFunction(p,t,T)), p0)
I have no idea what ‘t’ and ‘T’ are supposed to be, so if they are vectors, it will be necessary to loop through the individual values and estimate ‘Parms’ at each of them. In this code, ‘Parms’ is a (2x1) vector, with ‘a=p(1)’ and ‘b=p(2)’.

Janna Hinchliff on 11 Dec 2019
I have done the calculation to some extent using this - t is the vector given above and T is t_i+1-t_i, which is also a vector. I can then get a set of Parms values for each value of t, however, it still isn't obvious how I then take this and work out what a and b should be for the full data set, as I have a set of values of a and b for each time step.
Star Strider on 11 Dec 2019
The ‘Parms’ vector at each value of ‘t’ and ‘T’ are the parameters:
a = Parms(1)
b = Parms(2)