How can I solve a set of equations using a function handle?
4 vues (au cours des 30 derniers jours)
Afficher commentaires plus anciens
Janna Hinchliff
le 11 Déc 2019
Commenté : Star Strider
le 11 Déc 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?
0 commentaires
Réponse acceptée
Star Strider
le 11 Déc 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)’.
2 commentaires
Star Strider
le 11 Déc 2019
The ‘Parms’ vector at each value of ‘t’ and ‘T’ are the parameters:
a = Parms(1)
b = Parms(2)
Plus de réponses (0)
Voir également
Catégories
En savoir plus sur Least Squares dans Help Center et File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!