Linear fit through loglog plots.

3 Ansichten (letzte 30 Tage)
Kevin Osborn
Kevin Osborn am 1 Okt. 2021
Beantwortet: Kevin Osborn am 1 Okt. 2021
With the iteration data below find the best linear fit in the formula above and determine the slope, α (order of convergence), and the value of λ. How would I go about coding this in MatLab? I think I need to use loglog plots, but I'm not exactly sure how to go about this. I have used Newton's Method of approximation as shown below.
f = @(x) 2*exp(-2*x) + 4*sin(x) - 2*cos(2*x);
fp = @(x) 4*(-exp(-2*x) + sin(2*x) + cos(x));
x0 = 0.3;
N = 10;
tol = 1E-6;
x(1) = x0;
n = 2;
nfinal = N + 1;
while (n <= N + 1)
fe = f(x(n - 1));
fpe = fp(x(n - 1));
x(n) = x(n - 1) - fe/fpe;
if (abs(fe) <= tol)
nfinal = n;
break;
end
n = n + 1;
end
figure, plot(0:nfinal - 1,x(1:nfinal),'o-')
title('Newton''s Method')
xlabel('Iterations')
ylabel('Root Approximation')
table([1:length(x)]', x', 'VariableNames', {'Iteration' 'Root Approximation'})
x(n) = x(n - 1) - fe/fpe;
fprintf('Iteraion Root Approximation Tolerance Level\n')
fprintf('%3d: %20g %20g\n', n, x(n), abs(fe));
Thank you in advance for your help.
  1 Kommentar
Kevin Osborn
Kevin Osborn am 1 Okt. 2021
I figured it out actually! Just needed to work on it a bit longer.

Melden Sie sich an, um zu kommentieren.

Akzeptierte Antwort

Kevin Osborn
Kevin Osborn am 1 Okt. 2021
This is how I plotted it, before cleaning up the graph to make it more pretty and what not.
y = log(abs(x(3:end) - x(2:end-1)));
x = log(abs(x(2:end-1) - x(1:end-2)));
p = polyfit(x,y,1);
alpha = p(1);
lambda = exp(p(2));
scatter(x,y)
hold on
plot(x, alpha*x + log(lambda))
hold off
legend('Data','Fitted line','location','best')
title(['\alpha = ',num2str(alpha),' \lambda = ',num2str(lambda)])

Weitere Antworten (0)

Kategorien

Mehr zu Biotech and Pharmaceutical finden Sie in Help Center und File Exchange

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by