Effacer les filtres
Effacer les filtres

how to plot the equation

1 vue (au cours des 30 derniers jours)
shiv gaur
shiv gaur le 21 Fév 2022
Commenté : Image Analyst le 21 Fév 2022
x(0)=1;
y(0)=0
t=1:15;
for n=1:15
x(n+1)=(x(n-1)-y(n))/n+1;
y(n+1)=(x(n)+y(n)-t)/n+1
plot(t,x)
end
  2 commentaires
shiv gaur
shiv gaur le 21 Fév 2022
x(0) is the initial value you take as x0
Image Analyst
Image Analyst le 21 Fév 2022
Yeah, I figured that, but what I meant by my answer below is you're supposed to do
x(1) = 1;
y(1) = 0;

Connectez-vous pour commenter.

Réponse acceptée

Image Analyst
Image Analyst le 21 Fév 2022
Modifié(e) : Image Analyst le 21 Fév 2022
Indexing starts at 1, not 0. Also, check your formulas - they look weird.
For example in
y(n+1)=(x(n)+y(n)-t)/n+1
you need t(n), not just t which is the entire array.
y(n+1)=(x(n)+y(n)-t(n))/n+1
  4 commentaires
shiv gaur
shiv gaur le 21 Fév 2022
it is
Image Analyst
Image Analyst le 21 Fév 2022
OK, so it's all working now that you're starting your vectors at index 1 instead of 0, and changed t to t(n)? If so, can you accept this answer because I pointed out the problems? If not, post your current code so we can continue to work on it. This is what I got. Let me know if it's the same as what you have.
x = zeros(1, 16);
y = zeros(1, 16);
x(1)=1;
% y(1)=0
t=1:15;
for n = 2: length(x) - 1
fprintf('n = %d.\n', n);
x(n+1) = (x(n-1)-y(n)) / n + 1; % Note only n is in the denominator, not (n+1)
y(n+1) = (x(n)+y(n)-t(n)) / n + 1;
end
n = 2. n = 3. n = 4. n = 5. n = 6. n = 7. n = 8. n = 9. n = 10. n = 11. n = 12. n = 13. n = 14. n = 15.
plot(t, x(1:end-1), 'b.-', 'LineWidth', 2, 'MarkerSize', 30);
hold on;
plot(t, y(1:end-1), 'r.-', 'LineWidth', 2, 'MarkerSize', 30);
grid on;
xlabel('t', 'FontSize', 20);
ylabel('x or y', 'FontSize', 20);
title('x or y VS. t', 'FontSize', 20);
legend('x', 'y')

Connectez-vous pour commenter.

Plus de réponses (0)

Catégories

En savoir plus sur MATLAB dans Help Center et File Exchange

Tags

Produits


Version

R2021b

Community Treasure Hunt

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

Start Hunting!

Translated by