Effacer les filtres
Effacer les filtres

How do I avoid the mistake?

4 vues (au cours des 30 derniers jours)
Alisa-Oleksandra Kotliarova
Alisa-Oleksandra Kotliarova le 11 Déc 2023
I have the following function:
function [x,y,k] = Hord(fun,a,b,tolx,toly)
ya=feval(fun,a);
k=0;
while 1
k=k+1;
x=(a*fun(b)-b*fun(a))./(fun(b)-fun(a));
y=feval(fun,x);
if(x(k)-x(k-1)<=tolx)&(abs(y)<=toly), break, end
if y*ya>0; a=x; ya=y;
else b=x;
end
end
Then my code is:
clc, clear all, close all, format short
x1=linspace(-pi/2,pi/2);%Заданий інтервал значень;
y1=cos(x1+0.3)-x1.^2;%Функція 1;
figure(1), plot(x1,y1,[-pi/2,pi/2], [0,0]),grid on, hold on%Графік;
f1=(@(x1) cos(x1+0.3)-x1.^2);%Анонімне задання функції 1;
format long
[X1,Y1] = Hord(f1, 0, pi/2, 1e-3, eps);%Перший додатний корінь;
X1,Y1%Виведення кореня;
plot(X1,Y1,'o','r--')%На графіку;
It should output the graph with a root, however, it only gives:
Array indices must be positive integers or logical values.
Error in Hord (line 9)
if(x(k)-x(k-1)<=tolx)&(abs(y)<=toly), break, end
Error in Lab7_var2 (line 36)
[X1,Y1] = Hord(f1, 0, pi/2, 1e-3, eps);%Перший додатний корінь;
How can I correct it?

Réponse acceptée

Torsten
Torsten le 11 Déc 2023
Modifié(e) : Torsten le 11 Déc 2023
x in your code is a scalar, not a vector. Thus something like x(k) does not exist.
You could work with xold and x where xold would be the x-value of the previous iteration.
  1 commentaire
Alisa-Oleksandra Kotliarova
Alisa-Oleksandra Kotliarova le 13 Déc 2023
Thanks, that helped a lot!

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Image Analyst
Image Analyst le 13 Déc 2023

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