secant method exceeds the number of array

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Faisal Al-Wazir
Faisal Al-Wazir 2022 年 10 月 2 日
編集済み: Torsten 2022 年 10 月 2 日
hi, i'm trying to solve this question but i don't know why i'm getting error every time i run it
here is the qustion part c
clear
clc
fun=@(x) x^3-6*x^2+11*x-6.1;
x=3.5;
c_factor=0.001;
d=@(x) 3*x^2-12*x+11;
for i=1:100
x(i+2)=x(i)-(fun(x(i))*((x(i+1)-x(i))/(fun(x(i+1))-fun(x(i)))));
e(i)=abs((x(i+1)-x(i))/x(i));
if e(i)<c_factor
break
end
end
Index exceeds the number of array elements. Index must not exceed 1.
fprintf('The root of given equation is: %.4f\n', x(i))

採用された回答

Torsten
Torsten 2022 年 10 月 2 日
編集済み: Torsten 2022 年 10 月 2 日
x(i+2)=x(i)-(fun(x(i))*((x(i+1)-x(i))/(fun(x(i+1))-fun(x(i)))));
For i = 1, the evaluation of the right-hand side needs x(1) and x(2).
You didn't define them before the loop. Set
x(1) = 2.5;
x(2) = 3.5;
instead of
x=3.5;

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