function [ A ] = cal( a,b,v )%a,b表示區間，v是精度

i=1;

x = (a+b)/2;

A=[i x];

t = x-(x^3-x-1)/(3*x^2-1);%迭代函式

while(abs(t-x)>v)

i=i+1;

x = t;

A = [A;i x];

t = x-(x^3-x-1)/(3*x^2-1);%迭代函式

end

A = [A;i+1 t];

end

執行結果：

>> format long;

>> cal(1,2,0.00001)

ans =

1.000000000000000 1.500000000000000

2.000000000000000 1.347826086956522

3.000000000000000 1.325200398950907

4.000000000000000 1.324718173999054

5.000000000000000 1.324717957244790