用MATLAB的迭代法求解x^3-x-1=0在x0=1.5附近的一个根?

2025-04-01 04:38:11
推荐回答(2个)
回答1:

 %牛顿迭代法 解方程y=x.^3-x-1
  x=1.5;
  format long;
  x1=x-func1_1(x)/func1_1_1(x);
  if(abs(x1)<1.5)
  delt=abs(x1-x);
  else
  delt=abs((x1-x)/x1);
  end
  while(delt>1e-6|abs(func1_1(x))>1e-6)
  x=x1; x1=x-func1_1(x)/func1_1_1(x);
  if(abs(x1)<1.5)
  delt=abs(x1-x);
  else
  delt=abs((x1-x)/x1);
  end
  if func1_1(x1)==0
  break
  end
  end

  disp('解方程y=x.^3-x-1,牛顿迭代法结果')
  x1

  编辑函数

  function y=func1_1(x)
  y=x.^3-x-1;

  function y=func1_1_1(x)
  y=3*x.^2-1

  % 对分法
  delta=10e-5;
  a=1;
  b=2;
  fa=func2_1(a);
  fb=func2_1(b);
  n=1;
  while(1)
  if(fa*fb>0)
  break;
  end
  x=(a+b)/2;
  fx=func2_1(x);
  if(abs(fx)  break;
  elseif(fa*fx<0)
  b=x;fb=fx;
  else
  a=x;fa=fx;
  end
  if(b-a  break;
  end
  n=n+1;
  end
  disp('对分法结果');
  x

  % 调用roots函数求解结果并作图
  x=linspace(-5,5);
  y=x.^2-x-1;
  plot(x,y)
  p=[1 -1 -1];
  disp('调用roots函数求解')
  x=roots(p)

  编辑函数

  function y=func2_1(x)
  y=x.^2-x-1;

  function y=func2_1_1(x)
  y=2*x-1

回答2:

x=1.5;
m=0.00000001
N=500000;
for k=1:N
y=x-(x^3-x-1)/(3*x^2-1);
if (abs(x-y))break;
end
x=y;
end
x
x^3-x-1
k
采用的是牛顿迭代法 x是解 x^3-x-1如果十分接近0证明解正确 k是迭代次数 可以通过调整m的值使精度继续提高

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