二分法、不动点迭代、牛顿法求根实验

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问题：

结果：

解\方法	二分法	不动点迭代	牛顿迭代
近似解	1.3688	1.9332	1.3688
迭代次数	14	10000	3

一、二分法

matlab代码

clear all;clc
f=@(x)x^3+2*x^2+10*x-20;
a=0;
b=2;
if(f(a)*f(b)<0)
  a0=a;
      b0=b;
      else
      error('a,b处同号')
end

iter = 0;
eps=0.0001;

while((b0-a0)/2)>=eps
     iter=iter+1;
     c=(a0+b0)/2;
     if f(c)==0
        solution=c;
        break
     elseif f(c)*f(a0)<0
         b0=c;
     else
         a0=c;
     end
end
[c iter]

python代码

def f(x):
    return x**3+2*x**2+10*x-20
a = 0
b = 2
if f(a)*f(b) < 0:
    a0 = a
    b0 = b
else:
    print('a,b同号')

iter = 0
eps=0.0001

while((b0-a0)/2)>=eps:
     iter=iter+1
     c=(a0+b0)/2
     if f(c)==0:
        solution=c
        break
     elif f(c)*f(a0)<0:
         b0=c
     else:
         a0=c

print(c,iter)

二、不动点迭代

matlab代码

clear all;clc
iter = 0;
eps=0.0001;
x0=1.2;
N=10000
% phi=@(x)(-2*x^2-10*x+20)^(1/3); 
% phi=@(x)((-x^3-10*x+20)*0.5)^(1/2); 
phi=@(x)((-x^3-2*x^2+20)/10); 
x1=phi(x0)
while abs(x0-x1)>=eps & iter<N
    x0=x1
    x1=phi(x0);
    iter=iter+1;
end
if iter==N
      warning('迭代次数达到允许上限')
end
format short g
[x1 iter]

这个构造的phi()好像不太收敛
python代码

def phi(x):
    return (20-x**3-2*x**2)/10

#def phi(x):
#    return ((-x**3-10*x+20)*0.5)**(0.5)
# phi=@(x)((-x^3-10*x+20)*0.5)^(1/2);

eps = 1e-4
N=10000
x0=1.2
iter=0
x1=phi(x0)

while abs(x0-x1)>=eps and iter<N:
    iter+=1
    x0=x1
    x1=phi(x0)
    if iter==N:
        print('迭代次数达到允许上限')
        break

print(x1,iter)

三、牛顿法

matlab代码

clear all;clc
iter = 0;
eps=0.0001;
x0=2;

df=@(x)3*x^2+4*x+10;
f=@(x)x^3+2*x^2+10*x-20;

while abs(x0-(x0-f(x0)/df(x0)))>=eps 
   x0=(x0-f(x0)/df(x0));
   iter=iter+1;
end
[x0 iter]

python代码

def df(x):
    return 3*x**2+4*x+10

def f(x):
    return x**3+2*x**2+10*x-20

iter = 0
eps=0.0001
x0=2

while abs(x0-(x0-f(x0)/df(x0)))>=eps:
   x0=(x0-f(x0)/df(x0))
   iter=iter+1

print(x0,iter)