我编的程序是：

NewtonIterate.m

function s=NewtonIterate(x,eps)
  % Newton迭代法求非线性方程组
  % x为迭代初值，eps为允许误差值
  if nargin==1
        eps=1.0e-6;
  elseif nargin<1
        error
        return
  end
  x1=fx1(x); %非线性方程组
  x2=-dfx1(x); %非线性方程组导数
  x3=pinv(x2);
  x0=x3*x1';
  while norm(x0)>=eps %循环迭代
      x=x0'+x;
      x1=fx1(x);
      x2=-dfx1(x);
      x3=pinv(x2);
      x0=x3*x1';
  end
  s=x0'+x;
  return
     

fx1.m

function y=fx1(x)
  syms t
  E=714711.9864;
  D=955030;
  s=0.32531;
  e=2.71828;
  y(1)=[x(1)/sqrt(2*pi)]*int(e^(-0.5*t^2),-inf,[log(x(1)/D)+r+0.5*x(2)*x(2)]/x(2))-D*e^(-r)/sqrt(2*pi)*int(e^(-0.5*t^2),-inf,[log(x(1)/D)+r-0.5*x(2)*x(2)]/x(2))-E
  y(2)=x(1)*x(2)*int(e^(-0.5*t^2),-inf,[log(x(1)/D)+r+0.5*x(2)*x(2)]/x(2))-Es

dfx1.m

function y=dfx1(x)
   syms t
   e=2.71828;
   E=714711.9864;
   s=0.32531;
   y(1)=[1/sqrt(2*pi)]*int(e^(-0.5*t^2),-inf,[log(x(1)/D)+r+0.5*x(2)*x(2)]/x(2))+[1/sqrt(2*pi)]*[1/(x(1)*x(2))]*e^{-0.5*[log(x(1)/D)+r+0.5*x(2)*x(2)]^2/(x(2)*x(2))}-D*e^(-r)*{1/(x(1)*x(2))}*e^{-0.5*[log(x(1)/D)+r-0.5*x(2)*x(2)]^2/(x(2)*x(2))}*[1/sqrt(2*pi)];
   y(2)=[x(1)/sqrt(2*pi)]*[-(log(x(1)/D)+r)/(x(2)*x(2))+0.5]*e^{-0.5*[log(x(1)/D)+r+0.5*x(2)*x(2)]^2/(x(2)*x(2))}-D*e^(-r)*[1/sqrt(2*pi)]*[-(log(x(1)/D)+r)/(x(2)*x(2))-0.5]*e^{-0.5*[log(x(1)/D)+r-0.5*x(2)*x(2)]^2/(x(2)*x(2))};
   y(3)=x(2)*int(e^(-0.5*t^2),-inf,[log(x(1)/D)+r+0.5*x(2)*x(2)]/x(2))+x(1)*x(2)*[1/sqrt(2*pi)]*[1/(x(1)*x(2))]*e^{-0.5*[log(x(1)/D)+r+0.5*x(2)*x(2)]^2/(x(2)*x(2))};
   y(4)=x(1)*[1/sqrt(2*pi)]*int(e^(-0.5*t^2),-inf,[log(x(1)/D)+r+0.5*x(2)*x(2)]/x(2))+x(1)*x(2)*[1/sqrt(2*pi)]*[-(log(x(1)/D)+r)/x(2)*x(2)+0.5]*e^{-0.5*[log(x(1)/D)+r+0.5*x(2)*x(2)]^2/(x(2)*x(2))};
   y=[y(1)*y(2);y(3)*y(4)];

请高手指出错误，不吝赐教！非常感谢！

可能是此方程组较难解吧，怎么没其他人回复呢？

fx1.m中r没有定义啊

谢谢你，tabuce。

我修改后，程序如下：

fx1.m

function y=fx1(x)
  syms t
    e=2.71828;
    E=714711.9864;
    s=0.32531;
    r=0.02;
    D=955030;
  y(1)=[x(1)/sqrt(2*pi)]*int(e^(-0.5*t^2),-inf,[log(x(1)/D)+r+0.5*x(2)*x(2)]/x(2))-D*e^(-r)/sqrt(2*pi)*int(e^(-0.5*t^2),-inf,[log(x(1)/D)+r-0.5*x(2)*x(2)]/x(2))-E
  y(2)=x(1)*x(2)*int(e^(-0.5*t^2),-inf,[log(x(1)/D)+r+0.5*x(2)*x(2)]/x(2))-Es

dfx1.m

function y=dfx1(x)
   syms t
    e=2.71828;
    E=714711.9864;
    s=0.32531;
    r=0.02;
    D=955030;
   y(1)=[1/sqrt(2*pi)]*int(e^(-0.5*t^2),-inf,[log(x(1)/D)+r+0.5*x(2)*x(2)]/x(2))+[1/sqrt(2*pi)]*[1/(x(1)*x(2))]*e^{-0.5*[log(x(1)/D)+r+0.5*x(2)*x(2)]^2/(x(2)*x(2))}-D*e^(-r)*{1/(x(1)*x(2))}*e^{-0.5*[log(x(1)/D)+r-0.5*x(2)*x(2)]^2/(x(2)*x(2))}*[1/sqrt(2*pi)];
   y(2)=[x(1)/sqrt(2*pi)]*[-(log(x(1)/D)+r)/(x(2)*x(2))+0.5]*e^{-0.5*[log(x(1)/D)+r+0.5*x(2)*x(2)]^2/(x(2)*x(2))}-D*e^(-r)*[1/sqrt(2*pi)]*[-(log(x(1)/D)+r)/(x(2)*x(2))-0.5]*e^{-0.5*[log(x(1)/D)+r-0.5*x(2)*x(2)]^2/(x(2)*x(2))};
   y(3)=x(2)*int(e^(-0.5*t^2),-inf,[log(x(1)/D)+r+0.5*x(2)*x(2)]/x(2))+x(1)*x(2)*[1/sqrt(2*pi)]*[1/(x(1)*x(2))]*e^{-0.5*[log(x(1)/D)+r+0.5*x(2)*x(2)]^2/(x(2)*x(2))};
   y(4)=x(1)*[1/sqrt(2*pi)]*int(e^(-0.5*t^2),-inf,[log(x(1)/D)+r+0.5*x(2)*x(2)]/x(2))+x(1)*x(2)*[1/sqrt(2*pi)]*[-(log(x(1)/D)+r)/x(2)*x(2)+0.5]*e^{-0.5*[log(x(1)/D)+r+0.5*x(2)*x(2)]^2/(x(2)*x(2))};
   y=[y(1)*y(2);y(3)*y(4)];

运行NewtonIterate([0 0])，结果是：
error_msg : socket 4280 listen failed (err=10013).

??? Maple initialization failed.

Error in ==> mapleinit at 47
maplemex(maplelib,0);

Error in ==> maple at 104
[result,status] = maplemex(statement);

Error in ==> sym.maple at 85
[result,status] = maple(statement);

Error in ==> sym.mpower at 17
   B = maple(A,'^',p);

Error in ==> fx1 at 8
  y(1)=[x(1)/sqrt(2*pi)]*int(e^(-0.5*t^2),-inf,[log(x(1)/D)+r+0.5*x(2)*x(2)]/x(2))-D*e^(-r)/sqrt(2*pi)*int(e^(-0.5*t^2),-inf,[log(x(1)/D)+r-0.5*x(2)*x(2)]/x(2))-E

Error in ==> NewtonIterate at 10
  x1=fx1(x); %非线性方程组

还是不行，请不吝赐教，多谢！

怎么没有高手来回答呢？

期望中。。。。