matlab - MATLAB ERROR Feval需要一个函数句柄作为第一个参数

Question

我有此代码(*)，当我这样做时:

»syms x
»newton_raphson({((5400.*(1 + x)*0^360) - (1450000.*x.*(1 + x).^360))}, diff(((5400.*(1 + x)*0^360) - (1450000.*x.*(1 + x).^360)),1), 0.001, eps, 5, 0.1)

出现此错误:

Error using feval Argument must contain a string or function_handle.

Error in newton_raphson (line 10) fz = feval(f,z(1));

如何解决此错误？

(*)

 function [raiz, zn, fz, i] = newton_raphson(f, flinha, x0, eps, iter_max, debug)
   if nargin < 4
      eps = 1e-6;
   end
   if nargin < 5
      iter_max = 1e3;
   end

   z(1) = x0;
   fz = feval(f,z(1));
   fzlinha = feval(flinha,z(1));
   if (nargin > 5 && debug > 0)
      fprintf(2,'i=%d z=%23.18G fz=%G fzlinha=%G\n',0,x0,fz,fzlinha);
   end
   for i = 1:iter_max
      if abs(fzlinha) == 0 % f'(x0) equal zero 
         disp('O valor da derivada em Xi não pode ser zero');
         z(i+1) = x0;
         return
      end
      z(i+1) = x0 - fz / fzlinha;
      fz = feval(f,z(i+1));
      fzlinha = feval(flinha,z(i+1));
      dif = abs(z(i+1) - x0);
      if (nargin > 5 && debug > 0)
         fprintf(2,'i=%d z=%23.18G fz=%G fzlinha=%G dif=%E\n',i,z(i+1),fz,fzlinha,dif);
      end
      if dif < eps
         break;
      elseif i == iter_max
         disp('Foi excedido o número máximo de iterações (iter_max)'); 
         break
      end 
      x0=z(i+1);
   end
   zn = z';
   raiz = z(i+1);
end`

Answer 1

您正在将symbolic expression传递给旨在评估anonymous function，function handle的函数，或通过name indicated by a string将 feval 传递给Matlab路径上的函数。如果您希望Matlab为您做区分，您可以先使用符号表达式，然后通过 matlabFunction 将它们转换为匿名函数，如下所示

syms x f Df
%
% Symbolic expressions
f  = (5400.*(1 + x)*0^360) - (1450000.*x.*(1 + x).^360); 
Df = diff(f,x); 
%
%  Convert to anonymous functions
f  = matlabFunction(f ,'Vars',x); 
Df = matlabFunction(Df,'Vars',x);