fzero problem when doing integration using quadl.

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Lin LI 2012-1-6

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https://ww2.mathworks.cn/matlabcentral/answers/25482-fzero-problem-when-doing-integration-using-quadl

1. I am trying to do an integration using quadl, so I need to define an inline function first. But this inline function is somewhat complex. I need to use the function fzero to define it and this causes a lot of errors. I dont know how to fix it.

2. If there is no fzero, it would be very easy, I have to use fzero to find the value of cc and then to do integration. If anyone can help, thanks a lot.

 dd=quadl(@kk,1,5);
function kk=kk(y)
aa=y.^2;
bb=@(w) w^5-2*w-4-aa;
cc=fzero(bb,3);
kk=cc+1;

The error is the following:

??? Operands to the and && operators must be convertible to logical scalar values.

Error in ==> fzero at 333 elseif ~isfinite(fx) ~isreal(fx)

Error in ==> kk at 4 cc=fzero(bb,3);

Error in ==> quadl at 70 y = feval(f,x,varargin{:}); y = y(:).';

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采纳的回答

Andrew Newell 2012-1-6

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在 MATLAB Online 中打开

You get those error messages because quadl expects a function that can input a vector but fzero can only solve for a scalar input. Replace fzero by fsolve.

EDIT: You also need to change the initial guess for fsolve to a vector:

function kk=kk(y)
bb=@(w) w.^5-2*w-4-y.^2;
cc=fsolve(bb,3*ones(size(y)));
kk=cc+1;

But don't expect the integration to work then. You're finding one of the five roots of a quintic equation, and fsolve could jump discontinuously as you change y.^2. Getting the correct answer may require some careful mathematics.

EDIT 2: If you run the following code

y = 1:.05:5;
polyRoots = zeros(5,length(y));
for ii=1:length(y)
  polyRoots(:,ii) = roots([1,0,0,0,-2,-4-y(ii).^2]);
end
polyRoots(abs(imag(polyRoots))>10*eps)=NaN;
plot(y,polyRoots,'.')