Undefined function 'abs' for input arguments of type 'function_handle'

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Meva 2015-3-15

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评论： John D'Errico 2015-3-15

Hi guys! I am getting this error when I try :

warning ('off','all');
beta1=0.2;
alphabar=-0.1;
theta4range=-1:0.1:4;
h4range=0:0.1:1;
c13range=-1:0.1:1;
% c23=-c13;
rmass=1.0;
rmomi= 2.0;
eps=0.01;
hassolution=0;
for theta4=theta4range
  for h4=h4range
      for c13=c13range
      syms x;
      H0=@(x) 1/2*(1-beta1.*x); 
      H14 =@(x)(-h4-theta4.*(x-1./2)+beta1);
      H24 =@(x)(h4 + theta4.*(x-1./2));
      u13=@(x)1./H0.* -4.* integral(H14,0,x) +c13;
      u23=@(x)1./H0.* -4.* integral(H24,0,x) -c13;
      p12 = @(x) -3.* integral(u13,0,x);
      p22 = @(x) -3.* integral(u23,0,x);
      kutta = @(x) 3.* integral(u23 - u13,0,1)+beta1.*alphabar;
      lateral = @(x) integral(p22 - p12,0,1)-12.*rmass.*h4;
      angular = @(x) integral((x-1./2).*(p22 - p12),0,1)-12.*rmomi.*theta4;
        if abs(kutta)<eps & abs(lateral)<eps & abs(angular)<eps
          c13
          c23
          theta4
          h4
          hassolution=1;
        end
      end
    end
  end
  if ~hassolution
    disp('No solution');
  end

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Answer 1

John D'Errico 2015-3-15

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https://ww2.mathworks.cn/matlabcentral/answers/183261-undefined-function-abs-for-input-arguments-of-type-function_handle#answer_171414

编辑：John D'Errico 2015-3-15

It is clearly time for you to learn about functions and how to call them. Start reading the basic tutorials.

Anyway, READ the error message. What is kutta? For that matter, what is lateral? And what is angular?

Answer: They are all function handles. They are NOT numbers. What is the absolute value of a number? That I can do. What is the absolute value of a function handle? Meaningless.

There is a difference between the function handle kutta as you have defined it, and the VALUE of that function on some input like kutta(x), where x is a variable.

By the way, why do you think you need to define the variable x as a symbolic one BEFORE you defined those function handles? There is no need to do so. That line of code was just a waste of CPU cycles.

What I don't know is if you actually want to do symbolic computation on those expressions, or if you want to do numerical computations. If your goal is symbolic computation, then all of those function handles were the wrong thing to do.

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Meva 2015-3-15

编辑：Meva 2015-3-15

在 MATLAB Online 中打开

well I have a code from my colleaque that is why I have defined x as a symbolic variable and all the other functions. His code is such that :

% script for solving non-linear equations using brute force approach
warning ('off','all');
theta0range=0:0.1:4;
h0range=0:0.1:1;
c1range=0:0.1:4;
eps=0.01;
hassolution=0;
for theta0=theta0range
  for h0=h0range
    for c1=c1range
      firstexp=c1^2-2*c1^2*h0-c1^2*theta0-h0^2-h0*theta0-theta0^2/4+2*c1*h0^2+2*h0*c1*theta0+c1*theta0^2/2;
      syms x;
      fun1=@(x)(1-c1).^2./(1-h0-(x-1./2).*theta0-0.1.*sin(pi.*x)).^2;
      fun2=@(x)c1.^2./(h0+(x-1./2).*theta0-0.1.*sin(pi.*x)).^2;
      fun3=@(x)(x-1./2).*((1-c1).^2./(1-h0-(x-1./2).*theta0-0.1.*sin(pi.*x)).^2-c1.^2./(h0+(x-1./2).*theta0-0.1.*sin(pi.*x)).^2);
      secondexp=0.5*integral(fun1,0,1)-0.5*integral(fun2,0,1);
      thirdexp=0.5*integral(fun3,0,1);
      if abs(firstexp)<eps & abs(secondexp)<eps & abs(thirdexp)<eps
        theta0
        h0
        c1
        hassolution=1;
      end
    end
  end
end
if ~hassolution
  disp('No solution');
end

As far as I know, this code works and I modified this code to my problem. Any suggestion?

My x is varied from 0 to 1 with the increment of 0.01.

Jan 2015-3-15

Warnings are extremely useful when running Matlab. Therefore warning ('off','all') is a very bad idea.

John D'Errico 2015-3-15

Jan makes a superb point. Warnings are there for a good reason. Disabling them is the equivalent to closing your eyes and your ears, and pretending that nothing is wrong. I see nothing. I hear nothing...

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Undefined function 'abs' for input arguments of type 'function_handle'

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Undefined function 'abs' for input arguments of type 'function_handle'

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