How do I fix the 'Invalid use of operator' Error in my anonymous function bisect code?

function [root,fx,ea,iter]=bisect( @(x)x^7+3*x-1;0;1;0;3 )
% bisect: root location zeroes
% [root,fx,ea,iter]=bisect(func,xl,xu,es,maxit,p1,p2,...):
% uses bisection method to find the root of func
% input:
% func = name of function
% xl, xu = lower and upper guesses
% es = desired relative error (default = 0.0001%)
% maxit = maximum allowable iterations (default = 50)
% p1,p2,... = additional parameters used by func
% output:
% root = real root
% fx = function value at root
% ea = approximate relative error (%)
% iter = number of iterations
if nargin<3,error('at least 3 input arguments required'),end
test = func(xl)*func(xu);
if test>0,error('no sign change'),end
if nargin<4|isempty(es), es=0.0000001;end
if nargin<5|isempty(maxit), maxit=3;end
iter = 0; xr = xl; ea = 100;
while (1)
 xrold = xr;
 xr = (xl + xu)/2;
 iter = iter + 1;
 if xr ~= 0,ea = abs((xr - xrold)/xr) * 100;end
 test = func(xl)*func(xr);
 if test < 0
 xu = xr;
 elseif test > 0
 xl = xr;
 else
 ea = 0;
 end
 if ea <= es | iter >= maxit,break,end
end
root = xr; fx = func(xr);

Then it says 
"Error: ... Line: 1 Column: 36
Invalid use of operator." while underlining the '@' and last parantheses ')' for bisect in red. I'm not sure why

Answer 1

[Alan Stevens]

Your syntax isn't correct. Try

f = @(x)x^7+3*x-1;
xl = 0; xu = 1; es = 10^-4; maxit = 30;

[root,fx,ea,iter]=bisect(f ,xl,xu,es,maxit );
disp(root)

function [root,fx,ea,iter]=bisect(f ,xl,xu,es,maxit )
% bisect: root location zeroes
% [root,fx,ea,iter]=bisect(func,xl,xu,es,maxit,p1,p2,...):
% uses bisection method to find the root of func
% input:
% func = name of function
% xl, xu = lower and upper guesses
% es = desired relative error (default = 0.0001%)
% maxit = maximum allowable iterations (default = 50)
% p1,p2,... = additional parameters used by func
% output:
% root = real root
% fx = function value at root
% ea = approximate relative error (%)
% iter = number of iterations
if nargin<3,error('at least 3 input arguments required'),end
test = f(xl)*f(xu);
if test>0,error('no sign change'),end
if nargin<4||isempty(es), es=0.0000001;end
if nargin<5||isempty(maxit), maxit=3;end
iter = 0; xr = xl; ea = 100;
while (1)
 xrold = xr;
 xr = (xl + xu)/2;
 iter = iter + 1;
 if xr ~= 0,ea = abs((xr - xrold)/xr) * 100;end
 test = f(xl)*f(xr);
 if test < 0
 xu = xr;
 elseif test > 0
 xl = xr;
 else
 ea = 0;
 end
 if ea <= es || iter >= maxit,break,end
end
root = xr; fx = f(xr);
end

[Ymbar Polanco]

Thank you!

Answer 2

[Steven Lord]

function [root,fx,ea,iter]=bisect( @(x)x^7+3*x-1;0;1;0;3 )

The values @(x)x^7+3*x-1, 0, 1, 0, and 3 are the values you should pass into bisect when you call bisect.

They are not the input argument names that you should specify when you define bisect. Those names need to be valid variable names to which MATLAB will assign the values with which the user of this function calls the function.

As you've written it, if MATLAB could somehow get past that invalid definition, when it reached this line:

test = func(xl)*func(xu);

the variable func would not exist.

If instead you had defined bisect as:

function [root, fx, ea, iter] = bisect(func, xl,xu,es,maxit,varargin)

then calling it like:

[root, fx, ea, iter] = bisect(@(x) x^7+3*x-1, 0, 1, 0, 3)

would define func as the anonymous function inside the workspace in which bisect operates, xl to be 0, etc.

[Ymbar Polanco]

Thank you, this was very helpful!