Conversion to function_handle from double is not possible.

f=@(x) ((1/3)*(x-2)^2)-3

function [r,k] = RegulaFalsi_Mod(f,a,b,kmax,tol)

if nargin <5

isempty(tol), tol = 1e-4;

end

if nargin<4

isempty(kmax), kmax = 20;

end

c=zeros(1,kmax);

if f(a)*f(b)>0

r = 'failure';

return

end

disp (' k a b')

for k = 1:kmax

c(k) = (a*f(b) - b*f(a))/(f(b)-f(a));

if f(c(k)) == 0

return

end

for k = 1:3

if f(3) == f(k)

f(a) = (1/k)*f(a);

elseif f(3) ~= f(k)

continue

for k=1:6

if c(6) == c(k)

f(a) = (1/(2*k))*f(a);

elseif c(6)~=c(k)

continue

end

end

end

end

end

fprintf('%2i, %11.6f, %11.6f\n',k,a,b)

if f(b)*f(c(k))>0

b=c(k);

else a = c(k);

end

c(k+1) = (a*f(b)-b*f(a))/(f(b)-f(a));

if abs(c(k+1)-c(k)) < tol

r = c(k+1);

return

end

end

Error Message

Conversion to function_handle from double is not possible.

Error in RegulaFalsi_Mod (line 21)

f(a) = (1/k)*f(a);

[darova]

Maybe it is becase your function handle is not in the function body? Why it is outside?

[John D'Errico]

f is passed in as an argument. That is not the issue.

Answer 1

[John D'Errico]

LOOK AT THE ERROR MESSAGE. Think about what it tells you.

Error in RegulaFalsi_Mod (line 21)
        f(a) = (1/k)*f(a);

What re you trying to do there? f is a function handle! The right hand side of that expression is a double precision number.

You cannot assign something to an existing function handle as you tried to do.

Essentially, you cannot use f as both a function handle (which it is) and a variable where you will store information. And worse, even if you could do that, you cannot use a floating point number as an index into a vector or any variable. You CANNOT assign something to f(a).

And that is the gist of your problem.