Undefined operand in function_handle

When I enter ' H(3) ' in the command line, the following error is shown:

Undefined operator '-' for input arguments of type 'function_handle'.

Error in

Trial_H>@(m)d1+s*(cos(gama)+sum(cos(gama+(pi/2-phi_i))))+((2*r)-omega)*sin(B_j(m))+(r-omega)*sum(sin((1:m-1)*pi-sum(B_j(1:m-1))))

Here is the code:

% Variables
L = 10;
R = 1;
r = R/(m+1);
S = sqrt(2);
s = S/(m+1);
d1 = 1;
d2 = 0.25;
omega = R/(3*(m+1));
gama = atan(R/d1);

% Functions
phi_i = @(ii, m) atan( (d1+(ii*d2)./(m+1) )./(r/3) );

B_j = @(ii, m) 2*phi_i(ii, m);

H = @(m) d1 + s*( cos(gama) + sum(cos(gama+(pi/2-phi_i(1:m)))) ) + ( (2*r)-omega)*sin(B_j(m) ) + (r-omega)*sum(sin((1:m-1)*pi - sum(B_j(1:m-1))));

Answer 1

[Tommy]

Where you call phi_i or B_j within the definition of H, you need to supply the arguments ii and m. For example,

..(pi/2-phi_i)..

is problematic because MATLAB thinks you are trying to subtract the function phi_i from pi/2. If you pass arguments,

..(pi/2-phi_i(1,m))..

for example, MATLAB will instead subtract the output of phi_i from pi/2.

[JAYEN PILLAY]

I input avalue for the variable m.

I also want to substract phi from pi/2

[Tommy]

Run this from the command window:

>> B_j(3)
Not enough input arguments.

Error in answers>@(ii,m)2*phi_i(ii,m)

The function B_j takes two inputs, ii and m, and calls phi_i(ii, m). If you call B_j(3), you are only giving B_j the first input, ii. It does not know what m is, and when it tries to call phi_i, it cannot. You need to pass both arguments into B_j:

>> B_j(1,3)

ans =

    2.9851

Now run this from the command window:

>> 2+phi_i
Undefined operator '+' for input arguments of type
'function_handle'.

phi_i is a function. You cannot add or subtract anything to/from it. You can, however, do this:

>> 2+phi_i(1,3)

ans =

    3.4925
    

Fill all of your function calls with both arguments and you shouldn't get any errors.

[JAYEN PILLAY]

It's there

B_j = @(ii, m) 2*phi_i(ii, m)

and for the others too

[Tommy]

That is where you define B_j. You call B_j twice within H() (and you call phi_i once):

H = @(m) d1 + s*( cos(gama) + sum(cos(gama+(pi/2- ...
    phi_i ... HERE
    ))) ) + ( (2*r)-omega)*sin( ...
    B_j(m) ... HERE
    ) + (r-omega)*sum(sin((1:m-1)*pi - sum( ...
    B_j(1:m-1) ... HERE
    )));

All three of those lines need to be changed, as described above.

Something like this would work:

H = @(m) d1 + s*( cos(gama) + sum(cos(gama+(pi/2- ...
    phi_i(1,m) ... HERE
    ))) ) + ( (2*r)-omega)*sin( ...
    B_j(1,m) ... HERE
    ) + (r-omega)*sum(sin((1:m-1)*pi - sum( ...
    B_j(1,1:m-1) ... HERE
    )));

I have used 1 for ii in all three lines, but I do not know if that is the value for ii that you want to use.

[JAYEN PILLAY]

Yeah exactly I am using ii = 1. But, for the first B_j I'm using 'm' only

[Tommy]

You cannot use m only, because your definition of B_j

B_j = @(ii, m) 2*phi_i(ii, m)

requires both ii and m.

[Steven Lord]

Calling B_j, which requires two inputs, with just one would be like me asking you "The first number is 7. What is the sum of the two numbers?" There's not enough information to answer the question.