Vector function is only returning a scalar output

I have the function:

function [ theta4 ] = ThetaFour( L, theta2 )

%given equations, finding polynomials for the quad formula
k1 = (L(1)/L(2));
k2 = (L(1)/L(4));
k3 = ((L(1)^2) + (L(2)^2) - (L(3)^2) + (L(4))^2) / (2*L(2)*L(4));

a = cosd(theta2) - k1 -k2*cosd(theta2) + k3;
b = -2*sind(theta2);
c = k1 - (k2 + 1) * cosd(theta2) + k3;

%finding final angle using the quadratic function function created for
%projectile motion script
theta4 = 2*atand(Quadratic(a,b,c,-1))

end

• L is a vector [0.1313, 0.0475, 0.0880, 0.0960]
• theta 2 is a vector [0:90]

During this process: a, b, and c all become vectors of length 90

My Quadratic function is:

function Roots = Quadratic( a, b, c, plusOrMinus )
%Uses the quadratic formula to solve for roots

if ((b .^ 2)-(4 .* a .* c) >= 0)
    if (plusOrMinus == 1)
        Roots = (-b + sqrt( (b .^2 - 4.* a.* c) ) ) / (2 .* a);
    else
        Roots = (-b - sqrt( (b .^2 - 4.* a.* c) ) ) / (2 .* a);
    end
else
    error('Coefficiants yield a complex root. b^2 - 4ac must be 0 or greater.')

end

a, b, c should be going in as vectors, while plusOrMinus is -1, returning a vector

Changing a, b, or c to a scalar produces a vector, but all three as vectors produces a scalar.

For example, Quadratic(-.5, b, c, -1) returns a full vector as does Quadratic(a, 10, c, -1) while Quadratic(a, b, c, -1) is just returning one value.

I would like to know how to get Quadratic(a, b, c, -1) to return a vector. Thank you in advance.

[Matt J]

If a, b, and c are vectors then so is the discriminant (b .^ 2)-(4 .* a .* c). Currently your function returns an error message unless all the discriminants are >0. Is this what you want?

[Adam Vanek]

That is correct. In the case above, a,b,c are all positive numbers, so no error message occurs.

[Matt J]

That is not sufficient for a no-error condition. If a=b=c=ones(N,1) an error would occur.

Answer 1

[Matt J]

function Roots = Quadratic( a, b, c, plusOrMinus )

    Roots = (-b + plusOrMinus.*sqrt( (b .^2 - 4.* a.* c) ) ) ./ (2 .* a);

     Roots(imag(Roots)~=0)=nan;

[Guillaume]

And to explain why that works when the original code didn't

The problem is all down to the if. If a, b or c is a vector, then the test expression in the if, b.^2 - 4.*a.*c >= 0 is a logical vector.

if, when given a vector, only executes the true branch if all the elements of the vector are true. The statement is only executed only once. if does not loop over the elements of the vector, choosing a branch for each. If you want to do that you have to do the looping yourself.

What Matt does is avoid the if entirely.

[Matt J]

The problem is all down to the if.

No, the reason that a scalar was returned is that the element-wise division operator './' was not being used. Although, I do wonder if the OP really wants a vector argument to if, see see my Comment above.

[Adam Vanek]

Thank you Matt, I am glad you both answered my question as well as expounded upon the arguments in my if statement.