Is it possible to solve this equation without symbolics?

Hi all, I am doing research involving numerical computations. Obviously symbolics are expensive to compute. So I am trying to compute something without using symbolics. Here is my sample code:

F1=1725;
F2=228;
F6=76;
syms F0;
s_12=[-0.875*F0;-0.625*F0; 0.21651*F0];
temp1=((s_12(1,1))^2/(F1)^2)+((s_12(2,1))^2/(F2)^2)+((s_12(3,1))^2/(F6)^2)-((s_12(1,1)*(s_12(2,1)))/(F1)^2);
a=double(solve(temp1==1,F0));

Can anyone suggest me a way to solve this equation without using symbolics? Thanks in advance.

Answer 1

[Star Strider]

Another method, using the equations you provided:

F1=1725;
F2=228;
F6=76;
s_12= @(F0) [-0.875*F0;-0.625*F0; 0.21651*F0];
temp1= @(s_12) ((s_12(1,1))^2/(F1)^2)+((s_12(2,1))^2/(F2)^2)+((s_12(3,1))^2/(F6)^2)-((s_12(1,1)*(s_12(2,1)))/(F1)^2);

fcn = @(F0) temp1(s_12(F0))-1;

a = [];
for k1 = [-500 500]
    S0 = fzero(fcn,k1);
    a = [S0; a];
end

The for loop uses two different starting values to get both roots, stored in vector a.

[Amit Kumar]

Thanks. I am looking for a generalized method and solution you provided can work.

[Amit Kumar]

In the last for loop, why are you storing a in vector a? a=[S0;a]. Can you explain this part?

[Star Strider]

I solved it for S0 and saved all the results in vector a because there are at least two solutions. In this case, I chose two starting points (-500, +500) and saved all (in this situation, two) solutions in array a. It’s just easier that way, since fzero returns one root for each starting point (at least in my experience). I chose to return the results in a because it matches your code.

[Amit Kumar]

Thanks a lot for a wonderful solution!!

[Star Strider]

My pleasure!

Answer 2

[Roger Stafford]

 K = (-0.875/F1)^2+(-0.625/F2)^2+(0.21651/F6)^2-(-0.875)*(-0.625)/(F1)^2;
 F0 = sqrt(1/K);

or (Two solutions)

 F0 = -sqrt(1/K);