# Is it possible to solve this equation without symbolics?

3 views (last 30 days)
Amit Kumar on 12 Mar 2014
Commented: Star Strider on 14 Mar 2014
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.

Star Strider on 13 Mar 2014
Edited: Star Strider on 13 Mar 2014
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.
Star Strider on 14 Mar 2014
My pleasure!

Roger Stafford on 13 Mar 2014
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);

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