a multivariate quadratic function
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I'm not asking for a full code, but a way that makes sense.
find a multivariate equation that solves, please no codes
%want to solve using a 2x2 matrix.
f(0,1)=6, f(1,0)=6, f(1,1)=3, f(-1,-1)=7, f(1,2)=2. f(2,1)-=6
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Alan Stevens
2021-2-4
编辑：Alan Stevens
2021-2-4
Here's one way. It uses an initial guess for each element of P, q and r, then combined them into a single column vector which is used by fminsearch.
f = @(x,P,q,r) x'*P*x + q'*x + r;
% initial guesses (only three elements guessed for P as it's a symmetric matrix)
P0 =[1; 0; 1];
q0 = [1; 1];
r0 = 1;
Pqr0 =[P0(:); q0(:); r0]; % initial guesses combined into a single column vector
Pqr = fminsearch(@(Pqr) Pqrfn(Pqr,f), Pqr0); % fminsearch tries to find values of Pqr that minimise
% the value returned from Pqrfn
% reconstruct P, q and r
P = [Pqr(1) Pqr(2); Pqr(2) Pqr(3)];
q = Pqr(4:5);
r = Pqr(6);
% display results
disp(P)
disp(q)
disp(r)
% Check results
d(1) = f([0;1],P,q,r);
d(2) = f([1;0],P,q,r);
d(3) = f([1;1],P,q,r);
d(4) = f([-1;-1],P,q,r);
d(5) = f([1;2],P,q,r);
d(6) = f([2;1],P,q,r);
disp(d)
function F = Pqrfn(Pqr,f)
% construct P, q and r from Pqr
P = [Pqr(1) Pqr(2); Pqr(2) Pqr(3)];
q = Pqr(4:5);
r = Pqr(6);
% Calculate differences of function values from desired values
d(1) = f([0;1],P,q,r) - 6;
d(2) = f([1;0],P,q,r) - 6;
d(3) = f([1;1],P,q,r) - 3;
d(4) = f([-1;-1],P,q,r) - 7;
d(5) = f([1;2],P,q,r) - 2;
d(6) = f([2;1],P,q,r) - 6;
% return the norm of the differences
F = norm(d);
end
If you do
format long
before displaying the results, you will see that they are not exact, but are good to four decimal places.
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