Q3: Define the linear functional J: H(2) R by 1(v) = a(v. v) - L(v) Let u be the unique weak solution to a(u,v) = L(v) in H() and suppose that a(...) is a symmetric bilinear form on H(2) prove that 1- u is minimizer. 2- u is unique. 3- The minimizer J(u,) can be rewritten under algebraic form u Au-ub. J(u)=u'Au- Where A. b are repictively the stiffence matrix and the load vector

Compute the discriminant D(x, y) of the function. f(x, y) = x³ + y² − 6x − 2y² + 1 (Express numbers in exact form. Use symbolic notation and fractions where needed.) D(x, y): = Which of these points are saddle points? (-√2, 1) (-√2,0) (√2, -1) (√2,0) (-√2, -1) (√2, 1) Which of these points are local minima? (-√√2, -1) (√2, -1) (√2, 1) (-√2,0) (-√2, 1) (√2,0)

Fit a quadratic function of the form f(t) = co + cit + czť² to the data points (0, 1), (1, –7), (2,7), (3,3), using least squares. f(t) =

Let q = - 2xy - y² + 2xz+2yz + z² be a quadratic form on R³ viewed as a polynomial in 3 variables. Find a linear change of variables tou, v, w that puts q into the canonical form in `Sylvester's law of inertia'!. What are the values of the associated indices s, t? Select one: O We let u = √3(x − y + 2), v = x+y, w = (x - y)/2 to find that q = u²+². Hence s = 2, t = 0 in Sylvester's law of intertia. O We let u = x - 2y, v = z+y, w = √3y to find that q=u²v² w². Hence s = 1,t = 2 in Sylvester's law of intertia. O we let u = x, v= √2(x+y), w = x+y+z to find that q = u²v² + w²2. Hence s = 2, t = 1 in Sylvester's law of intertia. O None of the others apply O The quadratic form does not obey the condition to be diagonalisable over R. This is because the minimal polynomial of the corresponding matrix is not a product of distinct linear factors. Hence Sylvester's law of inertia does not apply. By convention, we set s = t = ∞ when this happens.