Question 1. Consider the average cost function (Q) = e-Q+Q2. Show that the cost function is strictly convex. Find the minimum of the average cost c: find the value of Q which minimizes c. (For the algorithm described below, use a starting value of xo = {). (Some background information:) First observe that we can write a Taylor series approximation of c: c(Q) + c(Q.) + c'(Q.)(Q Qo) +1" (Q.)(Q Q.) When Q. is close to the value of Q that minimizes c, then minimizing c(Q) should give approximately the same answer as minimizing c(Q.) + c'(Q.)(Q Qo) +3." (Q.)(Q Q.). If we minimize the latter expression by differentiating with respect to Q, we get the first order condition: d' (Q.) +c" (Q.)(Q-Q.) = 0 Solving for Q gives Q=Qo - c This is the basis of an iteration: guess an initial Qo, this gives Q1 = Qo With Q1 we can obtain another estimate Q2 = 21 and with Q2 we can obtain Q3, and so on. At the nth iteration Qn+1 = Qn The process converges when the change from Qn to Qn+1 becomes small. In this case we end up with (approx- imately) On = n+1. Looking at the iteration rule (n) Qn+1 = Qn - with Qn+1 = Qn yields 0= Qn+1 - Qn = -( Qn) Thus (cm)] = 0, and since [c" (Qn)] # 0, we get c'(Qn) = 0. Thus, since c is conver, Qn minimizes c. Question 1. Consider the average cost function (Q) = e-Q+Q2. Show that the cost function is strictly convex. Find the minimum of the average cost c: find the value of Q which minimizes c. (For the algorithm described below, use a starting value of xo = {). (Some background information:) First observe that we can write a Taylor series approximation of c: c(Q) + c(Q.) + c'(Q.)(Q Qo) +1" (Q.)(Q Q.) When Q. is close to the value of Q that minimizes c, then minimizing c(Q) should give approximately the same answer as minimizing c(Q.) + c'(Q.)(Q Qo) +3." (Q.)(Q Q.). If we minimize the latter expression by differentiating with respect to Q, we get the first order condition: d' (Q.) +c" (Q.)(Q-Q.) = 0 Solving for Q gives Q=Qo - c This is the basis of an iteration: guess an initial Qo, this gives Q1 = Qo With Q1 we can obtain another estimate Q2 = 21 and with Q2 we can obtain Q3, and so on. At the nth iteration Qn+1 = Qn The process converges when the change from Qn to Qn+1 becomes small. In this case we end up with (approx- imately) On = n+1. Looking at the iteration rule (n) Qn+1 = Qn - with Qn+1 = Qn yields 0= Qn+1 - Qn = -( Qn) Thus (cm)] = 0, and since [c" (Qn)] # 0, we get c'(Qn) = 0. Thus, since c is conver, Qn minimizes c