[MUSIC] Now we'll proceed with the topic how to identify global extrema. The thing is that there are no clear-cut methods for this, but we'll consider two problems. We'll provide ideas which we need to follow dealing with this optimization problem. So usually we start with solving a problem of optimization which will give us the set of critical points. And we need to identify them using known techniques. But in order to answer the question whether the found local extrema are global ones, we need to do some extra investigation. So we start with a problem in two-dimensional case. There is a function z in the form of. We exclude the 0 point because the logarithm of 0 doesn't exist at the same time. This is a removable discontinuity. So (0,0) is a removable, Discontinuity point. Because the limit of z when x and y turns 0 is 0. It's quite easy to show. Now, the problem is to find global extrema, or this function z, and as always we start with the first order conditions. It's quite convenient to change cartesian variables x, y into polar variables, the radius r and the angle phi. So from (x,y) We move to (r,phi) using two formulas, Where r takes the non-negative values and phi the angle ranges from 0, to 2 pi. After the change of variables, the system can be written again. And it becomes, Since r is strictly positive here, we can cross it out. And the system becomes a bit simpler. If I take out sin phi in the first equation, And cos phi in the second. Now it's clear that if, r = 1 then logarithm r squared = 0. And for the values, 0, pi over 2, pi, 3 over 2, pi, the system is satisfied and easily seen. Because for instance, if you choose 0 for phi, sin turns 0. At the same time sin is here, also turns 0. If you choose pi over 2, then cos is 0 and cos here is also 0, so we got four points. And all these four points belong to the unit circle. Let's draw it. So these are the points when plotting, Them. Can we tell whether these critical points are points of extremum of the function? And we can do it without even checking the Hessian matrix because it's clear that they are not points of extrema. Where it's clear, let's consider the very first one, this one, point 1. If we move slightly up in that direction. So y goes up and x is still 1. So from point (1,0), we move to the point (1,epsilon), where epsilon is greater than 0. Let's find the value of the function at this new point. X is still 1, Y is epsilon. We get a positive value. But if we move down. So this time from we move to (1,-epsilon). We'll get a negative result. So this is clearly a saddle point. This analysis can be repeated again and again at any of these four points. Now, let's proceed and finish solution of this system. Now, when we exclude all these four points, we can cross out sin phi and cos phi. So after crossing out sin and cos, we need to add up two equations. When we add them up, we get, This being 0, then, This is 1. And we get That gives us the value of r, 1 over the square root of e. But now we need to take this value and plug in two of the equations again. What do we get then? We get then a system. It's not a system, it's a collection. Either cos squared phi = one-half or sin squared phi = one-half. And after solving, we'll get another four critical points. This time the 5 values are pi over 4, three-fourths pi, five-fourths pi, and the final, seven-fourths pi. [MUSIC]