[PDF] Lecture12: Global extrema - Harvard University

Lecture 12: Global extrema - Harvard University

boundary points, then pick the largest Global maxima or minima do not need to exist however The function f(x) = x2 has a global minimum at x= 0 but no global maximum The function f(x) = x3 has no global extremum at all We can however look at global maxima on nite intervals

Lecture12: Global extrema - Harvard University

then pick the largest Global maxima or minima do not need to exist The function f(x) = x2 has a global minimum at x = 0 but no global maximum The function f(x) = x3 has no global extremum at all We can however look at global maxima on finite intervals 1 Find the global maximum of f(x) = x2 on the interval [−1,2] Solution

Global and Local Extrema - Open Computing Facility

local and global cases, it is important to be cognizant of the domain over which the function is defined That which is an extremum on one domain may very well not be over a new domain, and vice versa Before delving further, let us give the formal definitions of these various extrema Global Extrema

Linear approximation Global extrema

value) Note that a global extremum is also a lo-cal extremum but a local extremum might not be a global extremum Extreme Value Theorem: If a function is continuous on a closed interval (an interval which includes the endpoints) then the function has a global maximum and a global minimum on the interval Knowing that there is an extreme value

Extrema for Functions of Two Variables

global extrema 2 There is no point in doing a second derivative test for a global extremum problem A local minimum cannot be a global maximum, but it need not be a global minimum, so identifying a candidate point as a local minimum does not determine the global minimum Global extrema are determined by comparing the function values for the

On Global Extremum Seeking In The Presence Of Local Extrema

We analyze global extremum seeking in the presence of local extrema for a simple scalar extremum seeking feedback scheme Sufficient conditions are given under which it is possible to tune the controller parameters to achieve convergence to an arbitrarily small neighborhood of the global extremum from an arbitrarily large set of initial conditions

Lecture 17: Derivatives and extrema

X at x =3/4 a global minimum is a˛ained; X at x =−1 and x =1 relative maxima are a˛ained; X at x =0 we do not have a local extremum at all Since f(−1)=3 > 1 =f(1), f has a global maximum at x =−1

Extremum Seeking Control: Convergence Analysis

extremum seeking is achieved if the system is initialized close to the extremum We introduced a simplified adaptive scheme in [17] where it was shown under slightly stronger conditions that non-local (even semi-global) extremum seeking is achieved if the controller is tuned appropriately Moreover, by using the

Local Extrema - Math

18B Local Extrema 2 Definition Let S be the domain of f such that c is an element of S Then, 1) f(c) is a local maximum value of f if there exists an interval (a,b) containing c such that f(c) is the maximum value of f on (a,b)∩S

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Math 1A: introduction to functions and calculus Oliver Knill, 2012Lecture 12: Global extremaIn this lecture we are interested in the points where a function is maximal overall. Theseglobal

extremacan occur at critical points offor at the boundary of the domain, wherefis defined. A pointpis called aglobal maximumoffiff(p)≥f(x) for allx. A pointpis called aglobal minimumoffiff(p)≤f(x) for allx. How do we find global maxima? We just make a list of all local extrema and boundary points, then pick the largest. Global maxima or minima do not need to exist. The functionf(x) =x2 has a global minimum atx= 0 but no global maximum. The functionf(x) =x3has no global extremum at all. We can however look at global maxima on finite intervals. 1 Find the global maximum off(x) =x2on the interval [-1,2].Solution. We look for local extrema at critical points and at the boundary. Then we compare all these extrema to find the maximum or minimum. The critical points arex= 0. The boundary points are-1,2. Comparing the valuesf(-1) = 1,f(0) = 0 andf(4) = 4 shows thatfhas a

global maximum at 2 and a global minimum at 0.Extreme value theoremA continuous functionfon a finite interval [a,b] attains

a global maximum and a global minimum. Here is the argument: Because the function is continuous, the image of the interval [a,b] is a closed interval [c,d].1There is a point such thatf(x) =c, which is a global minimum and a point wheref(x) =dwhich is a global maximum. Note that the global maximum or minimum can also also on the boundaryor points where the derivative dos not exist. 2 Find the global maximum and minimum of the functionf(x) =|x|. The function has no absolute maximum as it goes to infinity forx→ ∞. The function has a global minimum atx= 0 but the function is not differentiable there. The pointx= 0 is a point which does not belong to the domain off?.3 Asoda canis a cylinder of volumeπr2h. The surface area 2πrh+2πr2measures the amount of material used to manufacture the can. Assume the surface area is 2π, we can solve the equation forh= (1-r2)/r= 1/r-r Solution:The volume isf(r) =π(r-r3). Find the can with maximal volume:f?(r) =π-3r2π= 0 showing r= 1/⎷

3. This leads toh= 2/⎷

3.

1This statement needs more justification but is intuitive enough thatwe can accept it.

1 2 4 Take a card of 2×2 inches. If we cut out 4 squares of equal side lengthxat the corners, we can fold up the paper to a tray with width (2-2x) length (2-2x) and heightx. For whichx?[0,1] is the tray volume maximal? SolutionThe volume isf(x) = (2-2x)(2-2x)x. To find the maximum, we need to compare the critical points which is atx= 1/3 and the boundary pointsx= 0 and x= 1.5

Find the global maxima and minima of the function

f(x) = 3|x| -x3on the interval [-1,2].

Solution.Forx >0 the function is 3x-x3which can

be differentiated. The derivative 3-3x2is zero atx= 1.

Forx <0 the function is-3x-x3. The derivative is

-3-x2and has no root. The only critical points are 1. There is also the pointx= 0 which is not in the domain where we can differentiate the function. We have to deal with this point separately. We also have to look at the boundary pointsx=-1 andx= 2. Making a list of function values atx=-1,x= 0,x= 1,x= 2 gives the maximum. ?1.0 ?0.5 0.5 1.0 1.5 2.0 ?2?11234 3

Homework1

Find the global maxima and minima of the functionf(x) = (x-2)2on the interval [-1,4].2 Find the global maximum and minimum of the functionf(x) = 2x3-3x2-36xon the interval [-4,4]3 A candy manufacturer builds spherical candies. Its effectivenessisA(r)-V(r), where A(r) is the surface area andV(r) the volume of a candy of radiusr. Find the radius, wheref(r) =A(r)-V(r) has a global maximum forr≥0. 4 A ladder of length 1 is one side at a wall and on one side at the floor. First verify that the distance from the ladder to the corner isf(x) = sin(x)cos(x). Find the anglex for whichf(x) is maximal. x f?x? 5 a) The functionS(x) =-xlog(x) is called theentropy function. Find the prob- ability 0< x≤1 which maximizes entropy. important principle in all science is that nature tries to maximize entropy. In some sense we compute here the number of maximal entropy. 4 b) We can write 1/xx=e-xlog(x). Find the positive valuex, wherex-xhas a local maximum. 2 Entropy has been introduced by Bolzman. It is important in physics and chemistry.

2We have used the identityab=eblog(a)

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