12.1 Taylor Polynomials In Two-Dimensions: Nonlinear Approximations of F(x,y) Polynomials in Two-Variables. Taylor Polynomials and Approximations.

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1 Week 12 Lecture Overheads Math2080 Week 12 Page Taylor Polynomials In Two-Dimensions: Nonlinear Approximations of F(x,y) Polynomials in Two-Variables. Taylor Polynomials and Approximations Maximum and Minimum of Multivariate functions Maximum, minimum, and critical points. A Second Partial Derivative Test for Extreme Values. Optimization: Determining Maximum and Minimum Values of z = F(x, y) Optimization with Constraints. 1. The Method of Lagrange Multipliers. 2. Why the Lagrange Method works.

2 Lecture Overheads Math2080 Week 12 Page 2 Polynomials in two variables. For example Polynomials in two variables, say x and y, are linear combinations of terms of the form x n y m, where n and m are positive integers or zero. P(x, y) = y + x 2 y - 3x 2 y 4 or P(x, y) = x - y + xy The degree or "order" of a term x n y m is the sum of the exponents: n + m. The order of a polynomial is the highest order of its individual terms.

3 Lecture Overheads Math2080 Week 12 Page 3 A more general form of a polynomial, referred to as a "shifted" form, involves powers of the differences x - x 0 and y - y 0, where (x 0, y 0 ) is a specific reference point. The general term of a shifted polynomial has the form c i,j (x - x 0 ) i (y - y 0 ) j General Polynomial Term of degree n = i + j with coefficient c i,j The general shifted polynomial of degree N can thus be expressed using a double summation in the form N n0 n i0 P(x, y) = [ c i,n - i (x - x 0 ) i (y - y 0 ) n - i ]

4 The Lecture Overheads Math2080 Week 12 Page 4 N th -order Taylor Polynomial expansions of a function F(x, y) about a point (x 0, y 0 ) is the polynomial function P N (x,y) and with the coefficients c i,n-i chosen so that P N (x,y) matches the function F(x, y) at the point (x 0, y 0 ) all of the partial derivatives of F and P N of order N at the point (x 0, y 0 ) also match exactly.

5 Lecture Overheads Math2080 Week 12 Page 5 The functions match: P N (x 0, y 0 ) = F (x 0, y 0 ) Their first partial derivatives match: x y x P N (x 0, y 0 ) = F (x 0, y 0 ) y P N (x 0, y 0 ) = F (x 0, y 0 )

6 Lecture Overheads Math2080 Week 12 Page 6 Their second partial derivatives match: 2 P N (x 0, y 0 ) = F (x 0, y 0 ) x x 2 P N (x 0, y 0 ) = F (x 0, y 0 ) y x y y 2 2 P N (x 0, y 0 ) = F (x 0, y 0 ) x y

7 Lecture Overheads Math2080 Week 12 Page 7 The mixed partial derivatives of the Taylor Polynomial and the Function agree at (x 0, y 0 ) for all n N and i n: n x i y ni n P N (x 0, y 0 ) = F (x 0, y 0 ) x i y ni

8 Lecture Overheads Math2080 Week 12 Page 8 Assuming the function F(x, y) has partial derivatives of all orders, some of the mixed partial derivatives will be identical. For instance F xxyxy is a 5 th -order mixed partial derivative. It is formed by taking the partial derivative first with respect to x, then x again, then y, then x, and finally with respect to y. y x y x x { [ ( ( F(x, y) ) ) ] } How many derivatives with respect to x? How many derivatives with respect to y?

9 Lecture Overheads Math2080 Week 12 Page 9 What is another 5 th order partial derivative that is the same as F xxyxy? And another? What is the simpler way to write F yyxxx? x y F(x, y)

10 Lecture Overheads Math2080 Week 12 Page 10 How many 5 th -order partial derivatives are there that are the same as F xxyxy? Consequently, the Taylor polynomial will have terms that agree with F xxyxy

11 Lecture Overheads Math2080 Week 12 Page 11 Consider the problem of having an n th -order partial derivative with i of the derivatives being with respect to x. How many are there? Think of having i-x s that you have to place into n-slots. The number of ways this can be done is denoted using the mathematical notation for combinations: C(n,i) = the number of combinations of n terms taken i at a time C(n,i) = = n! i! (ni)!

12 Lecture Overheads Math2080 Week 12 Page 12 Think of how you can arrange in 5 slots 3-x s and 2- y s. Just put the x s in first, then we worry about placing the y s the remaining slots.

13 Think! Lecture Overheads Math2080 Week 12 Page 13 How many slots you can put one of the x s in. Because the x s are all the same, (not different colors) we can not distinguish between (1) putting the first x in slot 2 and the second x in slot 4 and (2) putting the first x in slot 4 and the second x in slot 2 To avoid counting such arrangements multiple times we must divide by the number of ways we could achieve a given order.

14 Lecture Overheads Math2080 Week 12 Page 14 That is each time we must ask How many x s do you have to choose from? The number of choices must be divided by this number of x s that we could place there. Then, in how many slots can you put the second x? But, remember to divide by how many x s you have to choose from for the second x? Then how many slots are available to place the third x in? Now, how many x s do you have to choose from?

15 Lecture Overheads Math2080 Week 12 Page 15 Now consider the y s. How many slots you can put one of the y s in. But remember, how many y s do you have to choose from? Then how many can you put the second y in? How many y s do you have to choose from? Thus, the possible number of ways to have a 5 th order partial derivative with 3-x and 2-y derivatives is

16 Lecture Overheads Math2080 Week 12 Page 16 1 xxxyy 2 xxyyx 3 xxyxy 4 xyxxy 5 xyxyx 6 xyyxx 7 yxxxy 8 yxxyx 9 yxyxx 10 yyxxx

17 Lecture Overheads Math2080 Week 12 Page 17 What is this about? Why do we want to know how many ways you can have a 5 th -order partial derivative with 3-x and 2-y derivatives? Because, all of these are the same, provided F(x,y) is repeatedly differentiable. Hence, when writing out the Taylor expansion of F(x,y) instead of having a separate term for each different 5 th -order partial derivatives, these can all be combined into one term with the coefficient C(5,3) F xxxyy (x 0, y 0 ) / 5! Thus one of the 5 th order terms in the Taylor expansion will be 10 F xxxyy (x 0,y 0 ) (xx 5! 0 ) 3 (yy 0 ) 2

18 Lecture Overheads Math2080 Week 12 Page 18 The N th -order Taylor Polynomial is thus N n0 N n0 n i0 c(n,i) n! P N (x, y) = [ (x - x 0 ) i (y - y 0 ) n - i ] 1 n! (xx 0 ) x (yy 0 ) y = [ F(x 0,y 0 )] The binomial theorem is used to expand the n th general derivative, like a power: n 1 n! (xx 0 ) x (yy 0 ) y n F(x 0, y 0 ) n i0 c(n,i) n! n F(x 0.y 0 ) = (x - x 0 ) i (y - y 0 ) n - i x i y ni

19 Lecture Overheads Math2080 Week 12 Page 19 The binomial expansion of (a+b) n : n(n1) (a+b) n = a n + n(a n-1 b + a n-2 b c(n,i) a i b n-i n(ab n-1 + b n

20 Lecture Overheads Math2080 Week 12 Page 20 z y x

21 Lecture Overheads Math2080 Week 12 Page 21 The graph of a function z = F(x,y) viewed with lines parallel to the x and y axis z = sin 2 (π(x-0.5)) sin(πy/2) z x y

22 Lecture Overheads Math2080 Week 12 Page 22 z x y

23 Lecture Overheads Math2080 Week 12 Page 23 Extrema of multivariate functions. z = F(x,y) Maximum. A function z = F(x,y) has a local maximum at a point (x 0,y 0 ) if, for all points (x,y) within a sufficiently small circle centered at (x 0,y 0 ), F(x,y) F(x 0,y 0 ). In this case, the value z 0 = F(x 0,y 0 ) is a local maximum value of F(x,y).

24 Lecture Overheads Math2080 Week 12 Page 24 Minimum. A function z = F(x,y) has a local minimum at a point (x 0,y 0 ) if, for all points (x,y) within a sufficiently small circle centered at (x 0,y 0 ), F(x,y) F(x 0,y 0 ). In this case, the value z 0 = F(x 0,y 0 ) is a local minimum value of F(x,y).

25 Lecture Overheads Math2080 Week 12 Page 25 A small open circle centered at (x 0,y 0 ) of radius δ, is the open ball B δ (x 0,y 0 ) = { (x, y) (x - x 0 ) 2 + (y - y 0 ) 2 < δ } A set D is closed if its compliment is open, i.e., if (x 1,y 1 ) D then there is an open neighborhood of (x 1,y 1 ) that does not intersect D: there exist a radius δ such that B δ (x 1,y 1 ) D = φ the empty set.

26 Lecture Overheads Math2080 Week 12 Page 26 DEFINITION OF STATIONARY AND CRITICAL POINTS. x A function F(x,y) is stationary at a point (x 0,y 0 ) if y F(x 0,y 0 ) = 0 and F(x 0,y 0 ) = 0. In this case, (x 0,y 0 ) is called a stationary point. The critical points of a function F(x,y) on a region D are: i) all stationary points for F in D; ii) all points (x,y) D for which either F x (x,y) or F y (x,y) does not exist; iii) points on the boundary of D.

27 Lecture Overheads Math2080 Week 12 Page 27 Theorem. Continuous functions can only have local maximums or minimums on a closed region D at critical points in D. If F(x, y) is continuous and D is closed then the only points at which extrema values can occur are critical points.

28 Lecture Overheads Math2080 Week 12 Page 28 THE SECOND PARTIAL DERIVATIVE TEST. Let (x 0,y 0 ) be a stationary point of a function F(x,y), so that x Set y F(x 0,y 0 ) = 0 and F(x 0,y 0 ) = 0. A = F xx (x 0,y 0 ) B = F xy (x 0,y 0 ) C = F yy (x 0,y 0 ) and define the discriminate D as Then D = B 2 - AC.

29 Lecture Overheads Math2080 Week 12 Page 29 i) If D > 0 the point (x 0,y 0 ) is a saddle point and the graph of F(x,y) does not have a maximum or a minimum at (x 0,y 0 ). ii) If D < 0 then (x 0,y 0 ) is a local extreme point; a) F(x,y) has a (local) maximum at (x 0,y 0 ) if A < 0 or C < 0; b) F(x,y) has a (local) minimum at (x 0,y 0 ) if A > 0 or C > 0. iii) If D = 0 no conclusion is drawn. The function F(x,y) may or may not have an extreme value at (x 0,y 0 ).

30 Lecture Overheads Math2080 Week 12 Page 30 z y x z = x 2 + y 2

31 Lecture Overheads Math2080 Week 12 Page 31 z = - x 2 - y 2 z x y x = 0 x = 2

32 Lecture Overheads Math2080 Week 12 Page 32 z = x2 - y 2 z x = -2 y x x = 0

33 Identify Lecture Overheads Math2080 Week 12 Page 33 the critical points of the given function and apply the Second Partial Derivative Test to characterize the stationary points. a) z = 2xy 2-8x - 8y b) z = xy + x 2 + y 2-4y c) z = x y 1/3 - x - 2y

34 Lecture Overheads Math2080 Week 12 Page 34 The following are all the same equation z = x[ (x - 1) 2 + (y - 2) 2-9] z = x(x 2-2x + y 2-4y - 4) z = x 3-2x 2 + xy 2-4yx - 4x From the first form we see that z is x times a function in [.. ] whose graph is a conical cone opening upward, having a minimum of -9 at (1,2). Multiplying by x skews this cone upward for x > 0 but negatively downward for x < 0.

35 Lecture Overheads Math2080 Week 12 Page 35 The first partial derivatives of z = x 3-2x 2 + xy 2-4yx - 4x are z x = 3x 2-4x + y 2-4y - 4 z y = 2xy - 4x Critical Points: z y = 0 ; 2xy - 4x = 2x(y - 2) = 0 hence x = 0 or y = 2 Then,. z x = 0 ; If x = 0, z x = y 2-4y - 4 = 0 gives the roots y = 2 ± 8 1/2. So two stationary points are (x E1, y E1 ) = (0, /2 ) and (x E2, y E2 ) = (0,2-8 1/2 ) If y = 2, then z x = 3x 2-4x - 8 = 0 gives x = 2/3(1 ±3 1/2 ). Hence two more stationary points are (x E3, y E3 ) = (2/3(1+ 3 1/2 ), 2) and (x E4, y E4 ) = (2/3(1-3 1/2 ), 2)

36 Lecture Overheads Math2080 Week 12 Page 36 The graph of z x = 0 in the x-y plane is the graph of the ellipse 3(x - 2/3) 2 + (y - 2) 2 = 28/3 The critical points are the points where the lines x = 0 (the y-axis) and y = 2 intersect this ellipse. y x

37 Lecture Overheads Math2080 Week 12 Page 37 The second partial derivatives of z = x 3-2x 2 + xy 2-4yx - 4x are: A = z xx = 6x - 4 B = z xy = 2y - 4 C = z yy = 2x

38 Lecture Overheads Math2080 Week 12 Page 38 This is a table of the calculations for the Second Partial derivative test. Stationary point (x E1, y E1 ) (x E2, y E2 ) (x E3, y E3 ) (x E4, y E4 ) A 0 0 4(3 1/2-4(3 1/2 B 2(8 1/2-2(8 1/2 0 0 C 0 0 4/3(1+ 3 1/2 ) Pos 4/3(1-3 1/2 ) D = B 2 - AC /3 1/2 (1+ 3 1/2 ) Neg 16/3 1/2 (1-3 1/2 ) < 0 2 nd Partial Test conclusion Saddle point Saddle point Minimum point as A > 0 Maximum point as A < 0

39 Lecture Overheads Math2080 Week 12 Page 39 The Method of Lagrange Multipliers The problem is to optimize the objective function z = F(x,y) with the constraint g(x, y) = 0 We use a method that utilizes a new variable, called a "multiplier", that is traditionally denoted by the lower case Greek lambda, λ. In this method, the calculus optimization problem is transformed into an algebra problem that involves the simultaneous solution of three equations in the three unknowns, x, y, and the parameter λ. The optimal function values occur at the points determined by the roots x and y. The value of λ is not utilized in computing the optimal function value but is usually essential to find the roots x and y. It is called the "Lagrange multiplier method" after Joseph Lagrange ( ), who pioneered the

40 Lecture Overheads Math2080 Week 12 Page 40 branch of mathematics called "calculus of variations" THE METHOD OF LAGRANGE MULTIPLIERS. The local optimum values of a differentiable function F(x,y), subject to the differentiable constraint g(x,y) = 0, occur at the points (x,y) that are the roots of the Lagrange system of Equations: F x (x,y) = λ g x (x,y) F y (x,y) = λ g y (x,y) g(x,y) = 0 for som constant Lagrange Parameter, λ.

41 Lecture Overheads Math2080 Week 12 Page 41 The Lagrange Method is deduced from three observations that show how the constrained problem is really a one dimensional problem whose solution is obtained via multivariable techniques. The key is to think of the constraint as a curve, like a hiking trail on a map, specified by a parameter, t. A point on the curve associated with each time value indicates the hiker's position on a trail at time t. The Lagrange system is derived by assuming that such a parametric description of the curve g(x, y) = 0 exists, but it does not actually require that a specific parametric function is known. The key that ties the following observations together is the fact that at a highest (or lowest) point on the trail the direction of the trail must be perpendicular to the direction that points uphill or downhill, i.e. in the direction of steepest assent or descent. This allows us to relate the partial derivatives of g and F.

42 Lecture Overheads Math2080 Week 12 Page 42 Constraint Curve g(x,y) = 0 g(x,y) = x^2-3xy + (y/2)^2 + y^3 y g(x,y) Tangent Vector _r (t) Observation I. First, we introduce a parametric representation of the "trail", which is the curve g(x, y) = 0 (see the graph on the following page). The exact form of this parameterization is not important! We will simply assume there is some vector-valued function r(t) that describes the curve g(x, y) = 0, i.e., that every point (x, y) on this curve can be represented as r(t) = (x(t), y(t)) _r(t) x Considered as a parametric curve and as a level curve. g(x, y) = x 2-3xy + (y/2) 2 + y 3 = 0.

43 Lecture Overheads Math2080 Week 12 Page 43 In Chapter 11, Section 11.1, we introduced the tangent vector to such curves and established that at any point r(t) the tangent vector is given by T = r (t). At the point r(t) two vectors have been sketched, a tangent vector r and the gradient vector g(x, y).

44 Lecture Overheads Math2080 Week 12 Page 44 Observation II. Next, observe that the curve g(x, y) = 0 is in fact a level curve of the surface that is the graph of z = g(x, y); the level curve with z = 0. The importance of this observation is the fact, established in Chapter 11, Section 11.3, that the directional derivative of z = g(x, y) in the direction of the level curve is zero.

45 Lecture Overheads Math2080 Week 12 Page 45 Thus, at any point (x, y) = r(t) on the constraint curve, the directional derivative in the direction T is zero: T D T g(x, y) = g(x, y)! = 0 T As T = r (t), this means that the dot product of r (t) with the gradient of g equals zero, g(x, y) r (t) = 0, which implies that r is perpendicular to g.

46 Lecture Overheads Math2080 Week 12 Page 46 Observation III. The third observation is that if a point (x, y) = r(t) corresponds to an extreme value of F(x, y) then the directional derivative of F(x, y) in the direction of the path g must be zero. This means that r (t) D r (t) F(x, y) = F(x, y)! = 0. r (t) But this also indicates that F(x, y)!r (t) = 0, which means the gradient vector F is perpendicular to the tangent vector r.

47 Lecture Overheads Math2080 Week 12 Page 47 Combining the last two observations we see that F(x, y) and g(x, y) are perpendicular to the same vector r (t) (which we do not need to know!) Consequently, the two vectors F(x, y) and g must be parallel. This means that there must be a constant, which is denoted by λ, such that F(x, y) = λ g(x, y) The constant λ is the Lagrange Multiplier. The x and y-component equations of this vector equation provide the first two equations of the Lagrange System. The third Lagrange equation is provided by the original constraint: g(x, y) = 0.

48 Lecture Overheads Math2080 Week 12 Page 48 Example. Find the points on the circle x 2 + y 2 = 1 at which f(x,y) = xy has it maximum and minimum values. Here: g(x,y) = f x = f y = g x = g y = The Lagrange system is f x = λg x f y = λg y This gives λ = = Substituting into g = 0 gives Critical points: