Defining a function: f[x,y ]:= x 2 +y 2. Take care with the elements in red since they are different from the standard mathematical notation.

Transcription

1 Chapter 1 Functions Exercises 1,, 3, 4, 5, 6 Mathematica useful commands Defining a function: f[x,y ]:= x +y Take care with the elements in red since they are different from the standard mathematical notation Graphic representation for a domain given by inequalities Ejemplo 1 A { 1 x + y < x + y 0 Is a domain that consists of the points x, y R such that satisfy both inequalities To represent it we use To represent the boundary we use RegionPlot[1 x +y < && x+y 0,{x,-5,5},{y,-5,5}] ContourPlot[{1==x +y, x +y ==, x+y==0},{x,-5,5},{y,-5,5}] Some definitions For a domain D R n : p D is an interior point whenever we can find a ball p Bp, r D That is to say, there is at least a ball centered at p completely included in D p D is a boundary point if for every ball centered at p, Bp, r, we can find q 1, q Bp, r such that q 1 Bp, r and q Bp, r That is to say, every ball around p contains points inside and outside D D = {p D : p is an interior point of D} is the set of all interior points of D We also denote it IntD D = {p R n : p is a boundary point of D} is the boundary set of D that consist of all boundary points of D We also denote it FrD 1

2 D = D D is the closure set of D D is said to be open whenever D = D or D D = D is said to be closed whenever D = D or D D D is bounded whenever we can find a ball such that D Bp, r D is said to be compact whenever D is bounded and closed D is said to be convex whenever for all p, q D we have [p, q] D, where [p, q] denotes de segment from p to q D is said to be arc-connected whenever for all p, q D we have that there is a continuous curve c : [a, b] R D such that ca = p and cb = q That is to say, any two points inside D can be connected by means of a continuous curve completely inside D Exercises 7 15, 18, 19 Use polar change or slopes to study all directions at a time Many times it is useful to study the curves in the directions 1, 0, 0, 1 and 1, 1 and, of course, the ones proposed in the exercise as suggestions Some useful infinitesimal expressions: e x 1 x e x 1 + x sinx x 1 cosx 1 x cosx 1 1 x log1 + x x n x x n Ejemplo lim x,y 0,0 x y sinx y = sinx x sinx y x y = = lim x,y 0,0 x + y = 0 x y lim x,y 0,0 x y = lim x yx + y x,y 0,0 x y Exercises 3, 4, 6 For suitable multivariate functions f and g the Chain Rule Theorem establishes the following formula for the Jacobian matrix of a composite function: f g p = f gp g p Ejemplo 3 Consider the functions f : R R 3 fx, y = x, xy, y and g : R 3 R gu, v, w = u + uv + w, uw + vw Compute f g 1, 0, 0 and g f 0, 1 by means of the Chain Rule Theorem

3 We have that x 0 f x, y = y x and g 1 + v u 1 u, v, w = w w u + v 0 y For the compositions we only need to apply the formula of the Chain Rule Theorem 0 f g 1, 0, 0 = f g1, 0, 0 g 1, 0, 0 = f 1, 0 g 1, 0, 0 = = =1,0 g f 1, 0 = g f1, 0 f 1, 0 = g 1, 0, 0 f 1, 0 = =1,0, = Exercises 35, 36, 40, 43, 50 For a function F : R n+m R m F X, Y = F 1 X, Y,, F m X, Y with variables X = x 1,, x n and Y = y 1,, y m, consider the system of equations F X, Y = 0 and a point p = p x1,, p xn, p y1,, p ym =, p Y which is solution of the system, that is to say, such p Y that F p = F, p Y = 0 The problem here is to study whether it is possible to solve the system for variables Y in terms of variables X around the point p In other words we want to know if we can find formulas to compute y 1,, y m in terms of x 1,, x n, y 1,, y m = y 1 x 1,, x n,, y m x 1,, x n Y Y X for short we write Y = Y X, such that, when we substitute these formulas in the system, it is satisfied, Y X F X {}}{ x 1,, x n, y 1 x 1,, x n,, y m x 1,, x n = 0 or more short F X, Y X = 0 is a solution of the system Since p is a solution of the system, we have that F {}}{ p x1,, p xn, Y {}}{ p y1,, p ym p Y p=,p Y = 0 and accordingly Y px = p Y Therefore, for the formulas Y X that we have to find, initially we only know Y = p Y, that is to say, the value at one point X = The question is to determine if we can compute them for more possible values of X, not only for X =, over a complete ball B, r around with any r > 0 The Implicit Function Theorem allows us to give an answer to the problem in the following way: 3

4 1 Compute the Jacobian matrix F p = X p partial derivatives for X The system can be solved arroun p whenever Y p 0 Y p partial derivatives for Y Then, every y 1,, y m can be computed by means of a formula in terms of x 1,, x n so that Y : B, r R m Y X = y 1 x 1,, x n,, y 1 x 1,, x n is a multivariate function defined in the ball B, r around Maybe we cannot obtain the explicit formulas for Y X but the Implicit Function Theorem tell us that its Jacobian matrix at, x 1 Y = y m x 1 we can also denote Y not = Y X, can be computed as x n y m x n 1 Y = Y p X p Ejemplo 4 Determine if the system for the variables x 1, x, x 3, y 1, y, { x 1 + x + x 3 + sinx 3 y 1 y = 5 x 1 x x 3 + y 1 + e xy1y = can be solved for the variables y 1 and y around the point p = 0,, 1, 1, 0 and compute y 0,, 1 In this case we have X = x 1, x, x 3 and Y = y 1, y Take the function F x 1, x, x 3, y 1, y = x 1 + x + x 3 + sinx 3 y 1 y 5, x 1 x x 3 + y 1 + e xy1y X Y and then the system can be written as F X, Y = 0 It is easy to see that the given point p = 0,, 1, 1, 0 = =p Y therefore, = 0,, 1 and p Y = 1, 0 is a solution since F p = F, p Y = F 0,, 1, 1, 0 = 0, 0 = 0 The Jacobian matrix for a general point is F x 1, x, x 3, y 1, y = x 1 x 1 + y 1 y cosx 3 y 1 y x 3 y cosx 3 y 1 y x 3 y 1 cosx 3 y 1 y x x 3 x 1 x 3 + y 1 y e xy1y x 1 x 1 + x y e xy1y x y 1 e xy1y X x 1, x, x 3, y 1, y Y x 1, x, x 3, y 1, y 4

5 and for the point p, F 0,, 1, 1, 0 = X p 1 Puesto que Y p 0 1 = = variables Y = y 1, y can be solved in terms of X = x 1, x, x Y p For = 0,, 1 we can compute the Jacobian matriz of Y X = y 1 x 1, x, x 3, y x 1, x, x 3 by means of the formula Y = 1 Y p X p = = = Therefore, In particular, p Y X = x 1 y x 1 x y x y = y =