Transcript Slide 1
Local Linear Approximation for Functions of Several Variables Functions of One Variable • When we zoom in on a “sufficiently nice” function of one variable, we see a straight line. Functions of two Variables Functions of two Variables Functions of two Variables Functions of two Variables Functions of two Variables When we zoom in on a “sufficiently nice” function of two variables, we see a plane. f ( x, y ) 5 x sin( y ) y cos( x) 2 Describing the Tangent Plane • To describe a tangent line we need a single number--the slope. • What information do we need to describe this plane? • Besides the point (a,b), we need two numbers: the partials of f in the x- and ydirections. T( a ,b ) ( x, y ) f ( a, b) y (a,b) f x ( a, b) Equation? f (a, b) f (a, b) ( x a) ( y b) f ( a, b) x y Describing the Tangent Plane T( a ,b ) ( x, y ) f (a, b) f (a, b) ( x a) ( y b) f ( a, b) x y We can also write this equation in vector form. Write x = (x,y), p = (a,b), and f (p) Tp (x) f (p) (x p) f (p) Dot product! f (a, b) f (a, b) , x y Gradient Vector! General Linear Approximations In the expression Tp (x) f (p) x f (p) we can think of f (p) As a scalar function on 2: x f (p) x . This function is linear. For a function general vector-valued function F: m n and for a point p in we want a linear function Lp : F(x) Lp (x) F(p) m n m that satisfies for x "close" to p. Why don’t we just subsume F(p) into Lp? Linear--- in the linear algebraic sense. General Linear Approximations In the expression p (x) f (p) x f (p) we can think of f (p) As a scalar function on 2: x f (p) x . This function is linear. For a function general vector-valued function F: m n and for a point p in we want a linear function Lp : F(x) Lp (x) F(p) m n m that satisfies for x "close" to p. Note that the expression Lp (x-p) is not a product. It is the function Lp acting on the vector (x-p). To understand Differentiability in Vector Fields We must understand Linear Vector Fields: Linear Transformations from m to n Linear Functions A function L is said to be linear provided that L ( x y ) L ( x) L ( y ) and L ( x ) L ( x ) Note that L(0) = 0, since L(x) = L (x+0) = L(x)+L(0). For a function L: m → are very prescriptive. n, these requirements Linear Functions It is not very difficult to show that if L: m →n is linear, then L is of the form: L(x) L( x1 , x2 , x3 , , xm ) a11 x1 a12 x2 a x a x 22 2 21 1 a31 x1 a32 x2 an1 x1 an 2 x2 a13 x3 a23 x3 a33 x3 an 3 x3 a1m xm a2 m xm a3m xm anm xm where the aij’s are real numbers for j = 1, 2, . . . m and i =1, 2, . . ., n. Linear Functions Or to write this another way. . . L(x) L( x1 , x2 , x3 , a11 a12 a 21 a22 a31 a32 an1 an 2 Αx , xm ) a13 a23 a33 an 3 a1m x1 a2 m x2 a3m x3 anm xm In other words, every linear function L acts just like leftmultiplication by a matrix. Though they are different, we cheerfully confuse the function L with the matrix A that represents it! (We feel free to use the same notation to denote them both except where it is important to distinguish between the function and the matrix.) One more idea . . . a11 x1 a x 21 1 A(x) a31 x1 am1 x1 a12 x2 a13 x3 a22 x2 a32 x2 a23 x3 a33 x3 am3 x3 am 2 x2 a1n xn a2 n xn a3n xn amn xn Suppose that A = (A1 , A2 , . . ., An) . Then for 1 j n Aj(x) = aj1 x1+aj 2 x2+. . . +aj n xn What is the partial of Aj with respect to xi? One more idea . . . a11 x1 a x 21 1 A(x) a31 x1 am1 x1 a12 x2 a13 x3 a22 x2 a32 x2 a23 x3 a33 x3 am3 x3 am 2 x2 a1n xn a2 n xn a3n xn amn xn Suppose that A = (A1 , A2 , . . ., An) . Then for 1 j n Aj(x) = aj1 x1+aj 2 x2+. . . +aj n xn The partials of the Aj’s are the entries in the matrix that represents A! Local Linear Approximation Lp (x) Αx where A is the m n matrix a11 a 21 A a31 an1 a12 a13 a22 a23 a32 a33 an 2 an 3 a1m a2 m a3m anm For x "close" to p we have F(x) Lp (x) F(p) For all x we have F(x) Lp (x) F(p) E(x) where E(x) is the error committed by Lp in approximating F. Local Linear Approximation Suppose that F: m → n is given by coordinate functions F=(F1, F2 , . . ., Fn) and all the partial derivatives of F exist at p in m and are continuous at p then . . . there is a matrix Lp such that F can be approximated locally near p by the affine function Lp (x p) F(p) What can we say about the relationship between the matrix DF(p) and the coordinate functions F1, F2, F3, . . ., Fn ? Quite a lot, actually. . . Lp will be denoted by DF(p) and will be called the Derivative of F at p or the Jacobian matrix of F at p. A Deep Idea to take on Faith I ask you to believe that for all i and j with 1 i n and 1 j m L j xi (p) Fj xi (p) This should not be too hard. Why? Think about tangent lines, think about tangent planes. Considering now the matrix formulation, what is the partial of Lj with respect to xi? The Derivative of F at p (sometimes called the Jacobian Matrix of F at p) F1 F1 F1 ( p ) ( p ) (p) x x x 2 3 1 F3 F2 F2 ( p ) ( p ) (p) x x x 2 3 1 D F(p) J F (p) F3 F3 F3 ( p ) ( p ) (p) x x x 2 3 1 Notice two Fm common Fm Fm the (p) (p) (p) x x x nomeclatures for the 2 3 1 derivative of a vector – valued function. F1 (p) xn F2 (p) xn F3 (p) xn Fm (p) xn In Summary. . . If F is a “reasonably well How should behaved” vector field we around thethink pointabout p, wethis function E(x)? can form the Jacobian Matrix DF(p). For x "close" to p we have F(x) DF(p)(x p) F(p) For all x, we have F(x)=DF(p) (x-p)+F(p)+E(x) where E(x) is the error committed by DF(p) +F(p) in approximating F(x). E(x) for One-Variable Functions E(x) measures the vertical distance between f (x) and Lp(x) But E(x)→0 is not enough, even for functions of one variable! ( x, Lp ( x)) ( x, Lp ( x)) E ( x) E ( x) ( p, f ( p)) ( x, f ( x)) What happens to E(x) as x approaches p? ( p, f ( p)) ( x, f ( x)) Definition of the Derivative A vector-valued function F=(F1, F2 , . . ., Fn) is differentiable at p in m provided that 1) All partial derivatives of the component functions F1, F2 , . . ., Fn of F exist on an open set containing p, and 2) The error E(x) committed by DF(p)(x-p)+F(p) in approximating F(x) satisfies E(x) lim 0 x p x p (Book writes this as lim xp F(x) DF(p)(x p) F(p) xp 0 .)