# Scalar Valued Functions of Several Variables; the Gradient Vector

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1 Scalar Valued Functions of Several Variables; the Gradient Vector Scalar Valued Functions vector) valued function of n variables: Let us consider a scalar (i.e., numerical, rather than y = φ(x) = φ(x 1, x 2,..., x n ), X D, where the domain D is a region in R n. By this we mean that D is an open, connected set. The term open means that each point X 0 D is the center of a ball N(X 0, r) = { X X X0 < r }, for some r > 0, which still lies in D. The term connected, as we will use it here, means that any two points in D can be connected by a curve lying in D. The partial derivatives of the function φ(x) at a point X 0 D are the numbers, which we denote by (X 0 ), k = 1, 2,..., n, obtained by differentiating the function φ(x) with respect to the variable x k at the point X 0, treating the other variables, x 1,..., x k 1, x k+1,..., x n, as constant quantities. Taking the components of X 0 to be x 0,k, k = 1, 2,..., n, these partial derivatives exist just in case the corresponding limits of difference quotients φ(x 0,1,..., x 0,k 1, x k, x 0,k+1,..., x 0,n ) φ(x 0,1,..., x 0,k 1, x 0,k, x 0,k+1,..., x 0,n ) lim x k x 0,k x k x 0,k exist; the limit values are the partial derivatives we have indicated. As we allow the point X 0 to range throughout the region D the relationship thus engendered between X 0 and the partial derivative (X 0 ) defines a function; replacing X 0 by the symbol X, we denote this function by (X). It is also common to see this function written as φ xk (X) = φ xk (x 1, x 2,..., x n ). If each of these functions is continuous as a function of X D then we say that φ(x) is continuously differentiable in the region D. In the work to follow we will assume continuous differentiability unless we specifically indicate otherwise. Most functions of several variables commonly encountered have continuous partial 1

2 derivatives if we stay away from singularities. Thus the function f(x, y) = cos(x + y) sin(x y) is well defined and continuous except along the lines x y = k π, k an integer. The partial derivative functions are f sin(x + y) (x, y) = sin(x y) f sin(x + y) (x, y) = y sin(x y) + cos(x + y) sin 2 (x y) cos(x + y) sin 2 (x y) cos(x y), cos(x y). Each of these are also continuous functions of x, y except along the lines x y = k π, k an integer. Example Where the Partial Derivatives are Defined but not Continuous We consider the function defined by f(x, y) = x 3 + y 3 x 2 + y 2, (x, y) (0, 0), f(0, 0) = 0. Since this is defined in a non-standard way at (0, 0), we should check on continuity at that point. If (x, y) (0, 0) lies in the square x b, y b, b > 0, we have x 3 + y 3 x 2 + y 2 x 3 + y 3 b (x 2 + y 2 ) = b. x 2 + y 2 x 2 + y 2 Thus as b 0, forcing (x, y) (0, 0), the value of f(x, y) tends to f(0, 0) = 0 and we conclude f(x, y) is continuous at (0, 0). On the x-axis f(x, 0) x so f (x, 0) 1; similarly on the y-axis. So the partial derivatives are defined on the axes; in particular they are defined and both equal to 1 at (0, 0). But for (x, y) (0, 0) we can compute ( x 3 + y 3 ) = 3x 2 x 2 + y 2 x 2 + y x 3 + y 3 2 (x 2 + y 2 ) 2 2x. Along a line y = a x this gives f (x, a x) = a 2(1 + a 3 ) 2 (1 + a 2 ). 2

3 Taking a = 1 we obtain f (x, x) 3/2 4/4 = 1/2 so, as we approach (0, 0) along this line f has a different value than if we approach the origin along either of the coordinate axes. We conclude f argument shows that f y is not continuous at the origin. A similar is also not continuous at the origin. The Gradient If we combine the partial derivatives (X 0 ), k = 1, 2,..., n into an n-dimensional vector, in the obvious order, we obtain the gradient (vector) of the function φ(x) at the point X 0. A commonly used notation for the gradient at X 0 is φ(x 0 ). Again letting X 0 range throughout D, and replacing X 0 by X we obtain the vector function In R 3 we can write = φ(x) = φ(x 1, x 2,..., x n ) = (x, y, z)i + y φ(x) = ( (x, y, z), y ( (X), (X),..., ) (X). 1 2 n ) (x, y, z), (x, y, z) z (x, y, z)j + (x, y, z)k, i = (100), j = (010), k = (001). z In this way we obtain a vector field in the region D, since the dimension of φ(x) is n, the same as the dimension of the independent vector variable X. Example 1 If, in R 2, we define φ(x, y) = x 2 y + y3 3, then the corresponding gradient field is ( ) φ(x, y) = (x2 y + y3 3 ), y (x2 y + y3 3 ) = ( 2xy, x 2 + y 2). The gradient vector at the point (1, 2), for example, is (4, 5). A Notational Convention Vectors can be written in row form: (x 1, x 2,..., x n ), or in column form: x 1 x 2.. x n 3

4 These vectors are said to be transposes of each other; if the row vector is designated as X, then the corresponding column vector is designated as X, and vice versa. If X and Y are n-dimensional column and row vectors, respectively, it is common to write x 1 X Y = Y (X) = (y 1 y 2 x 2... y n ). = n y k x k. k=1 x n If W is an n-dimensional row vector then W X n k=1 w k x k. Gradients are usually written as row vectors; thus the preceding implies that φ(x 0 )X n k=1 (X 0 )x k. We will use the notation repeatedly in the sequel. First Order Linear Approximation to a Scalar Function If φ(x) is a continuously differentiable function of X in the region D which contains the point X 0 the gradient φ(x 0 ) can be used to approximate values of the function φ(x) at points X D lying close to X 0. The formula for this approximation relationship is φ(x) φ(x 0 ) + φ(x 0 ) (X X 0 ) = φ(x 0 ) + n k=1 (X 0 ) (x k x 0,k ). The right hand side defines a linear function of the vector variable X, which we call the linear approximation to φ(x) at the point X 0. We will denote this linear function by L φ,x0 (X). This relationship can be illustrated for n = 2, φ = φ(x, y) by noting that in this case the graph of z = L φ,x0,y 0 (x, y) is a plane in R 3 which is tangent to the graph of z = φ(x, y) at the point (x 0, y 0, φ(x 0, y 0 ). 4

5 Example 2 Let w = x 2 + y 3 + z 4. Taking the base point X 0 = (x 0, y 0, z 0 ) = (1, 1, 1) we have w 0 φ(1, 1, 1) = 3. Now 2 φ(x, y, z) = (2x 3y 2 4z 3 ); φ(1, 1, 1) = 3. 4 Accordingly, the linear approximation to φ(x, y, z) as given by the above formula is given by the function x 1 w = 3 + (2 3 4 ) y 1 = 3 + 2(x 1) + 3(y 1) + 4(z 1) z 1 = 6 + 2x + 3y + 4z. The linear approximation can be used to estimate the value of φ at points X near the base point X 0 (with some degree of error corresponding to the term o ( X X 0 ) ). Thus, in the example just studied, if we take (x, y, z) = (1.1,.9, 1.05), the linear approximation gives the value L φ,(1,1,1) (1.1,.9, 1.05) = 6 + 2(1.1) + 3(.9) + 4(1.05) = 3.1, whereas the actual value is φ(1.1,.9, 1.05) = (1.1) 2 + (.9) 3 + (1.05) 4 = Basis for the Approximation; Error We want to see why the linear approximation, involving the gradient vector, works as it does, and to obtain a bound on the error incurred in use of this approximation. For clarity and brevity we carry out the calculations for the case n = 2; the argument in a larger number of variables is essentially identical. Let φ(x, y) be continuously differentiable in a region D, let X 0 = (x 0, y 0 ) be a point in D and let N(X 0, r) be a disc of positive radius, r, centered at X 0, which still lies in D. Let X = (x, y) be a point in N(X 0, r). We consider the difference φ(x) φ(x 0 ) = φ(x, y) φ(x 0, y 0 ) = φ(x, y) φ(x, y 0 ) + φ(x, y 0 ) φ(x 0, y 0 ). 5

6 Applying the Mean Value Theorem twice, we have φ(x, y) φ(x, y 0 ) + φ(x, y 0 ) φ(x 0, y 0 ) = y (x, η)(y y 0) + (ξ, y 0)(x x 0 ), where ξ lies between x 0 and x and η lies between y 0 and y. Then, clearly, + φ(x, y) φ(x 0, y 0 ) = y (x 0, y 0 )(y y 0 ) + (x 0, y 0 )(x x 0 ) ( ) ( (x, η) y y (x 0, y 0 ) (y y 0 ) + (ξ, y 0) ) (x 0, y 0 ) (x x 0 ) where γ(x 0, X) is the row vector = φ(x 0, y 0 )(X X 0 ) + γ(x 0, X)(X X 0 ), γ(x 0, X) = ( y (x, η) y (x 0, y 0 ) (ξ, y 0) (x 0, y 0 )). Applying the Schwarz inequality to the error term we have γ(x 0, X)(X X 0 ) γ(x 0, X) X X 0. Because φ(x) is continuously differentiable, lim γ(x 0, X) = X X 0 lim ( X X y 0 (x, η) y (x 0, y 0 ) (ξ, y 0) (x 0, y 0 )) = 0. This means that γ(x 0, X)(X X 0 ) γ(x 0, X) X X 0 has the property γ(x 0, X)(X X 0 ) lim X X 0 X X 0 = 0, a relationship which we express by saying that φ(x) = φ(x 0 ) + φ(x 0 )(X X 0 ) + o( X X 0 ); i.e., the error in the approximation tends to zero, as X X 0 0, more rapidly than any multiple of X X 0. It should be noted that the error estimate just obtained depends critically on the continuity of the partial derivatives of the function φ(x). Without that property one can show, with appropriate examples, that such an estimate need not hold. The 6

7 approximation relationship holds in exactly the same way in R n, for a general positive integer n, as we have shown it to hold for n = 2. We conclude with the following version of the Chain Rule. Proposition Suppose X(t) is a continuously differentiable curve in R n and φ(x) is a continuously differentiable function of X D R n. Then d dt φ(x(t)) = φ(x(t))x (t). Fix a value of t, say t 0. Then X(t) = X(t 0 ) + X (t 0 )(t t 0 ) + o( t t 0 ) as t t 0. Using the formula for the first order approximation to φ(x) at X = X(t 0 ) we find that φ(x(t)) = φ(x(t 0 )) + φ(x(t 0 ))(X(t) X(t 0 )) + o( X(t) X(t 0 ) ) = φ(x(t 0 )) + φ(x(t 0 ))(X (t 0 )(t t 0 ) + o( t t 0 ) X(t 0 )) +o ( φ(x(t 0 ))(X (t 0 )(t t 0 ) + o( t t 0 ) ) = φ(x(t 0 ) + φ(x(t 0 ))(X (t 0 )(t t 0 ) + o( t t 0 ). Dividing by t t 0 we then have φ(x(t)) φ(x(t 0 )) t t 0 = φ(x(t 0 ))X (t 0 ) + o( t t 0 ) t t 0. Letting t t 0 the difference quotient on the left approaches d dt φ(x(t)) t=t0 while the expression at the right approaches φ(x(t 0 ))X (t 0 ). Thus d dt φ(x(t)) = φ(x(t 0 ))X (t 0 ). t=t0 Since t 0 could be any value of t, the result follows. 7

8 Application: The Directional Derivative Let φ(x) be a continuously differentiable scalar valued function of the n-vector variable X. Let X 0 be a point in its domain and let U be a unit vector. Then we can construct the line through X 0 in the direction of U via X(t) = X 0 + t U, t 0. Clearly X(0) = X 0. We form the composite function φ(x(t)) and differentiate with respect to t using the chain rule: d dt φ(x 0 + t U) = φ(x 0 + t U) d dt (X 0 + t U) = φ(x 0 + t U)U. This is the rate of change of φ along the line described by X(t). Taking t = 0, the rate of change of φ in the direction of this line, i.e., in the direction of U, at X 0 = X(0), is seen to be φ(x 0 ) U. This is called the directional derivative of φ in the direction of U at the point X 0. The notation φ U U denote this quantity. Example is often used to For the function φ(x, y, z) = x 2 + 4y 2 + 9z 2 the directional derivative in the direction given by U = ( 1/ 6, 2/ 6, 1/ 6) at the point X 0 = (1, 1, 1) is / 6 2/ 6 1/ 6 = 1 (2( 1) + 8(2) + 18( 1)) = Properties of the Directional Derivative the dot, or inner product we have From the corresponding property of φ(x 0 )U = φ(x 0 ) U cos θ = φ(x 0 ) cosθ, where θ is the angle between the gradient vector φ(x 0 ) and U. If the gradient vector φ(x 0 ) 0 it is clear that this quantity reaches its maximum, φ(x 0 ), when θ = 0, in which case U = U + φ(x 0), and reaches its minimum, φ(x φ(x 0 ) 0), when θ = ±π, in which case U = U φ(x 0) φ(x 0. The direction corresponding to ) U + is called the steepest ascent direction for φ at X 0 while the direction corresponding to U is called the steepest descent direction. 8

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