Section 3.4. is itself a function from R n to R. The partial derivatives of f

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1 The Calculus of Functions of Several Variables Section 3.4 Second-Order Approximations In one-variable calculus, Taylor polynomials provide a natural way to extend best affine approximations to higher-order polynomial approximations. It is possible to generalize these ideas to scalar-valued functions of two or more variables, but the theory rapidly becomes involved technical. In this section we will be content merely to point the way with a discussion of second-degree Taylor polynomials. Even at this level, it is best to leave full explanations for a course in advanced calculus. Higher-order derivatives The first step is to introduce higher order derivatives. If f : R n R has partial derivatives which exist on an open set U, then, for any i = 1, 2, 3,..., n, f x i is itself a function from R n to R. The partial derivatives of f x i, if they exist, are called second-order partial derivatives f of f. We may denote the partial derivative of x i with respect to x j, j = 1, 2, 3,..., evaluated at a point x, by either 2 x j x i fx), or f xi x j x), or D xi x j fx). Note the order in which the variables are written; it is possible that differentiating first with respect to x i second with respect x j will yield a different result than if the order were reversed. If j = i, we will write 2 fx) for 2 x 2 x i i x i fx). It is, of course, possible to extend this notation to third, fourth, higher-order derivatives. so Suppose fx, y) = x 2 y 3x sin2y). Then f x x, y) = 2xy 3 sin2y) f y x, y) = x 2 6x cos2y), f xx x, y) = 2y, f xy x, y) = 2x 6 cos2y), f yy x, y) = 12x sin2y), f yx x, y) = 2x 6 cos2y). Note that, in this example, f xy x, y) = f yx x, y). For an example of a third-order derivative, f yxy x, y) = 12 sin2y). 1 Copyright c by Dan Sloughter 2001

2 2 Second-Order Approximations Section 3.4 Also, so Suppose w = xy 2 z 3 4xy logz). Then, for example, w y x = ) w = y x y y2 z 3 4y logz)) = 2yz 3 4 logz) w z 2 = z w x y = x ) w = 3xy 2 z 2 4xy ) = 6xy 2 z + 4xy z z z z 2. ) w = y x 2xyz3 4x logz)) = 2yz 3 4 logz), w y x = 2 w x y. In both of our examples we have seen instances where mixed second partial derivatives, that is, second-order partial derivatives with respect to two different variables, taken in different orders are equal. This is not always the case, but does follow if we assume that both of the mixed partial derivatives in question are continuous. Definition We say a function f : R n R is C 2 on an open set U if f xj x i is continuous on U for each i = 1, 2,..., n j = 1, 2,..., n. Theorem If f is C 2 on an open ball containing a point c, then for i = 1, 2,..., n j = 1, 2,..., n. x j x i fc) = 2 x i x j fc) Although we have the tools to verify this result, we will leave the justification for a more advanced course. We shall see that it is convenient to use a matrix to arrange the second partial derivatives of a function f. If f : R n R, there are n 2 second partial derivatives this matrix will be n n. Definition Suppose the second-order partial derivatives of f : R n R all exist at the point c. We call the n n matrix x 2 fc) fc) fc) fc) 1 x 2 x 1 x 3 x 1 x n x 1 fc) x 1 x 2 x 2 fc) fc) fc) 2 x 3 x 2 x n x 2 Hfc) = fc) fc) x 1 x 3 x 2 x 3 x 2 fc) fc) 3.4.1) 3 x n x fc) fc) fc) fc) x 1 x n x 2 x n x 3 x n the Hessian of f at c. x 2 n

3 Section 3.4 Second-Order Approximations 3 Put another way, the Hessian of f at c is the n n matrix whose ith row is f xi c). Suppose fx, y) = x 2 y 3x sin2y). Then, using our results from above, [ ] fxx x, y) f Hfx, y) = xy x, y) 2y 2x 6 cosy) =. f yx x, y) f yy x, y) 2x 6 cos2y) 12x sin2y) Thus, for example, Hf2, 0) = Suppose f : R n R is C 2 on an open ball B 2 c, r) let h = h 1, h 2 ) be a point with h < r. If we define ϕ : R R by ϕt) = fc + th), then ϕ0) = fc) ϕ1) = fc+h). From the one-variable calculus version of Taylor s theorem, we know that ϕ1) = ϕ0) + ϕ 0) ϕ s), 3.4.2) where s is a real number between 0 1. Using the chain rule, we have ϕ t) = fc + th) d dt c + th) = fc + th) h = f xc + th)h 1 + f y c + th)h ) where we have used the notation ϕ t) = h 1 f x c + th) h + h 2 f y c + th) h = h 1 f x c + th) + h 2 f y c + th) h fxx c + th) f = [ h 1 h 2 ] xy c + th) h1 f yx c + th) f yy c + th) = h T Hfc + th)h, 3.4.4) h = [ h1 h 2 ] h T = [ h 1 h 2 ], the latter being called the transpose of h see Problem 12 of Section 1.6). Hence ϕ 0) = fc) h 3.4.5) h 2 ϕ s) = 1 2 ht Hfc + sh)h, 3.4.6) so, substituting into 3.4.2), we have fc + h) = ϕ1) = fc) + fc) h ht Hfc + sh)h ) This result, a version of Taylor s theorem, is easily generalized to higher dimensions.

4 4 Second-Order Approximations Section 3.4 Theorem Suppose f : R n R is C 2 on an open ball B n c, r) let h be a point with h < r. Then there exists a real number s between 0 1 such that fc + h) = fc) + fc) h ht Hfc + sh)h ) If we let x = c + h evaluate the Hessian at c, 3.4.8) becomes a polynomial approximation for f. Definition If f : R n R is C 2 on an open ball about the point c, then we call P 2 x) = fc) + fc) x c) x c)t Hfc)x c) 3.4.9) the second-order Taylor polynomial for f at c. compute To find the second-order Taylor polynomial for fx, y) = e 2x+y at 0, 0), we from which it follows that Then fx, y) = 2e 2x+y, e 2x+y ) [ 4e 2x+y 2e Hfx, y) = 2x+y 2e 2x+y f0, 0) = 2, 1) Hf0, 0) = e 2x+y P 2 x, y) = f0, 0) + f0, 0) x, y) + 1 x 2 [ x y ] Hf0, 0) y = 1 + 2, 1) x, y) x 2 [ x y ] 2 1 y = 1 2x + y = 1 4x 2y 2 [ x y ] 2x + y = 1 2x + y x2 2xy 2xy + y 2 ) = 1 2x + y + 2x 2 2xy y2. ], Symmetric matrices Note that if f : R 2 R is C 2 on an open ball about the point c, then the entry in the ith row jth column of Hfc) is equal to the entry in the jth row ith column of Hfc) since x j x i fc) = 2 x i x j fc).

5 Section 3.4 Second-Order Approximations 5 Definition We call a matrix [a ij ] with the property that a ij = a ji for all i j a symmetric matrix. The matrices are both symmetric, while the matrices are not symmetric The Hessian of any C 2 scalar valued function is a symmetric matrix. For example, the Hessian of fx, y) = e 2x+y, namely, is symmetric for any value of x, y). [ 4e 2x+y 2e Hfx, y) = 2x+y 2e 2x+y e 2x+y Given an n n symmetric matrix M, the function q : R n R defined by qx) = x T Mx is a quadratic polynomial. When M is the Hessian of some function f, this is the form of the quadratic term in the second-order Taylor polynomial for f. In the next section it will be important to be able to determine when this term is positive for all x 0 or negative for all x 0. ], Definition Let M be an n n symmetric matrix define q : R n R by qx) = x T Mx. We say M is positive definite if qx) > 0 for all x 0 in R n, negative definite if qx) < 0 for all x 0 in R n, indefinite if there exists an x 0 for which qx) > 0 an x 0 for which qx) < 0. Otherwise, we say M is nondefinite.

6 6 Second-Order Approximations Section 3.4 In general it is not easy to determine to which of these categories a given symmetric matrix belongs. However, the important special case of 2 2 matrices is straightforward. Consider a b b c let qx, y) = [ x y ] M If a 0, then we may complete the square in ) to obtain qx, y) = a x 2 + 2b ) a xy + cy 2 = a x + b ) ) 2 a y b2 a 2 y2 + cy 2 = a x + b ) 2 a y + = a x + ba ) 2 y + = a x = ax 2 + 2bxy + cy ) y c b2 a ac b2 y 2 a ) y 2 x + b ) 2 a y + detm) y ). a Now suppose detm) > 0. Then from ) we see that qx, y) > 0 for all x, y) 0, 0) if a > 0 qx, y) < 0 for all x, y) 0, 0) if a < 0. That is, M is positive definite if a > 0 negative definite if a < 0. If detm) < 0, then q1, 0) q b a, 1) will have opposite signs, so M is indefinite. Finally, suppose detm) = 0. Then qx, y) = a x + b ) 2 a y, so qx, y) = 0 when x = b ay. Moreover, qx, y) has the same sign as a for all other values of x, y). Hence in this case M is nondefinite. Similar analyses for the case a = 0 give us the following result. Theorem Suppose a b. b c If detm) > 0, then M is positive definite if a > 0 negative definite if a < 0. detm) < 0, then M is indefinite. If detm) = 0, then M is nondefinite. The matrix is positive definite since detm) = 5 > 0 2 > 0. If

7 Section 3.4 Second-Order Approximations 7 The matrix is negative definite since detm) = 7 > 0 2 < 0. The matrix is indefinite since detm) = 7 < 0. The matrix is nondefinite since detm) = 0. In the next section we will see how these ideas help us identify local extreme values for scalar valued functions of two variables. Problems 1. Let fx, y) = x 3 y 2 4x 2 e 3y. Find the following. a) x y fx, y) b) fx, y) y x c) x 2 fx, y) d) 3 fx, y) x y x e) 3 x y 2 fx, y) f) 3 fx, y) y3 g) f yy x, y) h) f yxy x, y) xy 2. Let fx, y, z) = x 2 + y 2. Find the following. + z2 a) z x fx, y, z) b) fx, y, z) y z c) z 2 fx, y, z) d) 3 fx, y, z) x y z e) f zyx x, y, z) f) f yyy x, y, z) 3. Find the Hessian of each of the following functions. a) fx, y) = 3x 2 y 4xy 3 b) gx, y) = 4e x cos3y) c) gx, y, z) = 4xy 2 z 3 d) fx, y, z) = logx 2 + y 2 + z 2 ) 4. Find the second-order Taylor polynomial for each of the following at the point c. a) fx, y) = xe y, c = 0, 0) b) gx, y) = x sinx + y), c = 0, 0)

8 8 Second-Order Approximations Section 3.4 c) fx, y) = 1 x + y, c = 1, 1) d) gx, y, z) = ex 2y+3z, c = 0, 0, 0) 5. Classify each of the following symmetric 2 2 matrices as either positive definite, negative definite, indefinite, or nondefinite a) b) c) d) e) f) Let M be an n n symmetric nondefinite matrix define q : R n R by qx) = x T Mx. Explain why 1) there exists a vector a 0 such that qa) = 0 2) either qx) 0 for all x in R n or qx) 0 for all x in R n. 7. Suppose f : R n R is C 2 on an open ball B n c, r), fc) = 0, Hfx) is positive definite for all x in B n c, r). Show that fc) < fx) for all x in B n c, r). What would happen if Hfx) were negative definite for all x in B n c, r)? What does this say in the case n = 1? 8. Let a) Show that f x 0, y) = y for all y. b) Show that f y x, 0) = x for all x. c) Show that f yx 0, 0) f xy 0, 0). d) Is f C 2? xyx 2 y 2 ) fx, y) = x 2 + y 2, if x, y) 0, 0), 0, if x, y) = 0, 0).