� 11. Higher derivatives We first record a very useful: Theorem 11.1. Let A ⊂ Rn be an open subset. Let f : A −→ Rm and g : A −→ Rm be two functions and suppose that P ∈ A. Let λ ∈ A be a scalar. If f and g are differentiable at P , then (1) f +g is differentiable at P and D(f +g)(P )= Df(P )+Dg(P ). (2) λf is differentiable at P and D(λf)(P )= λD(f)(P ).· Now suppose that m =1. (3) fg is differentiable at P and D(fg)(P )= D(f)(P )g(P )+f(P )D(g)(P ). (4) If g(P )=0, then fg is differentiable at P and D(f/g)(P )= D(f)(P )g(P ) − f(P )D(g)(P ) . g2(P ) If the partial derivatives of f and g exist and are continuous, then (11.1) follows from the well-known single variable case. One can prove the general case of (11.1), by hand (basically lots of �’s and δ’s). Howevver perhaps the best way to prove (11.1) is to use the chain rule, proved in the next section. What about higher derivatives? Definition 11.2. Let A ⊂ Rn be an open set and let f : A −→ R be a function. The kth order partial derivative of f, with respect to the variables xi1 , xi2 , ... xik is the iterated derivative ∂kf ∂∂ ∂∂f (P )= ( (... () ... ))(P ). ∂xik ∂xik−1 ...∂xi2 ∂xi1 ∂xik ∂xik−1 ∂xi2 ∂xi1 We will also use the notation fxik xik−1 ...xi2 xi1 (P ). Example 11.3. Let f : R2 −→ R be the function f(x,t)= e−at cos x. Then ∂∂ fxx(x,t)= ((e−at cos x))∂x∂x∂ =(−e−at sin x)∂x= −e−at cos x. 1 On the other hand, ∂∂ fxt(x,t)= ((e−at cos x))∂x∂t∂ =(−ae−at cos x)∂x= ae−at sin x. Similarly, ∂∂ ftx(x,t)= ((e−at cos x))∂t∂x∂ =(−e−at sin x)∂t= ae−at sin x. Note that ft(x,t)= −ae−at cos x. It follows that f(x,t) is a solution to the Heat equation: ∂2f ∂f a = . ∂x2 ∂t Definition 11.4. Let A ⊂ Rn be an open subset and let f : A −→ Rm be a function. We say that f is of class Ck if all kth partial derivatives exist and are continuous. We say that f is of class C∞ (aka smooth) if f is of class Ck for all k. In lecture 10 we saw that if f is C1, then it is differentiable. Theorem 11.5. Let A ⊂ Rn be an open subset and let f : A −→ Rm be a function. If f is C2, then ∂2f∂2f = ,∂xi∂xj ∂xj ∂xi for all 1 ≤ i,j ≤ n. The proof uses the Mean Value Theorem. Suppose we are given A ⊂ R an open subset and a function f : A −→ R of class C1 . The objective is to find a solution to the equation f(x)=0. Newton’s method proceeds as follows. Start with some x0 ∈ A. The best linear approximation to f(x) in a neighbourhood of x0 is given by f(x0)+ f�(x0)(x − x0). 2 � If f�(x0) = 0, then the linear equation f(x0)+ f�(x0)(x − x0)=0, has the unique solution, f(x0) x1 = x0 − . f�(x0) Now just keep going (assuming that f�(xi) is never zero), f(x0) x1 = x0 − f�(x0) f(x1) x2 = x1 − f�(x1) .. ...= . xn = xn−1 − ff�((xxnn−−11)) . Claim 11.6. Suppose that x∞ = limn→∞ xn exists and f�(x∞) == 0�. Then f(x∞)=0. Proof of (11.6). Indeed, we have xn = xn−1 − ff�((xxnn−−11)) . Take the limit as n goes to ∞ of both sides: f(x∞) x= x,∞∞ − f�(x∞)we we used the fact that f and f� are continuous and f�(x∞) = 0. But �then f(x∞)=0, as claimed. � Suppose that A ⊂ Rn is open and f : A −→ Rn is a function. Suppoos that f is C1 (that is, suppose each of the coordinate functions f1,f2,...,fn is C1). The objective is to find a solution to the equation f(P )= �0. Start with any point P0 ∈ A. The best linear approximation to f at P0 is given by f(P0)+ Df(P0)−−→PP0. 3 � � Assume that Df(P0) is an invertible matrix, that is, assume that det Df(P0) �= 0. Then the inverse matrix Df(P0)−1 exists and the unique solution to the linear equation f(P0)+ Df(P0)−−→PP0 = �0, is given by P1 = P0 − Df(P0)−1f(P0). Notice that matrix multiplication is not commutative, so that there is a difference between Df(P0)−1f(P0) and f(P0)Df(P0)−1 . If possible, we get a sequence of solutions, P1 = P0 − Df(P0)−1f(P0) P2 = P1 − Df(P1)−1f(P1) .. ...= . Pn = Pn−1 − Df(Pn−1)−1f(Pn−1). Suppose that the limit P∞ = limn→∞ Pn exists and that Df(P∞) is invertible. As before, if we take the limit of both sides, this implies that f(P∞)= �0. Let us try a concrete example. Example 11.7. Solve x 2 + y 2 =1 23 y = x. First we write down an appropriate function, f : R2 −→ R2, given by f(x,y)=(x2 + y2 − 1,y2 − x3). Then we are looking for a point P such that f(P ) = (0, 0). Then ��2x 2yDf(P )= 2 . −3x2y The determinant of this matrix is 4xy +6x 2 y =2xy(2 + 3x). Now if we are given a 2 × 2 matrix ab cd, 4 � � � � then we may write down the inverse by hand, 1 d −b . ad − bc −ca So �� 12y −2yDf(P )−1 =2xy(2 + 3x)3x2 2x So, ��� � Df(P )−1f(P )= 1 2y 2 −2yx2 + 2 y2 −3 1 2xy(2 + 3x)3x2xy− x12x2y − 2y +2x3y= 2xy(2 + 3x) x4 +3x2y2 − 3x2 +2xy2 One nice thing about this method is that it is quite easy to implement on a computer. Here is what happens if we start with (x0,y0) = (5, 2), (x0,y0) = (5.00000000000000, 2.00000000000000) (x1,y1) = (3.24705882352941, −0.617647058823529) (x2,y2) = (2.09875150983980, 1.37996311951634) (x3,y3) = (1.37227480405610, 0.561220968705054) (x4,y4) = (0.959201654346683, 0.503839504009063) (x5,y5) = (0.787655203525685, 0.657830227357845) (x6,y6) = (0.755918792660404, 0.655438554539110), and if we start with (x0,y0) = (5, 5), (x0,y0) = (5.00000000000000, 5.00000000000000) (x1,y1) = (3.24705882352941, 1.85294117647059) (x2,y2) = (2.09875150983980, 0.363541705259258) (x3,y3) = (1.37227480405610, −0.306989760884339) (x4,y4) = (0.959201654346683, −0.561589294711320) (x5,y5) = (0.787655203525685, −0.644964218428458) (x6,y6) = (0.755918792660404, −0.655519172668858). One can sketch the two curves and check that these give reasonable solutions. One can also check that (x6,y6) lie close to the two given curves, by computing x62 + y62 − 1 and y62 − x63 . 5 MIT OpenCourseWarehttp://ocw.mit.edu 18.022 Calculus of Several Variables Fall 2010 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.