THE IMPLICIT FUNCTION THEOREM. g ( x 0 2,x 0 3,..., x 0 n)

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1 THE IMPLICIT FUNCTION THEOREM 11 Statement of the theorem 1 A SIMPLE VERSION OF THE IMPLICIT FUNCTION THEOREM Theorem 1 (Simple Implicit Function Theorem) Suppose that φ is a real-valued functions defined on a domain D and continuously differentiable on an open set D 1 D R n, ( x 1,x 2,, x n) D 1, and Further suppose that φ ( x 1,x 2,, x n) (1) φ ( x 1,x 2,, x n) (2) Then there exists a neighborhood V δ (x 2,x 3,, x n) D 1, an open set W R 1 containing x 1 and a real valued function ψ 1 : V W, continuously differentiable on V, such that ( ) x 1 ψ 1 x 2,x 3,, x n φ ( ) (3) ψ 1 (x 2,x 3,, x n),x 2,x 3,, x n Furthermore for k 2,, n, ψ 1 (x 2,x 3,, x n) φ(ψ 1(x 2,x 3,, x n ),x 2,x 3,, x n ) x k φ(ψ 1(x 2,x 3,, x n ),x 2,x 3,, x n ) φ(x ) φ(x ) (4) We can prove this last statement as follows Define g:v R n as g ( x 2,x 3,, x n) ψ 1 (x 2,, x n) x 2 x 3 x n (5) Then (φ g)(x 2,, x n ) for all (x 2,, x n ) V Thus by the chain rule Date: October 1, 25 1

2 2 THE IMPLICIT FUNCTION THEOREM D (φ g)(x 2,, x n) Dφ ( g(x 2,, x n) ) Dg(x 2, x n) For any k 2, 3,, n, we then have Dφ ( x 1,x 2,, xn) Dg(x 2, x n) x 1 ψ 1 (x x 2,, x n) 2 x Dφ x 2 3 D x 3 x n x n ψ 1 ψ 1 ψ x 3 1 x n [ ] 1 φ, φ,, φ x n 1 [ ] 1 (6) φ(x 1,x 2,,x n ) ψ 1 (x 2,,x n ) + φ(x 1,x 2,,x n ) ψ 1(x 2,,x n) φ(x 1,x 2,,x n ) φ(x 1,x 2,,x n ) We can, of course, solve for any other x k ψ k ( x 1,x 2,,x k 1,x k+1,,x n), as a function of the other x s rather than x 1 as a function of ( x 2,x 3,x n) 12 Examples 121 Production function Consider a production function given by (7) Write this equation implicitly as y f(x 1,x 2, x n ) (8) φ(x 1,x 2,, x n,y) f(x 1,x 2, x n ) y (9) where φ is now a function of n+1 variables instead of n variables Assume that φ is continuously differentiable and the Jacobian matrix has rank 1 ie, φ f, j 1, 2,,n (1) Given that the implicit function theorem holds, we can solve equation 9 for x k as a function of y and the other x s ie Thus it will be true that or that x k ψ k (x 1,x 2,, x k 1,x k+1,, y) (11) φ(y, x 1,x 2,, x k 1,x k,x k+1,, x n ) (12)

3 THE IMPLICIT FUNCTION THEOREM 3 y f(x 1,x 2,, x k 1,x k,x k+1,, x n ) (13) Differentiating the identity in equation 13 with respect to x j will give or Or we can obtain this directly as f + f f (14) RT S MRS (15) f φ(y, x 1,x 2,, x k 1,ψ k,x k+1,, x n ) ψ k φ(y, x 1,x 2,, x k 1,ψ k,x k+1,, x n ) Consider the specific example φ φ f f MRS (16) y f (x 1,x 2 ) x 2 1 2x 1 x 2 + x 1 x 3 2 To find the partial derivative of x 2 with respect to x 1 we find the two partials of f as follows f f 2 x 1 2 x 2 + x x 1 x x 1 2 x 2 2 x 1 x x 1 x x Utility function u u (x 1,x 2 ) To find the partial derivative of x 2 with respect to x 1 we find the two partials of U as follows u u 123 Another utility function Consider a utility function given by This can be written implicitly as u x x x f (u, x 1,x 2,x 3 ) u x x x 1 6 3

4 4 THE IMPLICIT FUNCTION THEOREM Now consider some of the partial derivatives arising from this implicit function First consider the marginal rate of substitution of x 1 for x 2 f f ( 1 3 ) x x 3 2 x ( 1 4 ) 3 x 4 1 x x x 1 3 x 2 Now consider the marginal rate of substitution of x 2 for x 3 Now consider the marginal utility of x 2 x 3 u f x 3 f ( 1 6 ) x x x ( 1 3 ) x x x x 2 2 x 3 f f u ( 1 3 ) x x 3 2 x x x x THE GENERAL IMPLICIT FUNCTION THEOREM WITH P INDEPENDENT VARIABLES AND M IMPLICIT EQUATIONS 21 Motivation for implicit function theorem We are often interested in solving implicit systems of equations for m variables, say x 1,x 2,,x m in terms of m+p variables where there are a minimum of m equations in the system We typically label the variables x m+1,x m+2,,x m+p,y 1,y 2,,y p We are frequently interested in the derivatives xi where it is implicit that all other x k and all y l are held constant The conditions guaranteeing that we can solve for m of the variables in terms of p variables along with a formula for computing derivatives are given by the implicit function theorem One motivation for the implicit function theorem is that we can eliminate m variables from a constrained optimization problem using the constraint equations In this way the constrained problem is converted to an unconstrained problem and we can use the results on unconstrained problems to determine a solution 22 Description of a system of equations Suppose that we have m equations depending on m + p variables (parameters) written in implicit form as follows

5 THE IMPLICIT FUNCTION THEOREM 5 φ 1 (x 1,x 2,,x m,y 1,y 2,,y p ) φ 2 (x 1,x 2,,x m,y 1,y 2,,y p ) φ m (x 1,x 2,,x m,y 1,y 2,,y p ) (17) For example with m 2 and p 3, we might have φ 1 (x 1,x 2,p,w 1,w 2 ) (4) px1 6 x2 2 w 1 φ 2 (x 1,x 2,p,w 1,w 2 ) (2) px1 4 x2 8 w 2 where p, w 1 and w 2 are the independent variables y 1,y 2,andy 3 (18) 23 Jacobian matrix of the system The Jacobian matrix of the system in 17 is defined as matrix of first partials as follows J φ m φ m This matrix will be of rank m if its determinant is not zero x m x m φ m x m (19) 24 Statement of implicit function theorem Theorem 2 (Implicit Function Theorem) Suppose that φ i are real-valued functions defined on a domain D and continuously differentiable on an open set D 1 D R m+p, where p > and φ 1 (x 1,x 2,, x m,y 1,y 2,, y p) φ 2 (x 1,x 2,, x m,y 1,y 2,, y p) (2) We often write equation 2 as follows φ m (x 1,x 2,, x m,y 1,y 2,, y p) (x,y ) D 1 φ i (x,y ), i 1, 2,, m, and (x,y ) D 1 (21) [ ] φi(x Assume the Jacobian matrix,y ) has rank m Then there exists a neighborhood N δ (x,y ) D 1, an open set D 2 R p containing y and real valued functions ψ k, k 1, 2,, m, continuously differentiable on D 2, such that the following conditions are satisfied:

6 6 THE IMPLICIT FUNCTION THEOREM x 1 ψ 1 (y ) x 2 ψ 2 (y ) (22) For every y D 2, we have x m ψ m (y ) φ i (ψ 1 (y), ψ 2 (y),, ψ m (y), y 1,y 2,, y p ), i 1, 2,, m or (23) φ i (ψ(y),y), i 1, 2,, m We also have that for all (x,y) N δ (x,y ), the Jacobian matrix [ ] φi(x, y) the partial derivatives of ψ(y) are the solutions of the set of linear equations has rank m Furthermore for y D 2, m k1 m k1 m k1 (ψ(y),y) (ψ(y),y) φ m (ψ(y),y) ψ k (y) ψ k (y) ψ k (y) We can write equation 24 in the following alternative manner (ψ(y),y) (ψ(y),y) φ m(ψ(y),y) (24) m k1 φ i (ψ(y),y) φ m ψ k (y) or perhaps most usefully as the following matrix equation φ 1 x m φ 2 x m φ m This of course implies ψ 1(y) ψ 2(y) ψ m(y) φ m x m φ m φ i(ψ(y),y) i 1, 2,, m (25) ψ 1(y) ψ 2(y) ψ m(y) φ m x m x m φ m x m 1 (ψ(y),y) (ψ(y),y) φ m(ψ(y),y) (ψ(y),y) (ψ(y),y) φ m(ψ(y),y) (26) (27)

7 THE IMPLICIT FUNCTION THEOREM 7 We can expand equation 27 to cover the derivatives with respect to the other y l by expanding the matrix equation ψ 1 (y) 1 ψ 2 (y) 1 ψ m(y) 1 ψ 1 (y) 2 ψ 2 (y) 2 ψ m(y) 2 ψ 1 (y) p ψ 2 (y) p ψ m(y) p φ m φ m 1 (ψ(y),y) x m 1 (ψ(y),y) x m 1 φ m φ m (ψ(y),y) x m 1 (ψ(y),y) 2 (ψ(y),y) 2 φ m(ψ(y),y) 2 (ψ(y),y) p (ψ(y),y) p φ m(ψ(y),y) p (28) 25 Some intuition for derivatives computed using the implicit function theorem To see the intuition of equation 24, take the total derivative of φ i in equation 23 with respect to y j as follows φ i ψ 1 ψ 1 φ i (ψ 1 (y), ψ 2 (y),,ψ m (y),y) + φ i ψ 2 ψ φ i ψ m and then move φi equations to obtain m equations in the m partial derivatives, xi ψ m + φ i to the right hand side of the equation Then perform a similar task for the other For the case of only one implicit equation 24 reduces to m k1 φ(ψ(y), y) ψi (29) ψ k (y) φ(ψ(y), y) (3) If there is only one equation, we can solve for only one of the x variables in terms of the other x variables and the p y variables and obtain only one implicit derivative which can be rewritten as φ(ψ(y), y) ψ k (y) φ(ψ(y), y) (31) ψ k (y) This is more or less the same as equation 4 φ(ψ(y), y) φ(ψ(y), y) (32) If there are only two variables, x 1 and x 2 where x 2 is now like y 1, we obtain φ(ψ 1 (x 2 ),x 2 ) ψ 1 (x 2 ) φ(ψ 1(x 2 ),x 2 ) ψ 1(x 2 ) φ(ψ 1(x 2),x 2) φ(ψ 1(x 2),x 2) which is like the example in section 121 where φ takes the place of f (33) 26 Examples

8 8 THE IMPLICIT FUNCTION THEOREM 261 One implicit equation with three variables φ(x 1,x 2,y ) (34) The implicit function theorem says that we can solve equation 34 for x 1 as a function of x 2 and y, ie, and that The theorem then says that x 1 ψ 1 (x 2,y ) (35) φ(ψ 1 (x 2,y), x 2,y) (36) 262 Production function example φ(ψ 1 (x 2,y), x 2,y) ψ 1 φ(ψ 1(x 2,y), x 2,y) (x 2,y) (x 2,y) φ(x 1,x 2,y ) y f(x 1,x 2) The theorem says that we can solve the equation for x 1 φ(ψ 1(x 2,y), x 2,y) φ(ψ 1(x 2,y), x 2,y) φ(ψ1(x2,y),x2,y) φ(ψ 1(x 2,y),x 2,y) (37) (38) It is also true that x 1 ψ 1 (x 2,y ) (39) φ(ψ 1 (x 2,y), x 2,y) y f(ψ 1 (x 2,y),x 2 ) Now compute the relevant derivatives (4) The theorem then says that φ(ψ 1 (x 2,y), x 2,y) f(ψ 1(x 2,y),x 2 ) φ(ψ 1 (x 2,y), x 2,y) f(ψ 1(x 2,y),x 2 ) (x 2,y) [ φ(ψ1(x 2,y),x 2,y) φ(ψ 1(x 2,y),x 2,y) [ f(ψ 1(x 2,y),x 2) f(ψ1(x2,y),x2) f(ψ 1(x 2,y),x 2) f(ψ 1(x 2,y),x 2) ] ] (41) (42)

9 THE IMPLICIT FUNCTION THEOREM 9 27 General example with two equations and three variables Consider the following system of equations The Jacobian is given by φ 1 (x 1,x 2,y) 3x 1 +2x 2 +4y φ 2 (x 1,x 2,y) 4x 1 + x 2 + y φ [ φ1 1 ] [ ] We can solve system 43 for x 1 and x 2 as functions of y Move y to the right hand side in each equation (43) (44) 3x 1 +2x 2 4y (45a) 4x 1 + x 2 y (45b) Now solve equation 45b for x 2 x 2 y 4x 1 (46) Substitute the solution to equation 46 into equation 45a and simplify 3x 1 +2( y 4x 1 ) 4y 3x 1 2y 8x 1 5x 1 4y 2y (47) x y ψ 1(y) Substitute the solution to equation 47 into equation 46 and simplify x 2 y 4 [ ] 2 5 y x y 8 5 y (48) 13 5 y ψ 2(y) If we substitute these expressions for x 1 and x 2 into equation 43 we obtain and ( ) [ ] φ 1 y, 5 5 y, y ) 3 5 y +2 [ 135 ] y 6 5 y 26 5 y y 2 5 y y +4y (49)

10 1 THE IMPLICIT FUNCTION THEOREM ( ) [ ] φ 2 y, 5 5 y, y ) 4 5 y + [ 13 ] 5 y + y Furthermore 8 5 y 13 5 y y 13 5 y 13 5 y (5) ψ 1 ψ We can solve for these partial derivatives using equation 24 as follows (51) ψ 1 + ψ 2 ψ 1 + ψ 2 (52a) (52b) Now substitute in the derivatives of φ 1 and φ 2 with respect to x 1,x 2, and y 3 ψ 1 +2 ψ 2 4 ψ 1 +1 ψ 2 4 (53a) 1 (53b) Solve equation 53b for ψ2 ψ ψ 1 Now substitute the answer from equation 54 into equation 53a (54) 3 ψ ( ψ ) ψ ψ 1 5 ψ 1 If we substitute equation 55 into equation 54 we obtain ψ (55)

11 THE IMPLICIT FUNCTION THEOREM 11 ψ 2 ψ ψ 1 ( ) We could also do this by inverting the matrix (56) 271 Profit maximization example 1 Consider the system in equation 18 We can solve this system of m implicit equations for all m of the x variables as functions of the p independent (y) variables Specifically, we can solve the two equations for x 1 and x 2 as functions ( of ) p, w 1, and w 2, ie, x 1 ψ 1 (p, w 1,w 2 ) and x 2 ψ 2 (p, w 1,w 2 ) We can also find x1 x2 and, that is ψi, from the following two equations derived from equation 18 which is repeated here φ 1 (x 1,x 2,p,w 1,w 2 ) (4) px1 6 x2 2 w 1 φ 2 (x 1,x 2,p,w 1,w 2 ) (2) px1 4 x2 8 w 2 [ ] ( 24) px 16 1 x [ (8) px1 6 x 2 8 ] [ (8) px 6 1 x2 8 ] + [ ( 16) px1 4 x 2 18 ] We can write this in matrix form as [ ( 24) px 16 1 x2 2 (8) px1 6 x2 8 (8) px1 6 x2 8 ( 16) px1 4 x 2 18 ][ ] x1 (4) x 6 1 x 2 2 (2) x4 1 x2 8 [ ] (4) x 6 1 x 2 2 (2) x 4 1 x 2 8 (57) (58) 272 Profit maximization example 2 Let the production function for a firm by given by Profit for the firm is given by y 14x 1 +11x 2 x 2 1 x 2 2 (59) π py w 1 x 1 w 2 x 2 p (14x 1 +11x 2 x 2 1 x 2 (6) 2) w 1 x 1 w 2 x 2 The first order conditions for profit maximization imply that π 14px 1 +11px 2 px 2 1 px 2 2 w 1 x 1 w 2 x 2 π φ 1 14p 2 px 1 w 1 π φ 1 11p 2 px 2 w 2 We can solve the first equation for x 1 as follows (61)

12 12 THE IMPLICIT FUNCTION THEOREM π 14p 2 px 1 w 1 2 px 1 14p w 1 x 1 14 p w 1 2 p 7 w 1 2 p In a similar manner we can find x 2 from the second equation π 11p 2 px 2 w 2 2 px 2 11p w 2 x 2 11 p w 2 2 p 55 w 2 2 p We can find the derivatives of x 1 and x 2 with respect to p, w 1 and w 2 directly as follows: (62) (63) x 1 x w 1 p w 2 p w 1 p 2 w p 1 w 2 (64) 1 2 w 2 p 2 w p 1 w 2 We can also find these derivatives using the implicit function theorem The two implicit equations are φ 1 (x 1,x 2,p,w 1,w 2 ) 14p 2 px 1 w 1 φ 2 (x 1,x 2 p, w 1,w 2 ) 11p 2 px 2 w 2 (65) First we check the Jacobian of the system It is obtained by differentiating φ 1 and φ 2 with respect to x 1 and x 2 as follows [ ] φ1 [ ] x J 1 2 p (66) 2p The determinant is 4p 2 which is positive Now we have two equations we can solve using the implicit function theorem The theorem says

13 THE IMPLICIT FUNCTION THEOREM 13 m k1 φ i (g (y), y) For the case of two equations we obtain g k (y) φ i(ψ(y), y),i 1, 2, (67) + + We can write this in matrix form as follows Solving for x1 [ φ1 x2 and we obtain [ ] x1 [ φ1 ][ x1 ] [ ] φ2 ] 1 [ ] φ2 We can compute the various partial derivatives of φ as follows (68) (69) (7) φ 1 (x 1,x 2,p,w 1,w 2 ) 14p 2 px 1 w 1 2 p 14 2 x 1 φ 2 (x 1,x 2 p, w 1,w 2 ) 11p 2 px 2 w 2 2 p 11 2 x 2 Now writing out the system we obtain for the case at hand we obtain [ x1 ] [ φ1 Now substitute in the elements of the inverse matrix [ x1 If we then invert the matrix we obtain ] ] 1 [ ] φ2 [ 2p ] 1 [ ] 2 x1 14 2p 2 x 2 11 (71) (72) (73)

14 14 THE IMPLICIT FUNCTION THEOREM Given that we obtain [ x1 ] x 1 [ 1 2p 1 2p ][ ] 2 x x 2 11 [ ] (74) 7 x1 p 55 x 2 p w 1 p 1 x w 2 p 1 (75) 7 x 1 p 7 (7 1 2 w 1 p 1 ) p 1 2 w 1 p 2 55 x 2 p 55 ( w 2 p 1 ) p (76) which is the same as before 1 2 w 2 p Profit maximization example 3 Verify the implicit function theorem derivatives x1 for the profit maximization example 2 w 1, w 2, w 1, x2 w Profit maximization example 4 A firm sells its output into a perfectly competitive market and faces a fixed price p It hires labor in a competitive labor market at a wage w, and rents capital in a competitive capital market at rental rate r The production is f(l, K) The production function is strictly concave The firm seeks to maximize its profits which are π pf(l, K) wl rk (77) The first-order conditions for profit maximization are π L pf L (L,K ) w π K pf K (L,K ) r This gives two implicit equations for K and L The second order conditions are (78) 2 π L 2 (L,K ) <, We can compute these derivatives as 2 π L 2 2 π K 2 [ 2 ] 2 π > at (L,K ) (79) L K

15 THE IMPLICIT FUNCTION THEOREM 15 2 π (pf L (L,K ) w ) L 2 L pf LL 2 π K (pf K (L,K ) r ) 2 L pf KK (8) 2 π L K (pf L (L,K ) w ) K pf KL The first of the second order conditions is satisfied by concavity of f We then write the second condition as D pf LL pf LK pf KL pf KK > p 2 f LL f LK > (81) f KL f KK p 2 (f LL f KK f KL f LK ) > The expression is positive or at least non-negative because f is assumed to be strictly concave Now we wish to determine the effects on input demands, L and K, of changes in the input prices Using the implicit function theorem in finding the partial derivatives with respect to w, we obtain for the first equation L π LL w + π K LK w + π Lw L pf LL w + pf K KL w 1 For the second equation we obtain (82) L π KL w + π K KK w + π Kw L pf KL w + pf K (83) KK w We can write this in matrix form as [ ][ ] [ ] pfll pf LK L / w 1 pf KL pf KK K (84) / w Using Cramer s rule, we then have the comparative-statics results L w 1 pf LK pf KK pf KK D D K w pf (85) LL 1 pf KL pf KL D D The sign of the first is negative because f KK < and D > from the second-order conditions Thus, the demand curve for labor has a negative slope However, in order to sign the effect of a change in the wage rate on the demand for capital, we need to know the sign of f KL, the effect of a change in the labor input

16 16 THE IMPLICIT FUNCTION THEOREM on the marginal product of capital We can derive the effects of a change in the rental rate of capital in a similar way 275 Profit maximization example 5 Let the production function be Cobb-Douglas of the form y L α K β Find the comparative-static effects L / w and K / w Profits are given by π pf(l, K) wl rk pl α K β wl rk The first-order conditions are for profit maximization are π L αpl α 1 K β w π K βpl α K β 1 r This gives two implicit equations for K and L The second order conditions are (86) (87) 2 π L 2 (L,K ) <, We can compute these derivatives as 2 π L 2 2 π K 2 [ 2 ] 2 π > at (L,K ) (88) L K 2 π L 2 α (α 1) pl α 2 K β 2 π K 2 β (β 1) plα K β 2 (89) 2 π L K αβplα 1 K β 1 The first of the second order conditions is satisfied as long as α, β<1 We can write the second condition as D α (α 1) plα 2 K β αβpl α 1 K β 1 αβpl α 1 K β 1 β (β 1) pl α K β 2 > (9) We can simplify D as follows D α (α 1) plα 2 K β αβpl α 1 K β 1 αβpl α 1 K β 1 β (β 1) pl α K β 2 The condition is then that p 2 α (α 1) L α 2 K β αβ L α 1 K β 1 αβ L α 1 K β 1 β (β 1) L α K β 2 p [ 2 αβ(α 1) (β 1) L α 2 K β L α K β 2 α 2 β 2 L 2α 2 2β K 2] p [ 2 αβl α 2 K β L α K β 2 ((α 1) (β 1) αβ ) ] p [ 2 αβl α 2 K β L α K β 2 ( αβ β α +1 αβ ) ] p [ 2 αβl α 2 K β L α K β 2 (1 α β ) ] (91) D p 2 αβl 2α 2 K 2β 2 (1 α β) > (92) This will be true if α+β <1

17 THE IMPLICIT FUNCTION THEOREM 17 Now we wish to determine the effects on input demands, L and K, of changes in the input prices Using the implicit function theorem in finding the partial derivatives with respect to w, we obtain for the first equation α (α 1) pl α 2 K β L w For the second equation we obtain π LL L w + π LK K w + π Lw + αβplα 1 K β 1 K w 1 (93) L π KL w αβpl α 1 K β 1 L + π KK We can write this in matrix form as [ α (α 1) pl α 2 K β αβpl α 1 K β 1 K w + π Kw w + β (β 1) plα K β 2 αβpl α 1 K β 1 β (β 1) pl α K β 2 Using Cramer s rule, we then have the comparative-statics results L 1 αβplα 1 K β 1 w β (β 1) pl α K β 2 D ][ L / w K / w ] β (β 1) plα K β 2 D (96) K α (α 1) plα 2 K β 1 w αβpl α 1 K β 1 αβplα 1 K β 1 D D The sign of the first of is negative because pβ(β 1) L α K β 2 < (β <1), and D > by the second order conditions The cross partial derivative K w is also less than zero because pαβlα 1 K β 1 > and D > So, for the case of a two-input Cobb-Douglas production function, an increase in the wage unambiguously reduces the demand for capital 3 TANGENTS TO φ(x 1, x 2,,)c AND PROPERTIES OF THE GRADIENT 31 Direction numbers A direction in R 2 is determined by an ordered pair of two numbers (a,b), not both zero, called direction numbers The direction corresponds to all lines parallel to the line through the origin (,) and the point (a,b) The direction numbers (a,b) and the direction numbers (ra,rb) determine the same direction, for any nonzero r A direction in R n is determined by an ordered n-tuple of numbers ( x 1,x 2,,xn), not all zero, called direction numbers The direction corresponds to all lines parallel to the line through the origin (,,,) and the point ( ( ( ) x 1,x 2,,xn) The direction numbers x 1,x 2,,xn) and the direction numbers rx 1,rx 2,,rx n determine the same direction, for any nonzero r If pick an r in the following way [ 1 ] (94) (95) then r 1 (x 1 ) 2 +(x 2 )2 + +(x n) 2 1 x (97)

18 18 THE IMPLICIT FUNCTION THEOREM cos γ 1 x 1 (x 1 ) 2 +(x 2 )2 + +(x n) 2 cos γ 2 x 2 (x 1 ) 2 +(x 2 )2 + +(x n) 2 (98) cos γ n x n (x 1 ) 2 +(x 2 )2 + +(x n )2 where γ j is the angle of the vector running through the origin and the point ( x 1,x 2,,x n) with the xj axis With this choice of r, the direction numbers (ra,rb) are just the cosines of the angles that the line makes with the positive x 1,x 2, and x n axes These angles γ 1, γ 2, are called the direction angles, and their cosines (cos[γ 1 ], cos[γ 2 ],) are called the direction cosines of that direction See figure 1 They satisfy cos 2 [γ 1 ] + cos 2 [γ 2 ]+ + cos 2 [γ n ] 1 FIGURE 1 Direction numbers as cosines of vector angles with the respective axes x x 1 1 x 1 x 3 x 2 x 3 5,, x 1 From equation 98, we can also write x ( x 1,x 2,,x ) ( n x cos[γ 1 ], x cos[γ 2 ],, x ) cos[γ n x (cos[γ 1 ], cos[γ 2 ],,cos[γ n ]) Therefore (99)

19 THE IMPLICIT FUNCTION THEOREM 19 1 x x (cos[γ 1 ], cos[γ 2 ],,cos[γ n ]) (1) which says that the direction cosines of x are the components of the unit vector in the direction of x Theorem 3 Ifθ is the angle between the vectors a abd b, then a b a b cos θ (11) 32 Planes in R n 321 Planes through the origin A plane through the origin in R n is given implicitly by the equation p x p 1 x 1 + p 2 x p n x n (12) 322 More general planes in R n A plane in R n through the point ( x 1,x 2,,x n) is given implicitly by the equation p (x x ) (13) p 1 (x 1 x 1 )+p 2(x 2 x 2 )+ + p n(x n x n ) We say that the vector p is orthogonal to the vector ( x x ) or x 1 x 1 [ ] x 2 x 2 p1 p 2 p n (14) x n x n The coefficients [p 1,p 2,,p n ] are the direction numbers of a line L passing through the point x or alternatively they are the coordinates of a point on a line through the origin that is parallel to L If p 1 is equal to -1, we can write equation 14 as follows x 1 x 1 p 2(x 2 x 2 )+p 3(x 2 x 3 )+ + p n(x n x n ) (15) 33 The general equation for a tangent hyperplane Theorem 4 Suppose D 1 R n is open, ( x 1,x 2,,x n) D 1, φ : D 1 R 1 is continuously differentiable, and Dφ ( x 1,x 2,,x n) Suppose φ ( x 1,x 2,,x n) c Then consider the level surface of the function φ (x 1,x 2,,x n ) φ ( x 1,x 2,,x n) c Denote this level service by M φ 1 ({c}) which consists all all values of (x 1,x 2,,x n ) such that φ(x 1,x 2,,x n ) c Then the tangent hyperplane at ( x 1,x 2,,x n) of M is given by T x M {x R n : Dφ(x )(x x )} (16) that is φ(x ) is nomal to the tangent hyperplane

20 2 THE IMPLICIT FUNCTION THEOREM Proof Because Dφ(x ), we may assume without loss of generality that φ (x ) Apply the implicit function theorem (Theorem 1) to the function φ c We then know that the level set can be expressed as a graph of the function x 1 ψ 1 (x 2,x 3,,x n ) for some continuous function ψ 1 Now the tangent plane to the graph x 1 ψ 1 (x 2,x 3,,x n ) at the point x [ψ 1 (x 2,x 3,,x n ),x 2,x 3,x n ] is the graph of the gradient ( ) Dψ 1 (x 2,x 3,,x n) translated so that it passes through x Using equation 15 where ψ1 x 2,x 3,,x n are the direction numbers we obtain n x 1 x ψ 1 ( ) 1 x,x 3,,x n (xk x k) (17) k k2 Now make the substitution from equation 4 in equation 17 to obtain x 1 x 1 Rearrange equation 18 to obtain n k2 34 Example Consider a function n k2 n k2 φ(ψ 1(x 2,x 3,, x n ),x 2,x 3,, x n ) φ(ψ 1(x 2,x 3,, x n ),x 2,x 3,, x n ) φ(x ) φ(x ) (x k x k) (18) (x k x k ) φ(x ) (x k x k)+ φ(x ) (x 1 x 1) Dφ(x )(x x ) (19) y f (x 1,x 2,,x n ) (11) evaluated at the point ( x 1,x 2,,x n ) y f (x 1,x 2,,x n) (111) The equation of the hyperplane tangent to the surface at this point is given by (y y ) f (x 1 x x 1)+ f (x 2 x 2 )+ + f (x n x 1 x n) n f (x 1 x x 1)+ f (x 2 x 2 )+ + f (x n x 1 x n) ( y y ) n f(x 1,x 2,,x n ) f(x 1,x 2,,x n)+ f (x 1 x x 1)+ f (x 2 x 1 x 2)+ + f (x n x 2 x n) n where the partial derivatives f x i are evaluated at ( x 1,x 2,,x n ) We can also write this as (112) f (x) f (x ) f (x )(x x ) (113) Compare this to the tangent equation with one variable y f (x ) f (x )(x x ) y f (x )+f (x )(x x ) and the tangent plane with two variables (114)

21 THE IMPLICIT FUNCTION THEOREM 21 y f(x 1,x 2) f (x 1 x 1)+ f (x 2 x 2) y f(x 1,x 2)+ f (x 1 x 1)+ f (x 2 x 2) 35 Vector functions of a real variable Let x 1,x 2,,x n be functions of a variable t defined on an interval I and write x(t) (x 1 (t),x 2 (t),,x n (t)) (116) The function t x(t) is a transformation from R to R n and is called a vector function of a real variable As x runs through I, x(t) traces out a set of points in n-space called a curve In particular, if we put x 1 (t) x 1 + ta 1,x 2 (t) x 2 + ta 2,,x n (t) x n + ta n (117) the resulting curve is a straight line in n-space It passes through the point x (x 1,x 2,,x n ) at t, and it is in the direction of the vector a (a 1,a 2,,a n ) We can define the derivative of x(t) as dx dt ẋ (t) ( dx1 (t) dt, dx 2(t) dt,, dx n(t) dt If K is a curve in n-space traced out by x(t), then ẋ (t) can be interpreted as a vector tangent to K at the point t 36 Tangent vectors Consider the function r(t) (x 1 (t), x 2 (t),,x n (t))and its derivative ṙ (t) (ẋ 1 (t), ẋ 2 (t),,ẋ n (t)) This function traces out a curve in R n We can call the curve K For fixed t, r (t + h) r (t) ṙ (t) lim (119) h h If ṙ (t), then for t + h close enough to t, the vector r(t+h) - r(t) will not be zero As h tends to, the quantity r(t+h) - r(t) will come closer and closer to serving as a direction vector for the tangent to the curve at the point P This can be seen in figure 2 It may be tempting to take this difference as approximation of the direction of the tangent and then take the limit lim [ r (t + h) r (t)] (12) h and call the limit the direction vector for the tangent But the limit is zero and the zero vector has no direction Instead we use the a vector that for small h has a greater length, that is we use ) (115) (118) r (t + h) r (t) h For any real number h, this vector is parallel to r(t+h) - r(t) Therefore its limit r (t + h) r (t) ṙ (t) lim h h can be taken as a direction vector for the tangent line (121) (122)

22 22 THE IMPLICIT FUNCTION THEOREM FIGURE 2 Tangent to a Curve 37 Tangent lines, level curves and gradients Consider the equation f (x) f (x 1,x 2,,x n ) c (123) which defines a level curve for the function f Let x (x 1,x 2,,x n ) be a point on the surface and let x (t) (x 1 (t), x 2 (t),,x n (t)) represent a differentiable curve K lying on the surface and passing through x attt Because K lies on the surface, f [x(t)] f [x 1 (t), x 2 (t),,,, x n (t)] c for all t Now differentiate equation 123 with respect to t f(x) t + f(x) t + + f(x) t f (x ) ẋ (t ) Because the vector ẋ (t) (ẋ 1 (t), ẋ 2 (t),, ẋ n (t))has the same direction as the tangent to the curve Katx, the gradient of f is orthogonal to the curve K at the point x Theorem 5 Suppose f(x 1,x 2,,x n ) is continuous and differentiable in a domain A and suppose that x(x 1,x 2,,x n ) A The gradient (124)

23 THE IMPLICIT FUNCTION THEOREM 23 ( f f, f,, f ) x n has the following properties at points x where f(x) (i) f(x) is orthogonal to the level surfaces f(x 1,x 2,,x n )c (ii) f(x) points in the direction of the steepest increase on f (iii) f(x) is the value of the directional derivative in the direction of steepest increase

24 24 THE IMPLICIT FUNCTION THEOREM REFERENCES [1] Hadley, G Linear Algebra Reading, MA: Addison-Wesley, 1961 [2] Sydsaeter, Knut Topics in Mathematical Analysis for Economists New York: Academic Press, 1981