6. Implicit functions If a function is given by the expression y = f (x), i.e. the value of the dependent variable y is given by the formula in the independent variable x, we say the function f is given explicitly ( f is an explicit function). For instance, the linear function y = 7x + 3 is an explicit function. If the dependence of y on x is given by an equation in both x and y, for instance x2 + y2 = 1, we say the function y(x) is given implicitly (y is an implicit function). However, not every equation in x and y defines an implicit function y(x), for instance x2 + y2 + 1 = 0. 6.1 Implicit functions of one variable Definition 6.1.1 An equation F(x;y) = 0 defines an implicit function y = f (x) on a neighbourhood of a point (x0;y0) if 1. F(x0;y0) = 0, 2. 9d > 0;e > 0 such that 8x 2 Od (x0) 9!y 2 Oe (y0) : F(x;y) = 0: R The function y = f (x) is implicitly defined by the equation F(x;y) = 0 in the neighbourhood of the point (x0;y0). Thus, • f is a function of one real variable, • f is defined for all x 2 (x0 − d;x0 + d) for a d > 0, • the values y = f (x) lie within the interval (y0 − e;y0 + e) for an e > 0, • for an x 2 (x0 − d;x0 + d) there exists exactly one value y = f (x) 2 (y0 − e;y0 + e) such that the equality F(x;y) = 0 holds, i.e. the solution to the equation F(x;y) = 0 within the rectangle (x0 − d;x0 + d) × (y0 − e;y0 + e) is the set f(x; f (x)) : x 2 (x0 − d;x0 + d)g and thus F(x; f (x)) = 0 on (x0 − d;x0 + d); • f (x0) = y0. 2 k Theorem 6.1.1 Let G ⊆ R be an open set and let F 2 C (G) for k 2 N. Let (x0;y0) 2 G be a point such that ¶F F(x ;y ) = 0 and (x ;y ) 6= 0: 0 0 ¶y 0 0 Then the equation F(x;y) = 0 defines an implicit function y = f (x) on a neighbourhood of a point (x0;y0). k Moreover, f 2 C (x0 − d;x0 + d) for a d > 0. 66 Chapter 6. Implicit functions R • The existence and continuity of higher order partial derivatives of F implies the existence and k continuity of higher order derivatives of f . More precisely, if F 2C ((x0 −d;x0 +d)×(y0 −e;y0 +e)) k then f 2 C (x0 − d;x0 + d). • The existence of a point (x0;y0) such that F(x0;y0) = 0 is necessary for the existence of an implicit function y = f (x). It specifies a point lying on the graph of f . If such (x0;y0) did not exist, there would not exist a function y = f (x) implicitly defined by F(x;y) = 0, i.e. satisfying F(x; f (x)) = 0. • The graph of implicitly defined function is a subset of the contour line of the graph of F(x;y) ¶F corresponding to the cutting plane z = 0. The condition (x ;y ) 6= 0 implies this contour line is the ¶y 0 0 graph of a function in x in a neighbourhood of the point (x0;y0). Hence, the equation F(x;y) = 0 has, except the solution (x0;y0), infinitely many solutions of the form (x; f (x)). The set of these solutions is the graph of the implicit function f . Figure 6.1: Relation between the graphs of f and F Theorem 6.1.1 is referred to as the theorem on the existence of implicit functions. It does not say anything about how large d and e in Definition 6.1.1 could be, the explicit expression of the implicit function is neither specified. Often it is even impossible to determine explicit expressions of implicit functions. Example 6.1 Let us consider the equation F(x;y) = x2 + y2 − 1 = 0 ¶F describing a circle. Because F(0;1) = 0, the point (0;1) lies on the circle. Because (0;1) = 2 6= 0, the ¶y equation F(x;y) = 0 defines an implicit function y = f (x) in the neighbourhood of (0;1). Because the equation F(x;y) = x2 + y2 − 1 = 0 is simple, we can even derive the explicit expression of the implicit function to be p y = f (x) = 1 − x2: ¶F The following four examples will demonstrate the necessity of the assumption (x ;y ) 6= 0 in Theorem ¶y 0 0 6.1.1 to be able to conclude the existence of an implicit function. ¶F Example 6.2 The point (x ;y ) = (0;0) satisfies the equation F(x;y) = x2 + y2 = 0. Further, (x ;y ) = 0 0 ¶y 0 0 2y0 = 0. Therefore, the assumptions of Theorem 6.1.1 are not satisfied and we cannot conclude anything about the existence of an implicit function in a neighbourhood of (0;0). However, because the equation F(x;y) = x2 + y2 = 0 is satisfied only for the point (0;0), there does not exist an implicit function defined in the neighbourhood of (0;0). Example 6.3 The equation F(x;y) = x2 + 2y2 − 1 = 0 6.1 Implicit functions of one variable 67 2 ¶F defines an ellipse in R . The point (x0;y0) = (1;0) is such that F(1;0) = 0 and (x0;y0) = 4y0 = 0, thus the ¶y assumptions of Theorem 6.1.1 are not satisfied. The equation F(x;y) = 0 does not define an implicit function in a neighbourhood of the point (1;0) from two reasons: 1. There does not exist y 2 R such that F(x;y) = 0 for x > 1. 2. For every x 2 (1 − d;1), where 0 < d < 2, there are two values r r 1 − x2 1 − x2 y = and y = − 1 2 2 2 such that both (x;y1) and (x;y2) satisfy the equation F(x;y) = 0. Hence, in arbitrarily small neighbourhood q q 1−x2 1−x2 of (x0;y0) there are always two distinct points (x;y1 = 2 ) and (x;y2 = − 2 ) satisfying the equation F(x;y) = 0. Example 6.4 The set of all points satisfying the equation F(x;y) = y2 − x2 = 0 is the set of points of the lines y = x and y = −x. For every x close to 0 there exist two values y1 = x and y2 = −x close to 0 such that F(x;y1;2) = 0. Therefore, the equation y2 − x2 = 0 does not define an implicit function in the neighbourhood of (0;0). Note ¶F that F(0;0) = 0 = (0;0). ¶y However, the equation y2 − x2 = 0 defines the implicit function y = x for instance in the neighbourhood ¶F of the point (x ;y ) = (2;2). Note that F(2;2) = 0 and (2;2) = 4 6= 0 which means the assumptions of 0 0 ¶y Theorem 6.1.1 are satisfied. Example 6.5 Let us consider the equation F(x;y) = y3 − x = 0: Theorem 6.1.1 cannot be applied to decide the existence of an implicit function in the neighbourhood of (0;0) ¶F since F(0;0) = 0 and (0;0) = 0. However, the equation F(x;y) = 0 is equivalent to ¶y p y = 3 x p and thus the equation y3 − x = 0 defines the implicit function y = 3 x in the neighbourhood of (0;0) (even in 2 R ). If the equation F(x;y) = 0 defines an implicit function y = f (x) in the neighbourhood of (x0;y0), then certainly y0 = f (x0). In general, without further theory, nothing else could be said about the function values of f around x0. In the example below we will apply the Newton’s method to approximate values of an implicit function. Example 6.6 Let us consider the equation x3 + y3 − 3xy − 3 = 0: This equation is satisfied for example by the point (x0;y0) = (1;2). Let us denote F(x;y) = x3 + y3 − 3xy − 3: Then ¶F ¶F = 3y2 − 3x and (x ;y ) = 9 6= 0: ¶y ¶y 0 0 68 Chapter 6. Implicit functions The given equation therefore implicitly defines a function y = f (x) in the neighbourhood of (x0;y0). We know that f (1) = 2. Further, let us assume f is defined on O0:11(1) = (0:89;1:11). What would be the function value of f at x = 0:9? By substituting x = 0:9 into the equation x3 + y3 − 3xy − 3 = 0 one derives y3 − 2:7y − 2:271 = 0: This equation has to have exactly one solution y = f (0:9) in the neighbourhood of the point y0 = 2. The first approximation of this solution by the Newton’s method is : y = f (0:9) = 1:963783: The theorem below shows us how to differentiate implicit functions. The computed derivatives can be used to decide monotonicity, convexity and concavity of implicit functions, to construct Taylor polynomials to approximate implicit functions, to determine their tangents and differentials. 2 k Theorem 6.1.2 Let G ⊆ R be an open set and let F 2 C (G) for k 2 N. Let (x0;y0) 2 G be a point such that ¶F F(x ;y ) = 0 and (x ;y ) 6= 0: 0 0 ¶y 0 0 Then the derivative of an implicit function y = f (x) defined by the equation F(x;y) = 0 in a neighbourhood of (x0;y0) is given as ¶F (x; f (x)) f 0(x) = − ¶x for x 2 (x − d;x + d) for some d > 0: (6.1) ¶F 0 0 ¶y (x; f (x)) Proof.