Chapter 8: Taylor s theorem and L Hospital s rule Theorem: [Inverse Mapping Theorem] Suppose that a < b and f : [a, b] R. Given that f (x) > 0 for all x (a, b) then f 1 is differentiable on (f(a), f(b)) and (f 1 ) = 1/(f f 1 ). Proof: 1. Let y 0 (f(a), f(b)). Since f is strictly increasing there is an x 0 (a, b) with f(x 0 ) = y 0. This is so if and only if f 1 (y 0 ) = x 0. 2. Define a function H : [a, b] R by, x x 0 f(x H(x) := 0 ) 1 f (f 1 (y 0 )) x x 0 ; x = x 0. 3. Then Figure 23: Motivation for the definition of H. H(x) = 1/f (x 0 ) = H(x 0 ) = H(f 1 (y 0 )) and so H is continuous at x 0 = f 1 (y 0 ). x x 0 4. Moreover from the earlier work on continuity for monotonic functions we know that f 1 is continuous on [f(a), f(b)]. Therefore, ( ) 1/f (x 0 ) = H(x 0 ) = H(f 1 (y 0 )) = H f 1 (y) y y 0 = y y0 H(f 1 (y)) = y y0 f 1 (y) f 1 (y 0 ) y y 0 = (f 1 ) (y 0 ) 54
Example: = x 2 + 1 on (0, 1) has inverse f 1 (y) = y 1 on (1, 2) and (f 1 ) (y) = 1 2 y 1 = 1 2x. Remark: The hypothesis that f (x) > 0 for all x (a, b) is essential. In fact if f is strictly increasing and differentiable on (a, b) but f (x 0 ) = 0 for some x 0 (a, b) then the inverse, f 1 is not differentiable at f(x 0 ). Indeed, if f 1 were differentiable at f(x 0 ) then by the chain rule we would have, 1 = (f 1 f) (x 0 ) = (f 1 ) (f(x 0 )) f (x 0 ) = 0; which is impossible. Therefore, f 1 cannot be differentiable at f(x 0 ). The function := x 3 is strictly increasing and differentiable on R however x 3 x (the inverse of f) is not differentiable at f(0) = 0. Figure 24: Differentiable function with a non-differentiable inverse. Theorem: [Cauchy s Mean Value Theorem]. Suppose that a < b and f : [a, b] R and g : [a, b] R are continuous on [a, b] and differentiable on (a, b). If g (x) 0 for all x (a, b) then there exists a point x 0 (a, b) such that f(b) f(a) g(b) g(a) = f (x 0 ) g (x 0 ). Proof: Consider the auxiliary function h : [a, b] R defined by the 3 3 determinant, g(x) 1 h(x) := f(a) g(a) 1 f(b) g(b) 1. Now h is continuous on [a, b] and differentiable on (a, b) and h(a) = h(b) = 0 so by Rolle s theorem there exists a point x 0 (a, b) such that h (x 0 ) = 0; that is, f (x 0 ) g (x 0 ) 0 f(a) g(a) 1 f(b) g(b) 1 = 0. 55
ie: f (x 0 ) [g(b) g(a)] = g (x 0 ) [f(b) f(a)]. Now since g (x) 0 for all x (a, b), Rolle s theorem tells us that g(b) g(a) 0 and so the result follows. Remark: Cauchy s mean value theorem has a geometric interpretation. If we consider the curve defined by the parametric equation α(t) := (f(t), g(t)), t [a, b]. Then the conclusion of the theorem is that there exists a point (f(x 0 ), g(x 0 )) on the curve such that the slope, g (x 0 )/f (x 0 ) of the tangent line to the curve at that point is equal to the slope of the line segment joining the end points of the curve. Figure 25: Cauchy s Mean Value Theorem interpreted. The next theorem has sometimes been said to be the most important in Calculus or Analysis. We use the notation for higher derivatives, f (0) (x) =, f (1) (x) = f (x) and, in general for n N, f (n+1) (x) = (f (n) ) (x). If f (n) is differentiable on an interval, then f (n+1) exists on the interval. Theorem: [Taylor s Theorem] Suppose that a < b and f : [a, b] R. If f (n) is continuous on [a, b] and differentiable on (a, b) then for each x (a, b] there exists a point ζ (a, x) such that = P n (x) + R n (x), where P n (x) := f(a) + f (k) (x a)k (a) and R n (x) := f (n+1) (ζ) (x a)n+1. (n + 1)! Note: The conclusion of the Mean Value Theorem can be written = f(a) + f (1) (ζ)(x a) 1 = P 0 (x) + R 0 (x), so can be regarded as Taylor s Theorem of order 0. Proof: 1. Fix x 0 (a, b] and define M R by, f(x 0 ) = f(a) + f (k) (a) (x 0 a) k + M (x 0 a) n+1 (n + 1)! We need to show that there exists a ζ (a, x 0 ) such that f (n+1) (ζ) = M. 56
2. Consider the auxiliary function g : [a, x 0 ] R defined by, g(x) := f(x 0 ) + + f (k) (x) (x 0 x) k + M (x 0 x) n+1. (n + 1)! 3. Now g is continuous on [a, x 0 ] and differentiable on (a, x 0 ) and g(a) = g(x 0 ) = 0. Therefore, by Rolle s theorem there exists a point ζ (a, x 0 ) such that g (ζ) = 0. 4. But g (x) = f (x) + = (x 0 x) n {f (k+1) (x) (x 0 x) k {f (n+1) (x) M} for all x (a, x 0 ). Therefore, f (n+1) (ζ) = M. This completes the proof. f (k) (x) (x } 0 x) k 1 M (x 0 x) n (k 1)! In Taylor s theorem the polynomial P n (x) is called the n-th degree Taylor polynomial for f at a and R n (x) is called the Lagrange remainder. Suppose that a < b and f : [a, b] R. If for each fixed x [a, b], R n(x) = 0. n Then for each x [a, b], the series of powers (a so-called power series), converges to. n=0 f (n) (x a)n (a) Functions with this property are common, e.g. e x, sin(x), polynomials, rational functions. They are called real analytic. L Hospital s Theorems Recall that if x x0 = L 1 and x x0 g(x) = L 2, then x x 0 g(x) = L 1, provided L 2 0. L 2 We now look into the case when L 1 = L 2 = 0 (there s no point in considering the case L 1 0 and L 2 = 0 since the it will be ± ). Theorem: Suppose that a < b and f : [a, b] R and g : [a, b] R. If f(x 0 ) = g(x 0 ) = 0, g (x 0 ) 0 and both f (x 0 ) and g (x 0 ) exist at some point x 0 (a, b), then x x 0 g(x) = f (x 0 ) g (x 0 ). 57
Figure 26: Convergence of Taylor polynomials to a real analytic function. Proof: The trick in this proof is to multiply and divide by (x x 0 ). x x 0 g(x) f(x 0 ) = x x0 g(x) g(x 0 ) ( ) f(x0 ) = x x0 x x 0 = f (x 0 )/g (x 0 ). ( x x0 ) g(x) g(x 0 ) Theorem: Suppose that a < b and f : [a, b] R and g : [a, b] R are continuous on [a, b]. Let x 0 be any point in (a, b) such that f(x 0 ) = g(x 0 ) = 0 and g (x) 0 for all x x 0. Then, x x 0 g(x) = f (x), whenever the it on the right exists x x 0 g (x) Proof: By Cauchy s mean value theorem there exists for each x (a, b) a point ζ x between x 0 and x such that Then g(x) = f(x 0) g(x) g(x 0 ) = f (ζ x ) g (ζ x ). x x 0 g(x) = f(x 0 ) x x 0 g(x) g(x 0 ) = f (x) x x 0 g (x), because when x tends to x 0, ζ x tends to x 0. Notes: 1. The previous theorem is also true when = ± and g(x) = ± x x0 x x0 and may also be extended to the case when x 0 is replaced by ±. 58
2. It can also be iterated by replacing f by f and g by g e.t.c for higher order derivatives. If f(x 0 ) = f (x 0 ) = g(x 0 ) = g (x 0 ) and the it of the ratio of the second derivatives exists then x x 0 g(x) = f (x) x x 0 g (x) = f (x) x x 0 g (x). Exercises 1. Calculate the derivative of the function f : R R defined by, := x 3. 2. Consider the function f : (0, ) R defined by, := log e (x). Show that the inverse of f exists and is differentiable. Moreover, show that (f 1 ) (x) = f 1 (x) for all x R. Note: the function f 1 is usually called the exponential function. 3. Let f : (0, ) R be defined by, := x log e (1 + 1/x). (a) Calculate x. Hint: Consider the derivative of the function g : (0, ) R defined by, g(x) := log e (x) at x = 1. (b) Show that n n log e (1 + 1/n) = 1. (c) By using the fact that the exponential function, x e x, is continuous show that n (1 + 1/n)n = e. 4. Show that the function f : [0, π/2] R defined by, := sin(x) has a differentiable inverse. Moreover show that (sin 1 ) (x) = 1/ 1 x 2. 5. Let f : [a, b] R be continuous on [a, b] and differentiable on (a, b). Show that f is a constant function if, and only if, f (x) 0 on (a, b). 6. Let f : R R differentiable. Show that if f : R R is increasing on R then f is continuous on R. Hint: use the fact that an increasing function is continuous if, and only if, it satisfies the intermediate value property, (see the section on continuity). 7. Let f : R R and let x 0 R. If f exists and is continuous on R and (i) f (x 0 ) = 0; (ii) f (x 0 ) > 0. Show that f has a local minimum at x 0. Hint: Consider the 1st order Taylor s expansion of f around x 0, (with remainder). 8. (Taylor s Theorem) Suppose that a < b and f : [a, b] R. If f (n) is continuous on [a, b] and differentiable on (a, b) then for each x 0 (a, b] there exists a point ζ (a, x 0 ) such that f(x 0 ) = P n (x 0 ) + R n (x 0 ), where P n (x 0 ) := f(a) + f (k) (a) (x 0 a) k and R n (x 0 ) := f (n+1) (ζ) (x 0 ζ) n (x 0 a) Hint: Consider the auxiliary function g : [a, b] R defined by, g(x) := f(x 0 ) + + f (k) (x) (x 0 x) k + M (x 0 x). 59
9. Let I := [a, b] and let f : I R be differentiable on I. Suppose that f(a) < 0 < f(b) and that there exist m, M such that 0 < m f (x) M for all x I. Let x 1 I be arbitrary and define x n+1 := x n f(x n )/M for all n N. Show that the sequence (x n : n N) is well defined and converges to the unique zero r I of f. Hint: if φ(x) := x /M, show that 0 φ (x) 1 m/m < 1 and that φ([a, b]) [a, b]. 10. Calculate the following its. tan 1 (x) sin(x) (a) ; (b) x 0 x x 0 x ; (c) x 0 e x 1 1 cos(x) ; (d). x x 0 x 2 60