, applyingl Hospital s Rule again x 0 2 cos(x) xsinx

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1 Lecture 3 We give a couple examples of using L Hospital s Rule: Example 3.. [ (a) Compute x 0 sin(x) x. To put this into a form for L Hospital s Rule we first put it over a common denominator [ x 0 sin(x) ] x sin(x) = x x 0 x sin(x) cos(x) = x 0 sin(x) + x cos(x), applying L Hospital s Rule once sin(x) =, applyingl Hospital s Rule again x 0 2 cos(x) xsinx = 0. Note, the justification of the use of L Hospital s rule twice is thought of backwards: After applying it the second time, the it of the ratio of (second) derivatives exist, which means the it of the ratios of first derivatives exist, thereby justifying the first use of the rule. (b) Compute x 0 (cos(x)) x 2. To put this in the form of L Hospital s Rule, we take let g(x) = ln(cos(x)) x 2. Using the properties of logarithms, we have g(x) = ln(cos(x)) x 2. Since the ln function and its inverse, exp, are continuous on their domains, we can compute the it as x 0 of g(x) and then apply the exponential function: ln(cos(x)) x 0 x 2 = x 0 sin(x) cos(x) 2x = x 0 sin(x) 2x, by L Hospital s Rule cos(x) = 2 where we have used the known its for sin(x)/x and cos(x). Taking exponentials, we find (cos(x)) x 2 = exp(g(x)) = exp( /2). x 0 x 0 Example 3.2. Here we answer problem 30.6 from the text. The function f is assumed to be differentiable on an interval (c, ) and [f(x)+f (x)] = L, for finite L. Taking a what me worry approach, we use the hint given and find f(x)e x f(x) = e x f(x)e x + f (x)e x = e x, assuming L Hospital s Rule = [f(x) + f (x)] = L.

2 The use of L Hospital s Rule will be justified if we know that f(x)e x = + (or, for then we could apply the argument to g = f). For g(x) = f(x)e x, we find that g (x) = f(x)e x + f (x)e x = (f(x) + f (x))e x. (3.) If L 0, say L > 0, then g (x) = + and there is N > 0 so that x > N implies g (x) >. Then g(x) g(n) = g (c)(x N) > x N, x > N = g(x) =. (If L < 0, we get the it in a similar way.) If L = 0, then it really seems puzzling. But, suppose we set f(x) = f(x) +. Then f(x) + f (x) = f(x) + f (x) + = ( f(x) + f (x)) = L + =. We can apply what we have proved to f to get f(x) = = f(x) = 0, and f (x) = f (x) = 0. We finish up our theorems about differentiation with the theorem about inverse functions. Theorem 3.3. (Derivatives of Inverses) Let f be a - continuous function on an open interval I and let J = f(i). If f is differentiable at x 0 I, and if f (x 0 ) 0, then f is differentiable at y 0 = f(x 0 ) and (f ) (y 0 ) = f (x 0 ). Proof. Note that J is an open interval containing y 0 (theorem from last semester). We have that f(x) f(x 0 ) = f (x 0 ) 0. x x 0 x x 0 Now by the - property, f(x) f(x 0 ) for x x 0. Thus, x x 0 x x 0 f(x) f(x 0 ) = f (x 0 ). By the definition of the it, this means, given ε > 0, there is a δ = δ ε > 0 so that 0 < x x 0 < δ = x x 0 f(x) f(x 0 ) f (x 0 ) < ε. Now, f continuous at x 0 implies g = f is continuous at y 0 (last semester). Hence, given that δ ε > 0, there is an η > 0 so that 0 < y y 0 < η = g(y) g(y 0 ) = g(y) g(f(x 0 )) = g(y) x 0 < δ ε. 2

3 Taking these two statements together, we have 0 < y y 0 < η = g(y) x 0 f(g(y)) f(x 0 ) f (x 0 ) < ε. But Hence, which shows g(y) x 0 f(g(y)) f(x 0 ) = g(y) g(y 0). y y 0 0 < y y 0 < η = g(y) g(y 0 ) y y 0 f (x 0 ) < ε, g(y) g(y 0 ) = y y 0 y y 0 f (x 0 ). Taylor s Theorem A function defined by a power series with radius R, f(x) = a k x k, x < R, has derivatives at all orders of x <. That is, for any natural number n, f (n) (x) = k(k ) (k n + )a k x k n, x < R. k=n Therefore, we find that f (n) (0) = a n. That is, the coefficients are given in terms of the value of the derivatives of the function at x = 0. A power series can be of the form f(x) = a k (x x 0 ) k, for any fixed number x 0. This is said to be a power series expansion around x 0 and the radius of convergence (found in the usual way using the nth root of a n or the ratio of consecutive a n ) determines that the power series converges for x x 0 < R (this is simply by a change of variable x x x 0 ). Again, f will have derivatives of any order given by f (n) (x) = k(k ) (k n + )a k (x x 0 ) k n, x x 0 < R. k=n This time, we find the relation a n = f (n) (x 0 ). 3

4 This means that a function f which is given by a power series expansion around x 0 can be written as f(x) = f(x 0 ) + k= f (k) (x 0 ) (x x 0 ) k, x x 0 < R. k! Thus, the function is determined completely by its value and the values of its derivatives at a single point! The partial sums, S n, of this series provide an approximation to the function that gets better as n gets large. Taylor s Theorem will say that we have a similar type of approximation process even if we don t have infinitely many derivatives. However, we don t necessarily have a iting process that converges to the function, but we do have various estimates for the error, or remainder. Theorem 3.4. (Taylor s Theorem) Suppose that (i) f (n ) exists and is continuous on [a, b] and (ii) f (n) exists on (a, b). Then for all x (a, b) f(x) = f(a) + f (a) (x a) f (n ) (a)! (n )! (x a)n + R n (f, a, x), where the remainder R n (f, a, x) is given by R n (f, a, x) = f (n) (c) (x a) n, for some c, a < c < x. Proof. Fix x (a, b). Let M (M = M x ) be the unique solution of f(x) = f(a) + f (a)! (x a) f (n ) (a) (x a) n + M n )! (x a)n. (3.2) This is related to a polynomial interpolation problem. Find a polynomial p(t) by choosing coefficients for the powers, (t a),! (t a)2,..., 2! to fit the interpolation conditions (t a)n, p(t) = n (t a) k c k, k! p (k) (a) = f (k) (a), k = 0,..., n ; p(x) = f(x). (3.3) It is easy to check that by choosing c k = f (k) (a), k = 0,..., n, the first equation is satisfied. Then the M chosen above is the c n that finally determines p(t). 4

5 Consider the function g(t) = f(t) p(t), t [a, b], where p is the polynomial satisfying (3.3). Then, g, g,..., g (n ) are continuous on [a, x], g (n) exists on (a, x) and g(a) = g(x) = 0. Hence, by Rolle s theorem there is a c, a < c < x, such that g (c ) = 0. Since g (a) = g (c ) = 0, by the choice of p(t), we can apply Rolle s Theorem once more to obtain c 2, a < c 2 < c, so that g (c 2 ) = 0. Since g (a) = g (a) =... = g (n ) (a) = 0, we may continue this process inductively applying Rolle s Theorem at each stage to obtain points c k, a < c k < c k, with g (k) (c k ) = 0 for k =, 2,..., n. For c = c n, we have g (n) (c) = f (n) (c) M = 0. Solving for M and substituting this into (3.2) gives the theorem. 5