8.7 Taylor s Inequality Math 2300 Section 005 Calculus II. f(x) = ln(1 + x) f(0) = 0

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1 8.7 Taylor s Inequality Math 00 Section 005 Calculus II Name: ANSWER KEY Taylor s Inequality: If f (n+) is continuous and f (n+) < M between the center a and some point x, then f(x) T n (x) M x a n+ (n + )! Note: T n (x) is the n th degree Taylor polynomial for f(x) centered at a.. (a) Find the Taylor series for f(x) = ln( + x) centered at a = 0. f(x) = ln( + x) f(0) = 0 f (x) = ( + x) f (0) = f (x) = ( + x) f (0) = f (x) = ( + x) f (0) = f (4) (x) = ( + x) 4 f (4) (0) =! f (5) (x) = 4 ( + x) 5 f (5) (0) = 4! f (n) (x) = ( ) n (n )!( + x) n f(n)(0) = ( ) n (n )! n (n )! T (x) = ( ) x n = ( ) n! n= n= n xn n (b) If we use the series in part (a) to estimate ln 4 with n = 5, how accurate will the estimate be? Find M: We want M to be the max of f (6) (x) for 0 x. The 6 th derivative is f (6) (x) = 5! ( + x) 6. Since f (6) 5! (x) = is strictly decreasing, its max occurs at the left-most x-value ( + x) 6 5! in the interval. Thus M = ( + 0) 6 = 5! Substituting n = 5, x =, and the value for M back into Taylor s inequality, f() T 5 () 5! 6! 6 = 6 6 =.5

2 8.7 Taylor s Inequality Math 00 Section 005 Calculus II (c) Why is the error bound in part (b) so bad? is very far from the center a = 0! In fact, is not even within the interval of convergence! (d) Show that the series in part (a) converges to ln( + x) on the interval (, ) by showing that the limit of the error term is zero. Find M: From part (a), we know f (n+) (x) = ( )n n! and we want M to be an upper bound ( + x) n+ of f (n+) (x) on the given interval. For any value of n, this function is a decreasing function of n! x, so the max occurs at the left endpoint, x = /. So M = (/) n+ = n! n+. Plug in M and maximize error term: f(x) T n (x) n+ n! (n + )! x n+ = n+ n + x n+ On the interval x, the right-hand side of the inequality above is maximized for x = (or x =, doesn t matter). So our inequality now says f(x) T n (x) n+ n + n+ = n +. Finally, lim = 0. This means that as n increases (in other words, as we use higher degree n n + Taylor polynomials to approximate f), the error goes to zero. This means the Taylor series converges to f(x) = ln( + x).

3 8.7 Taylor s Inequality Math 00 Section 005 Calculus II. Suppose the function g(x) has derivatives of all orders and the Maclaurin series for g is n=0 ( ) n xn+ (a) Does the series for g evaluated at x = converge? n +. ( ) n ( )n+ n + = ( ) n. This is an alternating series, so we apply the alternating (n + ) n+ n=0 n=0 series test. Let b n =. Since lim (n + ) n+ b n = lim = 0, the series for g at n n (n + ) n+ x = converges by the alternating series test. (b) Approximate g g ( ) ( ) ( ) = ( ) using the first two nonzero terms of the series. (c) Show that the approximation above differs from g ( ) by less than 00. Taylor s Inequality states that the error from using the two-term approximation is less than or equal to the maximum value of the third term for values of x between the center x = 0 and the estimation point x =. In other words, error x5 7 for 0 x The maximum value of the Taylor error estimate occurs at x =, so error ( )5 7 = 7 = 4 Since Taylor s Inequality states that the error in the approximation is less than, the difference 4 between the estimate and the exact value is certainly less than 00.

4 8.7 Taylor s Inequality Math 00 Section 005 Calculus II. Consider the function f(x) = ln x. (a) Find the third-degree Taylor polynomial for f(x) centered at x =. f(x) = ln x f() = ln f (x) = x f () = f (x) = x f () = 9 f (x) = x f () = 7 The general formula for the third-degree Taylor polynomial is T (x) = f(a) + f (a)(x a) + f (a) (x a) + f () (a) (x a), so the third-degree Taylor polynomial for ln x is!! T (x) = ln + (x ) 8 (x ) + (x ) 8 (b) Use Taylor s Inequality to estimate the accuracy of the approximation f(x) T (x) for x 4. Taylor s Inequality states that f(x) T (x) M x 4 4! First we find M: Ṫhe fourth derivative of f(x) is f (4) (x) = 6. On the interval x 4, the x4 maximum for f (4) (x) is achieved when x is the smallest. Thus the max is M = 6 4 = 8 Now we use Taylor s Inequality: f(x) T (x) 8 ( ) x 4. The max on x 4 is 4! 64 4

5 8.7 Taylor s Inequality Math 00 Section 005 Calculus II 4. The table below gives some values of h and its derivatives. x h(x) h (x) h (x) h () (x) h (4) (x) (a) Write the first-degree Taylor polynomial for h about x = and use it to approximate h(.9). Is this approximation greater than or less than h(.9)? Explain your reasoning. The general formula for a first-degree Taylor polynomial is T (x) = f(a) + f (a)(x a). So the first-degree approximation for h about x = is T (x) = h() + h ()(x ) = (x ) We use the approximation to estimate h(.9): h(.9) T (.9) = (.9 ) = 67. Since h () = 488, which is positive. This means the graph of h(x) is concave up at x =, so the linear approximation lies below h(x). Thus 67. is an underestimate for the true value of h(.9). (b) Write the third-degree Taylor polynomial for h about x = and use it to approximate h(.9). The third-degree Taylor polynomial is T (x) = f(a)+f (a)(x a)+ f (a)! (x a) + f () (a) (x a)! So the third-degree approximation is T (x) = (x ) (x ) (x ) 6 Approximating h(.9) gives h(.9) T (.9) = (.) (.) + 6 (.) = (c) Use Taylor s Inequality to show that the third-degree Taylor polynomial for h about x = approximates h(.9) with error less than 0 4. Taylor s Inequality states that h(x) T (x) M 4! (x )4. First we find M: M should be an upper bound on h (4) (x) for values of x between the center x = and the estimation point x =.9. The best information we have is that h (4) () = 584 9, so we should pick this value for M Now Taylor s Inequality gives error 4! (x )4 = In other words, Taylor s Inequality shows the error from using T (x) to estimate h(.9) is less than.0007, which is certainly less than

6 8.7 Taylor s Inequality Math 00 Section 005 Calculus II 5. Let f(x) = sin x. (a) Approximate f(x) = sin x with a fourth-degree Taylor polynomial about a = π 6. f(x) = sin(x) f (x) = cos(x) f (x) = sin(x) f (x) = cos(x) f (4) (x) = sin(x) ( π ) f = ( 6 f π ) = ( 6 f π ) = ( 6 f π ) = ( 6 f (4) π ) = 6 So T 4 (x) = + ( x π ) ( x π ) ( x π ) ( + x π ) 4 6! 6! 6 4! 6 (b) Estimate the accuracy of the approximation f(x) T 4 (x) when x lies in the interval 0 x π. For 0 x π, Taylor s Inequality states that f(x) T 4(x) M x π 5. 5! 6 First we find M: f (5) (x) = cos x. On the interval 0 x π, the maximum value of cos x is. So pick M =. Then f(x) T 4 (x) x π 5. When 0 x π x 5! 6, the quantity π < π 6 6, so f(x) T 4 (x) x π So the error is at most ( π ) = π