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Formulas for the Remainder Term in Taylor Series

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Formulas for the Remainder Term in Taylor Series Formulas for the Remainder Term in Taylor Series In Section 8.7 we considered functions f with derivatives of all orders and their Taylor series ϱ ͑n͒͑ ͒ f a n ͚ ͑x Ϫ a͒ n෇0 n! The nth partial sum of this Taylor series is the nth-degree Taylor polynomial of f at a: f Ј͑a͒ f Љ͑a͒ f ͑n͒͑a͒ T ͑x͒ ෇ f ͑a͒ ϩ ͑x Ϫ a͒ ϩ ͑x Ϫ a͒2 ϩ иииϩ ͑x Ϫ a͒n n 1! 2! n! We can write f ͑x͒ ෇ Tn͑x͒ ϩ Rn͑x͒ where Rn͑x͒ is the remainder of the Taylor series. We know that f is equal to the sum of its Taylor series on the interval Խ x Ϫ a Խ Ͻ R if we can show that lim n l ϱ Rn͑x͒ ෇ 0 for Խ x Ϫ a Խ Ͻ R. Here we derive formulas for the remainder term Rn͑x͒. The first such formula involves an integral. ͑ ϩ ͒ 1 Theorem If f n 1 is continuous on an open interval I that contains a, and x is in I, then x 1 n ͑nϩ1͒ Rn͑x͒ ෇ y ͑x Ϫ t͒ f ͑t͒ dt n! a Proof We use mathematical induction. For n ෇ 1, R1͑x͒ ෇ f ͑x͒ Ϫ T1͑x͒ ෇ f ͑x͒ Ϫ f ͑a͒ Ϫ f Ј͑a͒͑x Ϫ a͒ xx ͑ Ϫ ͒ Љ͑ ͒ and the integral in the theorem is a x t f t dt. To evaluate this integral we integrate by parts with u ෇ x Ϫ t and dv ෇ f Љ͑t͒ dt, so du ෇ Ϫdt and v ෇ f Ј͑t͒. Thus x t෇x x y ͑x Ϫ t͒ f Љ͑t͒ dt ෇ ͑x Ϫ t͒ f Ј͑t͔͒t෇a ϩ y f Ј͑t͒ dt a a ෇ 0 Ϫ ͑x Ϫ a͒ f Ј͑a͒ ϩ f ͑x͒ Ϫ f ͑a͒ (by FTC 2) ෇ f ͑x͒ Ϫ f ͑a͒ Ϫ f Ј͑a͒͑x Ϫ a͒ ෇ R1͑x͒ The theorem is therefore proved for n ෇ 1. Now we suppose that Theorem 1 is true for n ෇ k, that is, x 1 k ͑kϩ1͒ Rk͑x͒ ෇ y ͑x Ϫ t͒ f ͑t͒ dt k! a We want to show that it’s true for n ෇ k ϩ 1, that is x 1 kϩ1 ͑kϩ2͒ Rkϩ1͑x͒ ෇ y ͑x Ϫ t͒ f ͑t͒ dt ͑k ϩ 1͒! a 1 2 ■ FORMULAS FOR THE REMAINDER TERM IN TAYLOR SERIES Again we use integration by parts, this time with u ෇ ͑x Ϫ t͒kϩ1 and dv ෇ f ͑kϩ2͒͑t͒. Then du ෇ Ϫ͑k ϩ 1͒͑x Ϫ t͒k dt and v ෇ f ͑kϩ1͒͑t͒, so x 1 kϩ ͑kϩ ͒ y ͑x Ϫ t͒ 1 f 2 ͑t͒ dt ͑k ϩ 1͒! a t෇x 1 k ϩ 1 x ෇ ͑ Ϫ ͒kϩ1 ͑kϩ1͒͑ ͒ͬ ϩ y ͑ Ϫ ͒k ͑kϩ1͒͑ ͒ x t f t x t f t dt ͑ ϩ ͒ ͑ ϩ ͒ a k 1 ! t෇a k 1 ! 1 kϩ ͑kϩ ͒ 1 x k ͑kϩ ͒ ෇ 0 Ϫ ͑x Ϫ a͒ 1 f 1 ͑a͒ ϩ y ͑x Ϫ t͒ f 1 ͑t͒ dt ͑k ϩ 1͒! k! a f ͑kϩ1͒͑a͒ ෇ Ϫ ͑x Ϫ a͒kϩ1 ϩ R ͑x͒ ͑k ϩ 1͒! k f ͑kϩ1͒͑a͒ ෇ f ͑x͒ Ϫ T ͑x͒ Ϫ ͑x Ϫ a͒kϩ1 k ͑k ϩ 1͒! ෇ f ͑x͒ Ϫ Tkϩ1͑x͒ ෇ Rkϩ1͑x͒ Therefore, (1) is true for n ෇ k ϩ 1 when it is true for n ෇ k. Thus, by mathematical induction, it is true for all n. To illustrate Theorem 1 we use it to solve Example 4 in Section 8.7. EXAMPLE 1 Find the Maclaurin series for sin x and prove that it represents sin x for all x. SOLUTION We arrange our computation in two columns as follows: f ͑x͒ ෇ sin x f ͑0͒ ෇ 0 f Ј͑x͒ ෇ cos x f Ј͑0͒ ෇ 1 f Љ͑x͒ ෇ Ϫsin x f Љ͑0͒ ෇ 0 f ٞ͑x͒ ෇ Ϫcos x f ٞ͑0͒ ෇ Ϫ1 f ͑4͒͑x͒ ෇ sin x f ͑4͒͑0͒ ෇ 0 Since the derivatives repeat in a cycle of four, we can write the Maclaurin series as follows: f Ј͑0͒ f Љ͑0͒ f ٞ͑0͒ f ͑0͒ ϩ x ϩ x 2 ϩ x 3 ϩ иии 1! 2! 3! 3 5 7 ϱ 2nϩ1 x x x n x ෇ x Ϫ ϩ Ϫ ϩиии෇ ͚ ͑Ϫ1͒ 3! 5! 7! n෇0 ͑2n ϩ 1͒! With a ෇ 0 in Theorem 1, we have x 1 n ͑nϩ1͒ Rn͑x͒ ෇ y ͑x Ϫ t͒ f ͑t͒ dt n! 0 FORMULAS FOR THE REMAINDER TERM IN TAYLOR SERIES ■ 3 Since f ͑nϩ1͒͑t͒ is Ϯsin t or Ϯcos t, we know that Խ f ͑nϩ1͒͑t͒ Խ ഛ 1 for all t. We use the fact that, for a ഛ b, b b Խ y f ͑t͒ dt Խ ഛ y Խ f ͑t͒ Խ dt a a Thus, for x Ͼ 0, 1 x 1 x ͑ ͒ ෇ ͑ Ϫ ͒n ͑nϩ1͒͑ ͒ ഛ ͑ Ϫ ͒n ͑nϩ1͒͑ ͒ Խ Rn x Խ Խ y x t f t dt Խ y x t Խ f t Խ dt n! 0 n! 0 nϩ1 nϩ1 x 1 n 1 x x ഛ y ͑x Ϫ t͒ dt ෇ ෇ n! 0 n! n ϩ 1 ͑n ϩ 1͒! For x Ͻ 0 we can write 1 0 n ͑nϩ1͒ Rn͑x͒ ෇ Ϫ y ͑x Ϫ t͒ f ͑t͒ dt n! x so nϩ1 1 0 1 0 ͑Ϫx͒ ͑ ͒ ഛ Ϫ n ͑nϩ1͒͑ ͒ ഛ ͑ Ϫ ͒n ෇ Խ Rn x Խ Խ y Խ x t Խ Խ f t Խ dt y t x dt n! x n! x ͑n ϩ 1͒! Thus, in any case, we have Խ x Խnϩ1 Խ Rn͑x͒ Խ ഛ ͑n ϩ 1͒! The right side of this inequality approaches 0 as n l ϱ (see Equation 8.7.10), so Խ Rn͑x͒ Խ l 0by the Squeeze Theorem. It follows that Rn͑x͒ l 0as n l ϱ, so sin x is equal to the sum of its Maclaurin series. For some purposes the integral formula in Theorem 1 is awkward to work with, so we are going to establish another formula for the remainder term. To that end we need to prove the following generalization of the Mean Value Theorem for Integrals (see Section 6.4). 2 Weighted Mean Value Theorem for Integrals If f and t are continuous on ͓a, b͔ and t does not change sign in ͓a, b͔, then there exists a number c in ͓a, b͔ such that b b y f ͑x͒t͑x͒ dx ෇ f ͑c͒ y t͑x͒ dx a a Proof Because t doesn’t change sign,either t͑x͒ ജ 0 or t͑x͒ ഛ 0 for a ഛ x ഛ b. For the sake of definiteness, let’s assume that t͑x͒ ജ 0. By the Extreme Value Theorem (4.2.3),f has an absolute minimum value m and an absolute maximum value M, so m ഛ f ͑x͒ ഛ M for a ഛ x ഛ b. Since t͑x͒ ജ 0, we have mt͑x͒ ഛ f ͑x͒t͑x͒ ഛ Mt͑x͒ a ഛ x ഛ b 4 ■ FORMULAS FOR THE REMAINDER TERM IN TAYLOR SERIES and so b b b 3 m y t͑x͒ dx ഛ y f ͑x͒t͑x͒ dx ഛ M y t͑x͒ dx a a a xb t͑ ͒ ෇ xb ͑ ͒t͑ ͒ ෇ If a x dx 0, these inequalities show that a f x x dx 0 and so Theorem 2 is xb t͑ ͒ true because both sides of the equation are 0. If a x dx 0, it must be positive and xb t͑ ͒ we can divide by a x dx in (3): xb f ͑x͒t͑x͒ dx m ഛ a ഛ M xb t͑ ͒ a x dx Then, by the Intermediate Value Theorem (2.4.10), there exists a number c in ͓a, b͔ such that xb f ͑x͒t͑x͒ dx ͑ ͒ ෇ a b ͑ ͒t͑ ͒ ෇ ͑ ͒ b t͑ ͒ f c b and so y f x x dx f c y x dx x t͑ ͒ a a a x dx ͑ ϩ ͒ 4 Theorem If f n 1 is continuous on an open interval I that contains a, and x is in I, then there exists a number c between a and x such that f ͑nϩ1͒͑c͒ R ͑x͒ ෇ ͑x Ϫ a͒nϩ1 n ͑n ϩ 1͒! Proof The function t͑t͒ ෇ ͑x Ϫ t͒n doesn’t change sign in the interval from a to x, so the Weighted Mean Value Theorem for Integrals gives a number c between a and x such that x x n ͑nϩ1͒ ͑nϩ1͒ n y ͑x Ϫ t͒ f ͑t͒ dt ෇ f ͑c͒ y ͑x Ϫ t͒ dt a a t෇x ͑x Ϫ t͒nϩ1 ͑x Ϫ a͒nϩ1 ෇ Ϫ ͑nϩ1͒͑ ͒ ͬ ෇ ͑nϩ1͒͑ ͒ f c ϩ f c ϩ n 1 t෇a n 1 Then, by Theorem 1, x 1 n ͑nϩ1͒ Rn͑x͒ ෇ y ͑x Ϫ t͒ f ͑t͒ dt n! a 1 ͑x Ϫ a͒nϩ1 f ͑nϩ1͒͑c͒ ෇ f ͑nϩ1͒͑c͒ ෇ ͑x Ϫ a͒nϩ1 n! n ϩ 1 ͑n ϩ 1͒! The formula for the remainder term in Theorem 4 is called Lagrange’s form of the remainder term. Notice that this expression f ͑nϩ1͒͑c͒ R ͑x͒ ෇ ͑x Ϫ a͒nϩ1 n ͑n ϩ 1͒! is very similar to the terms in the Taylor series except that f ͑nϩ1͒ is evaluated at c instead of at a. All we can say about the number c is that it lies somewhere between x and a. In the following example we show how to use Lagrange’s form of the remainder term as an alternative to the integral form in Example 1.
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