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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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    GEOL 595 - Mathematical Tools in Geology Lab Assignment # 12- Nov 13, 2008 (Due Nov 19) Name: Polynomials, Taylor Series, and the Temperature of the Earth How Hot is it ? A.Linear Temperature Gradients Near the surface of the Earth, borehole drilling is used to measure the temperature as a function of depth in the Earth’s crust, much like the 3D picture you see below. These types of measurements tell us that there is a constant gradient in temperture with depth. 1. Write the form of a linear equation for temperature (T) as a function of depth (z) that depends on the temperature gradient (G). Assume the temperature at the surface is known (To). Also draw a picture of how temperture might change with depth in a borehole, and label these variables. 1 2. Measurements in boreholes tell us that the average temperature gradient increases by 20◦C for every kilometer in depth (written: 20◦C km−1). This holds pretty well for the upper 100 km on continental crust. Determine the temperature below the surface at 45 km depth if the ◦ surface temperature, (To) is 10 C. 3. Extrapolate this observation of a linear temperature gradient to determine the temperature at the base of the mantle (core-mantle boundary) at 2700 km depth and also at the center of the Earth at 6378 km depth. Draw a graph (schematically - by hand) of T versus z and label each of these layers in the Earth’s interior. 2 4. The estimated temperature at the Earth’s core is 4300◦C.
  • Integration by Parts

    Integration by Parts

    3 Integration By Parts Formula ∫∫udv = uv − vdu I. Guidelines for Selecting u and dv: (There are always exceptions, but these are generally helpful.) “L-I-A-T-E” Choose ‘u’ to be the function that comes first in this list: L: Logrithmic Function I: Inverse Trig Function A: Algebraic Function T: Trig Function E: Exponential Function Example A: ∫ x3 ln x dx *Since lnx is a logarithmic function and x3 is an algebraic function, let: u = lnx (L comes before A in LIATE) dv = x3 dx 1 du = dx x x 4 v = x 3dx = ∫ 4 ∫∫x3 ln xdx = uv − vdu x 4 x 4 1 = (ln x) − dx 4 ∫ 4 x x 4 1 = (ln x) − x 3dx 4 4 ∫ x 4 1 x 4 = (ln x) − + C 4 4 4 x 4 x 4 = (ln x) − + C ANSWER 4 16 www.rit.edu/asc Page 1 of 7 Example B: ∫sin x ln(cos x) dx u = ln(cosx) (Logarithmic Function) dv = sinx dx (Trig Function [L comes before T in LIATE]) 1 du = (−sin x) dx = − tan x dx cos x v = ∫sin x dx = − cos x ∫sin x ln(cos x) dx = uv − ∫ vdu = (ln(cos x))(−cos x) − ∫ (−cos x)(− tan x)dx sin x = −cos x ln(cos x) − (cos x) dx ∫ cos x = −cos x ln(cos x) − ∫sin x dx = −cos x ln(cos x) + cos x + C ANSWER Example C: ∫sin −1 x dx *At first it appears that integration by parts does not apply, but let: u = sin −1 x (Inverse Trig Function) dv = 1 dx (Algebraic Function) 1 du = dx 1− x 2 v = ∫1dx = x ∫∫sin −1 x dx = uv − vdu 1 = (sin −1 x)(x) − x dx ∫ 2 1− x ⎛ 1 ⎞ = x sin −1 x − ⎜− ⎟ (1− x 2 ) −1/ 2 (−2x) dx ⎝ 2 ⎠∫ 1 = x sin −1 x + (1− x 2 )1/ 2 (2) + C 2 = x sin −1 x + 1− x 2 + C ANSWER www.rit.edu/asc Page 2 of 7 II.
  • Integration-By-Parts.Pdf

    Integration-By-Parts.Pdf

    Integration INTEGRATION BY PARTS Graham S McDonald A self-contained Tutorial Module for learning the technique of integration by parts ● Table of contents ● Begin Tutorial c 2003 [email protected] Table of contents 1. Theory 2. Usage 3. Exercises 4. Final solutions 5. Standard integrals 6. Tips on using solutions 7. Alternative notation Full worked solutions Section 1: Theory 3 1. Theory To differentiate a product of two functions of x, one uses the product rule: d dv du (uv) = u + v dx dx dx where u = u (x) and v = v (x) are two functions of x. A slight rearrangement of the product rule gives dv d du u = (uv) − v dx dx dx Now, integrating both sides with respect to x results in Z dv Z du u dx = uv − v dx dx dx This gives us a rule for integration, called INTEGRATION BY PARTS, that allows us to integrate many products of functions of x. We take one factor in this product to be u (this also appears on du the right-hand-side, along with dx ). The other factor is taken to dv be dx (on the right-hand-side only v appears – i.e. the other factor integrated with respect to x). Toc JJ II J I Back Section 2: Usage 4 2. Usage We highlight here four different types of products for which integration by parts can be used (as well as which factor to label u and which one dv to label dx ). These are: sin bx (i) R xn· or dx (ii) R xn·eaxdx cos bx ↑ ↑ ↑ ↑ dv dv u dx u dx sin bx (iii) R xr· ln (ax) dx (iv) R eax · or dx cos bx ↑ ↑ ↑ ↑ dv dv dx u u dx where a, b and r are given constants and n is a positive integer.