Can you please help me with problems 12, 32, 56 on the bottom of the list of pages. Thanks so much! :)

AT&T S 9:13 PM Q1 70% SECTION 10.7 Taylor Polynomials 593 Define the ath Taylor polynomial 7, of f centered at x = a as follows: To(x ) = f(a) + (@) ( x - a ) + ( @ ) 2! " (x - a) + ...+ (a), n! ( x - a ) " REMINDER k-factorial is the number (4 - 1)(k - 2). . . (2)(1). Thus, The first few Taylor polynomials are 1! = 1, 2! = (2)1 = 2 To(x) = f(a) 3! = (3)(2)1 = 6 Ti(x) = f(a) + f'(a)(x -a) question, we define 0! = 1. T2(x) = f(a) + S' (a)(x - a)+ 5 1"(a)(x -a)? Ty(x) = f(a) + S' (a)(x - a)+ 5 1"(a)(x - a) + - f"(a)(x -a) Note that To is a constant function equal to the value of f at a, and that 71 is the lin- carization of f at a. Note also that 7, is obtained from T,-, by adding on a term of degree n: To(x ) = In-1(x) + (@) n! - (x - a)" The next theorem justifies our definition of T,. THEOREM 1 The polynomial T,, centered at a agrees with f to order n at x = a, and it is the only polynomial of degree at most n with this property. The verification of Theorem 1 is left to the exercises (Exercises 76-77), but we'll illustrate the idea by checking that 72 agrees with f to order n = 2: Ti(x) = f(a) + f' (a)(x - a) + ; f"(a)(x -a). Tz(a) = f(a) T'(x) = f'(a) + f"(a)(x - a). T'a) = f'(a) T"(x) = f"(a). T,'(a) = f"(a) This shows that the value and the derivatives of order up to n = 2 at x = a are equal. Before proceeding to the examples, we write T, in summation notation: TH(x) = P(a) (x - a) By convention, we regard f as the zeroth derivative, and thus f() is f itself. When a = 0, T, is also called the ath Maclaurin polynomial. EXAMPLE 1 Maclaurin Polynomials for f(x) =e' Plot the third and fourth Maclaurin polynomials for f(x) = et . Compare with the linear approximation. Solution All higher derivatives coincide with f itself: f()(x) = e". Therefore. f(0) = f'(0) = f"(0) = f"(0) = f(*)(0) = e' = 1 The third Maclaurin polynomial (the case a = 0) is = f(0) + f'(0)x + ;1"(0)x]+ =1"(0) = 1+x+ +x2+4x3 1/9AT&T S 9:13 PM Q1 70% 594 CHAPTER 10 INFINITE SERIES We obtain 74(x) by adding the term of degree 4 to 73(x): TA(x) = 7)(x)+ -"(Or' = Itx+ ;x +-x+: 24 Figure 3 shows that 73 and Ta approximate f(x) == e' much more closely than the linear approximation 7, on an interval around a = 0. Higher-degree Maclaurin polynomials would provide even better approximations on larger intervals. EXAMPLE 2 For objects near the surface of the earth, to two decimal places the accel- eration due to gravity is g = 9.81 m/s'. For objects at higher altitudes, Newton's Law of FIGURE 3 Maclaurin polynomials for Gravitation says that the acceleration due to gravity is G(h) = 8 (1 + 270)2 where G(h) is in m/s', h is the altitude above the surface of the earth in km, and 6370 is the radius of the earth in km. (a) Find a power series representation of G as a function of h. For what values of h is the power series valid? (b) Use the third Maclaurin polynomial to approximate the acceleration due to gravity on an object at an altitude of 1000 km, and estimate the error in the approximation. Solution (a) We use the power series representation for _ from Example 6 in the previous section. Substituting 5370 for x, we obtain G(h) = & _ " (n + 1 )(-h)" - =9.81 19.61 29.4h2 39.2h3 6370" 6370 63702 63703 + ... The power series for _is valid for -1 0, so its maximum value on [1, 12 is 1f(9)(1)| = 6. Therefore, we may take K = 6. 4/9\fAT&T S 9:14 PM Q1 70% SM CHAPTER 10 INFINITE SERIES THEOREM 3 Taylor's Theorem Assume that fit+!) exists and is continuous. Then R. ( x ) = - ( x - u ) (+ ") (1 ) du 2 Proof Set Our goal is to show that Ry(x) = 1,(x). For n = 0. Ro(x) = f(x) - f(a) and the desired result is just a restatement of the Fundamental Theorem of Calculus: lo(x ) = [ f'()du = f(x) - f(a) = Ro(x) Exercise 70 reviews this proof for the special case n = 2 To prove the formula for n > 0. we apply Integration by Parts to /,(x) with h(u ) = - (x - u). g (1 ) = f() (u ) Then g'(u) = fleth)(u). and so In (x ) = ["how)s'(u)du = hug(u) - [" n'tug(u) du - (x - a) fin)(a) + In-1(x) This can be rewritten as In-1(x) = (@) n! -(x - a)" + In(x) Now. apply this relation n times, noting that /o(x) = f(x) - f(a): f(x) = fla) + lo(x) = f(a) +@ 1! ( x - a) + h(x ) = f(a) + @ (x - 0)+f (@ I! 2 ! ( x - 2 ) 2 + 12 ( x ) = f(a)+ 1! (* - a)+ ...+- @(x - a)" + In(x) n! This shows that f(x) = T,(x) + In(x) and hence I,,(x) = R,,(x). as desired. Proof Now, we can prove Theorem 2. Assume first that x 2 a. Then To establish the inequality in (3), we use the inequality If (x) - T,(x)1 = 1R.(x)1 = (x -1)" g(+1(u)du which is valid for all integrable functions. s [ 1x - ul du K -(x - 1)"+1 /x = - = KE n+ 1 (n + 1)! 6/9