Can you please help me with questions 40 and 42. For 40, you have to find the interval of convergence. Thank so much!

\fAT&T S 8:49 PM 01 42% SH CHAPTER 10 INFINITE SERIES EXAMPLE 4 Geometric Series Prove that 1-7= [2 for |xly Solution Substitute 2x for x in Eq. (2): n= 3 Expansion (2) is valid for |x| an(x - c)" has radius of convergence R > 0. Then F is differentiable on (c - R. c + R). Further- more, we can integrate and differentiate term by term. For x E (c - R.c + R). The proof of Theorem 2 is somewhat technical and is omitted. See Exercise 70 F'(x) = [ nd.(x - c)"-1 for a proof that F is continuous. [F(x)dx = A + E (A any constant) For both the derivative series and the integral series the radius of convergence also R. Theorem 2 is a powerful tool for working with power series. The next two example show how to use differentiation or antidifferentiation of power series representations functions to obtain power series for other funct 4 / 11\fAT&T S 8:49 PM 01 42% 516 CHAPTER 10 INFINITE SERIES Since the power series is an alternating series, the error in this a y roximation is less than the first omitted term, the term with (0.3)". Therefore. error = [tan (0.3) - S.(0.3) 1, the situation changes drastically since the power series diverges and the partial sums deviate sharply from tan"x. Power Series Solutions of Differential Equations Power series are a basic tool in the study of differential equations. To illustrate, consider the differential equation with initial condition y' = y. y(0) = 1 From Example 5 in Section 9.1, it follows that f (x) = e" is a solution to this Initial Value Problem. Here, we take a different approach and find a solution in the form of a power series, F(x) = _any". Ultimately, this approach will provide us with a power 1=0 series representation of f(x) = e". We have F(x) = _ and " = ao + aux + azx? + agx' +... #=0 F'(x) = End,1"-1 = a1 + 2a2x + 3agx? + 4agr) + ... To satisfy the differential equation, we must have F'(x) = F(x). and therefore. ag = a1. a1 = 2a2. az = 3a3, a3 = 404. .. . In other words, F'(x) = F(x) if an-1 = nan. or an = an-1 An equation of this type is called a recursion relation. It enables us to determine all of the coefficients a, successively from the first coefficient an, which may be chosen arbitranh For example, n = 1: 1 = 2: 2 2. 1 n = 3: a2 43 = do 3 3.2 3 .2 . 1 3! To obtain a general formula for a,, apply the recursion relation n times: an-1 an - 2 an - 3 ac Un = - n n(n - 1) n(n - 1)(n - 2) We conclude that F(x) = do *" 1=0 6 / 11\fI AT&T S 8:49 PM 01 42% 383 CHAPTER 10 INFINITE SERIES The first few terms on each side of this equation are -do + 0. x + 3azr + 8ayx + 15air* + ...= 0+0.x -dox ax - agrt- ... Matching up the coefficients of x", we find that -10 = 0, 3ay = -do. 803 = -01. 1504 = -02 8 In general, (n' - 1)d, = -d.-2, and this yields the recursion relation an = - an- 2 72 - 1 for n 2 2 9 Note that ap = 0 by Eq. (8). The recursion relation forces all of the even coefficients oz. as. as. ... to be zero: 42 = 72 50 42 = 0, and then 42 - 1 = 0 so a4 = 0. and so on As for the odd coefficients, a, may be chosen arbitrarily. Because F'(0) = a1, we set 1 = 1 to obtain a solution y = F(x) satisfying F'(0) = 1. Now. apply Eq. (9): n = 3: (} = - 32 - 1 32 - 1 a3 n =5: as = - 52- 1 (52 - 1)(32 - 1) as n =7: 07 = - 72 - 1 (72 - 1)(32 - 1)(52 - 1) This shows the general pattern of coefficients. To express the coefficients in a compact form, let n = 2k + 1. Then the denominator in the recursion relation (9) can be written m' - 1 = (24 + 1)2 - 1 = 4k2 +4k = 4k(k + 1) and az-1 021+1 = - 4k(k + 1) Applying this recursion relation k times, we obtain the closed formula 021+1 = (-1)* (AK(K + 1)) (-1) 4* K! (k + 1)! Thus, we obtain a power series representation of our solution: F(x) = > (-1) x24+1 1=0 4 * *! (K + 1 ) ! A straightforward application of the Ratio Test shows that F has an infinite radius of convergence. Therefore, F(x) is a solution of the Initial Value Problem for all r. 10.6 SUMMARY . A power series is an infinite series of the form F(x) = _ an(x - c)" 1=0 The constant c is called the center of F(x) 8 / 11AT&T S 8:49 PM 01 42% 590 CHAPTER 10 INFINITE SERIES (b) Another way to obtain a power series for h. . 's to square the power 17. 5 (20)! M series for f(x) in Exercise 40. By multiplying ter . by term. determine the (2n + 1)!' terms up to degree 12 in the resulting power series for (f(x)) and show that they match the terms in the power series for hix ) found in part (a). 19. DY 20. 45. Use the equalities - 74 4 2 1 -13-( - 4) 1 + (4 4 ) 21 5 12441 to show that for |x - 4) (-1)"+1_2"+1 39. f(x) = _ 40. f(x) = n(n + 1) 41. Differentiate the power series in Exercise 39 to obtain a power series (b) Evaluate at x = } to prove 3x3 for & (x) = 7 313 2 1. 2 . 22 2 . 3. 23 + 3-4- 24 4 -5-25 42. Differentiate the power series in Exercise 40 to obtain a power series 4x (c) Use a calculator to verify that the partial sum $4 approximates for g(x) =- (1 - x) left-hand side with an error no greater than the term as of the series 52. Prove that for (x|