cc) c, >o,, c, =c7 $"(x\ <o - S ~ Z

Transcription

1 J cc) c, >o,, c, =c7 $"(x\ <o - S ~ co w, c, Co, c, +"(>C\ 7 < x<a 6) e. Co, c,, el (

2 x (b), (c) The initial terms of J (x) up to n = 5 are a = -, al =--,an x3 x5 x7 x9 = - as = ' 8,43' a* =,474,56' xll and a5 = - The partial sums seem to 76,947, ' approximate Jl (x) well near the origin, but as x increases, we need to take a large number of terms to get a good approximation. 3 xsn so lim - an+l - n=l (3n-l)(3n)' n-3 a, -'~~~~!%(3n+)(3n+3) =O 4. (a) A(x) = + C a,, where a, = for all x, so the domain is R. -8 -

3 46 CHAPTER8 SERIES m m 6. f (x) = = (-9')" = C (-l)n 3n xzn. The series converges when -9' + 9x - (-9x) 7=o n=o when x < $, SO I = (-f, f). < I; that is, 3 x 5"+l 7. f(x)= = - - C (zr or equivalently, - - xn. The series converges when ( < ; x-5-5 that is, when x < 5, so I = (-5,5). ( -5 ) 5,= n=~ x m m 8. f (x) = - = x. = x C (-4)" = C (-)nnxn+. The series converges when -4 < ; that is, 4x+ - (-4) n=o n=o when x < $, SO I = (-$,a). x w xn+l = C (-l)nn=o gn+l The geometric series C [- (;) n=o so R = 3 and I = (-3,3). ' n' converges when 9 n=o x x x3 m x3n+. f(x) = = - ='E(a)=~w. The series converges when x3/a3 a3 - x3 a3 ' - x3/a3 a3 n=o n=o x3 < la3] * x < la\, so R = la and I = (-la\, lal) f (x) = - - x+x- (x+)(x-) x+ x- A = -. Taking x =, we get B =. Thus, < + A = A(x - ) + B(x + ). Taking x = -, we get We represented the given fi~nction as the sum of two geometric series; the first converges for x E (-,l) and the second converges for x E (-,). Thus, the sum converges for x E (-,l) = I A B - +-=. -- 'f(x)=3x+x-l (3x-l)(x+l) 3 x x+l 3x- x+l - ( x) =- The series C (-x)" converges for x E (-,l) and the series C (3)" converges for x E (-4, i), so their sum converges for x E (-5,;) = I. d - d m 3. (a) f (x) = ----i [C(-)" x~] [from Exercise 3 (I+X) dx +x n=o = - (-) = -z m = C (-l)n+lnxn-l [from Theorem (i)] = C (-l)n(n + l)xn with R =. n=l n=o In the last step, note that we decreased the initial value of the summation variable n by, and then increased each occurrence df n in the term by [also note that (-I)"+~ = (-I)"].

4 d (I+x)~ dx [(+~)~] =-?z (b) f(%) = - = (- (n + m 3 = -4 C (-l)"(n+ l)nxn-' = i C (-l)n(n + )(n + l)xn with R =. n=l n.=o x (c) f (5) = - = x. - lc SECTION 8.6 REPRESENTING FUNCTIONS AS POWER SERIES 46 [from part (a)] [n=o = x ( + x )~ ( +.- x (-l)n(n + )(n + l)xn [frornpart (b)] x)3 3 = i C (-l)"(n + )(n + )xn+ n=o n=o To write the power series with xn rather than xn+', we will decrease each occurrence of n. in the term by and increase the initial value of the summation variable by. This gives us x (-l)"(n)(n - l)zn. n= 4. (a) - = - - x (-l)nxn [geometric series with R =, so +x - ( x) n=o [C = since f () = In =, with R = (-)n-lxn+' ca (-l)nxn (b) f(x) =xln(l+x) =x [ C * l n l X n ] [by pan (a)] = =x- with R =. n=l n n=l n n.= n - m (-l)n-(x)n, m (-l)n-lxn (c) f (x) = ln(x + ) = x [by past (a>] = C with R =. n=l n=l Putting x =, we get C = In 5 The series converges for /5 < e x < 5, so R = 5. m m 6. We know that = x ( ~)~. - n=o (- ~ ) n=l ~ n=o Differentiating, we get = C "nxn-i = C n+ (n + l)xn, so with R = $. x3 nf lxn - n+lxn+3 n- f (XI = - =x3 C- -C- Or C jfi zn for x <. Thus, R = and I = (-,). (x - ) n=o n+ n=o n+ n=3 -- x 8. From Example 7, g(x) = arctan z = (-)"- Thus, n = ~ n + '

5 . f(x) = tan-'(x) = J & = J E (-)" ( 4 ~ ~ ) ~ SECTION 8.6 REPRESENTING FUNCTIONS AS POWER SERIES 463 dx = (-)"4"x" dx n=o n=o (-l)n4nxn+ (-) n n+l x n+l =C+ C - [f() = tan-' =, so C =. n=o n+ n=o n+ S3 = S', S5 = S6 The series converges when 4' x < $, so \ / S, = S R = $. Ifx = f i, then f(x) = C (-)"- n=o n+ m m f (x) = C ( -)"+ll, respectively. Both series - n=o n + and converge by the Alternating Series Test. As n increases, f I sn(x) approximates f better on the interval of convergence, which is [-$, $. - I J m M w t t8n+ converges - t8 - t8 n=o n=o n=o 8n+' - t8 3. -=t.-- - t C (t8)" = C pn+l J & d t = ~ + ~ - The series for - when It8 It <, SO R = for that series and also the series for t/(l - t8). By Theorem, the series for / dt also has R =. 3 tn ln( - t) tn n=l n t n=l dt=~-)-. n=l n 4. By Example 6,n(l - t) = - C - for It <, so - = - By Theorem, R =.. J, 5. By Example 7, tan-' x = C (-l)n- with R =, so n=o n+ x-tan-'x=x- 3: x3 x5 x7 x3 x5 x = cc (-I)"+'- xn+l and ( n=l n + x - tan-' x 53 m xn- = 'y(-~)~+l- n=l n+ ' J xn- = C + C (-I)"+'-. By Theorem, R =. n=l 4n - w n+l m x4n+3 6. By Example 7, tan-l(x) dx = dx=c+c(-)" with R =.?l=~ (n + )(4n + 3) J dx = f E (-)"~~" dx = c + co x5n+ n=o n=~ 5n+' T~US,. dx= x6 xl (.)~ (.)~' = The series is alternating, so if we use [ 6 6 = to six decimal places. the first two terms, the error is at most (.)/ =.9 x lo-'. So I =. - (.)~/6

6 464 CHAPTER8 SERIES xn 8. From Example 6, we know ln( - x) = - C -, so n=l n co x~~ w x4n+l Thus, n n= n(4n + ) ' /n(+ x4) dx = (-~)~+l- dx = c + (-I)"+' The series is alternating, so if we use the first three terms, the error is at most (.4)7/68 x.5 x lo-'. So I x (.4)~/5- (.4)'/8 + (.9)3/39 =.34 to six decimal places. 9. We substitute 3x for x in Example 7, and find that SO " x arctan(3x) dx = [g -- 3"x x7 37x x 5 35 x 7 63 x 9 I:' 87 The series is alternating, so if we use three terns, the error is at most - x 3.5 x lo-'. 63 x 9 9 x arctan(3x) dx % x 5 35 x %. 983 to six decimal places. So The series is alternating, so if we use only two terms, the error is at most x places, l3 - dxx x4 3 x lo3 7 x 7 3 x. 6. So, to six decimal x xn 3. Using the result of Example 6, In( - x) = - C -, with z = -., we have n=l n In. = ln[l - (-.l)] = The series is alternating, so if we use only the.... first four terms, the error is at most =.. So In. x x.953.

7 468 CHAPTER8 SERIES n f(n)(x) cosx -sinx f(n) () We use Equation 7 with f (x) = cosx. -cosx sinx cosx (-l)nxn If a, =, then (n)! = < for all x. So R = cm (Ratio Test). n 3 4 f(n)(x) sin.7: cosx -' sinx -3 COS~X 4 sin s f(,)(~) -3 f (")(o) = if n is even and f ')() = (-l)nn+, SO an+l xi lim - - lim = < for all x. I-m/ a, I n--(n+3)(n+) So R = CQ (Ratio Test). an+' gn+l lxln+l n! lim - lim n-- a, I = n-- [ (n + I)! 5" x n-- n + = < for all x - = lim - 5x SoR=cm. = < for all x I = lirn - x n-m n

8 47 CHAPTER8 SERIES 3. The general binomial series in (I 8) is x rn (-I)"-' (n - 3)xn =I+?+ c for x <, so R =. n= n. n! ii4) = ( + xlp4 = xn. The binomial coetlicient is 4. - ( + x ) ~ n =O - (I + x)\=o 6 Thus, (-l)%(n + l)(n + )(n + 3) xn for 5. - ( + x)" [( + x/)i3 8 8 n=o <, so R =. = :(I+ z) -3 = (;3) (q).. The binomial coeficieot is and /-xi < e x <, so R =. m xzn (-l)n(7r~)n (-)"7rZnx", R=m,=o (Zn)!,=o (n)! (n)! 7, cos x = C (-l)n - + f (x) = cos(7rx) = C 8. ex = - + f (x) = e-x/ = 7 XT - OC 5" (-)" (-l)nxn, R = %=o n! n=o = n=o n!

9 SECTION 8.7 TAYLOR AND MACLAURIN SERIES (a),- = ( + x)-'i4 = (:$) xn = - -x + x3+... n=o 4! 3! (b) / = - $a + Px 3-5x3 8 + s x / = / qm, so let x =.. The sum of the first four terms is then - i(.) + &(.)' - s (.)~ x.976. The fifth term is g (.)~ %.95, which does not affect the third decimal place of the sum, so we have l/ $'ii x.976. (Note that the third decimal place of the sum of the first three terms is affected by the fourth term, so we need to use more than three terms for the sum.) ('7) xn m n=o (n)! n=o 43. cosx = C (-)n- cos(x3) = C (-)n- - m x6n+l m %6n+ x cos(x3) = = (-l)n- i /scos(x3)dx=c+ X(-l)n with R = w. n =O (n)! (6n + ) (n)! sinx x"+' (-)" xzn, so J sin x Cc xr~+l OJ x x,o (n + l)!,b=o (n + dx=c+zo (n+)(n+)! I)! x = - C = 5 dx = 45. Using the series from Exercise 3 and substituting x3 for x, we get ' with R = oo. J 47. By Exercise 43, x cos(x3) dx = C + x (-)" x6n+,l=o (6~~+)(n)!'~O -.69, so L x cos(x3) dx = %.44 (correct to three decimal places) by the = -. 6! 4, Alternating Series Estimation Theorem. 48. From the table of Maclaurin series in this section, we see that 3 xn+l m xn+l tan-'s = C (-l)n- for x in [-l,] and sin x = (- )" for all real numbers x, so n =O n+ n=o (n + I)!

10 476 CHAPTER8 SERIES (.)'~ But (i + ;) = (.)'~ = 5. x lo-', so by the Alternating Series Estimation Theorem, (.)~ I = - =. 8 (correct to five decimal places). [Actually, the value is. 8, correct to seven decimal places.] 49. We first find a series representation for f (x) = (I + x)-'/~, and then substitute = - -x3 + -x6 - -x j d'it7 8 6 dx I = (.)- -(.)~, by the Alternating Series Estimation 8 Theorem, since 6 (.)~ z. 54 < lo-', which is the maximum desired error. Therefore, 5..5 rn (-l)n xn+ [(-)nxn+3 LIZ ~'e-~'dx = dx= ] =C I n and since the term n=o n! n=o n!(n + 3).n=o n!(n + 3)n+3 with n = is 79 =.354. n=o n!(n + 3)n <., we use - x - tan-' x x- (x-+x3+ix5-fx7+... ix3 - ix5 + fx lim = lim ) = lim s-o 3 x-o x3 - x3 - lim(+ - ix + 4x4 -...) = x- since power series are continuous functions. 3 - cosx - ( - &x+ $4 - Ax6+...) 5. lim = lim ~-ol+x-e~ x-o~+x-(~+x+~x~+$x~+$x~+~x~+~x~+~~~) &x- +x4 + Ax6 -...,+o -LX - x3 - i4 - ix5 - AX6 -...! 3! = lim - - 4! gx +g = lim x,~ -I -, - Lx - LX3-5x4 -...! 3! 4! 5! since power series are continuous functions '-