Math 163 REVIEW EXAM 3: SOLUTIONS

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1 Math 63 REVIEW EXAM 3: SOLUTIONS These otes do ot iclude solutios to the Cocept Check o p8. They also do t cotai complete solutios to the True-False problems o those pages. Please go over these problems o your ow, they may be exam problems. Geeral, p.8: True-False. False. (Ca you thik of a couterexample?). False. (Why?) 3. True. 4. True. (Explai why) 5. False. (Explai why) 6. True. (Explai why) 7. False. (Explai why) 8. True. (Explai) 9. False. (Explai why) 0. True. (Explai why). True.. True. (Explai why) 3. True. (Explai why) 4. False. (Ca you thik of a couterexample?) 5. False. (Ca you thik of a couterexample?) 6. True. (Why?) 7. True. (Why?) 8. True. (Why?) Sequeces. (a) lim a = (d) diverges sice lim a ( 9 (b) lim a 9 0 ) = 0 (c) diverges sice lim a = / = (usig L Hopital s Rule) si (e) lim = 0 by the Squeeze Theorem, sice si ad both the left ad right sides of this iequality approach 0 as ( ) (f) lim = 0 sice lim ( ) =0 ( 0) (g) lim = 0 sice! grows faster tha the expoetial 0 (ca you show this?)! (h) {a } = {3,, 3,, 3,,...} diverges sice it oscillates betwee 3 ad ad ever approaches a fiite limit + +/ (i) lim (l( +) l ) l l =l Series. (a) If a 0 ad f(x) is a decreasig fuctios such that f() =a, the = o a coverges if ad oly if o f(x) dx coverges. (b) Note: a > 0 for. Let f(x) = l x x. The f() =a ad f (x) = lx x < 0ifx> e,so f(x) is defiitely decreasig if x. Usig first substitutio, the itegratio by parts, we fid that the improper itegral l x x dx = l u [ ] R du e u (u +) eu R l R ( e R (R +)+e l (l + ))

2 R ( R + e R )+e l (l + ) ( R e R )+e l (l + ) = (l + ) is fiite, so it coverges (for the limit we used L Hopitals rule oce). By the itegral test, it follows that the sum = l coverges. Therefore the sum l also coverges.. Try to state these tests i your ow words. Make sure to clearly state all coditios. Compare with the statemets i the text. 3. (a) The series coverges (geometric series with r < ) ad (b) The series diverges (geometric series with r > ). (c) The series coverges (geometric series with r < ) ad ( ) + 3 =-3 ( ) = 3 ( ) = 3 ( ) = + (d) The series diverges by the direct compariso test, sice l 0 ad (e) This is a covergig telescopig series: S = + + as. (f) The series diverges sice lim = = diverges. ( +) = ( + ) = sice the partial sums + = ( + ) = e 0. (Show this) (g) This is a divergig telescopig series, sice the partial sums S = l+l( +) as. ( ) x (h) This geometric series coverges if (x )/ <, ie <x<3, ad i that case 3 = 6 3 x. (i) This geometric series coverges if x <, ie <x<, ad i that case (j) The series diverges sice lim l = = 0. (k) The series diverges sice lim (l) This geometric series coverges ( r = e < ) ad equals /( e ). ( x) = +x.! = =0. ( grows faster tha!, show this) (m) This series coverges ad equals e x4 sice it is the series for e y = y! evaluated at y = x4. () This telescopig series coverges ad equals 3π/4 + ta sice this is the limit of the partial sums S = ta + ta ta ( +) ta ( +)as. (o) This series coverges ad equals 3/ sice it is the series for cos y = ( ) y evaluated at y = π/6. ()! (p) This series coverges ad equals / sice it is the series for si y = ( ) y+ evaluated at y = π/4. ( + )! p83, : The series coverges (absolutely) sice lim + 3 = ad is a covergig p-series. p83, : The series diverges sice grows at the same rate as ad diverges.

3 p83, 3: The series coverges (absolutely) by the Ratio Test sice lim a + a = 5 <. p83, 4: The series coverges coditioally. It does ot coverge absolutely by the Limit Compariso Test sice + lim = ad diverges. It coverges by the Alteratig Series Test, sice a / = > 0, + / is decreasig (a + = + < + = a ) ad a 0as. p83, 6: The series diverges sice lim a =l 3 0 p83, 7: The series coverges absolutely by the Compariso Test sice cos 3 +(.) +(.) ad +(.) coverges. The latter follows from the Limit Compariso Test sice +(.) lim = ad (.) (.) = (. ) is a covergig geometric series. p83, 8: The series coverges (absolutely) by limit compariso to ( ) p83, 3: The series coverges coditioally. It does ot coverge absolutely sice It coverges by the Alteratig Series Test, sice a = > 0, is decreasig /3 (if f(x) =x /3 the f (x) = 3 x /3 < 0) ad a 0as. p83, 4: The series coverges absolutely sice is a covergig p-series. 3 p83, 5: The series coverges absolutely by the Ratio Test, sice lim a + a = 3 4 <. l /( ) / p83, 6: The series diverges sice lim a (Here, we used L Hopitals Rule) p83, 7: 5 = ( 4 5 ) =( 4 5 ) ( 4 5 ) =( 4 5 ) 4/5 =8 p83, 8: This is a telescopig series sice (+3) = 3 ( +3 N s N = 3 is a divergig p-series. /3 = =0 ) write out partial sums ad take limit: ( +3 )= 3 [ N 3 N + + N N+ + N N+3 ]= 3 [ N+ N+ N+3 ] a s N = N 3 [ ]= 8 p83, 30: = ( ) ( π/3) = cos( π/3) ()! p83, 34: The series coverges if < l x<, that is, /e<x<e p83, 35: For =7,a = < 0.000, ad the coefficiet of the 7th term is positive. Therefore s 7 < a <s , that is, < a < Hece, to 4 decimal places, the aswer is Power Series ad Taylor Series p83, 40: By the Ratio Test, this series coverges absolutely if lim a + a 3 =... x 5 N N+ ( +) = x 5 <,

4 ( ) that is, if x < 5 ad diverges if x > 5. If x = 5, the series coverges. If x = 5, the series coverges. Hece, the radius of covergece is r = 5 ad the iterval of covergece is x [ 5, 5]. p83, 4: By the Ratio Test, this series coverges absolutely if lim a + x + x + a =... = <, ( ) that is, if x + < 4( 6 <x<), ad diverges if x + > 4. If x = 6, the series coverges. If x =, the series diverges (harmoic series). Hece, the radius of covergece is r = 4 ad the iterval of covergece is x [ 6, ). p83, 4: By the Ratio Test, this series coverges absolutely if lim a + x + a =... =0<, +3 Therefore, the series coverges for all x. The radius of covergece is r = ad the iterval of covergece is x (, ). p83, 43: By the Ratio Test, this series coverges absolutely if lim a + +3 a =... x 3 +4 = x 3 <, that is, if x 3 < / (.5 <x<3.5), ad diverges if x 3 > /. If x =.5, ( ) the series coverges by the Alteratig Series Test sice b = > 0, is decreasig (b + = +4 < +3 = b ) ad b 0as.Ifx =3.5, the series Limit Compariso Test sice lim +3 / iterval of covergece is x [.5, 3.5). +3 diverges by the =. Hece, the radius of covergece is r =/ ad the p83, 45: f(x) = si x, f (x) = cos x, f (x) = si x, f (x) = cos x. f(π/6) = /, f (π/6) = 3/, f (π/6) = /, f (π/6) = 3/. Thus, f(x) =/+ 3/(x π/6) /( )(x π/6) 3/( 3!)(x π/6) (You could fid a codesed form of writig the series - see solutio i back of book - but o the exam you will most likely be asked to fid oly the first few terms of the series) p83, 46: f(x) = cos x, f (x) = si x, f (x) = cos x, f (x) = si x. f(π/3) = /, f (π/3) = 3/, f (π/3) = /, f (π/3) = 3/. Thus, f(x) =/ 3/(x π/6) /( )(x π/6) + 3/( 3!)(x π/6) p83, 47: x =+x + x + x 3 + x 4 + x 5...= x, for x <. +x = x + x x 3 + x 4 x = ( x), for x <, that is x <. x +x = x x 3 + x 4 x 5 + x 6 x = ( ) x +, for x <. p83, 49: l( x) = dx x = +x + x + x 3...dx= (x + x x Thus l( x) = + x3 3 + x4 4 p83, 50: e x =+x + x /+x 3 /3! + x 4 /4! + x 5 /5!...= x /! for all x. e x =+x +(x) /+(x) 3 /3! + (x) 4 /4! + (x) 5 /5!...= (x) /! for all x C), where C =0.

5 xe x = x +x +4x 3 /+8x 4 /3! + 6x 5 /4! + 3x 6 /5!...= x + /! for all x. p83, 5: si x = x x 3 /3! + x 5 /5! x 7 /7! +...= x+ /( + )! for all x. si x 4 = x 4 x /3! + x 0 /5! x 8 /7! +...= x4(+) /( + )! for all x. p83, 5: e x =+x + x /+x 3 /3! + x 4 /4! + x 5 /5!...= x /! 0 x = e l 0x = + (l 0x) + (l 0x) / + (l 0x) 3 /3! + (l 0x) 4 /4! + (l 0x) 5 /5!...= (l 0x) /! p83, 55: e x x dx = +x+x /+x 3 /3!+x 4 /4!+x 5 /5!... x dx = x ++x/+x /3! + x 3 /4! + x 4 /5!...dx = C +l x + x + x /4+x 3 /(3 3!) + x 4 /(4 4!) + x 5 /(5 5!)... p83, 56: f(x) = +x, f (x) =(/)( + x) /, f (x) = (/4)( + x) 3/, f (x) =(/4)(3/)( + x) 5/. f(0) =, f (0) = (/), f (0) = (/4), f (0)=(/4)(3/). Thus +x =+x/ x /8+x 3 / x4 =+x 4 / x 8 /8+x / x4 dx =[x + x 5 /0 x 9 /7 + x 3 /(3 6) +...] 0 0 =+/0 /7 + /(3 6) +... = error where error < /(3 6) < sice the series is alteratig (after first term). si x x p83, 59: lim x 0 x 3 x x 3 /3! + x 5 /5! x 7 /7! +... x x 3 /3! + x 5 /5! x 7 /7! +... x 0 x 3 x 0 x 3 x 0 ( /3! + x /5! x 4 /7! +...)= /6 p806, 5: f(x) =(+x) 3, f (x) = 3( + x) 4, f (x) =3 4( + x) 5, f (x) = 3 4 5( + x) 6. Oe ca see the patter, f ( + )! (x) =( ) ( + x) (+3) Therefore f ( + )! (0) = ( ) ad the Taylor Series for ( + x) 3 ( + )! x is ( ) =! ( ) ( + )( +)x By the Ratio Test, the series coverges if lim a + + 3)( +) a =... x ( = x <, ( + )( +) ad diverges if x >. Hece, the radius of covergece is r =. p806, : f(x) =+x + x, f (x)=+x, f (x) =,f () (x) =0for 3. f() = 7, f () = 5, f () =, f () () = 0 for 3. ad the Taylor Series for + x + x cetered at a = is 7 + 5(x ) + /(x ). This series coverges everywhere sice it is a fiite sum. Therefore + x + x =7+5(x ) + (x ) for all x. p806, 4: f(x) =lx, f (x) = x, f (x) = x, f (x) = x, f (4) (x) = 3 3 x. Patter: 4 f + ( )! (x) =( ) x,. Therefore f() = l, f + ( )! () = ( ),. + ( )! (x ) ad the Taylor Series for l x cetered at x = is l + ( )! ( x) =l 5

6 3. (a) You ca either do this by computig all the derivatives of f () (), or you ca maipulate the geometric series: (b) x = x = x = ( (x ) = d [ ] dx x = + x + ( x ) + (x ( + x ) + 3x 3 + 4x = ) )3 (x) = x ( x ) 4. (a) By computig the derivatives f () (0) ad substitutig ito the formula for the Taylor series about a = 0 we fid that f(x)=+5x +0x +0x 3 +5x 4 + x 5 (b) By computig the derivatives f () (0) ad substitutig ito the formula for the Taylor series about a = 0 we fid that f(x) = (x ) (x ) +0 (x ) 3 +5 (x ) 4 +(x ) 5 (c) (Chage this problem to read: Fid Taylor series at a = ). The fuctio is already give i powers of x +, so its Taylor series is (x +) 5 6

PLEASE MARK YOUR ANSWERS WITH AN X, not a circle! 1. (a) (b) (c) (d) (e) 3. (a) (b) (c) (d) (e) 5. (a) (b) (c) (d) (e) 7. (a) (b) (c) (d) (e)

PLEASE MARK YOUR ANSWERS WITH AN X, not a circle! 1. (a) (b) (c) (d) (e) 3. (a) (b) (c) (d) (e) 5. (a) (b) (c) (d) (e) 7. (a) (b) (c) (d) (e) Math 0560, Exam 3 November 6, 07 The Hoor Code is i effect for this examiatio. All work is to be your ow. No calculators. The exam lasts for hour ad 5 mi. Be sure that your ame is o every page i case pages

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