MTH 142 Exam 3 Spr 2011 Practice Problem Solutions 1

Transcription

1 MTH 42 Exam 3 Spr 20 Practice Problem Solutios No calculators will be permitted at the exam. 3. A pig-pog ball is lauched straight up, rises to a height of 5 feet, the falls back to the lauch poit ad bouces straight up agai. It cotiues to bouce, each time reachig a height 90% of the height reached o the previous bouce. Fid the total distace that the ball travels. The ball has goe 30 ft at the first retur to the lauch poit, the 2(5)(.9) more feet at the secod retur, 2(5)(.9)(.9) at the third retur, etc. The total distace is the (.9) + 30(.9) (.9) 3 + = 30.9 = 300feet. 3.2 Use the itegral test to determie the covergece of the followig series: 3/2 e Series coverges if ad oly the improper itegral x 3/2 dx coverges. x 3/2 dx = 2 < so series coverges. Series coverges if ad oly the improper itegral xe x dx coverges. xe x dx = 2/e < (itegrate by parts ad use L Hopital s rule) so series coverges. 3.3 Determie if the followig series coverge or diverge. Give your reasoig usig complete seteces. l 2! ( + 2)!

2 MTH 42 Exam 3 Spr 20 Practice Problem Solutios 2 Series coverges if ad oly the improper itegral l x dx coverges. x 2 Itegrate by parts ad use L Hopital s rule to see that this itegral coverges to so series coverges. Alterately, l x < x /2 whe x is large sice by l x L Hopital s rule lim x = 0, so l x so x /2 x 2 x 3/2 compariso with the itegral i problem 2 above l x x 2 dx is coverget by! (+2)! = (+2)(+) < 2 so give series coverges by compariso with p-series with p = 2 which is coverget. 3.4 For each of the followig items ad choose a correct coclusio ad reaso from amog the choices (R), (C), (I) below ad provide supportig computatio. For example, if you choose (R) calculate ad iterpret a suitable ratio. (R) Coverges by the ratio test. (I) Coverges by the itegral test. (C) Diverges by a p-series compariso l (R). The ratio a + a = <. (C). For > 3 l > so the series diverges by compariso with the p-series with p = Which of the followig correctly classifies the series gives a valid reaso? Circle your aswer. No other reaso required. e ad i) Diverges: lim a 0. ii) Coverges: sum of two coverget geometric series. iii) Coverges: Compariso test with geometric series with ratio 3/e.

3 MTH 42 Exam 3 Spr 20 Practice Problem Solutios 3 iv) Coverges: Compariso test with geometric series with ratio e/4. v) Diverges: costat multiple of the harmoic series. Which of the followig correctly classifies the series valid reaso? Circle your aswer. No other reaso required. i) Diverges: lim a 0. ii) Coverges: Ratio test. iii) Coverges: Compariso test with a p-series, p >. iv) Diverges: Ratio test. 3 ad gives a v) Diverges: Compariso or limit compariso with the harmoic series. Aswer: i) Aswer: v) 3.6 Determie if the followig series coverge or diverge. Give your reasoig usig complete seteces. ( ) + 2 ( ) + =2 l A alteratig series, but the terms do ot approach 0 (they approach ) so diverget. A alteratig series with terms decreasig to 0 so coverget. 3.7 Fid the iterval of covergece for the power series Fid the radius of covergece for! (2)! x (x 3) 2

4 MTH 42 Exam 3 Spr 20 Practice Problem Solutios 4 (+)2 + (x 3) + 2 (x 3) = 2 (x 3) + ( + )2 + (x 3) = 2 + x 3 x 3 2 as. x 3 < for x 3 < 2, that is, < x < 5 so series coverges 2 if < x < 5 ad diverges if x > 5 or x <. At x = 5 the series becomes the harmoic series which diverges, but at x = it becomes the alteratig series ( ) which coverges. The iterval of covergece is therefore [, 5). (+)! (2+2)! x+! (2)! x = + (2 + 2)(2 + ) x 0 < for every x whe. Thus the series coverges for all x ad the radius of covergece is. 3.8 Fid the Taylor series about x = π of cos(x). Your aswer should clearly idicate the patter of the terms. (Note the ceter, π.) Fid the Taylor series of l(x) at x =. With f(x) = cos(x), f(π) = cos π =, f (π) = si π = 0, f (π) = cos π = f (π) = si π = 0,f (π) = cos π = ad this patter repeats, so the Taylor series about π is =0 f () (π) (x π) = +! 2! (x π)2 4! (x π)4 + 6! (x π)6 With f(x) = l(x), f() = 0, f (x) = /x so f () =, f (x) = x 2 so f () =, f (x) = 2x 3 so f () = 2, f (x) = 3 2x 4 so f () = 3! ad i geeral f () (x) = ( ) ( )!. The, rememberig that f() = 0, f () () (x ) =! ( ) ( )! (x ) =! ( ) (x ) =

5 MTH 42 Exam 3 Spr 20 Practice Problem Solutios 5 (x ) 2 (x )2 + 3 (x )3 4 (x ) Give the Taylor polyomial at 0 of degree 4 for each of the followig fuctios. Use the easiest method you kow, icludig your kowledge of the Taylor series for si x, e x, ( + x) p, etc. f(x) = x 2 cos(x). g(x) = c) f(x) = 2x x 0 si(t 2 ) dt d) f(x) = ( + x) si(2x) so x 2 cos(x) = x 2 ( x2 2! + x4 4! x6 6! + ) = x2 x4 2! + x6 4! x8 6! + P 4 (x) = x 2 x4 2!. Usig the biomial series ( + x) p = + px + p(p ) x 2 + 2! with p = /2 ad replacig x by 2x we get p(p )(p 2) x 3 + 3! ( 2x) /2 = ( + ( 2x)) /2 = 2 ( 2x) + ( 2 )( 3 2 )( 2x)2 /2!+ ( 2 )( 3 2 )( 5 2 )( 2x)3 /3! + ( 2 )( 3 2 )( 5 2 )( 7 2 )( 2x)4 /4!

6 MTH 42 Exam 3 Spr 20 Practice Problem Solutios 6 so P 4 (x) = + x x x x4. c) si(t 2 ) = t 2 (t 2 ) 3 /3! + (t 2 ) 5 /5! = t 2 t 6 /3! + t 0 /5! so x 0 si(t2 ) dt = 3 x3 7 3! x7 + 5! x so P 4 (x) = 3 x3. d) si(2x) = (2x) (2x)3 3! + so ( + x) si(2x) = ( + x)[2x 8x3 6 + ] = 2x + 2x2 8x3 6 8x4 6 + hece P 4 (x) = 2x + 2x x3 4 3 x The graph of y = f(x) passes through the poit (0, 2) ad ear that poit the fuctio is decreasig ad cocave up. If the third degree Taylor polyomial of f ear 0 is P 2 (x) = a + bx + cx 2 what are the sigs of a, b ad c? If f(x) = e x2 fid f (8) (0) ad f (9) (0), that is, the 8th ad 9th derivative of f at 0. Suggestio: Cosider the Taylor series of f ad look at the coefficiets of x 8 ad x 9. a = f(0) = 2 is egative, b = f (0) < 0 sice f is decreasig, c = f (0)/2! 0 sice f is cocave up. e x2 = x 2 + x 4 /2! x 6 /3! + x 8 /4! x 0 /5! + x 2 /6! + Sice the coefficiet of x 8 i a Taylor series about 0 must be f (8) (0)/8! we have /4! = f (8) (0)/8! = f (8) (0) = 8!/4! = = 680. Similarly, f (9) (0)/9! = 0 sice the series has o odd degree terms so f (9) (0) = 0

7 MTH 42 Exam 3 Spr 20 Practice Problem Solutios 7 3. Suppose you estimate /e = e by P 5 (.0) where P 5 is the 5th degree Taylor polyomial of the fuctio e x at 0. Give a upper boud for the error i usig this estimate. Suppose you wat to use a Taylor polyomial at 0 to estimate cos(x) ad you wat this estimate to be accurate to withi.0 for all values of x betwee 0 ad π/2. Will the 4th degree Taylor polyomial be adequate? If ot, what degree Taylor polyomial should you use? Give your reasos. The error is M 0 6 /6! where M is a upper boud for the maximum absolute value of the sixth derivative of e x for x betwee 0 ad. Sice the sixth derivative is e x ad this fuctio has absolute value o the iterval [0,.0] we ca take M =. Thus the error is /6! = /720. The error usig the degree Taylor polyomial will be M x 0 + /(+ )! where M is a upper boud for the absolute value of the + -st derivative of cos x betwee 0 ad π/2. Sice all derivatives are either ± si(x) or ± cos(x) we ca use M =, so the error i usig P 4 (x) will be less tha (π/2) 5 /5!. A rough upper estimate of this is 2 5 /5! = 4/5 sice π/2 < 2, so P 4 (x) might ot be accurate eough. We wat to use large eough so that, say, 2 + /( + )! <.0. If we try = 9 we fid 2 0 /0! = 024 < /6! = /720 so = 9 will certaily do, i fact = ! will work.