MTH Lecture 2: Solutios to Practice Problems for Exam December 6, 999 (Vice Melfi) ***NOTE: I ve proofread these solutios several times, but you should still be wary for typographical (or worse) errors.. Compute the itegral Solutio: 2. Compute the itegral Solutio: 0 0 8 8 x dx. a x dx x dx a 0 8 a a 0 (/2)x2/] 8 (/2)( 8) 2/ ]= 6. a 0 [(/2)a2/ x 4 dx. b dx x4 b x dx 4 ] b b x b [ b ]=.. Determie if the followig series coverge or diverge. Justify your aswers completely. (a) ( )! =0 Solutio: This series diverges because the terms do t coverge to zero. 4 + (b) 8 2 Solutio: We ca compare this series with the series, which we kow coverges. 4 Sice ( 4 +) 8 + 4 ( 8 2) (/4 ) 8 2 =, the series coverges by the it compariso test. Most of these problems were kidly provided by Dr. Richeso.
(c) 5 2 Solutio: This series coverges, sice the expoet p = /2 of i the deomiator is greater tha. (d) (l ) 2 =2 Solutio: We compare this series with l Hopital s rule twice, / l() 2 / l() 2, which we kow diverges. Now usig 2l()(/) 2l() 2(/) 2 =, (e) so the series diverges by the it compariso test. +! =0 Solutio: Computig the ratio of the ( + )stadth terms gives ( +2)/( +)! ( +)/! = +2 ( +)( +). Sice this coverges to 0 as, the series coverges by the ratio test. (2x ) 4. Fid the iterval of covergece for the power series ( ). Solutio: The ratio of the absolute values of the ( +)stadth terms is ( ) + (2x ) + ( +) ( ) (2x ) = (2x ) +. Sice this coverges to 2x as, the series coverges absolutely (by the ratio test) whe 2x <. Solvig for x, the series coverges absolutely whe <x<2. Checkig ( ) 2 the edpoits of this iterval, whe x = the series is give by =,which we kow diverges. Whe x =2,theseriesisgiveby ( ), the alteratig harmoic series, which we kow coverges. So the iterval of covergece is (, 2]. 2
5. (a) What are the first four terms of the Maclauri series for f(x) =(+x) 4? What is the iterval of covergece for this series? Solutio: First compute the first three derivatives: f (x) = ; f (x) = ; 4( + x) /4 6( + x) 7/4 f (x) = Pluggig i x =0yields f (0) = 4 ; f (0) = 6 ; f (0) = 2 64. So the first four terms of the series are give by f(0) + f (0)x + f (0) x 2 + f (0) 2!! This series coverges for <x<. 2 64( + x) /4. x =+ 4 x 2 x2 + 7 28 x. (b) What are the first four terms of the Maclauri series for g(x) =(+2x 2 ) 4? What is the iterval of covergece for this series? Solutio: Wecajustplugi2x 2 for x i the above series. This yields + 4 (2x2 ) 2 (2x2 ) 2 + 7 28 (2x2 ) =+ 2 x2 8 x4 + 7 6 x6. This series coverges whe 2x 2 <. Solvig for x, the series coverges whe (/ 2) <x<(/ 2). 6. Compute the sum ( ) 2 5. Solutio: This is just 2 5 r = 2/5, ad so coverges to a/( r) = 2/7. ( 2 5). This is a geometric series with a = 2/5 ad 7. Write dow the first five terms of the Taylor series for l x cetered at x =. Solutio: The first four derivatives are f (x) = /x, f (x) = /x 2, f (x) =2/x, ad f (4) (x) = 6/x 4. Pluggig i x =yieldsf () =, f () =, f () = 2, ad f (4) () = 6. So the first five terms of the Taylor series about are 0 + ()(x ) (/2!)(x ) 2 +(2/!)(x ) (6/4!)(x ) 4. 8. Give the first four terms of the Maclauri series for f(x) =x cos(2x). Solutio: The first four terms of the Maclauri series for cos(x) are x 2 /2! + x 4 /4! x 6 /6!. Plug i 2x for x to get the first four terms of the Maclauri series for cos(2x): So the aswer is 4x 2 /2! + 6x 4 /4! 64x 6 /6!. x 4x 5 /2! + 6x 7 /4! 64x 9 /6!.
9. Give the decimal expasio of e accurate to 5 decimal places. Hit: use the Maclauri series for e x ad the fact that e = e. Solutio: We kow that the Maclauri series for e x is so where f(x) =e x =+x + x 2 /2! + x /! + x 4 /4! +..., f() = e =++/2! + /! + /4! +... =++/2! + +/!+R (), R () = f (+) (c) ( +)! () = e c ( +)! for some c i the iterval [0, ]. Sice this fuctio is positive ad icreasig i c, ittakes its maximum value o the iterval [0, ] at c =,sothat R () e ( +)! < ( +)!. Whe = 8, this yields R 8 () < /9! 0.000008, which is less tha.0000. So ++/2! + /! + /4! + /5! + /6! + /7! + /8! approximates e to withi 5 decimal places. 0. Does the series ( ) coverge absolutely? Does the series coverge? Solutio: It does ot coverge absolutely, sice the series of absolute values is (/ ), which we kow diverges. But the alteratig series test tells us that the origial series coverges, sice the / is always positive ad decreases to 0.. For each of the followig series covergece problems, determie whether the argumet give is correct. If ot, explai precisely why ot, ad if possible, determie the aswer (covergece or divergece) to the problem. (a) CLAIM: The series 2 +7 coverges, because 2 +7 =0. Solutio: This argumet is faulty. Covergece of the terms of a series to 0 is ot sufficiet to imply that the series is coverget. I fact, by usig the it compariso test with the diverget series, we fid that the origial series diverges. 4
0 + (b) CLAIM: The series diverges by the it compariso test, because ( +)( +2) whe we compare the series with the coverget series we fid that 2 (0 +)/(( +)( +2)) =0. / 2 Sice this is bigger tha, the series diverges. Solutio: This argumet is also faulty. The it compariso test says that as log as the it is a positive umber, the two series either both coverge or both diverge. Sice the it is 0, the two series above either both diverge or both coverge. Sice is coverget, both series coverge. 2 (c) CLAIM: The series 2 e coverges, because ( +) 2 /e + = 2 /e e. Sice this is less tha, the series coverges by the ratio test. Solutio: This is correct. (d) CLAIM: The series ( ) is absolutely coverget by the ratio test, because 4 +7 ( +)/(4( +) +7) /(4 +7) Solutio: This is a faulty argumet. Whe the ratio of successive terms coverges to, the ratio test is icoclusive. I this case we ca compare the series of absolute values to the coverget series to show via the it compariso test that the 2 origial series is absolutely coverget. =. 5