Math 181, Solutions to Review for Exam #2 Question 1: True/False. Determine whether the following statements about a series are True or False.

Transcription

1 Math 8, Solutios to Review for Exam #2 Questio : True/False. Determie whether the followig statemets about a series are True or False. X. The series a diverges if lim s 5.! False: The series coverges to If the sequece {a } coverges the the series P a coverges. False: Tae the harmoic series as a couter examle, the sequece { } coverges but the harmoic series P diverges.. The series X coverges.. True:. > covergesby-test.. The sequece a coverges. True: lim!. X 5. is a geometric series with r /5. 5 False: r 5. Questio 2: a) Absolutely coverget by ratio test. Tae lim a +! a b) Absolutely coverget. Use comariso test with /2 >. c) Not absolutely coverget sice we ca comare X + to X X let a + ad b a the lim! lim + <,ad-test with 2 /2 + 2 <! diverges by the -test ad /2 diverges,ad usig the Limit Comariso Test. More secifically, b adthusbothseries X + ad diverge. However, it is coditioally coverget by alteratig series test: Let b + decreasig sice b + ale b. Ad lastly lim! X, b is + ;sicethelimitis,wecausethe alteratig series test. Questio : a) This oe is a geometric series with a 2adr, ad it coverges to. 5 /5 Or you ca rewrite the series as: X 2 2+ X X X 2 X /5.

2 b) Rewritig ( +)( +2) +2 telescoig series. The th artial sum S usig artial fractios, we recogize this as a i the limit, everythig cacels out excet the 2. + We ca coclude that the series coverges to lim! S 2. c) The series diverges sice lim! a 6 bythedivergecetest. d) Diverges, use either direct comariso or Limit comariso to comare to do limit comariso: it ll be slightly easier. Use a + 2 ad b, the X. Let s a + lim lim! b! 2 lim 2 +! 2, X ad sice diverges (harmoic series or ad-test) by Limit Comariso both series coverge. Questio : a) Radius of covergece:. Iterval of covergece: [ 6, 2). To figure out where it coverges, use the ratio test with a (x+2). This gives! a + a (x +2) + lim! ( +) + (x +2) (x +2) lim! + x +2. It will coverge if this is less tha : x +2 <, which meas < (x +2)<, i.e. <x+2<, so 6 <x<2. We ca see from this that the legth of the iterval is 8, so the radius of covergece is. We still eed to figure out what haes at the two edoits where r adtheratiotestisicoclusive. At x 6, the series becomes X ( 6+2) X ( ) X ( ). This is the classic alteratig harmoic series, which coverges accordig to the alteratig series test. At x 2,theseriesbecomes X (2 + 2) X X,

3 which diverges by the -test. b) This time we re u agaist X 2 (x ) 2. For this sort of thig, you always wat to use the ratio test. The ratio is 2 + (x ) + 2! ( +) 2 2 (x ) lim 2(x ) 2 2 x.! ( +) 2 This is less tha if 2 x <, so < 2(x ) <, which meas /2 <x < /2 ad 5/2 <x<7/2. So the radius of covergece is /2. We still eed to chec the edoits, as usual. At 5/2, this is X 2 (5/2 ) 2 X which coverges by the alteratig series test. At 7/2, this is X 2 (7/2 ) X 2 2 ( /2) 2 2 (/2) 2 X ( ), 2 which coverges by the -test. So it coverges at both edoits. Puttig it all together: radius of covergece is, iterval of covergece is [, + ] Questio 5: First rewrite x x +2 x 2 ( x/2). Now, usig the ower series reresetatio we get the ower series reresetatio x X ( x/2) X x x 2 X X 2, ( ) 2 x. Fially, the desired ower series is obtaied by multilyig through by x 2 : x x +2 X ( ) 2 + x+. We still eed to figure out where this coverges. For that, we use the ratio test as usual.! ( ) + x (+)+ 2 (+)+ 2 + ( ) x + lim! x 2 x 2.

4 This is less tha if 2 <x<2. At the edoit 2, the series is P ( ) 2 +, while at 2 + 2it s P ( ) ( 2) +. Both of these diverge accordig to the divergece test. 2 + The radius of covergece is 2 ad the iterval of covergece is ( 2, 2). Questio 6: (a) e x +x + x2 + x of e x to fid the Maclauri series of x 2 e ( x2). First, substitute i x 2 for x: e x +x + x2 + x e x2 +( x 2 )+ ( x2 ) 2 x 2 + x + ( x2 )!! + x8 + ( x2 )! b) Itegrate o both sides above to get Z e x2 dx K + x x + x5 x 7 7! + x9 9. Questio 7: The Taylor olyomial of degree 2 about x is P 2 (x) f() + f ()(x ) + f () (x ) 2 2+ x 2 Substitutig x 5itotheolyomialwegetthearoximatio (x ) 2. 6 Questio 8: (a) Use the Maclauri series e x +x + x2 + x of e x to fid the Maclauri series of x 2 e ( x2). First, substitute i x 2 for x: e x +x + x2 + x e x2 +( x 2 )+ ( x2 ) 2 x 2 + x + ( x2 )!! + x8 + ( x2 )! Now, multily through by x 2. To achieve that, we eed to multily every term i the series by x 2. e x2 x 2 e x2 x 2 x 2 + x x + x6! + x8 x 8! + x!

5 (b) Use the first three o-zero terms of the series you foud i art (a) to aroximate the defiite itegral. We have Z x 2 e ( x2) Z x 2 x Z x 2 e ( x2) dx x + x6 dx x x7 5 + ( +) 2.2.