Math 116 Practice for Exam 3

Transcription

1 Math 6 Practice for Eam 3 Geerated April 4, 26 Name: SOLUTIONS Istructor: Sectio Number:. This eam has questios. Note that the problems are ot of equal difficulty, so you may wat to skip over ad retur to a problem o which you are stuck. 2. Do ot separate the pages of the eam. If ay pages do become separated, write your ame o them ad poit them out to your istructor whe you had i the eam. 3. Please read the istructios for each idividual eercise carefully. Oe of the skills beig tested o this eam is your ability to iterpret questios, so istructors will ot aswer questios about eam problems durig the eam. 4. Show a appropriate amout of work (icludig appropriate eplaatio) for each eercise so that the graders ca see ot oly the aswer but also how you obtaied it. Iclude uits i your aswers where appropriate. 5. You may use ay calculator ecept a TI-92 (or other calculator with a full alphaumeric keypad). However, you must show work for ay calculatio which we have leared how to do i this course. You are also allowed two sides of a 3 5 ote card. 6. If you use graphs or tables to obtai a aswer, be certai to iclude a eplaatio ad sketch of the graph, ad to write out the etries of the table that you use. 7. You must use the methods leared i this course to solve all problems. Semester Eam Problem Name Poits Score Witer Witer Fall Witer Fall Fall Witer Witer Fall Fall Total 8 Recommeded time (based o poits): 98 miutes

2 Math 6 / Fial (April 23, 2) page 5 3. [2 poits] For each of the followig series, determie the iterval of covergece ad write it o the space provided to the right of the series. Be sure to show all appropriate work to justify your aswer. a. [6 poits] ( ) ( 2) Solutio: By usig the ratio test we see that < () 2 (+) = 2. So the ceter is at = 2 ad the radius is. We eed to check the edpoits = ad = 3. We see it coverges whe = 3 by alteratig series, ad diverges at = by the p-test. b. [6 poits]! = Solutio: By usig the ratio test we get (+)! +! (+) =. So it has radius of covergece ad coverges oly at the ceter which is =. Uiversity of Michiga Departmet of Mathematics Witer, 2 Math 6 Eam 3 Problem 3 Solutio

3 Math 6 / Fial (April 9, 22) page 5. [8 poits] Cosider 4 (+) 2. a. [2 poits] Does the series coverge for = 2? Justify your aswer. Solutio: At = 2 4 (+) 2 = The series diverge sice + =. 4 (+) 4 = + b. [2 poits] Based oly o your aswer from part a, what ca you say about R, the radius of covergece of the series? Circle your aswer. Solutio: R = 2 R > 2 R < 2 R 2 R 2 sice it is possible for = 2 to be oe of the edpoits i the iterval of covergece. c. [4 poits] Fid the iterval of covergece of the series. Solutio: (+2) 2+2 = 2 (+) 2 4(+2) = (+) 2 If 2 <, the the series coverges. Hece Ratio test states that the series coverges if 2 < < 2. We eed to check the edpoits = ±2. We already checked 4 = 2. For = 2, 4 (+) 2 = 4 (+) 4 = + diverges by part a. The iterval of covergece of the series is 2 < < 2. Uiversity of Michiga Departmet of Mathematics Witer, 22 Math 6 Eam 3 Problem 5 Solutio

4 Math 6 / Fial (December 4, 22) page 7 5. [9 poits] Cosider the followig power series: 5 ( 4)+ a. [4 poits] Fid the radius of covergece of the power series. Show all your work. Solutio: so R = 5. (+) = < 4 < 5, 5 + = 5 4. b. [5 poits] For which values of does the series coverge absolutely? For which values of does it coverge coditioally? Solutio: Coverges absolutely iside radius: (, 9). Left edpoit: =, 5 ( 5)+ = ( ) +, coverges coditioally (alteratig harmoic series). Right edpoit: = 9, = , diverges. So, coverges coditioally for =, absolutely for < < 9. Uiversity of Michiga Departmet of Mathematics Fall, 22 Math 6 Eam 3 Problem 5 Solutio

5 Math 6 / Fial (April 26, 23) page 2 8. [8 poits] Cosider the power series 2 ( 5). I the followig questios, you eed to support your aswers by statig ad properly justifyig the use of the test(s) or facts you used to prove the covergece or divergece of the series. Show all your work. a. [2 poits] Does the series coverge or diverge at = 3? Solutio: At = 3, the series is series test, sice / is decreasig ad coverges to. ( ), which coverges by the alteratig b. [2 poits] What does your aswer from part (a) imply about the radius of covergece of the series? Solutio: Because it coverges at = 3, we kow that the radius of covergece R 2. c. [4 poits] Fid the iterval of covergece of the power series. Solutio: Usig the ratio test, we have = 5 = L, so the radius of covergece is 2. Now we have to check the edpoits. We kow from part (a) that it coverges at = 3. For = 7, we get, which diverges. Thus, the iterval of covergece is 3 < 7. Uiversity of Michiga Departmet of Mathematics Witer, 23 Math 6 Eam 3 Problem 8 Solutio

6 Math 6 / Fial (December 7, 23) page 7. [8 poits] Cosider the power series 2 3 ( 5). I the followig questios, support your aswers by statig ad properly justifyig ay test(s), facts ad computatios you use to prove covergece or divergece. Show all your work. a. [4 poits] Fid the radius of covergece of the power series. Solutio: 2 + 3(+) ( 5)+ 2 3 ( 5) = = 2 5 the R = 2 or R = (+) = + 2 = 2. Radius of covergece=.5 b. [4 poits] Fid the iterval of covergece of the power series. Make sure to cite all the tests you use to fid your aswer. Solutio: Testig the edpoits: 2 = 4.5: 3 (4.5 5) = ( ) coverges by alteratig series test. 3 2 = 5.5: 3 (5.5 5) = diverges by p-series test p =. 3 Iterval of covergece: 4.5 < 5.5. Uiversity of Michiga Departmet of Mathematics Fall, 23 Math 6 Eam 3 Problem 7 Solutio

7 Math 6 / Fial (December 2, 24) page 8. [7 poits] Cosider the power series (+2) 3. a. [2 poits] At which -value is the iterval of covergece of this power series cetered? Solutio: This power series is cetered o = 2. (+2) b. [5 poits] The radius of covergece for the power series 3 is 3. Fid the iterval of covergece for this power series. Thoroughly justify your aswer. Solutio: Sice the radius of covergece for this power series is 3 ad it is cetered o = 2, the iterval of covergece cotais the ope iterval ( 2 3, 2+3) = ( 5,). Now we oly eed to check the edpoits = 5 ad =. (+2) For = : 3 = diverges by the p-test with p = (this is the harmoic series). ( 5+2) ( ) For = 5: 3 = which coverges by the alteratig series test. Therefore, the iterval of covergece for this power series is [ 5, ). 9. [5 poits] Fid the radius of covergece for the power series Solutio: Let the -th term be deoted by a a + (2(+))! 2(+) (!) 2 = a ((+)!) 2 (2)! 2 = (2+2)(2+) 2 ((+) 2 (2)! (!) 22 Therefore, we ca use the ratio test: a + = (2+2)(2+) 2 a ((+) 2 = 42. So this series coverges for with 4 2 <, or rather with 2 < 4 radius of covergece is /2. which implies that the Uiversity of Michiga Departmet of Mathematics Fall, 24 Math 6 Eam 3 Problem 8 Solutio

8 Math 6 / Fial (April 23, 25) page 3 2. [4 poits] a. [2 poits] You are give that the power series C (+3) coverges whe = 6 ad diverges whe =. What are the largest ad smallest possible values for the radius of covergece R? Solutio: = 3 R 4 b. [2 poits] Give the eact value of the ifiite series ( ) = ! = Solutio: ( ) = 3e 3 usig the Taylor series for e cetered at =.! = 3. [5 poits] Determie whether the followig itegral coverges or diverges. If the itegral coverges, circle the word coverges. If the itegral diverges, circle diverges. I either case, you must show all your work ad idicate ay theorems you use. cos() 2 d CONVERGES DIVERGES Solutio: For, we have that cos() cos(). We also kow that cos() diverges. Therefore, cos() 2 d diverges by the compariso test. 2d Uiversity of Michiga Departmet of Mathematics Witer, 25 Math 6 Eam 3 Problem 2 Solutio

9 Math 6 / Fial (April 23, 25) page 5 5. [ poits] a. [5 poits] Determie the radius of covergece of the power series ( ) ( 5) 4 5 (6). Solutio: The above series will coverge for values such that by the ratio test. We have ad so the desired iequality holds if Thus, the radius of covergece is (+) 5 (6) (+) 5 (6) + < (6) = (6) <. This is equivalet to 5 < 2. The radius of covergece is 2. (+2) b. [5 poits] The power series 4 has radius of covergece. Determie the + = iterval of covergece for this power series. Solutio: For =, we get the series = We have that + The it compariso test the tells us that the series if the series = coverges. For =, we get the series coverges absolutely. = = 4 + =. coverges if ad oly 3 coverges. Sice is a p-series with p > it coverges, ad so 3 The iterval of covergece for = = ( ) (+2) 4. By the work show above, this series + (+2) 4 + is the The iterval of covergece is [, ]. Uiversity of Michiga Departmet of Mathematics Witer, 25 Math 6 Eam 3 Problem 5 Solutio

10 Math 6 / Fial (December 7, 25) DO NOT WRITE YOUR NAME ON THIS PAGE page 6 6. [4 poits] For each of the followig questios, circle the aswer which correctly completes the statemet. You do ot eed to show your work. l a. [2 poits] The itegral 3/2d coverges diverges Solutio: l /4 for sufficietly large values of, so l /4 3/2 3/2 = 5/4 evetually. Sice d coverges by the p-test (p = 5 5/4 4 > ), l d also coverges by direct 3/2 compariso. b. [2 poits] The itegral 2 + 3/2d coverges diverges Solutio: for all positive values of. Sice 2 + 3/2 = /2 d also coverges by direct compariso. 7. [6 poits] The power series d coverges by the p-test (p = 2 < ), ( ) ( ) has a radius of covergece of 5. For each of the edpoits of the iterval of covergece, fill i the first two blaks with the edpoit ad the series at that edpoit (i sigma otatio or by writig out the first 4 terms), ad the idicate whether the series coverges at that edpoit i the fial blak. You do ot eed to show your work. 5 Attheedpoit = -4,theseriesis ad that series diverges. Attheedpoit = 6,theseriesis ( ) ad that series coverges. Uiversity of Michiga Departmet of Mathematics Fall, 25 Math 6 Eam 3 Problem 7 Solutio

11 Math 6 / Fial (December 7, 25) DO NOT WRITE YOUR NAME ON THIS PAGE page 7 8. [9 poits] Cosider the fuctio g() defied by the power series g() = 2 (!) 2. (2)! = a. [6 poits] Fid the radius of covergece of the power series. You do ot eed to fid the iterval of covergece. Solutio: Applyig the ratio test ( 2 + ((+)!) 2 + (2(+))! ( 2 (!) 2 (2)! ) ) = = 2. 2(+) 2 (2+2)(2+) <, or < 2. Hece the radius of cover- Therefore the series coverges wheever 2 gece is 2. b. [3 poits] Use the first 3 ozero terms of the power series to estimate g() d. Solutio: Sice g() = + 2 2! + 22 (2!) 2 4! 2 + = , g() d d = (+ 2 3 )d = 4 3. Uiversity of Michiga Departmet of Mathematics Fall, 25 Math 6 Eam 3 Problem 8 Solutio