Math 06 Fall 04 Exam 3 December 0, 04 Determie if the series is coverget or diverget by makig a compariso (DCT or LCT) with a suitable b Fill i the blaks with your aswer For Coverget or Diverget write Coverget or C if the series to the left is coverget Otherwise write Diverget or D The write a sequece that could be used to make your compariso You do ot eed to show ay work Series Coverget or Diverget? b? (a) l() C 3/ We ca use the fact that l() d (evetually) Usig d / ad doig some algebra gives a l() / 3/ b (evetually) (b) 6 3 + 9 C ( ) 3 6 From algebra (bigger deomiator meas smaller fractio) we kow that 3 + 9 6 3 ad 6 3 + 9 6 for all Oly oe of these sequeces gives us a coverget series ad that is 9 6 9 ( ) 3 So a 6 3 + 9 ( ) b 3 (c) 3 + si() D We kow that si() for all The si() ad 3 + si() 5 The we get the right compariso from a 3 + si() b
Aswer the followig questios about the series by fillig i the blaks For Coverget or Diverget write Coverget or C if the series to the left is coverget Otherwise write Diverget or D For Test Used write a appropriate series test that tells you whether the series is coverget or diverget You ca use test abbreviatios You do ot eed to show ay work, although you may fid it helpful to write the sums usig series otatio Series Coverget or Diverget? Test Used? (d) +4 3 +5 4 +6 5 +7 + D TFD 6 We ca recogize that the umerator ad deomiator each icrease by oe for each successive term So this is the series + + Sice lim 0 the series is diverget by the Test For Divergece (TFD) The Itegral Test (IT) would also work here (e) 8 + 8 + C GST 3 Testig ratios of successive terms shows that r ad this is the geometric series 4 ( 8 4) Sice r < the series is coverget by the Geometric Series Test (GST) The Alteratig 4 Series Test (AST) would also work here (f) + + 3 + 4 + + D PST 5 This series is fairly easy to recogize as the P -series where p Sice p the series is diverget by the P -Series Test (PST) The Itegral Test (IT) would also work here
Fid the idicated Taylor Series Simplify your aswer (i) f(x) e 3x ; a 4 Let t x 4 t + 4 x e 3x e 3(t+4) e 3t+ e 3t e! (3t) e e 3 t e 3 (x 4)!! (ii) f(x) cos(x); a π Let t x π t + π x ( ( cos(x) cos t + π )) cos(t + π) cos(t) cos(π) si(t) si(π) cos(t) ( ) (t) cos(t) ()! ( ) + t ()! ( ) + ( x π ()! ) ( ) 3 (a) Determie if the series + 5 is absolutely coverget, coditioally coverget, or diverget Use the Root Test: lim a lim By the Root Test (RoT), the series ( + 5 ) lim + 5 5 < ( ) + 5 is absolutely coverget ( ) (b) What ca you say about lim + 5? Support your aswer with words ( ) ( ) We must have lim + 5 0 If ot, lim + 5 0 would imply the series ( ) + 5 is Diverget by the Test For Divergece (TFD) But by part (a), we kow the series is absolutely coverget (ad therefore coverget)
4 (a) Use techiques of itegratio (ot power series) to evaluate the itegral si(8 x) x dx Note that si(8 x) si(8x / ) dx dx Use the substitutio u 8 x 8x / x x / The du dx 4x / 4 So x 4 du dx ad x si(8 x) dx x 4 si(u) du 4 cos(u) + C 4 cos(8 x) + C (b) Now evaluate the itegral si(8 x) x dx usig power series Use the MacLauri Series for cosie Sice si(x) si(8 x) ( ) ( + )! (8 x) + ( ) ( + )! x+ we kow ( ) ( + )! (8x/ ) + ( ) ( + )! 8+ x +/ (Fu fact: Eve though this series has a fractio i the expoet, the result still coverges to the desired fuctio This is eve true for complex values of x!) The si( x) x x ( ) ( + )! 8+ x +/ x / We ca ow itegrate the power series ( ) ( + )! 8+ x +/ ( ) ( + )! 8+ x ( ) ( + )! 8+ x dx ( ) ( + )! 8+ x ( ) 8 + dx ( + )! ( + ) x+ + C
5 (a) Evaluate the itegral x l ( x) dx usig power series Use the MacLauri Series for the atural logarithm Sice l( x) x l( x) x ( ) + x+ + x+ + x+ we kow The + x+ dx + x + dx ( + )( ) x+3 + C (b) Now use techiques of itegratio (ot power series) to evaluate the itegral x l ( x) dx The product i the itegral suggests we should use parts: x l ( x) dx (x ) l( x) dx x l( x) x l( x) For the itegral du dx ad The it follows that x x x x l(x) + x (l( x)) dx x x x dx ( ) dx x du dx use the substitutio u x x u Sice we have dx x ( u) x dx u x dx l u u + u u + u du du u u + u du + C l x ( x) + ( x) + C So we get x l( x) dx x l( x) + l x ( x) + ( x) + C
6 (a) Fid a rage of x values for which the fuctio is defied, the state the radius of covergece (3x ) f(x) lim a + a lim (3x ) + + (3x ) 3x + lim 3x + Settig 3x < < 3x < 0 < 3x < 0 < x < 3 Sice the ceter is a 3 we must have R 3 (3x ) (b) Is the iput x 0 valid i the fuctio f(x)? Explai your aswer (3x ) For x 0 i the series we have (0 ) ( ) Here b Sice lim 0 ad b ( ) is decreasig we kow is coverget by the Alteratig Series Test (AST) So the iterval of covergece I icludes x 0 I other words, ( ) (3x + ) x 0 is valid iput ito the fuctio f(x)