Math 106 Fall 2014 Exam 3.1 December 10, 2014

Transcription

1 Math 06 Fall 0 Exam 3 December 0, 0 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) D We ca use the fact that l) d evetually) Usig d ad takig the reciprocal gives a l) b evetually) b) 3 + cos) D We kow that cos) for all The cos) ad 3 + cos) 5 The we get the right compariso from a 3 + cos) b c) + 8 C ) From algebra bigger deomiator meas smaller fractio) we kow that + 8 ad + 8 for all Oly oe of these sequeces gives us a coverget series ad that is 8 8 ) So a + 8 ) b

2 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) C GST 3 Testig ratios of successive terms shows that r ad this is the geometric series 8 ) Sice r < the series is coverget by the Geometric Series Test GST) The Alteratig Series Test AST) would also work here e) 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 f) D PST 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

3 Fid the idicated Taylor Series Simplify your aswer i) fx) lx); a Let t x t + x lx) lt + ) l l) + ) t ) l) + t ))) l) + l t )) ) x ) ) l) ) t + + ) ii) fx) six); a π Let t x π t + π x six) si t + π )) sit + π) sit) cosπ) + siπ) cost) sit) sit) ) t) + + )! ) + + t + + )! ) + + x π + )! ) + 3 ) 3 a) Determie if the series + is absolutely coverget, coditioally coverget, or diverget Use the Root Test: lim a lim By the Root Test RoT), the series 3 + ) 3 lim + 3 < 3 ) + is absolutely coverget 3 ) b) What ca you say about lim +? Support your aswer with words 3 ) 3 ) We must have lim + 0 If ot, lim + 0 would imply the series 3 ) + 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 cos6 x) x dx Note that cos6 x) cos6x / ) dx dx Use the substitutio u 6 x 6x / x x / The du dx 3x / 3 So x 3 du dx ad x cos6 x) dx x 3 cosu) du 3 siu) + C 3 si6 x) + C b) Now evaluate the itegral cos6 x) x dx usig power series Use the MacLauri Series for cosie Sice cosx) cos6 x) ) )! 6 x) ) )! x we kow ) )! 6x/ ) ) )! 6 x The cos x) x x ) )! 6 x x / ) )! 6 x ) )! 6 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!) We ca ow itegrate the power series ) )! 6 x / dx ) )! 6 x / dx ) )! 6 + /) x+/ + C

5 5 a) Evaluate the itegral 3x arctax) dx usig power series Use the MacLauri Series for arctaget Sice arctax) 3x arctax) 3x ) + x+ ) + x+ we kow ) 3 + x+3 The ) 3 + x+3 dx ) 3 + x +3 dx ) 3 + ) + ) x+ + C b) Now use techiques of itegratio ot power series) to evaluate the itegral 3x arctax) dx The product i the itegral suggests we should use parts: 3x arctax) d x 3 ) arctax) dx x 3 arctax) x 3 arctax)) dx x 3 arctax) x 3 + x dx x 3 x 3 arctax) + x dx x 3 For the itegral + x dx use the substitutio u x + Sice du dx x we have du x dx We also kow that u x + u x The x 3 + x dx x + x x dx u du u u du The it follows that x 3 + x dx u l u ) + C x + l x + ) + C So we get 3x arctax) dx x 3 arctax) x + l x + ) + C

6 6 a) Fid a rage of x values for which the fuctio is defied, the state the radius of covergece ) 3x + ) fx) lim a + a lim ) + 3x + ) + + ) 3x + ) 3x + lim Settig 3x + < < 3x + < < 3x < 0 3 < x < 0 Sice the ceter is a 3 we must have R 3 3x + + ) 3x + ) b) Is the iput x 0 valid i the fuctio fx)? Explai your aswer ) 3x + ) For x 0 i the series we have ) ) 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 fx)