NATIONAL UNIVERSITY OF SINGAPORE FACULTY OF SCIENCE SEMESTER 1 EXAMINATION ADVANCED CALCULUS II. November 2003 Time allowed :

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1 NATIONAL UNIVERSITY OF SINGAPORE FACULTY OF SCIENCE SEMESTER EXAMINATION MA08 ADVANCED CALCULUS II November 003 Time allowed : hours INSTRUCTIONS TO CANDIDATES This examiatio paper cosists of TWO ) sectios: Sectio A ad Sectio B It cotais a total of SEVEN 7) questios ad comprises FIVE 5) prited pages Aswer ALL questios i Sectio A Sectio A carries a total of 60 marks 3 Aswer o more tha TWO ) questios from Sectio B Each questio i Sectio B carries 0 marks 4 Cadidates may use calculators However, they should lay out systematically the various steps i the calculatios

2 SECTION A Aswer all the questios i this sectio Sectio A carries a total of 60 marks Questio [6 marks] For each of the followig sequeces, either fid the it or show that the it does ot exist { ) } l ) a) arcta si { } l ) 3 + 8! b)! + 9 c) d) { ) } + / { } 3 5 )! Solutio a) si arcta b) ) l ) arctasi 0) l ) 3 + 8! l ) 3 /! /! 00 /! /! c) ) + / ) / / + ) / ) d) Let a The! 3 5 )! 3 5 ) + ) + )! + + a + a a! 3 5 ) + / + /

3 Questio [6 marks] Determie the covergece or divergece of each of the followig series Justify your aswers a) b) c) d) ) ! ) ) si + Solutio a) Divergece Let a 6 ad let b The a 6 b / + 3/ + / 3 Sice diverges by p-series, a divergece by the it compariso test b) Covergece by the root test because 5 ) a / / < c) Covergece by the ratio test because a + a + )! ) )3 + 5)! / 3 + 5/ 3 < d) Divergece Let a si ) + ad let b The a > 0 ad a si ) si ) + + b + + x + si x x 0 x / + 3

4 4 Sice test diverges by p-series, a diverges by the it compariso Questio 3 [0 marks] Fid the radius of covergece of each of the followig power series Justify your aswer a) b) Solutio a) x )!) 3 3)! x The radius of covergece R x ) k a b) The radius of covergece R a + a +) 3 3+3)3+)3+) x ) [+)!] 3 3)! 3+3)!!) 3 +/) 3 3+3/)3+/)3+/) 7 Questio 4 [8 marks] a) Determie whether the followig sequece of fuctios coverges uiformly o the idicated itervals Justify your aswers F x) x x), x [0, ] b) Determie whether the followig series of fuctios coverge uiformly o the idicated itervals Justify your aswers

5 5 i) ii) k k cos kx k3 + x 3, ) k k + x x [0, ) x [0, ) Solutio a) Whe 0 < x <, the x x) x / x)) 0 because / x) > ad F 0) F ) 0 Thus the itig fuctio F x) F x) 0 for x [0, ] Sice F x) 0 for x [0, ], T sup 0 x F x) F x) sup 0 x F x) From F x) x) + x x) ) x) +)x) 0, x + or ad so T max 0 x F x) max Sice { ) } F 0), F, F ) + + [ T + ) ] + /+/) 0 + the sequece of fuctios coverges uiformly o [0, ] by the T -test cos kx b) i) Let f k x) k The 3 +x 3 f k x) cos kx k3 + x 3 k 3/ Sice k coverges by p-series, the series of fuctios f 3/ k x) k coverges uiformly by Weierstrass M-test b)ii) Let b k k+x The for x 0, b k is positive ad mootoe decreasig with b k 0 By the alteratig series test, k the series of fuctios k ) k k+x coverges poitwise o [0, ) By + )

6 6 alteratig series estimatio, 0 T sup x 0 k+ ) k k + x sup x x + Sice + 0, T 0 by the Squeeze theorem ad so the series of fuctios coverges uiformly o [0, )

7 SECTION B Aswer ot more tha TWO ) questios from this sectio Each questio i this sectio carries 0 marks 7 Questio 5 [0 marks] a) Evaluate aswer 0 ) x + x + cosx ) dx Justify your b) Fid the iterval of covergece of the power series x + ) + l Justify your aswer c) Let {a } be a coverget sequece Defie 5 b a + a + + a Does the sequece {b } coverge? Justify your aswer ) Solutio a) Let F x) cosx ) The, for 0 x, Sice 4 ) 5 0, x +x+ 5 ) 4 F x) 5 ) 4 5 F x) F x) 0 by the Squeeze Theorem for 0 x Now ) x ) + x + 4 T sup F x) F x) sup cosx ) 0 x 0 x 5 5 Sice 4 ) 5 0, T 0 by the Squeeze Theorem ad so ) x + x + cosx ) dx 0 5 ) x + x + cosx ) dx 0 dx

8 8 b) The power series x + ) + l has the radius of covergece R a + a l l x + ) + + l ++) l+) l+) l l + ) with x 0 Whe x, the series x + ) + l ) + l coverges by the alteratig series test because is positive ad + l mootoe decreasig with 0 Whe x + 0, the + l series x + ) + l + l diverges by it compariso test with ad the series x l x l + l l because l diverges by the itegral test because, let fx), the fx) is positive mootoe decreasig with fx) dx x l x ul x dx l du u + Hece the iterval of covergece is [, 0) c) Sice {a } coverges, {a } is bouded ad so there exists a positive umber M such that a M for all Let A a Give ay ɛ > 0, there exists N such that a A < ɛ

9 for > N Let N be the smallest iteger such that N > max{ N M+ A ), N ɛ } For > N, b A a A) + a A) + + a A) a A) + + a N A) a + A + + a N + A ad hece b A N M + A ) + a N + A) + + a A) < ɛ + ɛ ɛ + a N + A + + a A + N ɛ 9 Questio 6 [0 marks] a) Usig ay applicable method, fid the Taylor series of the fuctio fx) 4 at x 0 0, ad determie its 6 x radius of covergece b) Suppose that the series of fuctios f x) coverges o a iterval I Prove that the sequece of fuctios {f x)} coverges uiformly to 0 o I c) Let {a } ad {b } be bouded sequeces Show that a + b ) a + b Solutio a) By usig biomial series, fx) 6 x) 4 / x ) 4 / ) 4 x 8 for x < 8 Thus radius of covergece is 8 ) ) x ) 4 8

10 0 b) Let S x) k f k x) Sice f x) coverges uiformly o I, the sequece of partial sums {S x)} coverges uiformly o I By Cauchy Criterio, give ay ɛ > 0, there exists N such that S m x) S x) < ɛ for all x I ad, m > N I particular, f x) 0 S x) S x) < ɛ for > N + ad hece {f x)} coverges uiformly to 0 by defiitio c) For each m, we have a m + b m sup{a k k } + sup{b k k } Thus sup{a k k } + sup{b k k } is a upper boud of the set {a m + b m m } {a k + b k k } It follows that sup{a k + b k k } sup{a k k } + sup{b k k } because sup is the least upper boud, ad so, by lettig ted to ifiity, a + b ) sup{a k + b k k }) sup{a k k } + sup{b k k }) sup{a k k }) + sup{b k k }) a + b

11 Questio 7 [0 marks] a) Let fx) x 3 six 8 ) Fid f 48) 0) b) Let {a } be the sequece defied recursively by a a + + a a + Prove that {a } is coverget, ad fid its it c) Show that the series aa )a ) a + ) +! coverges for ay real umber a > Solutio a) By usig Taylor series, fx) x 3 six 8 ) x 3 ) k+ x 8 ) k k )! Compare with k fx) f0) + k By the uiqueess theorem of power series, f 48) 0) 0 f k) 0) k! ) k+ x 8k )+3 k )! b) First we prove that a by iductio This is true for Assume that a The + a a + a + The iductio is fiished ad so a for all Next we prove that {a } is mootoe icreasig Note that a 3 > a Assume that a a The ) a + a + a a + ) + a a + k a a a + )a + ) 0 By mootoe covergece theorem, {a } coverges Let A a The A a + + a ) + A a + A + ad so A A 0 It follows that A ± 5

12 Sice a, A 5 is rejected ad hece A + 5, that is, a + 5 c) Observe that this series is evetually alteratig Let ) a aa )a ) a + ) a! The a + a aa ) a + )a ) + )!! aa )a ) a + ) a + a + because a > Thus { a } is mootoe decreasig From a a + a + a +, l a l a + k k Sice l a+ k 0 for k a +, the series l a + k k k is evetually positive Sice l a+ x k l a + )x) x 0 x because a > ) ad the series series k k l k k x 0 ) a+ k k a+ > 0 a + )x a ) diverges, the evetually positive) k l a + k k diverges by it compariso test ad so l a l a + k + k or l a It follows that a 0

13 3 By alteratig series test, the series + a ) coverges