Extra+examples+_previous+assignment+02+solutions_

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Unformatted text preview: TUTURIAL LETTER 202. - ' Question 1 1. Use a proof by contradiction to prove that if p is a prime number then ﬂ is an irrational number. Solution Suppose ﬂ is a rational number. Then as observed on p. 43(H), ﬂ can be expressed as a rational in its lowest form. Hence f = E:- where m and n are integers n # 0 and n have no common factor greater than one. Now m2 = pn2 and so n22 is divisible by p. As p is a prime number, it follows that m is divisible by p. Thus there exists a k e Z such that m = kp. Then m2 = [€2,1522 and consequently ,arm2 = kzpz. It now follows that n2 = k2 p and we again deduce that n is divisible by p. Thus p is a common divisor of both m and n. This contradicts our assumption that the greatest common divisor of m and n is one. Question 2 2. Investigate the following for absolute and conditional convergence or divergence (a) few (m —r) r—l Solution Let Ian! = «M in + l) —n _ ~/nln+l —n t/n(n+1)+n __._...—.___ rr->oo 1 t/n(n+1)+n n(n+1)—rl2 l'm n = ——-——-—— = l ——_— "moot/nin+l)+n n-><>0t/Jv(n+l)+n' l = lim———= l n->oo 1+#+] 2 Since lim [anl # 0 we also have lim an gé 0 and it follows from the contrapositive of the vanishing . {7—)CO . ' ‘ n—mo condition that the serles IS divergent. 3 MAT213T/202/2 00 .3 (—1 "I (b) I; r -l ' Solution Let ng n; 1 Ianl= n2- 1 andbn=ﬁzg Then 3 . lanl . n: l :: E El? b” Jl£;f12-— ‘n . n2 = lim 2 n—Hmn -— , l = 11m i n-DOO] _. :5 = l > 0. Since 00 CD 1 Zn- = Z—l + 1 r=l r=2ri diverges (see 4.1.5H), 2—]; diverges and we have 2 Ia,.| diverges (use the second comparison test r—Z" (4.2.2(H)). Using the diagram on p.l [9(H) we now apply the alternating series test. First we see that lanl is monotone decreasing: Let 3 x5 forx> ]. f(x)= xz—l Then f’ (X) < 0 forx>1. Then f is decreasing on the interval (1, 00) and thus f (n + l) < f (17) Le. la” + ll < lanl for all n > 1. Also i _ , :1 11m |a,,| = 11m n—aoo IP9OCn2 —-] , 1 2 11m I I n—)oo n?— E = 0. 00 It follows from (4.2.4H) that Ear is conditionally convergent. r-.2 4 m 1 l'. (C) 2% )‘=1 Solution We use D’Alembert’s ratio test ,, l . l ” lim la + I = hm 01+ )‘e_ n—wo fanl n—aoo em“ :1 l , (n+1) = —lIm en—mo n l 1 = — lim (1 + -—) era—mo n _ l _ e < l sincee > 2. The given series is absolutely convergent. Question 3 Consider the function f given by x2—6x+5 ' if x > 1 f(x) = ;-' . x —5 If x 5 1 Solution We have f(1) = (I)2 — 5 = —4 and for x > 1 we have __ (x-sxx—n f(x) — T =x—5. Now I __ - f(x)—f(1) L“) 31.1,"? x—(l) 2— .. _ ___ “mix—M x—H‘ x—l _ 1‘ x2—] —.r—l)nlll—x—l . (x—1)(x+l) = 11m— x—>l‘ x—l = linl]_(x+l) =2. Also f() 10) I _ ' x _ f+(l) —xl-l»1111+ x—l = “mt-w x—>I+ x—l = Iimx_1 x—>l+x --1 1. Consequently ff (1) 75 f: (l) and thus f’ (1) does not exist. [See p. 169(H).] 3.2 Is f continuous at x = 1? Give a reason for your answer. Answer limf(x) = 1imx2—5=—4=f(l) x—)|— x—)|_ limf(x) = limx—5=—4=f(l) .\‘—)|+ x—>l+ It follows from 5.2.3(I—l) that f is continuous in the point x = 1. Question 4 4.1 Find the fourth degree Taylor polynomial of f (x) about a : 1 when f (x) = ﬂ. MAT2 l3T/202/2 Solution We need to ﬁnd —] T4.1f(x) ._ —_2!f(1)+f’(1)(x—l)+f"(1)(x )2 _ 3 “.1 +fm (1) (x3 '1) +fm (1) (X4 '4) Now f(x)=xi so f(1)=1 f’(x)=%x-%=2—J/§ fr(l)=% f”(x)= “5—711: —% f”(])=—% fm (x): %X_§ 7 fm (1)_ _% flu (x): _%x—§ fiu(1)= __[% f” (x): '—°~5x -% f" (c)_ EC— 3 Substitution give T4,|f(x) = %(x _1)_ Mr; n2 + it: n3 _ :_g(x_4!;ﬁ i+:(x—l)—:(x2!2—l) +:—6‘(x—1)3— .3—3(x—1)4 4.2 Determine R4,, f (x) for the ﬁinctioin in (i). In general we have RM f (x) = (3—1)]??— f "+1 (c) where c lies between a and x. Solution From the above we have (x - U5 5! _<x-1)5 E L _ 5! 32 c; 7 (x — 1)5 =256 c: ' f 5 (c) where c is between 1 and x. R4,1f(x) = Question 5 Let f(x)=—x+2 03x5]. ) Let P,r be the partition I O, :IN ,---,l}of[0,l]. l n 5.] Calculate L (P,,) and U (Pn) Solution Pn=|0, l n 2 1 , Z, - - - , 1]subdivides [0, 1] into n subintervals each of length g?- 2 ﬂx)=—x+2 -_—% 0% 5 i % 3 u n 1| n n n J"0 I} 3‘2 x3 *4 In Since m; = inf[f(x) :xj_. 5 x 5 x;} we have m,- = f(x,~) = f (:7) = -—n—1 + 2 (the minimum value is at the right end point of the interval) and since M.- =Sllplf(x) :xH 5x sxil we have (the maximum value is at the left end point of the interval). Then 2:12 _ 3112 — n — Zn2 311 — l MAT213T/202/2 =é(—(":')+2)% l” . 2 =;I§(—I+l)+;(n) -1». I =__I(£.(Ll))+.l_+2 2 n —n2 —n+2n+4n2 2H2 _ 3112 + 1n _ 2.712 _3n+l _ 2n 5.2 Hence prove that f is Riemann integrable on [0, 1] and evaluate f0] f (x) dx. Solution Wehave 3 1 3 +1 n— n < < < 211 _£_L[_ 2n and by letting n —> 00 it can be deduced that L S U : %. Hence f is Riemann-integrable on [0, l] with fo‘ f(x)dx =f0‘(—x+2)dx ...
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