Student’s Manual Essential Mathematics for Economic Analysis

73 Pages·2008·0.56 MB·English

Student’s Manual
Essential Mathematics for
Economic Analysis
rd
3 edition
Knut Sydsæter
Arne Strøm
Peter Hammond
For further supporting resources please visit:
www.pearsoned.co.uk/sydsaeter
Preface
Thisstudent’ssolutionsmanualaccompaniesEssentialMathematicsforEconomicAnalysis(3rdedition,FT
⊂⊃
PrenticeHall,2008). Itsmainpurposeistoprovidemoredetailedsolutionstotheproblemsmarked SM inthe
text. ThisManualshouldbeusedinconjunctionwiththeanswersinthetext. Insomefewcasesonlyapart
oftheproblemisdoneindetail,becausetherestfollowsthesamepattern. WearegratefultoCarrenPindiriri
forhelpwiththeproofreading. Wewouldappreciatesuggestionsforimprovementsfromourreadersaswell
ashelpinweedingoutinaccuraciesanderrors.
OsloandCoventry,July2008
KnutSydsæter ([email protected])
ArneStrøm ([email protected])
PeterHammond ([email protected])
Version: 9July2008
Contents
1 IntroductoryTopicsI:Algebra ........................................................ 1
2 IntroductoryTopicsII:Equations ..................................................... 4
3 IntroductoryTopicsIII:Miscellaneous ................................................. 6
4 FunctionsofOneVariable ........................................................... 8
5 PropertiesofFunctions ............................................................ 12
6 Differentiation ................................................................... 14
7 DerivativesinUse ................................................................ 18
8 Single-VariableOptimization........................................................ 22
9 Integration ...................................................................... 26
10 InterestRatesandPresentValues..................................................... 33
11 FunctionsofManyVariables ........................................................ 35
12 ToolsforComparativeStatics ....................................................... 37
13 MultivariableOptimization ......................................................... 42
14 ConstrainedOptimization .......................................................... 48
15 MatrixandVectorAlgebra .......................................................... 57
16 DeterminantsandInverseMatrices ................................................... 61
17 LinearProgramming .............................................................. 67
©KnutSydsæter,ArneStrøm,andPeterHammond2008
CHAPTER 1 INTRODUCTORY TOPICS I: ALGEBRA 1
Chapter 1 Introductory Topics I: Algebra
1.1
1. (a)True (b)False. −5issmallerthan−3,soonthenumberlineitistotheleftof−3. (SeeFig.1.1.1
inthebook.) (c)False. −13isaninteger,butnotanaturalnumber. (d)True. Anynaturalnumberis
rational. For example 5 = 5√/1. (e)√False, since 3.1415 = 31415/10000, the quotient of two integers.
(f)False. Counterexample: 2+(− 2) = 0. (g)True. (h)True.
1.3
9. (a) (2t−1)(t2−2t+1) = 2t(t2−2t+1)−(t2−2t+1) = 2t3−4t2+2t−t2+2t−1 = 2t3−5t2+4t−1
(b) (a+1)2 +(a−1)2 −2(a+1)(a−1) = a2 +2a+1+a2 −2a+1−2a2 +2 = 4
(c) (x + y + z)2 = (x + y + z)(x + y + z) = x(x + y + z) + y(x + y + z) + z(x + y + z) =
x2+xy+xz+yx+y2+yz+zx+zy+z2 = x2+y2+z2+2xy+2xz+2yz (d) (x−y−z)2 =
(x−y−z)(x−y−z) = x2−xy−xz−xy+y2−yz−xz−yz+z2,so(x+y+z)2−(x−y−z)2 = 4xy+4xz
13. (a) a2+4ab+4b2 = (a+2b)2usingthefirstquadraticidentity. (d) 9z2−16w2 = (3z−4w)(3z+4w),
accordingtothedifference-of-squaresformula. (e) −1x2 +2xy −5y2 = −1(x2 −10xy +25y2) =
5 5
−1(x − 5y)2 (f) a4 − b4 = (a2 − b2)(a2 + b2), using the difference-of-squares formula. Since
5
a2 −b2 = (a−b)(a+b),theanswerinthebookfollows.
1.4
1 1 x +2 x −2 x +2−x +2 4
5. (a) − = − = =
x −2 x +2 (x −2)(x +2) (x +2)(x −2) (x −2)(x +2) x2 −4
(b)Since4x +2 = 2(2x +1)and4x2 −1 = (2x +1)(2x −1),theLCDis2(2x +1)(2x −1). Then
6x +25 6x2 +x −2 (6x +25)(2x −1)−2(6x2 +x −2) 21(2x −1) 21
− = = =
4x +2 4x2 −1 2(2x +1)(2x −1) 2(2x +1)(2x −1) 2(2x +1)
18b2 a 18b2 −a(a−3b)+2(a2 −9b2) a(a+3b) a
(c) − +2 = = =
a2 −9b2 a+3b (a+3b)(a−3b) (a+3b)(a−3b) a−3b
1 1 1 a2 −4−a(a−2)+8a 2(5a−2) 5a−2
(d) − + = = =
8ab 8b(a+2) b(a2 −4) 8ab(a2 −4) 8ab(a2 −4) 4ab(a2 −4)
(cid:2) (cid:3)
2t −t2 5t 2t t(2−t) 3t −t(t −2) 3t −3t2
(e) · − = · = · =
t +2 t −2 t −2 t +2 t −2 t +2 t −2 t +2
(cid:4) (cid:5) (cid:4) (cid:5)
a 1− 1 a− 1 a 1− 1
(f) 2a = 2 = 4a−2,so2− 2a = 2−(4a−2) = 4−4a = 4(1−a)
0.25 1 0.25
4
2 1 2(x +1)+x −3x(x +1) 2−3x2
6. (a) + −3 = =
x x +1 x(x +1) x(x +1)
t t t(2t −1)−t(2t +1) −2t
(b) − = =
2t +1 2t −1 (2t +1)(2t −1) 4t2 −1
3x 4x 2x −1 3x(x −2)+4x(x +2)−(2x −1) 7x2 +1
(c) − − = =
x +2 2−x (x −2)(x +2) (x −2)(x +2) x2 −4
©KnutSydsæter,ArneStrøm,andPeterHammond2008
2 CHAPTER 1 INTRODUCTORY TOPICS I: ALGEBRA
(cid:2) (cid:3) (cid:2) (cid:3)
1 1 1 1 1 1 1 1
+ + xy − − ·x2y2
x y x y y +x x2 y2 x2 y2 y2 −x2
(d) = = = x +y (e) = (cid:2) (cid:3) =
1 1 1 1 1 1 1 x2 +y2
·xy + + ·x2y2
xy xy x2 y2 x2 y2
a(y −x) y −x
(f)Multiplythenumeratorandthedenominatorbyxy. Thenthefractionreducesto = .
a(y +x) y +x
(cid:4) (cid:5) (cid:4) (cid:5)
8. (a) 1 − 1 = 5 − 4 = 1 ,so 1 − 1 −2 = 1 −2 = 202 = 400
4 5 20 20 20 4 5 20
n n·n n2 n(n−1)−n2 −n
(b) n− = n− (cid:6) (cid:7) = n− = =
1 1 n−1 n−1 n−1
1− 1− ·n
n n
1 1 1 1 1 u
(c) Ifu = xp−q,then + = + = + = 1
1+xp−q 1+xq−p 1+u 1+1/u 1+u 1+u
(cid:2) (cid:3)
1 1
+ (x2 −1)
x −1 x2 −1 x +1+1 x +2 1
(d) (cid:2) (cid:3) = = =
2 x3−x −2x +2 (x +2)(x2 −2x +1) (x −1)2
x − (x2 −1)
x +1
1 1
−
1 1 x2 −(x +h)2 −2xh−h2 (x +h)2 x2 −2x −h
(e) − = = ,so =
(x +h)2 x2 x2(x +h)2 x2(x +h)2 h x2(x +h)2
10x2 2x
(f) Multiplyingdenominatorandnumeratorbyx2 −1 = (x +1)(x −1)yields =
5x(x −1) x −1
1.5
√ √ √ √
5. (Ne√eds some hints√.) Mult√iply the d√enominato√r and the num√erator by: (a) 7 − 5 (b) 5 − 3
(c) 3+2 (d)x y −y x (e) x +h+ x (f)1− x +1
7. Theanswerswilldependonthecalculatoryouuse.
12. (a) For x = 1 the left-hand side is 4 and the right-hand side is 2. (In fact, (2x)2 = 22x.) (b) Correct
because ap−q = ap/aq (c) Correct because a−p = 1/ap (d) For x = 1 it says 5 = 1/5, which is
absurd. (e) For x = y = 1, it says that a2 = 2a, which is usually wrong. (In fact, ax+y = axay.)
√ √ √ √ √
(f)2 x ·2 y = 2 x+ y,not2 xy.
1.6
3x +1 3x +1 3x +1−2(2x +4) −x −7
4. (a)2 < hasthesamesolutionsas −2 > 0,or > 0,or > 0
2x +4 2x +4 2x +4 2x +4
A sign diagram reveals that the inequality is satisfied for −7 < x < −2. A serious error is to multiply
theinequalityby2x +4, withoutassumingthat2x +4 > 0. Whenmultiplyingwith2x +4whenthis
number is negative, the inequality sign must be reversed. (It might be a good idea to test the inequality
forsomevaluesofx. Forexample,forx = 0itisnottrue. Whataboutx = −5?)
120 3 3(160−n)
(b)The inequality is equivalent to − ≤ 0, i.e. ≤ 0. A sign diagram reveals that the
n 4 4n
inequality is satisfied for n < 0 and for n ≥ 160. (Note that for n = 0 the inequality makes no sense.
Forn = 160,wehaveequality.)
©KnutSydsæter,ArneStrøm,andPeterHammond2008
CHAPTER 1 INTRODUCTORY TOPICS I: ALGEBRA 3
(c) Easy: g(g − 2) ≤ 0 etc. (d) Note that p2 − 4p + 4 = (p − 2)2, and the inequality reduces to
p+1
≥ 0. Thefractionmakesnosenseifp = 2. Theconclusionfollows.
(p−2)2
−n−2 −n−2−2n−8 −3n−10
(e)Theinequalityisequivalentto −2 ≥ 0,i.e. ≥ 0,or ≥ 0,etc.
n+4 n+4 n+4
(f)Seethetextanduseasigndiagram. (Don’tcancelx2. Ifyoudo,x = 0appearsasafalsesolution.)
5. (a)Useasigndiagram. (b)Theinequalityisnotsatisfiedforx = 1. Ifx (cid:7)= 1,itisobviouslysatisfied
for x + 4 > 0, i.e. x > −4 (because (x − 1)2 is positive when x (cid:7)= 1). (c) Use a sign diagram.
(d)Theinequalityisnotsatisfiedforx = 1/5. Ifx (cid:7)= 1/5,itisobviouslysatisfiedforx < 1.
(e)Useasigndiagram. ((5x −1)11 < 0ifx < 1/5,> 0ifx > 1/5.)
3x −1 3x −1 −(1+x2)
(f) > x +3, −(x +3) > 0, > 0,sox < 0. (1+x2 isalwayspositive.)
x x x
x −3 x −3 −2x(x +2)
(g) > 2x −1or −(2x −1) < 0or < 0. Thenuseasigndiagram.
x +3 x +3 x +3
(h)Usethehintandasigndiagram. (Actually,thisproblemandthenextcouldbepostponedtoSection2.3
ifyouhaveforgottenyourhighschoolalgebra.) (i)Usethehintandasigndiagram.
ReviewProblemsforChapter1
4. (a)(2x)4 = 24x4 = 16x4 (b)2−1−4−1 = 1/2−1/4 = 1/4,so(2−1−4−1)−1 = 4
(c)Cancelthecommonfactor4x2yz2. (d)−(−ab3)−3 = −(−1)−3a−3b−9 = a−3b−9,so
a5·a3·a−2 a6
[−(−ab3)−3(a6b6)2]3 = [a−3b−9a12b12]3 = [a9b3]3 = a27b9 (e) = = a3
a−3·a6 a3
(cid:8)(cid:6) (cid:7) (cid:9) (cid:8) (cid:9)
x 3 8 −3 x3·8 −3
(f) · = = (x5)−3 = x−15
2 x−2 8·x−2
√ (cid:4)√ √ (cid:5) √ √ √ √ √
8. Allarestra√ightforward,except(c),(g),and(h): (c)− 3 3− 6 = −3+ 3 6 = −3+ 3 3 2
= −3+3 2 (g) (1+x +x2 +x3)(1−x) = (1+x +x2 +x3)−(1+x +x2 +x3)x = 1−x4
(h)(1+x)4 = (1+x)2(1+x)2 = (1+2x +x2)(1+2x +x2)a.s.o.
11. (a)and(b)areeasy. (c)ax+ay+2x+2y = ax+2x+ay+2y = (a+2)x+(a+2)y = (a+2)(x+y)
(d)2x2−5yz+10xz−xy = 2x2+10xz−(xy+5yz) = 2x(x+5z)−y(x+5z) = (2x−y)(x+5z)
(e)p2−q2+p−q = (p−q)(p+q)+(p−q) = (p−q)(p+q+1) (f)Seetheanswerinthebook.
s s s(2s +1)−s(2s −1) 2s
15. (a) − = =
2s −1 2s +1 (2s −1)(2s +1) 4s2 −1
x 1−x 24 −x(x +3)−(1−x)(x −3)−24 −7(x +3) −7
(b) − − = = =
3−x x +3 x2 −9 (x −3)(x +3) (x −3)(x +3) x −3
y −x y −x 1
(c) Multiplyingnumeratoranddenominatorbyx2y2 yields, = =
y2 −x2 (y −x)(y +x) x +y
16. (a) Cancel the factor 25ab. (b) x2 −y2 = (x +y)(x −y). Cancel x +y. (c) The fraction can be
(2a−3b)2 2a−3b 4x −x3 x(2−x)(2+x) x(2+x)
written = . (d) = =
(2a−3b)(2a+3b) 2a+3b 4−4x +x2 (2−x)2 2−x
©KnutSydsæter,ArneStrøm,andPeterHammond2008
4 CHAPTER 2 INTRODUCTORY TOPICS II: EQUATIONS
Chapter 2 Introductory Topics II: Equations
2.1
3. (a)Wenotefirstthatx = −3andx = −4bothmaketheequationabsurd. Multiplyingtheequationby
the common denominator, (x +3)(x +4), yields (x −3)(x +4) = (x +3)(x −4), and thus x = 0.
(b) Multiplying by the common denominator (x − 3)(x + 3) yields 3(x + 3) − 2(x − 3) = 9, from
whichwegetx = −6. (c)Multiplyingbythecommondenominator15x (assumingthatx (cid:7)= 0,yields
18x2 −75 = 10x2 −15x +8x2,fromwhichwegetx = 5.
5. (a)Multiplyingbythecommondenominator12yields9y−3−4+4y+24 = 36y,andsoy = 17/23.
(b) Multiplying by 2x(x +2) yields 8(x +2)+6x = 2(2x +2)+7x, from which we find x = −4.
2−2z−z
(c)Multiplyingthenumeratorandthedenominatorinthefirstfractionby1−z,leadsto =
(1−z)(1+z)
6
. Multiplyingby(1−z2)(2z+1)yields(2−3z)(2z+1) = 6−6z2,andsoz = 4.
2z+1
(d) Expanding the parentheses we get p − 3 − 1 + p − 1 + p = −1. Multiplying by the common
4 8 4 12 3 3 3
denominator24givesanequationwiththesolutionp = 15/16.
2.2
(cid:2) (cid:3)
a+b a b 1 1 1
2. (a)Multiplybothsidesbyabxtoobtainb+a = 2abx. Hence,x = = + = + .
2ab 2ab 2ab 2 a b
(b)Multiplytheequationbycx +d toobtainax +b = cAx +dA,or(a −cA)x = dA−b,andthus
x = (dA−b)/(a−cA). (c)Multiplytheequationbyx1/2 toobtain 1p = wx1/2,thusx1/2 = p/2w,
√
2
so,bysquaringeachside,x = p2/4w2. (d)Multiplyeachsideby 1+x toobtain1+x +ax = 0,
sox = −1/(1+a). (e)x2 = b2/a2,sox = ±b/a. (f)Weseeimmediatelythatx = 0.
4. (a)αx −a = βx −b ⇐⇒ (α−β)x = a−b,sox = (a−b)/(α−β).
√
(b)Squaringeachsideof pq = 3q +5yieldspq = (3q +5)2,sop = (3q +5)2/q.
(c)Y = 94+0.2(Y −(20+0.5Y)) = 94+0.2Y −4−0.1Y,so0.9Y = 90,andthenY = 100.
(cid:4) (cid:5)
r
(d)Raiseeachsidetothe4thpower: K21 K = Q4,soK3 = 2wQ4/r,andhenceK = 2wQ4/r 1/3.
2w
(e)Multiplyingnumeratoranddenominatorintheleft-handfractionby4K1/2L3/4,(cid:6)leads(cid:7)to2L/K = r/w,
r
fromwhichwegetL = rK/2w. (f)Raiseeachsidetothe4thpower: 1 p4K−1 1 = r4. Itfollows
16 2w
thatK−1 = 32r3w/p4,soK = 1 p4r−3w−1.
32
1 1 1 T −t tT √
5. (a) = − = , so s = . (b) KLM = B + αL, so KLM = (B + αL)2, and
s t T tT T −t
so M = (B +αL)2/KL. (c) Multiplying each side by x −z yields, x −2y +xz = 4xy −4yz, or
(x+4y)z = 4xy−x+2y,andsoz = (4xy−x+2y)/(x+4y). (d)V = C−CT/N,soCT/N = C−V
andthusT = N(1−V/C).
2.3
5. (a) See the answer in the book. (b) If the first natural number is n, then the next is n + 1, so the
requirementisthatn2+(n+1)2 = 13,whichreducesto2n2+2n−12 = 0,i.e. n2+n−6 = 0. This
second-orderequationhasthesolutionsn = −3andn = 2,sothetwonumbersare2and3. (Ifweasked
forintegersolutions,wewouldhave−3and−2inaddition.)
©KnutSydsæter,ArneStrøm,andPeterHammond2008
CHAPTER 2 INTRODUCTORY TOPICS II: EQUATIONS 5
(c) If the shortest side is x, the other is x + 14, so according to Pythagoras’Theorem (see page 633
and draw a picture), x2 +(x +14)2 = (34)2, or x2 +14x −480 = 0. The solutions are x = 16 and
x = −30, so the shortest side is 16 cm and the longest is 30 cm. (d) If the usual driving speed is x
km/handtheusualtimespentist hours,thenxt = 80. 16minutesis16/60 = 4/15hours,sodrivingat
thespeedx +10fort −4/15hoursgives(x +10)(t −4/15) = 80. Fromthefirstequation,t = 80/x.
Inserting this into the second equation, we get (x +10)(80/x −4/15) = 80. Rearranging, we obtain
x2 +10x −3000 = 0,whosepositivesolutionisx = 50. Sohisusualdrivingspeedis50km/h.
2.4
4. (a) If the two numbers are x and y, then x +y = 52 and x −y = 26. Adding the two equations gives
2x = 78, so x = 39, and then y = 52−39 = 13. (b) Let the cost of one chair be $x and the cost of
onetable$y. Then5x +20y = 1800and2x +3y = 420. Solvingthissystemyieldsx = 120,y = 60.
(c)UnitsproducedofB:x. Thenx+ 1x = 3x unitsareproducedofA,and300· 3x+200x = 13000,
2 2 2
or650x = 13000,sox = 20. Thus,30ofqualityAand20ofqualityBshouldbeproduced.
(d)Ifsheinvests$x at15%and$y at20%,thenx +y = 1500and0.15x +0.2y = 275. Thesolution
isx = 8000andy = 2000.
2.5
2. (a)Thenumerator5+x2 isnever0,sotherearenosolutions. (b)Theequationisobviouslyequivalent
x2 +1+2x (x +1)2
to = 0, or = 0, so x = −1. (c) x = −1 is clearly no solution. Multiply
x2 +1 x2 +1
the equation by (x + 1)2/3. Then the denominator becomes x + 1 − 1x, which is 0 for x = −3/2.
3
(d)Multiplyingbyx −1andrearrangingyieldsx(2x −1) = 0,andsox = 0orx = 1/2.
3. (a)z = 0satisfiestheequation. Ifz (cid:7)= 0,cancelingz2 yieldsz−a = za+zb,orz(1−(a+b)) = a. If
a+b = 1wehaveacontradiction. Ifa+b (cid:7)= 1,z = a/(1−(a+b)). (b)Theequationisequivalent
to(1+λ)μ(x −y) = 0,soλ = −1,μ = 0,orx = y. (c)μ = ±1makestheequationmeaningless.
Multiplyingtheequationby1−μ2 yieldsλ(1−μ) = −λ,orλ(2−μ) = 0,soλ = 0orμ = 2.
(d)Theequationisequivalenttob(1+λ)(a−2) = 0,sob = 0,λ = −1,ora = 2.
ReviewProblemsforChapter2
2. SeeProblem2.1.3.
3. (a)x = 2(y −3)+y = 2y −2+y = 5y −2,or 5y = x +2,soy = 3(x +2).
3 3 3 3 5
(b)√ax −cx = b+d,or(a−c)x = b+d,sox = (b+d)/(a−c).
(c) L = Y /AK,sosquaringeachsideyieldsL = (Y /AK)2. (d)qy = m−px,soy = (m−px)/q.
0 0
(e)and(f): Seetheanswersinthetext.
5. (a)Multiplytheequationby5K1/2 toobtainK1/2 = 15L1/3. SquaringeachsidegivesK = 225L2/3.
(b)Raiseeachsidetothepower1/t toobtain1+r/100 = 21/t,andsor = 100(21/t −1).
(c)abxb−1 = p,soxb−1 = p/ab. Nowraiseeachsidetothepower1/(b−1).
0 0
(d)Raiseeachsidetothepower−ρ toget(1−λ)a−ρ+λb−ρ = c−ρ,orb−ρ = λ−1(c−ρ−(1−λ)a−ρ).
Nowraiseeachsidetothepower−1/ρ.
9. (a)Seetheanswerinthetext. (b)Letu = 1/x andv = 1/y. Thenthesystemreducesto3u+2v = 2,
2u−3v = 1/4,withsolutionu = 1/2,v = 1/4. Itfollowsthatx = 1/u = 2andy = 1/v = 4.
(c)Seetheanswerinthetext.
©KnutSydsæter,ArneStrøm,andPeterHammond2008
6 CHAPTER 3 INTRODUCTORY TOPICS II: MISCELLANEOUS
Chapter 3 Introductory Topics II: Miscellaneous
3.1
3. (a)–(d): Lookatthelasttermandreplacenbyk. Sumoverk from1ton. (e)Thecoefficientsarethe
powers3n forn = 1,2,3,4,5,sothegeneraltermis3nxn. (f)and(g)seeanswersinthetext.
(h) This is tricky. One has to see that each term is 198 larger that the previous term. (The problem is
relatedtothestoryaboutGaussonpage56.)
(cid:10) (cid:10)
7. (a) n ck2 = c·12 +c·22 +···+c·n2 = c(12 +22 +···+n2) = c n k2
k=1 k=1
(b)Wrongevenforn = 2: Theleft-handsideis(a +a )2 = a2 +2a a +a2,buttheright-handside
1 2 1 1 2 2
isa2 +a2. (c)Bothsidesequalb +b +···+b . (d)Bothsidesequal51+52 +53+54+55.
1 2 1 2 N
(e)Bothsidesequala2 +···+a2 . (f)Wrongevenforn = 2: Theleft-handsideisa +a /2,but
0,j n−1,j 1 2
theright-handsideis(1/k)(a +a ).
1 2
3.2
5. Onedoesnothavetousesummationsigns. Thesumisa+(a+d)+(a+2d)+···+(a+(n−1)d.
Therearenterms. Thesumofallthea’sisna. Therestisd(1+2+···+n−1). Thenuseformula(4).
3.3
(cid:10) (cid:10) (cid:4) (cid:5) (cid:10) (cid:11)(cid:4) (cid:5) (cid:4) (cid:5) (cid:4) (cid:5) (cid:12) (cid:4) (cid:5) (cid:4) (cid:5) (cid:4) (cid:5)
1. (a)Seethetext. (b) 2 4 rs 2 = 2 2s 2+ 3s 2+ 4s 2 = 2 2+ 3 2+ 4 2+
(cid:4) (cid:5) (cid:4) (cid:5) (cid:4) (cid:5) s=0 r=2 r+s s=0 2+s 3+s 4+s 3 4 5
4 2 + 6 2 + 8 2 = 5+ 3113
4 (cid:10) 5(cid:10) 6 (cid:10) 3600(cid:10)
(c) m n i·j2 = m i· n j2 = 1m(m+1)·1n(n+1)(2n+1) = 1 m(m+1)n(n+1)(2n+1),
i=1 j=1 i=1 j=1 2 6 12
whereweused(4)and(5).
(cid:2) (cid:3)
(cid:13)n (cid:13)m (cid:13)n
1 1 1
4. a¯ isthemeanofthea¯ ’sbecausea¯ = a = a¯ .
s rs s
n m n
s=1 r=1 s=1
Toprove(∗),notethatbecausea −a¯ isindependentofthesummationindexs,itisacommonfactor
(cid:10) rj (cid:10)
whenwesumovers,so m (a −a¯)(a −a¯) = (a −a¯) m (a −a¯)foreachr. Next,summing
s=1 rj sj rj s=1 sj
overr gives
(cid:14) (cid:15)(cid:14) (cid:15)
(cid:13)m (cid:13)m (cid:13)m (cid:13)m
(a −a¯)(a −a¯) = (a −a¯) (a −a¯) (∗∗)
rj sj rj sj
r=1 s=1 r=1 s=1
Usingthepropertiesofsumsandthedefinitionofa¯ ,wehave
j
(cid:13)m (cid:13)m (cid:13)m
(a −a¯) = a − a¯ = ma¯ −ma¯ = m(a¯ −a¯)
rj rj j j
r=1 r=1 r=1
(cid:10)
Similarly, replacing r with s as the index of summation, one also has m (a − a¯) = m(a¯ − a¯).
s=1 sj j
Substitutingthesevaluesinto(∗∗)thenconfirms(∗).
3.4
√ √ √
6. (a)If (i) x −4 = x +5−9,thenalso (ii)x −4 = ( x +5−9)2,whichwegetbysquaringboth
√
sidesin(i). Calculatingthesquareontheright-handsideof(ii)gives x +5 = 5,andsox +5 = 25,
©KnutSydsæter,ArneStrøm,andPeterHammond2008
CHAPTER 3 INTRODUCTORY TOPICS II: MISCELLANEOUS 7
i.e. x = 20. This shows that if x is a solution of (i), then x = 20. No other va√lue of x can satisfy (i).
But if we√check this solution, we find that with x = 20 the LHS of (i) becomes 16 = 4, and the RHS
becomes 25−9 = 5−9 = −4. ThustheLHSandtheRHSaredifferent. Thismeansthatequation(i)
actuallyhasnosolutionsatall. (Butnotethat42 = (−4)2, i.e.thesquareoftheLHSequalsthesquare
oftheRHS.Thatishowthespurioussolutionx = 20managedtosneakin.)
√ √
(b)Ifxisasolutionof(iii) x −4 = 9− x +5,thenjustasinpart(a)wefindthatxmustbeasolution
√ √ √
of(iv)x −4 = (9− x +5)2. Now,(9− x +5)2 = ( x +5−9)2,soequation(iv)isequivalent
toequation(ii)inpart(a). Thismeansthat(iv)hasexactlyonesolution, namelyx = 20. Insertingthis
valueofx intoequation(iii),wefindthatx = 20isasolutionof(iii).
Ageometricexplanationoftheresultscanbegivenwithreferencetothefollowingfigure.
y √
y =9− x+5
5
√
y = x−4
x
5 10 15 20 25
-5 √
y = x+5−9
Figure SM3.4.6
√
Weseethatthetwosolidcurvesinthefigurehavenopointincommon,thatis,theexpressions x −4
√ √ √
and x +5−9arenotequalforanyvalueofx. (Infact,thedifference x −4−( x +5−9)increases
with x, so there is no point of intersection farther to the right, either.) This explains why the equation
√ √
in(a)hasnosolution. Thedashedcurvey = 9− x +5,ontheotherhand,intersectsy = x +5for
x = 20(andonlythere),andthiscorrespondstothesolutioninpart(b).
Comment: In part (a) it was necessary to check the result, because the transition from (i) to (ii) is only
an implication, not an equivalence. Similarly, it was necessary to check the result in part (b), since the
transitionfrom(iii)to(iv)alsoisonlyanimplication—atleast,itisnotclearthatitisanequivalence.
(Afterwards, it turned out to be an equivalence, but we could not know that until we had solved the
equation.)
√
7. (a)Herewehave“iff”since 4 = 2. (b)Itiseasytoseebymeansofasigndiagramthatx(x+3) < 0
precisely when x lies in the open interval (−3,0). Therefore we have an implication from left to right
(that is, “only if”), but not in the other direction. (For example, if x = 10, then x(x + 3) = 130.)
(c) x2 < 9 ⇐⇒ −3 < x < 3, so x2 < 9 only if x < 3. If x = −5, for instance, we have x < 3 but
x2 > 9. Hence we cannot have “if” here. (d) x2 +1 is never 0, so we have “iff” here. (e) If x > 0,
thenx2 > 0,butx2 > 0alsowhenx < 0. (f)x4+y4 = 0 ⇐⇒ x = 0andy = 0. Ifx = 0and,say,
y = 1,thenx4+y4 = 1,sowecannothave“if”here.
9. (a)Ifx andy arenotbothnonnegative,atleatoneofthemmustbenegative,i.e. x < 0ory < 0.
(b)Ifnotallx aregreaterthanorequaltoa,atleastonex mustbelessthana. (c)Atleastoneofthem
is less than 5. (Would it be easier if the statement to negate were “Neither John nor Diana is less than
5yearsold”?) (d)–(f)Seetheanswersinthetext.
©KnutSydsæter,ArneStrøm,andPeterHammond2008
8 CHAPTER 4 FUNCTIONS OF ONE VARIABLE
3.7
3. Forn = 1,bothsidesare1/2. Suppose(∗)istrueforn = k. Thenthesumofthefirstk+1termsis
1 1 1 1 1 k 1
+ + +···+ + = +
1·2 2·3 3·4 k(k+1) (k+1)(k+2) k+1 (k+1)(k+2)
k 1 (k+1)2 k+1
But + = = ,whichis(∗)forn = k+1. Thus,byinduction,
k+1 (k+1)(k+2) (k+1)(k+2) k+2
(∗)istrueforalln.
4. Theclaimistrueforn = 1. Astheinductionhypothesis,supposek3 +(k+1)3 +(k+2)3 isdivisible
by 9. Note that (k + 1)3 + (k + 2)3 + (k + 3)3 = (k + 1)3 + (k + 2)3 + k3 + 9k2 + 27k + 27 =
k3+(k+1)3+(k+2)3+9(k2+3k+3). Thisisdivisibleby9becausetheinductionhypothesisimplies
thatthesumofthefirstthreetermsisdivisibleby9,whereasthelasttermisalsoobviouslydivisibleby9.
ReviewProblemsforChapter3
6. (b) ⇒ false (because x2 = 16 also has the solution x = −4), ⇐ true, because if x = 4, then x2 = 16.
(c) ⇒ true, ⇐ false because with y > −2 and x = 3, (x −3)2(y +2) = 0. (d) ⇒ and ⇐ both true,
since the equation x3 = 8 has the solution x = 2 and no others. (In the terminology of Section 6.3,
f(x) = x3 isstrictlyincreasing. SeeProblem6.3.3andseethegraphFig. 7,page88.)
9. ConsiderFig.A3.6.8,page643inthebook,andletn denotethenumberofstudentsinthesetmarkedS ,
k k
fork = 1,2,...,8. SetsA,B,andCrefertothosewhostudyEnglish,French,andSpanish,respectively.
Since 10 students take all three languages, n = 10. There are 15 who take French and Spanish, so
7
15 = n +n , and thus n = 5. Furthermore, 32 = n +n , so n = 22. Also, 110 = n +n , so
2 7 2 3 7 3 1 7
n = 100. Therestoftheinformationimpliesthat52 = n +n +n +n ,son = 52−5−22−10 = 15.
1 2 3 6 7 6
Moreover,220 = n +n +n +n ,son = 220−100−5−10 = 105. Finally,780 = n +n +n +n ,
1 2 5 7 5 1 3 4 7
son = 780−100−22−10 = 648. Theanswerstotheproblemsare:
4 (cid:10)
(a): n = 100 (b): n +n = 648+22 = 670 (c)1000− 8 n = 1000−905 = 95
1 3 4 i=1 i
Chapter 4 Functions of One Variable
4.2
1. (a) f(0√) = 02 +√1 = 1, f(−1) = (−1)2 + 1 = 2, f(1/2) = (1/2)2 + 1 = 1/4 + 1 = 5/4,
and f( 2) = ( 2)2 + 1 = 2 + 1 = 3. (b) (i) Since (−x)2 = x2, f(x) = f(−x) for all x.
(ii)f(x+1) = (x+1)2+1 = x2+2x+1+1 = x2+2x+2andf(x)+f(1) = x2+1+2 = x2+3. Thus
equalityholdsifandonlyifx2+2x+2 = x2+3,i.e.ifandonlyifx = 1/2. (iii)f√(2x) = (2x)√2+1 =
4x2 +1and2f(x) = 2x2 +2. Now,4x2 +1 = 2x2 +2 ⇔ x2 = 1/2 ⇔ x = ± 1/2 = ±1 2.
2
10. (a) No: f(2+1) = f(3) = 1√8,whereasf(2)+f(1) = 10. (b) Y√es: f(2+1) = f(2)+f(1) = −9.
(c) No: f(2+1) = f(3) = 3 ≈ 1.73,whereasf(2)+f(1) = 2+1 ≈ 2.41.
13. (a) Wemustrequire5−x ≥ 0,sox ≤ 5. (b) Thedenominatorx2 −x = x(x −1)mustbedifferent
from0,sox (cid:7)= 0andx (cid:7)= 1. (c) Tobeginwith,thedenominatormustbenonzero,sowemustrequire
x (cid:7)= 2 and x (cid:7)= −3. Moreover, since we can only take the square root of a nonnegative number, the
fraction(x−1)/(x−2)(x+3)mustbe≥ 0.AsigndiagramrevealsthatD = (−3,1]∪(2,∞). Note
f
inparticularthatthefunctionisdefinedwithvalue0atx = 1.
©KnutSydsæter,ArneStrøm,andPeterHammond2008