SOLUTIONS TO CALCULUS VOLUME 1 BY TOM APOSTOL. Fund Science!

236 Pages·2014·1.16 MB·English

SOLUTIONS TO CALCULUS VOLUME 1 BY TOM APOSTOL.
ERNESTYEUNG
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SOLUTIONSTO VOLUME1One-VariableCalculus,withanIntroductiontoLinearAlgebra
I2.5Exercises-Introductiontosettheory,Notationsfordesignatingsets,Subsets,Unions,intersections,complements.
Exercise10. Distributivelaws
LetX =A∩(B∪C),Y =(A∩B)∪(A∩C)
Supposex∈X
x∈Aandx∈(B∪C)=⇒x∈AandxisinatleastBorinC
thenxisinatleasteither(A∩B)or(A∩C)
x∈Y,X ⊆Y
Supposey ∈Y
yisatleastineither(A∩B)orA∩C
theny ∈AandeitherinBorC
y ∈X,Y ⊆X
X =Y
LetX =A∪(B∩C),Y =(A∪B)∩(A∪C)
Supposex∈X
thenxisatleasteitherinAorin(B∩C)
ifx∈A,x∈Y
ifx∈(B∩C),x∈Y x∈Y,X ⊆Y
Supposey ∈Y
thenyisatleastinAorinBandyisatleastinAorinC
ify ∈A,theny ∈X
ify ∈A∩Bory ∈A∪C,y ∈X (variouscarvingsoutofA,simply)
ify ∈(B∩C),y ∈X y ∈X,Y ⊆X
X =Y
1
Exercise11. Ifx∈A∪A,thenxisatleastinAorinA. Thenx∈A. SoA∪A⊆A. OfcourseA⊆A∪A.
Ifx∈A∩A,thenxisinAandinA. Thenx∈A. SoA∩A⊆A. OfcourseA⊆A∩A.
Exercise12. Letx∈A. y ∈A∪BifyisatleastinAorinB. xisinAsox∈A∪B. =⇒A⊆A∪B.
Suppose∃b∈Bandb∈/ A. b∈A∪Bbutb∈/ A. soA⊆A∪B.
Exercise13. Letx ∈ A∪∅, thenxisatleastinAorin∅. Ifx ∈ ∅, thenxisanullelement(notanelementatall). Then
actualelementsmustbeinA. =⇒A∪∅⊆A.
Letx∈A. Thenx∈A∪∅. A⊆A∪∅. =⇒A=A∪∅.
Exercise14. Fromdistributivity,A∪(A∩B)=(A∪A)∩(A∪B)=A∩(A∪B).
Ifx∈A∩(A∪B),x∈Aandx∈A∪B,i.e. x∈AandxisatleastinAorinB.
=⇒ x is in A and is in B or is not in B. Then x ∈ A. =⇒ A ∩ (A ∪ B) ⊆ A. Of course, A ⊆ A ∩ (A ∪ B).
=⇒A∩(A∪B)=A∪(A∩B)=A.
Exercise15. ∀a ∈ A,a ∈ C and ∀b ∈ B,b ∈ C. Consider x ∈ A∪B. x isat leastin A orin B. In eithercase, x ∈ C.
=⇒A∪B ⊆C.
Exercise16.
ifC ⊆AandC ⊆B, thenC ⊆A∩B
∀c∈C,c∈Aandc∈B
x∈A∩B,x∈Aandx∈B.Then∀c∈C,c∈A∩B. C ⊆A∩B
Exercise17.
(1)
ifA⊂BandB ⊂C then
∀a∈A,a∈B.∀b∈B,b∈C.
thensincea∈B,a∈C,∃c∈C suchthatc∈/ B.
∀a∈A,a∈Bsoa(cid:54)=c∀a.=⇒A⊂C
(2) IfA⊆B,B ⊆C,A⊆C since,∀a∈A,a∈B,∀b∈B,b∈C.Thensincea∈B,a∈C. A⊆C
(3) A⊂BandB ⊆C. B ⊂C orB =C. A⊂Bonly. ThenA⊂C.
(4) Yes,since∀a∈A,a∈B.
(5) No,sincex(cid:54)=A(setsaselementsaredifferentfromelements)
Exercise18. A−(B∩C)=(A−B)∪(A−C)
Supposex∈A−(B∩C)
thenx∈Aandx∈/ B∩C =⇒x∈/ B∩C
thenxisnotinevenatleastoneBorC
=⇒x∈(A−B)∪(A−C)
Supposex∈(A−B)∪(A−C)
thenxisatleastin(A−B)orin(A−C)=⇒xisatleastinAandnotinBorinAandnotinC
thenconsiderwhenoneofthecasesistrueandwhenbothcasesaretrue =⇒x∈A−(B∩C)
Exercise19.
(cid:91)
Supposex∈B− A
A∈F
(cid:91)
thenx∈B,x∈/ A
A∈F
(cid:91)
x∈/ A=⇒x∈/ A,∀A∈F
A∈F
(cid:92)
since∀A∈F,x∈B,x∈/ A, thenx∈ (B−A)
A∈F
2
(cid:92)
Supposex∈ (B−A)
A∈F
thenx∈B−A andx∈B−A and ...
1 2
then∀A∈F,x∈B,x∈/ A
thenx∈/ evenatleastoneA∈F
(cid:91)
=⇒x∈B− A
A∈F
(cid:92)
Supposex∈B− A
A∈F
(cid:92)
thenx∈/ A
A∈F
thenatmostx∈Afor∀A∈F butone
thenxisatleastinoneB−A
(cid:91)
=⇒x∈ (B−A)
A∈F
(cid:91)
Supposex∈ (B−A)
A∈F
thenxisatleastinoneB−A
thenforA∈F,x∈Bandx∈/ A
Consider∀A∈F
(cid:92)
=⇒ thenx∈B− A
A∈F
Exercise20.
(1) (ii)iscorrect.
Supposex∈(A−B)−C
thenx∈A−B,x∈/ C
thenx∈Aandx∈/ Bandx∈/ C
x∈/ Bandx∈/ C =⇒x∈/ evenatleastBorC
x∈A−(B∪C)
Supposex∈A−(B∪C)
thenx∈A,x∈/ (B∪C)
thenx∈Aandx∈/ Bandx∈/ C
=⇒x∈(A−B)−C
Toshowthat(i)issometimeswrong,
Supposey ∈A−(B−C)
y ∈Aandy ∈/ B−C
y ∈/ B−C
theny ∈/ Bory ∈C ory ∈/ C
(wheredoesthisleadto?)
Considerdirectly,
Supposex∈(A−B)∪C
thenxisatleastinA−BorinC
thenxisatleastinAand∈/ BorinC
Supposex=c∈C andc∈/ A
3
(2)
IfC ⊆A,
A−(B−C)=(A−B)∪C
I3.3Exercises-Thefieldaxioms. Thegoalseemstobetoabstracttheseso-calledrealnumbersintojustx’sandy’sthat
arepurelybuiltupontheseaxioms.
Exercise1. Thm. I.5. a(b−c)=ab−ac.
Lety =ab−ac;x=a(b−c)
Want: x=y
ac+y =ab(byThm. I.2,possibilityofsubtraction)
NotethatbyThm. I.3,a(b−c)=a(b+(−c))=ab+a(−c)(bydistributivityaxiom)
ac+x=ac+ab+a(−c)=a(c+(−c))+ab=a(0+b)=ab
ButthereexistsexactlyoneyorxbyThm. I.2. x=y.
Thm. I.6. 0·a=a·0=0.
0(a)=a(0)(bycommutativityaxiom)
Givenb∈Rand0∈R,∃exactlyone −bs.t. b−a=0
0(a)=(b+(−b))a=ab−ab=0(byThm. I.5. andThm. I.2)
Thm. I.7.
ab=ac
ByAxiom4,∃y ∈Rs.t. ay =1
sinceproductsareuniquelydetermined,yab=yac=⇒(ya)b=(ya)c=⇒1(b)=1(c)
=⇒b=c
Thm. I.8. PossibilityofDivision.
Givena,b,a(cid:54)=0,chooseysuchthatay =1.
Letx=yb.
ax=ayb=1(b)=b
Therefore, there exists at least one x such that ax = b. But by Thm. I.7, there exists only one x (since if az −b, and so
x=z).
Thm. I.9. Ifa(cid:54)=0,thenb/a=b(a−1).
b
Letx= forax=b
a
y =a−1foray =1
Want: x=by
Nowb(1)=b, soax=b=b(ay)=a(by)
=⇒x=by(byThm. I.7)
Thm. I.10. Ifa(cid:54)=0,then(a−1)−1 =a.
Now ab = 1 for b = a−1. But since b ∈ R and b (cid:54)= 0 (otherwise 1 = 0, contradiction), then using Thm. I.8 on b,
ab=b(a)=1; a=b−1.
Thm.I.11. Ifab=0,a=0orb=0.
ab=0=a(0)=⇒b=0orab=ba=b(0)=⇒a=0. (weusedThm. I.7,cancellationlawformultiplication)
Thm. I.12. Want: x=yifx=(−a)bandy =−(ab).
ab+y =0
ab+x=ab+(−a)b=b(a+(−a))=b(a−a)=b(0)=0
0isunique,soab+y =ab+ximpliesx=y(byThm. I.1)
Thm. I.13. Want: x+y =z,ifa=bx,c=dy,(ad+bc)=(bd)z.
(bd)(x+y)=bdx+bdy =ad+bc=(bd)z
Sousingb,d(cid:54)=0,whichisgiven,andThm. I.7,thenx+y =z.
4
Thm. I.14. Want: xy =zforbx=a,dy =c,ac=(bd)z.
(bd)(xy)=(bx)(dy)=ac=(bd)z
b,d(cid:54)=0,sobyThm. I.7,xy =z.
Thm.I.15. Want: x=yz,ifbx=a,dy =c,(bc)z =ad
(bc)z =b(dy)z =d(byz)=da
d(cid:54)=0sobyThm. I.7,byz =a,byz =abx
b(cid:54)=0sobyThm. I.7,yz =x
Exercise2. Consider0+z =0. ByThm. I.2,thereexistsexactlyonez,z =−0. ByAxiom4,z =0. 0=−0.
Exercise3. Consider1(z)z(1)=1. Thenz =1−1. ButbyAxiom4,thereexistsdistinct1suchthatz(1)=1,soz =1.
Exercise4. Supposethereexistsxsuchthat0x=1,but0x=0and0and1aredistinct,so zerohasnoreciprocal.
Exercise5. a+(−a)=0,0+0=0. Then
a+(−a)+b+(−b)=(a+b)+(−a)+(−b)=0
−(a+b)=−a+(−b)=−a−b
Exercise6. a+(−a)=0,b+(−b)=0,so
a+(−a)+b+(−b)=a+(−b)+(−a)+b=(a−b)+(−a)+b=0+0=0
−(a−b)=−a+b.
Exercise7.
(a−b)+(b−c)=a+(−b)+b+(−c)=a+(b+(−b))+(−c)=a−c
Exercise8.
(ab)x=1 (ab)−1 =x
a(bx)=1 a−1 =bx
b(ax)=1 b−1−ax
a−1b−1 =(abx)x=1(x)=(ab)−1
Exercise9. Want: x=y =z,if
a
z = a=zt b+t=0
−b
(−a)
y = by =u a+u=0
b
(cid:16)a(cid:17) (cid:16)a(cid:17)
x=− +x=v+x=0 vb=a
b b
a+(−a)=vb+by =b(v+y)=0
ifb(cid:54)=0,v+y =0, butv+x=0
byThm. I.1,x=y
b+t=0, thenz(b+t)=zb+zt=zb+a=z(0)=0
a+zb=0=⇒−a=zb=by
sinceb(cid:54)=0,z =ysox=y =z
Exercise10. Sinceb,d(cid:54)=0,Let
ad−bc
z = (bd)z =ad−bc bypreviousexerciseorThm. I.8,thepossibilityofdivision
bd
a
x= bx=a
b
−c
t= dt=−c(ByThm. I.3,weknowthatb−a=b+(−a))
d
5
dbx+bdt=(bd)(x+y)=ad−bc=(bd)z
b,d(cid:54)=0, sox+y =z
I3.5Exercises-Theorderaxioms.
Theorem1(I.18). Ifa<bandc>0thenac<bc
Theorem2(I.19). Ifa<bandc>0,thenac<bc
Theorem3(I.20). Ifa(cid:54)=0,thena2 >0
Theorem4(I.21). 1>0
Theorem5(I.22). Ifa<bandc<0,thenac>bc.
Theorem6(I.23). Ifa<band−a>−b. Inparticular,ifa<0,then−a>0.
Theorem7(I.24). Ifab>0,thenbothaandbarepositiveorbotharenegative.
Theorem8(I.25). Ifa<candb<d,thena+b<c+d.
Exercise1.
(1) ByThm. I.19,−c>0
a(−c)<b(−c)→−ac<−bc
−bc−(−ac)=ac−bc>0.Thenac>bc(bydefinitionof>)
(2)
a<b→a+0<b+0→a+b+(−b)<b+a+(−a)→(a+b)−b<(a+b)+(−a)
ByThm.I.18(a+b)+−(a+b)+(−b)<(a+b)−(a+b)+(−a)
−b<−a
(3)
Ifa=0orb=0,ab=0,but0≯0
Ifa>0,thenifb>0,ab>0(b)=0. Ifb<0,ab<0(b)=0. Soifa>0,thenb>0.
Ifa<0,thenifb>0,ab<0(b)=0. Ifb<0,ab>0(b)=0. Soifa<0,thenb<0.
(4)
a<csoa+b<c+b=b+c
b<dsob+c<d+c
ByTransitiveLaw,a+b<d+c
Exercise2. Ifx=0,x2 =0. 0+1=1(cid:54)=0. Sox(cid:54)=0.
Ifx(cid:54)=0,x2 >0, andbyThm. I.21,1>0
x2+1>0+0=0→x2+1(cid:54)=0
=⇒(cid:64)x∈Rsuchthatx2+1=0
Exercise3.
a<0,b<0,a+b<0+0=0(ByThm. I.25)
Exercise4. Considerax=1.
ax=1>0.ByThm. I.24,a,xarebothpositiveora,xarebothnegative
Exercise5. Definex,ysuchthatax=1,by =1. Wewantx>ywhenb>a.
xb−ax=xb−1>0=⇒bx>1=by
b>0sox>y
Exercise6.
6
Ifa=bandb=c, thena=c
Ifa=bandb<c, thena<c
Ifa<bandb=c, thena<c
Ifa<bandb<c, thena<c(bytransitivityoftheinequality)
=⇒a≤c
Exercise7. Ifa≤bandb≤c,thena≤c. Ifa=c,thenbypreviousproof,a=b.
Exercise8. Ifa≤bandb≤c,thena≤c. Ifa=c,thenbypreviousproof,a=b.
Exercise8. Ifaorbiszero,a2orb2 =0. ByThm. I.20,b2 ≥0ora2 ≥0,respectively.
Otherwise,ifneitherarezero,bytransitivity,a2+b2 >0.
Exercise9. Supposea≥x. Thena−x≥0.
Ifa∈Rso∃y ∈R,suchthata−y =0.
Considery+1∈R(byclosureunderaddition).
a−(y+1)=a−y−1=0−1<0Contradictionthata≥y+1
Exercise10.
Ifx=0, done.
x
Ifx>0,xisapositiverealnumber. Leth= .
2
x
=⇒ >xContradiction.
2
I 3.12 Exercises - Integers and rational numbers, Geometric interpretation of real numbers as points on a line,
Upperboundofaset,maximumelement,leastupperbound(supremum),Theleast-upper-boundaxiom(completeness
axiom),TheArchimedeanpropertyofthereal-numbersystem,Fundamentalpropertiesofthesuprenumandinfimum.
WeuseThmI.30,theArchimedeanpropertyofrealnumbers,alot.
Theorem9(I.30). Ifx>0andifyisanarbitraryrealnumber,thereexistsapositiveintegernsuchthatnx>y.
We will use the least upper-bound axiom (completeness axiom) alot for continuity and differentiation theorems later.
Apostol states it as an axiom; in real analysis, the existence of a sup for nonempty, bounded sets can be shown with an
algorithmtozoomintoasupwithmonotonicallyincreasingandmonotonicallydecreasingsequenceof“guesses”andshowing
itsdifferenceisaCauchysequence.
Axiom1(Leastupper-boundaxiom). EverynonemptysetS ofrealnumberswhichisboundedabovehasasuprenum;that
is,there’sarealnumberBs.t. B =supS.
Exercise1. 0<y−x.
=⇒n(y−x)>h>0,n∈Z+,harbitrary
y−x>h/n=⇒y >x+h/n>x
soletz =x+h/nDone.
Exercise2. x∈Rso∃n∈Z+suchthatn>x(Thm. I.29).
Setofnegativeintegersisunboundedbelowbecause
If∀m∈Z−,−x>−m,then−xisanupperboundonZ+. ContradictionofThm. I.29. =⇒∃m∈Zsuchthatm<x<n
Exercise3. UseArchimedianproperty.
x>0sofor1,∃n∈Z+suchthatnx>1,x> 1.
n
Exercise4. x is an arbitrary real number. By Thm. I.29 and well-ordering principle, there exists a smallest n+1 positive
integersuchthatx < n+1(considerthesetofallm+1 > xandsobywell-orderingprinciple,theremustbeasmallest
elementofthisspecificsetofpositiveintegers).
Ifx=nforsomepositiveintegern,done.
Otherwise,notethatifx<n,thenn+1couldn’thavebeenthesmallestelementsuchthatm>x. x>n.
Exercise5. Ifx=n,done. Otherwise,considerallm>x. Bywell-orderingprinciple,thereexistsasmallestelementnsuch
7
thatn>x.
Ifx+1<n,thenx<n−1,contradictingthefactthatnisthesmallestelementsuchthatx<n. Thusx+1>n.
Exercise6. y−x>0.
n(y−x)>h, harbitrary,n∈Z+
y >x+h/n=z >x
Sincehwasarbitrary,thereareinfinitelymanynumbersinbetweenx,y.
Exercise7. x= a ∈Q,y ∈/ Q.
b
a±by
x±y =
b
(cid:18) (cid:19)
a−mb
Ifa±bywasaninteger,saym,theny =± whichisrational. Contradiction.
b
ay ay
xy = =
b 1 b
n
Ifaywasaninteger,ay =n,y = ,butyisirrational. =⇒xyisirrational.
a
x
y
yisnotaninteger
Exercise8. Proofbycounterexamples. Wewantthatthesumorproductof2irrationalnumbersisnotalwaysirrational. Ify
isirrational,y+1isirrational,otherwise,ify+1∈Q, y ∈Qbyclosureunderaddition.
=⇒y+1−y =1
Likewise,y1 =1.
y
Exercise9.
y−x>0=⇒n(y−x)>k,n∈Z+,karbitrary. Choosektobeirrational. Thenk/nirrational.
k k
y > +x>x.Letz =x+ ,zirrational.
n n
Exercise10.
(1) Supposen=2m andn+1=2m .
1 2
1
2m +1=2m 2(m −m )=1 m −m = .Butm −m canonlybeaninteger.
1 2 1 2 1 2 2 1 2
(2) By the well-ordering principle, if x ∈ Z+ is neither even and odd, consider the set of all x. There must exist a
smallestelementx ofthisset. Butsincex ∈Z+,thentheremustexistan<xsuchthatn+1=x . nisevenor
0 0 0
oddsinceitdoesn’tbelongintheaboveset. Sox mustbeoddoreven. Contradiction.
0
(3)
(2m )(2m )=2(2m m )even
1 2 1 2
2m +2m =2(m +m )even
1 2 1 2
(2m +1)+(2m +1)=2(m +m +1)=⇒ sumoftwooddnumbersiseven
1 2 1 2
(n +1)(n +1)=n n +n +n+2+1=2(2m m )
1 2 1 2 1 1 2
2(2m m )−(n +n )−1odd,theproductoftwooddnumbersn ,n isodd
1 2 1 2 1 2
(4) Ifn2even,niseven,sinceforn=2m,(2m)2 =4m2 =2(2m2)iseven.
a2 =2b2. 2(b2)even. a2even,soaeven.
Ifaevena=2n.a2 =4n2
Ifbodd,b2odd. bhasnofactorsof2b2 (cid:54)=4n2
Thusbiseven.
8
(5) For p,Ifporqorbothareodd,thenwe’redone.
q
Else,whenp,qarebotheven,p=2lm,q =2np,m,podd.
p 2lm 2l−nm
= = andatleastmorpodd
q 2np p
Exercise11. a canbeputintoaformsuchthataorbatleastisoddbythepreviousexercise.
b
However,a2 =2b2,soaeven,beven,bythepreviousexercise,part(d)or4thpart. Thus a cannotberational.
b
Exercise12. ThesetofrationalnumberssatisfiestheArchimedeanpropertybutnottheleast-upper-boundproperty.
Since p ∈Q⊆R,np1 > p2 sinceifq ,q >0,
q q1 q2 1 2
np q q p
1 2 > 1 2 np q >q p
q q q q 1 2 1 2
1 2 1 2
nexistssince(p q ),(q p )∈R.
1 2 1 2
Thesetofrationalnumbersdoesnotsatisfytheleast-upper-boundproperty.
ConsideranonemptysetofrationalnumbersS boundedabovesothat∀x= r ∈S,x<b.
s
Supposex<b ,x<b ∀x∈S.
1 2
r r
<b <nb butlikewise <b <mb , n,m∈Z+
s 2 1 s 1 2
Soit’spossiblethatb >b ,butalsob >b .
1 2 2 1
I 4.4 Exercises - An example of a proof by mathematical induction, The principle of mathematical induction, The
well-orderingprinciple. Considerthese2proofs.
N +N +···+N =N2
N N−1
(cid:88) (cid:88)
(N −1)+(N −2)+···+(N −(N −1))+(N −N)=N2− j = j
j=1 j=1
N N
(cid:88) (cid:88) N(N +1)
N2+N =2 j =⇒ j =
2
j=1 j=1
Aninterestingpropertyisthat
n n
(cid:88) (cid:88)
S = j = (n+m−j)
j=m j=m
Sothat
N N m N
(cid:88) (cid:88) (cid:88) (cid:88) m(m+1) N(N +1)
j = j+ j = j+ =
2 2
j=1 j=m j=1 j=m
N
(cid:88) N(N +1)−m(m+1) (N −m)(N +m+1)
j = =
2 2
j=m
Anotherwaytoshowthisisthefollowing.
S = 1+ 2+ ···+ (N −2)+ (N −1)+ N
butS = N+ N −1+ ···+ 3+ 2+ 1
N(N +1)
2S =(N +1)N S =
2
9
Telescopingserieswillletyouget(cid:80)N j2andotherpowersofj.
j=1
N
(cid:88) N(N +1)
(2j−1)=2 −N =N2
2
j=1
N N N (cid:18) (cid:19)
(cid:88) (cid:88) (cid:88) N(N +1)
(j2−(j−1)2)= (j2−(j2−2j+1))= (2j−1)=2 −N =N2
2
j=1 j=1 j=1
N N N
(cid:88) (cid:88) (cid:88)
(j3−(j−1)3)=N3 = (j3−(j3−3j2+3j−1))= (3j2−3j+1)
j=1 j=1 j=1
(cid:88)N N(N +1) 2N3+2N −3N2−3N N(N +1)(2N +1) (cid:88)N
=⇒3 j2 =−3 +N =N3 =⇒ = = j2
2 2 6
j=1 j=1
N N N
(cid:88) (cid:88) (cid:88)
j4−(j−1)4 =N4 = j4−(j4−4j3+6j2−4j+1)= 4j3−6j2+4j−1=
j=1 j=1 j=1
N
(cid:88) N(N +1)(2N +1) N(N +1)
=4 j3−6 +4 −N =N4
6 2
j=1
N
(cid:88) 1 1
=⇒ j3 = (N4+N(N +1)(2N +1)−2N(N +1)+N)= (N4+(2N)N(N +1)−N(N +1)+N)
4 4
j=1
1 1 1(N(N +1))2
= (N4+2N3+2N2−N2−N +N)= N2(N2+2N +1)=
4 4 4 2
Exercise1. Inductionproof.
N+1 n
1(1+1) (cid:88) (cid:88) n(n+1) n(n+1)+2(n+1) (n+2)(n+1)
j = j+n+1= +n+1= =
2 2 2 2
j=1 j=1
Exercise6.
(1)
1 1 8k+8 (2k+3)2
A(k+1)=A(k)+k+1= (2k+1)2+k+1= (4k2+4k+1)+ =
8 8 8 8
(2) Then=1caseisn’ttrue.
(3)
(n+1)n n2+n n2+n+ 1
1+2+···+n= = < 4
2 2 2
(cid:18)2n+1(cid:19)2 1 (n+1/2)2 n2+n+1/4
and = =
2 2 2 2
Exercise7.
(1+x)2 >1+2x+2x2
1+2x+x2 >1+2x+2x2
0>x2 =⇒ Impossible
(1+x)3 =1+3x+3x2+x3 >1+3x+3x2
=⇒x3 >0
Bywell-orderingprinciple,wecouldarguethatn=3mustbethesmallestnumbersuchthat(1+x)n >1+2x+2x2. Or
wecouldfind,explicitly
n (cid:18) (cid:19) n (cid:18) (cid:19)
(cid:88) n n(n−1) (cid:88) n
(1+x)n = xj =1+nx+ x2+ xj
j 2 j
j=0 j=3
10