BEE1020—BasicMathematicalEconomicsDieterBalkenborg

Week15,LectureTuesday10/02/2004

DepartmentofEconomics

Exponentialandlogarithmicfunctions,Elasticities

UniversityofExeter

Introduction

—Exponentialfunctions:describesgrowthprocesseswithconstantgrowthrate

populationgrowth,growthofGDP,inflationetc...

—logarithm:inversefunction

—elasticities

ExponentialFunction

power

xy

basex

indexorexponenty

powerfunction:vary

xexponentialfunction:vary

yadmissiblevaluesfory:positiveintegers,integers,rationals,reelnumbers

problem:forgeneralythepowerxycanonlybedefinedforpositivex

05101520

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24

x

y=2:

f(x)=x2

01234

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24

x

y=−2:f(x)=x−2=

1 x2

0

0.20.40.60.811.21.41.61.822.2

12

34

5x

y=

1 2:f(x)=x1 2=√ x

012345

12

34

5x

y=−3 2:f(x)=x−3 2=

1x√x

2

approximateirrationalindex

ybyfraction

m n:

xy:=lim m n→yxm n.

02468

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12

y

x=3:

g(y)=3y

02468

-2-1

12

y

x=

1 3:g(y)=¡ 1 3¢ y

=3−

y

3

0.5

11.5

22.5

3x

0.6

0.8

11.2

1.4

y

1234

z=xy

x≥0

anexponentialfunctionax:

alwaysconvex,strictlypositivevalues.

a>1:increasingwithlim

x→−∞

ax=0andlim

x→∞ax=+∞.

0<a<1:decreasingwithlim

x→−∞

ax=+∞andlim

x→∞ax=0.

calculationalrulesforgeneralizedpowers:

as+

t=as a

tast=(a

s )t

(ab)

s=as b

s

4

but

(as )t6=a(s

t )

Compoundedinterestsandthenumbere.

PutP0>0(theprincipal)insavingsaccount

fixednominalannualinterestsrate

r>0

Interestspaidntimesduringtheyear

amount

Ptinyoursavingsaccountaftertyears:

formulaforcompoundedinterests

Pt=P0

¡ 1+r n

¢ ntr ninterestpaidperperiod

nttotalnumberofinterestpayments.

The(natural)exponentialfunction:

balanceinaccountafteroneyearifinterestspaidcontinuously:

exp(r)=lim

n→∞¡ 1+

r n

¢ n .5

Thetablebelowshowsthevalueof¡ 1+

r n

¢ n forvariousnandr:

r=5.4%

r=5.5%

r=100%

n=4

1.0551033751.0561448092.44140625

n=12

1.0553567521.05640786

2.61303529

n=364

1.0554803751.0565362252.714557303

n=8736

1.0554844261.0565404322.718126265

n=524160

1.0554845991.0565406122.718279235

n=314496001.0554846021.0565406132.718281796

n→+∞

1.0554846021.0565406152.718281828

Thenumber

eisdefinedas

e=exp(1)=lim

n→∞¡ 1+

1 n

¢ n .The‘naturalexponentialfunction’isindeedtheexponentialfunctionwithbase

e:exp(r)=er

6

“proof”forrationalr:

exp(r)=lim n→∞

³ 1+r n

´ n =lim

m→∞,n=rm

³ 1+r n

´ n=

limm→∞

µ 1+1 m

¶ rm=limm→∞

·µ 1+1 m

¶ m¸r

=

· lim m→∞µ 1+

1 m

¶ m¸r

=er

formulaforcontinuouslycompoundedinterests:

Pt=P0er

t .

7

Propertiesoftheexponentialfunction

1.e0=1,

2.e1=e

3.ex

>0forallx

4.d(ex)

dx=ex

Inparticular,exisstrictlyincreasingandconvex.

instantaneousgrowthrateofafunctiony=

f(x):

dydx

. ywhen

xisin-

creasedbyaexponentialfunctionhasconstantgrowthrate1.

5.Exponentialversuspolynomialgrowth:ForanypolynomialP(x)

limx→+∞

ex

P(x)=+∞

TakingP(x)asatheconstantpolynomialP(x)≡1onehas

limx→+∞ex=+∞

8

6.lim

x→−∞

ex=0

7.ea+b=eaeb

(ea)b=ea

b .

Inparticular1 ex=e−

x(because

e−xex=e0=1).

051015

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23

x

Thefunctiony=ex

Aquickerwaytocalculate

ex:istousetheformula

ex=1+x+x2 2!+x3 3!+...+

xn n!+...

9

Letf n(x)=¡ 1+

x n

¢ n .Intuitionforproperty3:¡ 1+

x n

¢ ispositivewhen

x>0orwhen

nlarge

comparedto|x|

.Then

f n(x)>0andsohence

ex>0.

Intuitionforproperty4:

dfn

dx=n³ 1+

x n

´ n−11 n=³ 1+

x n

´ n−1≈³ 1+

x n

´ n =f n(x)

fornverylargecomparedto|x|since1+

x nisthenverycloseto1.

Intuitionforproperty5:SupposeP(x)=

amxm−1+...hasdegreem.Ap-

proximateexby

f n(x)withnlargerthan

m.Then

limx→+∞

ex

P(x)≈

limx→+∞

¡ 1+x n

¢ nP(x)=

limx→+∞

¡ 1 n¢ nxn+...

amxm+...=

limx→+∞Cxn−m=+∞

10

Thelogarithmicfunction

naturallogarithmfunction x=ln(y)⇔

y=ex

-3-2-10234

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23

4x

Therulefordifferentiatinginversefunctionsyields:

dln(y)

dy

=dx dy=1 dy dx=1 ex=1 y

11

Propertiesofln(y):

1.ln(y)isonlydefinedforstrictlypositivey>0.

2.dln(y)

dy=

1 y.Inparticular,ln(y)isstrictlyincreasingandconcave.

3.ln(1)=0,ln(e)=1.

4.lim

y→0ln(y)=−∞

,lim

y→+∞ln(y)=+∞.

5.ln(ab)=ln(a)+ln(b),ln¡ ab¢

=bln(a).Inparticularln¡ 1 a¢ =

−ln(a).

Logarithmicdifferentiation:Combinedwiththechainruleoneobtains

thefollowingusefulformulawherey=g(x)isanydifferentiablefunction:

dln(g(x))

dx

=g0 (x)

g(x)

(1)

12

Differentiatinggeneralexponentialandlogarithmicfunctions

ln(x

y)=yln(x)generalpowersare:

xy=eyln(x)=eindex×ln(base).

Partialdifferentiationyields

∂xy

∂y=

eyln(x)ln(x)=xyln(x)

∂xy

∂x=

eyln(x)

µ y1 x¶ =

yxy1 x=yxy−1.

derivativeofanexponentialfunctiony=axis

dax

dx=ln(a)ax.

Theinstantaneousgrowthrateofy=axisln(a)

Thederivativeofapowerfunctiony=xbis

dxb

dx=bx

b−1

evenifbisirrational.

13

Theinversetotheexponentialfunctiony=ax(for

a>0,a6=1)iscalledthe

logarithmicfunctiontothebasicaandwrittenaslog a(y)

x=log a(y)⇔

y=ax.

Wehave

y=ax⇔

y=exln(a)⇔ln(y)=xln(a)⇔

x=ln(y)

ln(a)

solog ay=

ln(y)

ln(a)

log ayhashencethederivative

d(log

a(y))

dy

=1

ln(a)y

14

CompoundedInterests

Example:SupposeBankAofferstheannualnominalinterestrater A=5.5%

andpaysinterestsmonthly.BankBofferstheannualnominalinterestrate

r B=5.4%

andpaysinterestsdaily.Whichbankoffersthebetterdeal?

Solution:

r eff,A=³ 1+

r A 12

´ 12 −1=

µ 1+0.055

12

¶ 12 −1=5.64%

r eff,B=³ 1+

r B 364

´ 364 −1=

µ 1+0.054

364

¶ 364−1=5.55%

soBankAoffersbetterdeal.

15

Exponentialdecay

mostradioactivesubstancesdecayexponentially

sampleofinitialsize

Q0weightsQ(t)=Q0e−

ktattimet.

kmeasuresrateofdecay

half-lifeoftheradioactivesubstance:

Example:Showthataradioactivesubstancethatdecaysaccordingtothe

formulaQ(t)=Q0e−

kthasahalf-lifeoft̄=

ln2k.

Solution:findvaluet̄forwhichQ(t̄)=

1 2Q0,thatis

1 2Q0=Q0e−

kt̄ .

Divideby

Q0andtakenaturallogarithm:

ln1 2=−k

t̄.

Thusthehalf-lifeis

t̄=−ln1 2

k=ln2

k

16

Derivatives

Example:Findthederivativeofg(x)=xx.

Solution:Usinglogarithmicdifferentiationweobtain

g0 (x)

g(x)=

d(lnxx)

dx

=d¡ lnex

ln(x)¢

dx

=d(xln(x))

dx

=1×ln( x)+x×1 x=ln( x)+1

g0 (x)=(ln(x)+1)xx. 17

Thelogisticcurve

Thegraphofthefunctionoftheform

Q(t)=

B

1+Ae−

Bkt

whereA,B,karepositiveconstants,iscalledalogisticcurve.

describesgrowthprocesseswhenenvironmentalfactorsimposea“braking”effect

ontherateofgrowth.

Example:ShowthatthegrowthrateofthelogisticcurveQ(t)=

11+

e−tis

1−Q(t).

Solution:Wehave

Q0 (t)=−e−t(−1)

(1+e−

t )2=

e−t

(1+e−

t )2

Q0 (t)

Q(t)=

e−t

1+e−

t

1−Q(t)=1+e−

t−1

1+e−

t=

e−t

1+e−

t

18

Example:Publichealthrecordsindicatethat

tweeksaftertheoutbreakofa

certainformofinfluenza,approximatelyQ(t)=

201+19e−1.2tthousandpeoplehad

caughtthedisease.

a)Howmanypeoplehadthediseasewhenitbrokeout?Howmanyhadit

twoweekslater?

b)Atwhattimedoesthespreadoftheinfectionbegintodecline?

c)Ifthetrendcontinues,approximatelyhow

manypeoplewilleventually

contractthedisease?

Solution:a)Since

Q(0)=

201+19=1itfollowsthat1000peopleinitiallyhad

thedisease.When

t=2

Q(2)=

20

1+19e−2.4≈7.343

soabout7.343thousandpeoplehadcontractedthediseasebythesecondweek.

b)inflectionpointat

t̄≈2.5.For

t<

t̄convexandsothenumberofnewly

infectedincreasing.For

t>

t̄concaveandsothenumberofnewlyinfectedis

19

decreasing.

051015

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24

68

t

c)lim

t→+∞Q(t)=20,soroughly20000peoplecatchthediseaseontotal.

20

NOSUPPLEMENTARYCLASSTHISAFTERNOON

NEXTSUPPLEMENTARYCLASS:

FRIDAY20.02AT2P.M.IN

ROOMSCA!!!

Elasticities

Theown-priceelasticity

demandfunction

Qd=1000−P3.

Have

dQ

d

dP=−3

P2

currentpriceis$5,priceraisedbyapound:

quantitydemanddecreasesapproximatelyby

dQd

dP|P=5=3×52=75tons.

one-percentincreaseintheprice:increaseby5p=

1 20×$1

reducequantitydemandedbyapproximately3.75=

75 20tons.

demandat$5is1000−53=875

percentagedecreaseinquantitydemandedis3.75875≈0.0043=0.43%.

Thus1%

increaseinpricereducesquantitydemandedby0.43%.

Demandisinelasticatthisprice.

21

Moregenerally:

dQd

dPapproximatechangeindemandwhenthepriceincreases

byapound.

initialpriceisP,

increasebyonepoundisanincreaseby

100

Ppercent.

anincreaseofthepriceby1%

changesthequantitydemanded

byapproximatelyby

P 100×

dQd

dPtons.

percentagechangeinquantitydemandedisapproximately:

ped(P)=100

Qd×

P 100×dQ

d

dP=dQ

d

dP×

P Qd

Thisistheown-priceelasticityofdemand.

rewritethisformulaas

ped(P)=dQ

d

Qd÷dP P

where100dP Pisthepercentageincreaseinpriceand100dQd

Qdis(approximately)

theinducedpercentagechangeinquantity.

22

Inourexample

ped(P)=¡ −3P

2¢ ×

P Qd=−3

P3

1000−P3

Whenisdemandinelastic?

3P3

1000−P3<1

or

3P3<1000−P3

4P3<1000

P3<250

P<

3√ 250≈6.3

Exactlywhen

P=

3√ 250thereisunitelasticityandabovedemandiselastic.

23

Totalrevenue

Withthisdemand,totalrevenueofthemarketis

TR=PQ

d=P¡ 1000

−P3¢

Totalrevenueismaximizedwhen

dTR

dP=1000−4P

3=0

orP=

3√ 250,i.e.,exactlywhenthereisunitelasticity.

Since

d2TR

dP2=−12P

2<0,totalrevenuedecreasestotheleftandincreasesto

therightofthisprice.

24

Inversedemand

Theconsumerswilldemandaquantity

Qwhentheprice

Pissuchthat

Q=

Qd(P)=1000−P3

P3=1000−Q

Pd=

3p 1000−Q

inversedemandfunction.

25

Marginalrevenue

useinversedemandfunctiontoexpresstotalrevenueasafunctionofquan-

tity:

TR=PQ=

3p 1000−Q×Q=(1000−Q)1 3×Q

quantitydemandedisdecreasinginprice.

totalrevenueisincreasinginpricewhenitdecreasinginquantityand

viceversa.

Marginalrevenueisthechangeinrevenueifasmallunitmoreofthecom-

modityissoldonthemarket.

MR=dTR

dQ=−1 3

(1000−Q)−

2 3×Q+(1000−Q)1 3

marginalrevenueiszerowhen

(1000−Q)−

2 3×Q=3(1000−Q)1 3

Q=3( 1000−Q)=3000−3Q

4Q=3000

Q=750

atP=

3√ 250.Forlowerquantitiesitispositiveandforhigheronesnegative.

26

Ingeneral,marginalrevenueandown-priceelasticityarerelatedby

MR=P

µ 1+1

ped(P)¶

Thisissobecause

dTR

dP=d¡ PQ

d¢

dP

=Q

d+PdQ

d

dP

whereasbythechainrule

dTR

dP=dTR

dQ

dQ

d

dP

andso

MR=dTR

dQ=

µ Qd+PdQ

d

dP

¶ÁdQ

d

dP=

Qd

dQd

dP

+P=

PdQd

dP

P Qd

+P

27

Elasticitiesandlogarithms

data(x,y)=(lnP,lnQ).

dy

dQ=

1 Q,P=ex

dP dx=ex=P.Thechainruleappliedtwiceyields

dy

dx=

dy

dQ

dQ

dP

dP dx=1 Q

dQ

dPP=ped(P)

28

Otherelasticities.

demandforacommodityfunctionofownprice,incomeandotherprices.

Qd=100−p+2p∗ −

3y

pisthepriceofthecommodity,

p∗thepriceofanothercommodity

yisincome.

ownpriceelasticity:

∂Q

d

∂p

p Qd=−

p Qd

crosspriceelasticity:

∂Q

d

∂p∗

p∗ Qd=2p∗ Qd

incomeelasticity

∂Q

d

∂y

y Qd=−3

y Qd

29