BEE1024 Mathematics for Economists - Exeterpeople.exeter.ac.uk/dgbalken/ME08/week5b.pdf · 2008-03-25 · Objectives The Exponential Function Logarithmic functions Applications Elasticities

ObjectivesThe Exponential Function

Logarithmic functionsApplicationsElasticities

BEE1024 Mathematics for EconomistsExponential and logarithmic functions, Elasticities

Juliette Stephenson and Amr (Miro) AlgarhiAuthor: Dieter Balkenborg

Department of Economics, University of Exeter

Week 5

Balkenborg Exponential and logarithmic functions, Elasticities



1 Objectives

2 The Exponential FunctionDe�nitionProperties of exponential functionsCompounded interests and the number eProperties of the natural exponential function

3 Logarithmic functionsDe�nitionDi¤erentiating the natural logarithmProperties of the natural logarithmDi¤erentiating general exponential and logarithmic functions

4 ApplicationsCompounded InterestsExponential decayThe logistic curve




5 ElasticitiesOwn-price elasticityElasticities and logarithmsOther Elasticities




Objectives

Exponential functions: describes growth processes withconstant growth ratepopulation growth, growth of GDP, in�ation etc...

logarithm: the exponent required to produce a given numberinverse function, transforms multiplication into addition:10a � 10b = 10a+bLogarithmic di¤erentiation

Elasticities




Objectives



Elasticities




Objectives



Elasticities




De�nitionProperties of exponential functionsCompounded interests and the number eProperties of the natural exponential function

Gliederung1 Objectives2 The Exponential Function









The Exponential Function

power: xy

base: x

index or exponent: y

power function like x2: vary x

exponential function 2y : vary y

admissible values for y : positive integers, integers, rationals,real numbers

problem: for general y the power xy can only be de�ned forpositive x






power: xy

base: x











power: xy

base: x











power: xy

base: x











power: xy

base: x











power: xy

base: x











power: xy

base: x










Power functions

0

2468

1012141618202224

4 2 2 4x

y = 2: f (x) = x2

0

1

2

3

4

4 2 2 4x

y = �2:f (x) = x�2 = 1

x 2





Power functions

0

0.20.40.60.8

11.21.41.61.8

22.2

1 2 3 4 5x

y = 12 : f (x) = x

12 =

px

0

1

2

3

4

5

1 2 3 4 5x

y = � 32 :

f (x) = x�32 = 1

xpx





Exponential Functions

approximate irrational index y by fraction mn :

xy := limmn!y

xmn .

0

2

4

6

8

2 1 1 2y

x = 3: g (y) = 3y

0

2

4

6

8

2 1 1 2y

x = 13 : g (y) =

� 13

�y= 3�y





Exponential Functions

0.51.5 2 2.5 3

x0.6

0.81y

1

2

3

4

z = xy x � 0Balkenborg Exponential and logarithmic functions, Elasticities













Properties of exponential functions

An exponential function ax :

is strictly convex and has strictly positive values;

is for a > 1 strictly increasing with limx!�∞ ax = 0 andlimx!∞ ax = +∞;is for 0 < a < 1: decreasing with limx!�∞ ax = +∞ andlimx!∞ ax = 0.

Calculational rules for generalized powers:

as+t = asat ast = (as )t (ab)s = asbs

but(as )t 6= a(s t )






An exponential function ax :is strictly convex and has strictly positive values;











is for a > 1 strictly increasing with limx!�∞ ax = 0 andlimx!∞ ax = +∞;

is for 0 < a < 1: decreasing with limx!�∞ ax = +∞ andlimx!∞ ax = 0.



























Compounded interests and the number e

Put P0 > 0 (the principal) in savings account

�xed nominal annual interests rate r > 0

Interests paid n times during the year

amount Pt in your savings account after t years:

formula for compounded interests

Pt = P0�1+ r

n

�ntrn interest paid per period

nt total number of interest payments.











Pt = P0�1+ r

n













Pt = P0�1+ r

n













Pt = P0�1+ r

n













Pt = P0�1+ r

n

�nt

rn interest paid per period












Pt = P0�1+ r

n













Pt = P0�1+ r

n








The (natural) exponential function:balance in account after one year if interests paid continuously:

exp (r) = limn!∞�1+ r

n

�n.

The table below shows the value of�1+ r

n

�n for various n and r :r = 5.4% r = 5.5% r = 100%

n = 4 1.055103375 1.056144809 2.44140625n = 12 1.055356752 1.05640786 2.61303529n = 364 1.055480375 1.056536225 2.714557303n = 8736 1.055484426 1.056540432 2.718126265n = 524 160 1.055484599 1.056540612 2.718279235n = 31 449 600 1.055484602 1.056540613 2.718281796n! +∞ 1.055484602 1.056540615 2.718281828






The Euler number e is de�ned as

e = exp (1) = limn!∞�1+ 1

n

�n.

The �natural exponential function�is indeed the exponentialfunction with base e:

exp (r) = er

�proof� for rational r :

exp (r) = limn!∞

�1+

rn

�n= lim

m!∞,n=rm

�1+

rn

�n= lim

m!∞

�1+

1m

�rm= lim

m!∞

��1+

1m

�m�r=

�limm!∞

�1+

1m

�m�r= er






formula for continuously compounded interests:

Pt = P0ert .














Properties of the natural exponential function

1 e0 = 1,2 e1 = e3 ex > 0 for all x4

d (ex )dx = ex

In particular, ex is strictly increasing and convex.instantaneous growth rate of a function y = f (x): dy

dx

.y

when x is increased by a exponential function has constantgrowth rate 1.

5

ea+b = eaeb (ea)b = eab .

In particular 1ex = e

�x (because e�x ex = e0 = 1).





Let fn (x) =�1+ x

n

�n.Intuition for property 3:

�1+ x

n

�is positive when x > 0 or when n

large compared to jx j . Then fn (x) > 0 and so hence ex > 0.Intuition for property 4:

dfndx

= n�1+

xn

�n�1 1n=�1+

xn

�n�1��1+

xn

�n= fn (x)

for n very large compared to jx j since 1+ xn is then very close to 1.





Theorem

There is one and only one function y = f (x) which satis�es the�initial condition� f (0) = 1 and the �di¤erential equation�

dydx= y

and this is the exponential function f (x) = exp x = ex .





Properties of the natural exponential function

Exponential versus polynomial growth: For any polynomialP (x)

limx!+∞

ex

P (x)= +∞





Intuition: Suppose P (x) = amxm + . . . has degree m.Approximate ex by

�1+ x

n

�n with n larger than m. Thenlim

x!+∞

ex

P (x)� lim

x!+∞

�1+ x

n

�nP (x)

= limx!+∞

� 1n

�nxn + . . .

amxm + . . .=

limx!+∞

Cxn�m = +∞





A quicker way to calculate ex : is to use the formula

ex = 1+ x +x2

2!+x3

3!+ . . .+

xn

n!+ . . .




De�nitionDi¤erentiating the natural logarithmProperties of the natural logarithmDi¤erentiating general exponential and logarithmic functions










Logarithmic functions

Let a > 0. The logarithmic function loga x to the base a is de�nedas the inverse of the exponential function ay

y = loga x , ay = x

For instance, 1000 = 103, so log10 1000 = 3;18 = 2

�3, solog2

� 18

�= �3.





The natural logarithm

natural logarithm function

y = ln (x), x = ey

3

2

1

0

2

3

4

3 2 1 1 2 3 4x














Di¤erentiating the natural logarithm

x = e ln x

(ln x)0 =1e ln x

=1x= x�1

because by the chain rule

1 =dxdx= e ln x � (ln x)0 = x � (ln x)0














Properties of the natural logarithm

1 ln (y) is only de�ned for strictly positive y > 0.

2d ln(y )dy = 1

y . In particular, ln (y) is strictly increasing andconcave.

3 ln (1) = 0, ln (e) = 1.4 limy!0 ln (y) = �∞, limy!+∞ ln (y) = +∞.5 ln (ab) = ln (a) + ln (b), ln

�ab�= b ln (a). In particular

ln� 1a

�= � ln (a).

6 Logarithmic differentiation: Combined with the chainrule one obtains the following useful formula where y = g (x)is any di¤erentiable function:

d ln(g (x ))dx = g 0(x )

g (x ) (1)






1 ln (y) is only de�ned for strictly positive y > 0.2

d ln(y )dy = 1




ln� 1a

�= � ln (a).


d ln(g (x ))dx = g 0(x )

g (x ) (1)







d ln(y )dy = 1


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

4 limy!0 ln (y) = �∞, limy!+∞ ln (y) = +∞.5 ln (ab) = ln (a) + ln (b), ln


ln� 1a

�= � ln (a).


d ln(g (x ))dx = g 0(x )

g (x ) (1)







d ln(y )dy = 1


3 ln (1) = 0, ln (e) = 1.4 limy!0 ln (y) = �∞, limy!+∞ ln (y) = +∞.

5 ln (ab) = ln (a) + ln (b), ln�ab�= b ln (a). In particular

ln� 1a

�= � ln (a).


d ln(g (x ))dx = g 0(x )

g (x ) (1)







d ln(y )dy = 1




ln� 1a

�= � ln (a).


d ln(g (x ))dx = g 0(x )

g (x ) (1)







d ln(y )dy = 1




ln� 1a

�= � ln (a).


d ln(g (x ))dx = g 0(x )

g (x ) (1)





Theorem

The natural logarithm y (x) = ln x is the unique solution to thedi¤erential equation

dydx=1x

which satis�es the initial condition y (1) = 0.














Di¤erentiating general exponential and logarithmicfunctions

ln (xy ) = y ln (x)general powers can be rewritten as

xy = ey ln(x ) = e index�ln(base).

Partial di¤erentiation yields

∂xy

∂y= ey ln(x ) ln (x) = xy ln (x)

∂xy

∂x= ey ln(x )

�y1x

�= yxy

1x= yxy�1.





Derivative of an exponential function y = ax is

daxdx = ln (a) a

x .

The instantaneous growth rate of y = ax is ln (a)The derivative of a power function y = xb is

dx bdx = bx

b�1

even if b is irrational.





x = loga (y), y = ax .

We have

y = ax , y = ex ln(a) , ln (y) = x ln (a), x =ln (y)ln (a)

so

loga y =ln(y )ln(a)

loga y has hence the derivative

d (loga(y ))dy = 1

ln(a)y




Compounded InterestsExponential decayThe logistic curve










Compounded Interests

Example: Suppose Bank A o¤ers the annual nominal interest raterA = 5.5% and pays interests monthly. Bank B o¤ers the annualnominal interest rate rB = 5.4% and pays interests daily. Whichbank o¤ers the better deal?Solution:

reff ,A =�1+

rA12

�12� 1 =

�1+

0.05512

�12� 1 = 5.64%

reff ,B =�1+

rB364

�364� 1 =

�1+

0.054364

�364� 1 = 5.55%

so Bank A o¤ers better deal.














Exponential decay

most radioactive substances decay exponentiallysample of initial size Q0 weights Q (t) = Q0e�kt at time t.k measures rate of decay, �half-life� of the radioactive substance:Example: Show that a radioactive substance that decaysaccording to the formula Q (t) = Q0e�kt has a half-life of t̄ = ln 2

k .Solution: �nd value t̄ for which Q (t̄) = 1

2Q0, that is

12Q0 = Q0e�k t̄ .

Divide by Q0 and take natural logarithm:

ln12= �kt̄.

Thus the half-life is

t̄ =� ln 12k

=ln 2k





Di¤erentiation

Example: Find the derivative of g (x) = xx .Solution: Using logarithmic di¤erentiation we obtain

g 0 (x)g (x)

=d (ln xx )dx

=d�ln ex ln(x )

�dx

=d (x ln (x))

dx

= 1� ln (x) + x � 1x= ln (x) + 1

g 0 (x) = (ln (x) + 1) xx .














The logistic curve

The graph of the function of the form

Q (t) =B

1+ Ae�Bkt

where A, B, k are positive constants, is called a logistic curve.describes growth processes when environmental factors impose a�braking� e¤ect on the rate of growth.Example: Show that the growth rate of the logistic curveQ (t) = 1

1+e�t is 1�Q (t).





Solution: We have

Q 0 (t) =�e�t (�1)(1+ e�t )2

=e�t

(1+ e�t )2

Q 0 (t)Q (t)

=e�t

1+ e�t

1�Q (t) =1+ e�t � 11+ e�t

=e�t

1+ e�t





Theorem

The logistic curve y (x) = 11+e�x is the unique solution to the

di¤erential equationdydx= y (1� y)

which satis�es the initial condition y (0) = 12 .





Example

Public health records indicate that t weeks after the outbreak of acertain form of in�uenza, approximately Q (t) = 20

1+19e�1.2t

thousand people had caught the disease.a) How many people had the disease when it broke out? Howmany had it two weeks later?b) At what time does the spread of the infection begin to decline?c) If the trend continues, approximately how many people willeventually contract the disease?





Solution: a) Since Q (0) = 201+19 = 1 it follows that 1000 people

initially had the disease. When t = 2

Q (2) =20

1+ 19e�2.4� 7.343

so about 7.343 thousand people had contracted the disease by thesecond week.b) in�ection point at t̄ � 2.5. For t < t̄ convex and so the numberof newly infected increasing. For t > t̄ concave and so the numberof newly infected is decreasing.





0

5

10

15

4 2 2 4 6 8t

c) limt!+∞ Q (t) = 20, so roughly 20000 people catch the diseaseon total.




Own-price elasticityElasticities and logarithmsOther Elasticities










ElasticitiesOwn-price elasticity

demand function

Qd = 1000� P3dQd

dP= �3P2

current price is £ 5, price raised by a pound:quantity demand decreases approximately bydQ ddP jP=5 = 3� 5

2 = 75 tons.

one-percent increase in the price: increase by 5p= 120 � $1

reduce quantity demanded by approximately 3.75 = 7520 tons.

demand at £ 5 is 1000� 53 = 875percentage decrease in quantity demanded is 3.75875 �0.0043 = 0.43%.Thus 1% increase in price reduces quantity demanded by 0.43%.Demand is inelastic at this price.Balkenborg Exponential and logarithmic functions, Elasticities




More generally: dQd

dP approximate change in demand when theprice increases by a pound.initial price is P,increase by one pound is an increase by 100

P percent.an increase of the price by 1% changes the quantity demandedby approximately by P

100 �dQ ddP tons.

percentage change in quantity demanded is approximately:

ped (P) =100Qd

� P100

� dQd

dP=dQd

dP� PQd

This is the own-price elasticity of demand.





rewrite this formula as

ped (P) =dQd

Qd� dPP

where 100dPP is the percentage increase in price and 100dQd

Q d is(approximately) the induced percentage change in quantity.In our example

ped (P) =��3P2

�� PQd

= �3 P3

1000� P3





When is demand inelastic?

3P3

1000� P3 < 1

or

3P3 < 1000�P3 4P3 < 1000 P3 < 250 P < 3p250 � 6.3

Exactly when P = 3p250 there is unit elasticity and above demand

is elastic.





Constant Elasticity

A function has constant elasticity ε if and only if it is of the form

Q = αP ε

We have

dQdP

= αεP ε�1

dQdP

PQ

= αεP ε�1 PαP ε

= ε





Total Revenue

With this demand, total revenue of the market is

TR = PQd = P�1000� P3

�Total revenue is maximized when

dTRdP

= 1000� 4P3 = 0

or P = 3p250, i.e., exactly when there is unit elasticity.

Since d 2TRdP 2 = �12P

2 < 0, total revenue decreases to the left andincreases to the right of this price.





Inverse Demand

The consumers will demand a quantity Q when the price P is suchthat

Q = Qd (P) = 1000� P3

P3 = 1000�QPd = 3

p1000�Q

inverse demand function





Marginal Revenue

use inverse demand function to express total revenue as a functionof quantity:

TR = PQ = 3p1000�Q �Q = (1000�Q)

13 �Q

quantity demanded is decreasing in price.total revenue is increasing in price when it decreasing in quantityand vice versa.

Marginal revenue is the change in revenue if a small unit more ofthe commodity is sold on the market.

MR =dTRdQ

= �13(1000�Q)�

23 �Q + (1000�Q)

13





Marginal revenue is zero when

(1000�Q)�23 �Q = 3 (1000�Q)

13

Q = 3 (1000�Q) = 3000� 3Q4Q = 3000 Q = 750

at P = 3p250. For lower quantities it is positive and for higher

ones negative.





TheoremMarginal revenue and own-price elasticity are related by

MR = P�1+

1ped (P)

�





This is so because

dTRdP

=d�PQd

�dP

= Qd + PdQd

dP

whereas by the chain rule

dTRdP

=dTRdQ

dQd

dP

and so

MR =dTRdQ

=

�Qd + P

dQd

dP

��dQd

dP=Qd

dQ ddP

+P =P

dQ ddP

PQ d

+P














Elasticities and logarithms

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

dydx= ped (P)

dydQ =

1Q , P = e

x dPdx = e

x = P.The chain rule applied twice yields

dydx=dydQ

dQdP

dPdx=1QdQdPP = ped (P)














Other Elasticities

demand for a commodity function of own price, income and otherprices.

Qd = 100� p + 2p� � 3yp is the price of the commodity,p� the price of another commodityy is income.

own price elasticity:∂Qd

∂ppQd

= � pQd





Other Elasticities

demand for a commodity function of own price, income and otherprices.

Qd = 100� p + 2p� � 3ycross price elasticity:

∂Qd

∂p�p�

Qd= 2

p�

Qd

income elasticity∂Qd

∂yyQd

= �3 yQd


Documents

BEE1024 Mathematics for Economists - Exeterpeople.exeter.ac.uk/dgbalken/ME08/week5b.pdf · 2008-03-25 · Objectives The Exponential Function Logarithmic functions Applications Elasticities