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Mathe III Lecture 1 Mathe III Lecture 1

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Page 1: Mathe III Lecture 1 Mathe III Lecture 1. 2 WS 2005/6 Avner Shaked Mathe III Math III

Mathe IIILecture 1Mathe IIILecture 1

Page 2: Mathe III Lecture 1 Mathe III Lecture 1. 2 WS 2005/6 Avner Shaked Mathe III Math III

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WS 2005/6

Avner Shaked

Mathe III Math III

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Tutorien für Mathematik IIIIm WS 05/06

Tutor: ChongDae KIM

Mo. 11:00 Uhr - 12.30Uhr HS N. Mo. 12.30 Uhr - 14.00Uhr HS N.

Di. 9.30 Uhr - 11.00Uhr HS N.Di.13.30 Uhr - 15.00Uhr HS N.

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http://www.wiwi.uni-bonn.de/shaked/

Homepage addresswith PowerPoint Presentations:

http://www.wiwi.uni-bonn.de/shaked/

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Bibliography

• K. Sydsaeter, P.J. Hammond: Mathematics for Economic Analysis

Excellent,Comprehen

sive

• R. Sundaram: A First Course in Optimization Theory

• A. de la Fuente: Mathematical Methods and Model for Economists

• A. K. Dixit: Optimization in Economic Theory

Mathematical,covers less than Sydsaeter &

Hammond, more of dynamic programming

New, theoretical, good in dynamicsShort, concentrates on

Lagrange, Uncertainty & Dynamic Prog.

• A. C. Chiang: Elements of Dynamic Optimization

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Bibliography

• K. Sydsaeter, P.J. Hammond: Mathematics for Economic Analysis

• R. Sundaram: A First Course in Optimization Theory

• A. de la Fuente: Mathematical Methods and Model for Economists

• A. K. Dixit: Optimization in Economic Theory

• A. C. Chiang: Elements of Dynamic Optimization

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• Difference Equations

(Sydsaeter.& Hammond, Chapter 20, Old Edition)

• Differential Equations

(Sydsaeter.& Hammond, Chapter 21 Old Edition)

• Constrained Optimization

(Sydsaeter.& Hammond, Chapter 18)

• Uncertainty

(Dixit, Chapter 9)

• The Maximum Principle, Dynamic Programming

(Dixit, Chapters 10,11)• Calculus of Variations (Chiang, Part 2)

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Difference Equations

t t -1 t -2 t -kx = f(t, x , x , ..., x )

The state today is a function of the state yesterday

The state at time t is a function of the state at t-1

Or: The state at time t is a function of the states of the previous k periods: t-1, t-2, t-3…,t-k,

and possibly of the date t

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t t -1 t -2 t -kx = f(t, x , x , ..., x )The solution to the equation:

t = k, k + 1, k + 2, ..

is an infinite vector

0 1 2 k k+1 k+2(x , x , x , ...., x , x , x , ........., , )

satisfying the above equation for

t = k, k + 1, k + 2, ..

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Example:

t = 0,1,2, ...

t t -1x = 1+ (1+ r)x

Interest rate

saving

For a given x0:

1 0x = 1+ (1+ r)x

2 1x = 1+ (1+ r)x

01+ (= 1+ ( 1+1 r) r)x+12x = 1+ (1 r)x+

20= 1+ (1+ r)+ (1+ r) x

1 0x = 1+ (1+ r)x

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Example (cntd.):

t t -1x = 1+ (1+ r)x

2

01+ (1+ r)+ (1+ r)= 1+ (1+ xr)

23x = 1+ (1 r)x+

22 0x = 1+ (1+ r)+ (1+ r) x

2 30= 1+ (1+ r)+ (1+ r) + (1+ r) x

22 0x = 1+ (1+ r)+ (1+ r) x

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22 0x = 1+ (1+ r)+ (1+ r) x

2 33 0x = 1+ (1+ r)+ (1+ r) + (1+ r) x

2 3 44 0x = 1+ (1+ r)+ (1+ r) + (1+ r) + (1+ r) x

2 3 45

50

x = 1+ (1+ r)+ (1+ r) + (1+ r) + (1+ r)

+ (1+ r) x

2 t -1 tt 0x = 1+ (1+ r)+ (1+ r) + ...+ (1+ r) (1+ r) x

??

t……?

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60

nn

is an integer

60 60 60 = 60, = 30, = 20,

1 2 360 60

= 15, = 124 5

60

nn is an integer

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Mathematical Induction

A 1

n A n A n + 1

n A n

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Mathematical Induction

A 1

A 1 A 2

A 2 A 2 A 3

A 3

n A n

Etc. Etc. Etc.

Modus Ponens

(Abtrennregel)

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t t -1t x = 1+ (1+ r)x

0 1 t -1 tt 0t x = (1+ r) + (1+ r) + ...+ (1+ r) (1+ r) x

For t = 1

1

0+ 1+ r x

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t t -1t x = 1+ (1+ r)x

0 1 t -1 tt 0t x = (1+ r) + (1+ r) + ...+ (1+ r) (1+ r) x

if 0 n-1 nn 0x = (1+ r) + ...+ (1+ r) (1+ r) x

then 0 n n+1n+1 0x = (1+ r) + ...+ (1+ r) (1+ r) x

but n+1 nx = 1+ (1+ r)xhence

0 n-1 n

n+1 0x = 1+ (1+ r) (1+ r) + ...+ (1+ r) + (1+ r) x

?

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0 n-1 n

n+1 0x = 1+ (1+ r) (1+ r) + ...+ (1+ r) + (1+ r) x

0 n n+1n+1 0x = (1+ r) + (1+ r)+ ...+ (1+ r) + (1+ r) x

0 1 t -1 tt 0t x = (1+ r) + (1+ r) + ...+ (1+ r) (1+ r) x

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2 t -1 tt 0x = 1+ (1+ r)+ (1+ r) + ...+ (1+ r) (1+ r) x

€1 for 1 period

€1 for 2 periods

€1 for t-1 periods

€ x0 for

t periods

t

tt 0

1+ r - 1x = + (1+ r) x

r

t t -1x = 1+ (1+ r)x

The solution to the difference equation:

is:

Example (cntd.):

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First Order Difference Equations

t t -1x = f(t, x )

1 0t = 1, x = f(1, x )

2 1t = 2, x = f(2, x ) 0f 1= (2, , xf )1x

t 0x = f(t, f t - 1, f t - 2, f ......f 1, x )

0= f 3, f(2, f 1, x )

0= f(2, f 1, x )

3 2t = 3, x = f(3, x )2x

0= f(2, f 1, x )

etc. etc. etc.

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The difference equation xt =f(t , xt-1)has a unique solution with a given value x0 .

Theorem:

t 0x = f(t, f t - 1, f t - 2, f ......f 1, x )

i.e. For each value x0 there exists a unique vector , x1 , x2 , x3 ,……. satisfying the difference equation.

Existence & Uniqueness

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First Order Difference EquationsLinear Equations with Constant Coefficients

t t -1 tx = ax + brecall: t t -1x = 1+ (1+ r)x

1 0 1x = ax + b

2 1 2x = ax + b 2= a + b2

0 1 2= a x + ab + b3 2

3 0 1 2 3x = a x + a b + ab + b

t t -1 t -2 t -3 1 0t 0 1 2 3 t -1 tx = a x + a b + a b + a b + .....a b + a b

1x0 1ax + b 0a = 1

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t t -1 tx = ax + bt t -1 t -2 t -3 1 0

t 0 1 2 3 t -1 tx = a x + a b + a b + a b + .....a b + a b

t - jj

t

j=1

a btt 0x = a x +

t

t t - jt 0 j

j=1

x = a x + a b

t -1 t -2 t -3 1 01 2 3 t -1 ta b + a b + a b + .....a b + a b

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t t -1 tx = ax + b t

t t - jt 0 j

j=1

x = a x + a b

:assume tb b

t

t t - jt 0

j=1

x = a x + b a

tt 0

b bx = a x - +

1 - a 1 - a

t

t0

1 - aa x + b

1 - aa 1

When a = 1

t t

t - j

j=1 j=1

a 1 = t

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t tS = αY

t t t -1I = β Y -Y

t tS = I

National IncometY

Total SavingtS

Total InvestmenttI

Example: A Model of Growth

β > α > 0

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t tS = αY

t t t -1I = β Y -Y

t tS = I

t= αY

t t -1= β Y -Y

t t -1

βY = Y

β - α

t

t 0

βY = Y

β - α t

0= 1+ g Y

αg =

β - αt t -1

t -1

Y -Y=

Y

Proportional Growth Rate

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t t -1x = ax + b

tt 0

b bx = a x - +

1 - a 1 - a

Equilibrium & Stability

???

x* = ax* +bb

x* =1 - a

tt 0

b bx - = a x -

1 - a 1 - a

An EquilibriumA Stationary State

tt 0x - x* = a x - x*

tt 0

b b

1 - a 1 - ax - = a x -

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tt 0x - x* = a x - x*ta

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ta a0 1-1

t

1 < a

t2 1,2,4,8,16,32,64,128,256, .....

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ta

a0 1-1

t

a < -1

t-2 1, 2, 4, 8, 16, 32, 64, 128, 2+ - 56+ - + , .- + - + ....

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ta

a0 1-1

t

0 < a < 1

t1 1 1 1 1 1 1

1, , , , , , , .....2 2 4 8 16 32 64

0

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ta

a0 1-1

t

-1 < a < 0

t1 1 1 1 1 1 1

1, , , , , , , .....2 2 4 8 16 32 64

0

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tt 0x - x* = a x - x*

When the equation S a leis t ba < 1

x*

t

0x0

0 < a < 1

x* < x

tt

x x*

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tt 0x - x* = a x - x*

When the equation S a leis t ba < 1

x*

t

0x

tt

x x*

0

-1 < a < 0

x* < x

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tt 0x - x* = a x - x*

When the equation is n S a leot t ba > 1

x*

t

0x0

1 < a

x < x*0x < x*

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tt 0x - x* = a x - x*

When the equation is n S a leot t ba > 1

x*

t

0x0x

a < -1

< x*

a < -1

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tt 0x - x* = a x - x*

what if or a = 0 a = 1 ???

t = 1, 2 .t , .= ba x x*= 0

t 0x =a 1 x= + tb

,

t 0 0

t

t 0

x = x , 2x * -x

x - x*

= -1 x - x*

x* = b/2

a = -1