Post on 11-Jan-2016
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Mathe IIILecture 1Mathe IIILecture 1
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WS 2006/7
Avner Shaked
Mathe III Math III
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Mathematik IIIChong-Dae Kim
Donnerstag 9.15 – 10.45 Uhr – HS A Donnerstag 10.45 – 12.15 Uhr – HS A Donnerstag 12.15 – 13.45 Uhr – HS A Freitag 13.00 – 14.30 Uhr – HS G
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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
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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