Transcript PowerPoint

BASIC IDEA
Using a first-order Taylor series expansion
f  x 
f  x1   f  x0  
 x1  x0 
x x  x
0
f  x 
f  x1   f  x0  
 x1  x0 
x x  x
0
f  x1   f  x0 
f  x 
x x  x
0
 x0  x1
GENERAL FUNCTION
•
Take a general function
f  x   exp  0.50 x 2  0.75 x  2 
•
Suppose we want to solve for
f  x   exp  0.50 x 2  0.75 x  2   1
GENERAL FUNCTION
12.0
10.0
8.0
6.0
4.0
2.0
0.0
0.0
0.5
1.0
1.5
2.0
2.5
3.0
3.5
4.0
4.5
FIRST ORDER SETUP
•
Applying the first-order Taylor series expansion
f  x   exp  0.50 x 2  0.75 x  2   1
f  x 
  1.00 x  0.75 exp  0.50 x 2  0.75 x  2 
x
•
To find the solution we substitute for the desired value of the function
1 f  x1   f  2 
f  x 
x x  2
1  4.4817
 2  x1 
 2  2.6215
5.6021
ITERATIONS
X
2.0000
2.6215
2.8413
2.8844
2.8860
F(x)
4.4817
1.6989
1.0990
1.0034
1.0000
D f(x)
-5.60211
-3.17955
-2.29846
-2.14167
-2.13601
TWO EQUATIONS – TWO UNKNOWNS
•
Building a two equation system
f  x1 , x2   exp  0.50 x12  0.75 x1 x2  0.25 x22 
g  x1 , x2   exp  .75 x12  0.35 x1 x2  0.45 x22 
•
Extending the Taylor expansion
f  x1 , x2 
f  x1 , x2 
0
f  x , x   f  x1 , x2  
x1  x1  
x20  x2 


x1
x2
x x
x x
0
1
0
2
1
g  x10 , x20   g  x1 , x2  
1
2
2
g  x1 , x2 
g  x1 , x2 
0
0
x

x

x



1
1
2  x2 
x1
x2
x x
x x
1
1
2
2
MATRIX TAYLOR EXPANSION
•
In Matrix form
 f  x1 , x2 

 f  x10 , x20    f  x , x    x1
x1  x1
1
2



0
0
 g  x1 , x2    g  x1 , x2    g  x1 , x2 


 x
1
x1  x1


f  x1 , x2 

0

x2
x
 x1  
x2  x2   1

 0
g  x1 , x2 
  x2  x2  

x2
x2  x2 
JACOBIAN
•
Jacobian
 f  x 

x1

x  f  x  , g  x  
 g  x 

 x1
f  x  

x2 
g  x  

x2 
 1.00 x1  0.75 x2  f  x   0.75 x1  0.50 x2  f  x  



1.50
x

0.35
x
g
x
0.35
x

0.90
x
g
x

1
2  
1
2   

Iteration
 f  x 
  x10  x1    x
1


 x20  x2    g  x 

 
 x1
 f  x 

 x10   x1   x1
 0    
 x2   x2  g  x 

 x1
1
f  x  
0
0

x2    f  x1 , x2    f  x1 , x2   




g  x     g  x10 , x20    g  x1 , x2   


 
x2 
1
f  x  
0
0

x2    f  x1 , x2    f  x1 , x2   




g  x     g  x10 , x20    g  x1 , x2   


 
x2 
•
Suppose we want to find the solution for f(x) =0.95 and g(x) = 0.75, starting from
x1 = 0.50 and x2 = 0.50.
1
 x  1.0000   0.1250 0.1250    0.9500 1.0000    0.7146 




 





0.8086

0.4659

0.2224
0.7500
0.8086
0.3162
 
 
 
 

x  
0
1
0
2
1
 x10  0.7146  0.4282 0.3387    0.9500  0.8945   0.6464 




 0  





0.3162

0.6786

0.0233
0.7500
0.7055
0.3921
x
 
 
 
 

 2 
1
 x  0.6464  0.3327 0.2726    0.9500  0.9444   0.9499 




 





0.3921

0.6204

0.0944
0.7500
0.7454
0.7499
 
 
 
 

x  
0
1
0
2
PARAMETERS OF THE COBB-DOUGLAS
FUNCTION
•
Solving parameters


 
Y

Ax
1 x2

 

w
Ax
 
1
1 x2
Y  Ax1 x2    
x1
P
 
w
Ax
x
 2  1 2
x2
 P









•
X1=50, x2=65, P=3.00, w1=0.005, w2=0.004, Y=75.0



 75.0  A  50   65 



A  50   65 
 0.005
 3.00  
50



 0.004
A  50   65 


65
 3.00


 0
  
  0
 0 


