Fit three points to two coefficients
Use parameters c  c1 , c2  , an approximating function f A  c , h  , and three function values fi , hi
The value of the parameters for data set I is determined by minimizing
 f  c , hi   fi 
 c    A
 (1)
i
i 1 

2
3
2
The first derivatives with respect to c are
BK  c  
 2  c 
cK
3
 2
i 1
1

2
i
 f c, h   f 
A
i
f A  c , xi 
i
cK
(2)
The second derivatives are
AK , J  c  
 2  2  c0 
cK cJ
 2 f A  c , hi  
1  f A  c , hi  f A  c , hi 
 2 2 

  f A  c , hi   fi  

cJ
cJ ck 
i 1  i 
 cK
3
In general the second term in the bracket sums much smaller than the first and will be dropped.
Expand the first derivatives of 2, B1 and B2, about initial values c0.
BK  c   BK  c0     c j  c0, j  AK , J (4)
2
J 1
Set BK=0, multiply by AM1, K to find and sum over K to find
1
1
0   AMK
BK  c0     c j  c0, j   AMK
AK , J
2
2
2
K 1
J 1
K 1
1
0   AMK
BK  c0     c j  c0, j   M , J
2
2
K 1
J 1
(5)
2
1
cM  c0, M   AMK
BK  c0 
K 1
The standard deviation in a function F of the constants is
3  M F c  f   c  f  
 F  c  fi    2
 i   J i  2
  
 i    
 i





f

c

f
i 1
i

1
J

1
i
J
i




2
2
3
2
F
(6)
Errinfun.doc
(3)
cJ

fi
1B
  AJK
K
K
fi
  AJK 1
K
BK
fi
(7)
The coefficient of the partial of the second derivative array is exactly zero at the minimum of 2. The
partial of the second term with respect to fi is given by taking the partial of (2) with respect to fi.
BK
1 f  c , hi  
 2 2 A
 (8)
fi
cK
i
c  c
0
So that
cJ
1 f  c , hi 
 2 AJ , K 1 2 A
fi
cK
i
K
(9)
Insert (9) into (6) to find
 2 F 2 1 1 f A  c , xi   2
   
2 AJ , K
  (10)
 i  i cK  i
i 1  J 1 cI
K 1
2
3
2
F
Expand the sums
 2 F 2 1 1 f A  c , xi   2 F 2 1 1 f A  c , xi   2
   
2 AI , K 2
2 AJ , L 2
 
 i
cK
cL
i
i
i 1  I 1 cI
K 1
 J 1 cJ L1

3
2
F
(11)
Evaluate the sum on i first
2
 F2  2
I 1
K 1
J 1
L 1
F 1 F 1  3 2 1 f A  c , hi  f A  c , hi  
AI , K
AJ , L 2  i 4

cI
cJ
cK
cL
 i 1  i

(12)
The term inside the { } in (12) is very similar to the original AK J defined in(3). Assume that
 i  hi8
 i   i
(13)
With this assumption (1) at the minimum becomes
 f  c , hi   fi 
3 2 2
  cmin     A
3 (14)
 
i
3
i 1 

2
3
2
The (3 – 2)/3 corrects for fitting 3
data points to two constants, so that
 2   2  cmin 
M
 F2  2 2 
I 1
K 1
J 1
L 1
(15)
F 1 F 1
AI , K
AJ , L AKL
cI
cJ
(16)
The summon L yields δJK
  2
2
F
2
2
F
 c
I 1
K 1
J 1
I
AI,1K
F
 JK
cJ
F 1 F
 F2  2 2 
AI , K
cK
I 1 cI
(17)
2
K 1
For F=c1,
F
  I ,1 so that
cI
M
 c2  2 2   i ,1 AI,1K 1, K  2 2 A1,11
1
(18)
I 1
K 1
For three point Richardson extrapolation
Use εi = 1 for all 3 points
f A  h   c1  c2 h 6
f A  h 
c1
f A  h 
c2
2 f A  h
1
c12
h
0
2 f A  h
6
c2 2
0
(19)
2 f A  h
c2 c1
0
The lack of second derivatives allows starting from c = 0. In a single step(5), gives c. Equation (14) gives
the value of α. The parameter c1 is the extrapolated value of the integral. Inserting the values in (19)
into (2)
3
B1  c   2 fi
i 1
3
B2  c   2 h f
i 1
(20)
6
i i
Then inserting the values in(19) into (3)
3
A1,1  c   21
I 1
3
A1,2  c   2 hi6  A2,1 (21)AI
i 1
3
A2,2  c   2 hi12
i 1
 A B   D  B  AD  BC

 

C D  C A  CD  DC
det  A11 A22  A122
A111  A22 / det
A121   A12 / det  A211
A221  A11 / det
(23)
 AB  BA
CB  AD
(22)