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Estimating the error per point
fi 1  f  xi    m f '  xi  
fi  f  xi    i
fi 1  f  xi    p f '  xi  
 2m
f ''  xi    i 1
2
(1)

2
p
f ''  xi    i 1
2
Use the equation for fi to eliminate f(xi) from the other two
 2m
fi 1  fi   m f '  xi  
f ''  xi    i 1   i
2
(2)
 2p
fi 1  fi   p f '  xi  
f ''  xi    i 1   i
2
Multiply the top equation by p and the bottom by m and add the two equations
 m  2p   p  m2
 m  fi 1  fi    p  fi 1  fi  
f ''  xi    m  i 1   i    p  i 1   i  (3)
2
Assume that 2 f ''  xi    so that this term can be dropped. So that
m  fi 1  f i   p  fi 1  fi   m  i 1   i    p  i 1   i 
Square equation (4)
(4)
  m  fi 1  f i   p  fi 1  fi     m2   i21  2 i 1 i   i2   2 m  p  i 1 i 1   i  i 1   i 1    i2 
2
  2p   i21  2 i 1 i   i2 
(5)
Finally make a conceptual ensemble average over all possible values of  and note that
the cross terms are plus and minus averaging to zero while the squared terms are only +
so that
2
2
(6)
  mi  fi 1  fi    pi  fi 1  fi     mi
 i21   i2   2mi  pi  i2   2p  i21   i2 
Details on simulating errors are in SimErr.doc
A few details about this as an expectation value are in Expectation values.doc
Once this file has been constructed, this is just another curve fit problem. The normal
input of xi,fi,I fi,i2,3i2
On the first pass the error term is R,0,0,0. On the second pass it is given by the fit. The
poly is of order 3.
Minimization
Assume that
 i2  a  b fi  cfi 2 (7)
The constants a, b and c can be determined by minimizing
2/5
  f  f     f  f  2
pi
i 1
i 
i
  mi i 1
 2 2a  b f  f  c f 2  f 2
N 1 
 i 1 i   i 1 i 
 mi
2
  a, b, c    
2
i  2  2   a  b f  cf 
mi pi
i
i

 2p 2a  b  fi 1  fi   c  fi 21  fi 2 





2



 (8)





With respect to a, b, and c
Sort out the a, b, and c dependences
  f  f     f  f  2  a 2   2      2 
pi
i 1
i 
mi
pi mi
pi
i
  mi i 1

N 1


2
2
2
2
2
  a, b, c    b fi 1  mi  fi   mi  2 mi  pi   pi   fi 1  pi

i 2 

2
2
c fi 21  mi

 fi 2   mi
 2 mi  pi   2pi   fi 21 2pi






Define
Fdi    mi  fi 1  f i   pi  fi 1  fi  
2
ACi  2   mi
  pi  mi   2pi 
2
(10)
(11)
BCi  fi 1  2mi  fi   2mi  2 mi  pi   2pi   fi 1  2pi
(12)
2
2
CCi  fi 21  mi
 fi 2   mi
  mi  pi   2pi   fi 21 2pi
(13)
So that
N 1
2
 2  a, b, c     FC 2i  aACi  bBCi  cCCi  (14)
i 2
The partials with respect to a, b, and c are
 2  a, b, c  N 1
   Fdi  aACi  bBCi  cCCi ACi
a
i 2
 2  a, b, c 
b
N 1
   Fdi  aACi  bBCi  cCCi BCi
i 2
 2  a, b, c 
c
N 1
   Fdi  aACi  bBCi  cCCi CCi
i 2
Then define
N 1
FdA   Fdi ACi
i2
N 1
FdB   Fdi BCi
i 2
N 1
FdC   Fdi CCi
i 2
And also define
(16)
(15)
2
(9)
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N 1
AAS   ACi  ACi
i 2
N 1
ABS   ACi  BCi
i 2
N 1
ACS   ACi CCi
i 2
N 1
BAS   BCi  ACi  ABS
i2
N 1
BBS   BCi  BCi
i2
N 1
BCS   BCi  CCi
i 2
N 1
CAS   CCi ACi  ACS
i2
N 1
CBS   CCi BCi  BCS
i2
N 1
CCS   CCi CCi
(17)
i 2
So that setting the derivatives in (15) to zero the equations for a, b and c become
FdA  a  AAS  b  ABS  c  ACS
FdB  a  ABS  b  BBS  c  BCS
FdC  a  ACS  b  BCS  c  CCS
(18)
The code sminv from CHOLESKY.doc htm – or in nlfit\formp\robmin.for could be used
to solve (18), but it is also easily to manually do the eliminations. First note that since f’s
are traditionally between 0 and 100, the last terms dominate and can easily be rmoved
first. Thus begin by using the first line to eliminate c.
BCS
BCS 
BCS 


 a   ABS  AAS 
  b   BBS  ABS 

ACS
ACS 
ACS 


CCS
CCS 
CCS 


FdC  FdA 
 a   ACS  AAS 
  b   BCS  ABS 

ACS
ACS 
ACS 


FdB  FdA 
Define
BCS
ACS
CCS
FdChat  FdC  FdA 
ACS
And
FdBhat  FdB  FdA 
(20)
(19)
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BCS
ACS
CCS
ACShat  ACS  AAS 
ACS
And
BCS
BBShat  BBS  ABS 
ACS
CCS
BCShat  BCS  ABS 
ACS
So that (19) becomes
ABShat  ABS  AAS 
(21)
(22)
FdBhat  a  ABShat  b  BBShat
FdChat  a  ACShat  b  BCShat
Now eliminate b
FdBhat  a  ABShat  b  BBShat
BCShat
FdChat  FdBhat 
BBShat
a
BCShat
ACShat  ABShat 
BBShat
(23)
Then
FdChat  a  ACShat
BCShat
FdC  a  ACS  b  BCS
c
CCS
b
(24)
(25)
And a test equation
0  FdA  a  AAS  b  ABS  c  ACS
(26)
Simplification
The constant term only
Note that if only a is used to minimize 2 (9), that the solution is a=FdA/AAS
Linear only
If there is no c, the equations for minimizing 2 (18) become
FdA  a  AAS  b  ABS
(27)
FdB  a  ABS  b  BBS
FdB  FdA  BBS / ABS  a   ABS  AAS  BBS / ABS 
FdB  FdA  BBS / ABS
(29)
a
ABS  AAS  BBS / ABS
FdB  a  ABS
(30)
b
BBS
(28)
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B only
If there is only b - Poisson statistics err = (bfi) - , the equations for minimizing
 (9) become
2
FdB  b  BBS
b  FdB / BBS
c only
If there is only c - err = fi c ) - , the equations for minimizing 2 (9) become
FdC  c  CCS
c  FdC / CCS