The error estimate is from
2
  mi  fi 1  fi    pi  fi 1  fi     mi
 i21   i2   2mi  pi  i2   2p  i21   i2  (1)
2
This is supposed to average 62. A data set can be constructed in which this is 62 for
every value of i. Let
DO I=2,N-1,2
XM=(I-1)*delta
X=I*delta
F(I-1)=FUN(XM)-C
F(I)=FUN(X)+C
ENDDO
For this set of data, the left side of (1) is for
J=2,4,…

2  f ' j  2c    f ' j  2c 

2
 4c 2
J=3,5,…
2   f 'J  2c    f 'i  2c    4c 2
2
The equation for the error estimate has this average 62
Thus 16c2=62 or
c  3/ 8
sime.for – constant error – of 1
Figure 1 Data with half the points high and half low.
K:\public_html\Fittery\ErrorEstimation>ERREST
THE CODE ERREST IS DESIGNED TO FIT THE ERROR AS
ERR=SQRT(A+B*ABS(F)+C*F**2)
OPTIONS ARE
1 FIT A ONLY, B AND C ARE ZERO
2 FIT A AND B ONLY C = 0
3 FIT A, B AND C
4 FIT B ONLY
5 FIT C ONLY
ENTER THE DESIRED OPTION
3
ENTER THE NAME OF THE DATA FILE
TEST.OUT
A =
1.0000004064787330  limited only by machine precision
B = -9.1506823071757370D-007
C =
1.5539231141905710D-007
ERR= -1.6172424510685350D-011
OUTPUT WILL BE IN TEST.EST
Gauss random
Figure 2 Data with a Gaussian distribution of average 1
The value of any point in figure 2 is using (1) is
  mi  fi 1  fi    pi  f i 1  f i      mi   i 1   i    pi   i 1   i  
2
2
2
2
   mi 2   i 1   i   2 mi  pi   i 1   i   i 1   i    pi 2   i 1   i  


(2)
After discarding the cross terms which ensemble average to 0
2
  mi  fi 1  fi    pi  fi 1  fi    mi 2 i 12   mi 2 i 2  2 mi  pi  i 2   pi 2 i 12   pi 2 i 2 (3)
Over an ensemble average with a Gaussian distribution with width , Integral of a
Gaussian.doc, the odd terms integrate to zero and 2  2 so that
  mi  fi 1  fi    pi  f i 1  f i  
(4)
2
  mi 2 i 2   mi 2 i 12  2 mi  pi i 2   pi 2 i 12   pi 2 i 2
This for all delta’s equal 1 says that the average of the oscillating term squared is 62.
The average of the square of the term sumed in 2 is needed for the error estimate
2
2
  mi  fi 1  fi    pi  fi 1  f i      mi 2   i 1   i   2 mi  pi   i 1   i   i 1   i    pi 2   i 1   i  


4
  mi 4   i 1   i 



 4 mi 2  pi 2   i 1   i 2   i 1   i 2



  pi 4   i 1   i 4



 4 mi 2  mi  pi   i 1   i 2   i 1 i 1   i 1 i   i  i 1   i 2  


 2 2     2  2     2

mi
i 1
i
pi
i 1
i


 4 2               2

pi
mi pi
i 1
i
i 1
i
i 1
i


4
Holding only the even terms
  mi  fi 1  fi    pi  fi 1  fi    oddterms 
  mi 4   i 14  6 i 12  i 2   i 4 



 4 mi 2  pi 2   i 12  i 12   i 12  i 2   i 4 



  pi 4   i 14  6 i 12  i 2   i 4 



 4 mi 2  mi  pi   i 12  i 2  2 i 12  i 2   i 4 



 2 mi 2  pi 2   i 12  i 12   i 2  i 12   i 2  i 12   i 4  


 4 pi 2  mi  pi   i 12  i 2  2 i 12  i 2   i 4 



4
The squared terms average over a Gaussian distribution to 2 while the fourth powers
average to 24 Integral of a Gaussian.doc so that the ensemble average is
  mi  fi 1  fi    pi  fi 1  f i  
4
  mi 4  2  6  2   4 mi 2  pi 2 1  1  2 



  i 4   pi 4  2  6  2   4 mi 3  pi 1  2  2 



2
2
3
 2 mi  pi 1  1  1  2   4 pi  mi 1  2  2  
(5)
The distinction between i-1, i, and i+1 has been ignored in (5)
The error term is thus
2
  mi  fi 1  fi    pi  f i 1  f i  
4
   mi  f i 1  f i    pi  f i 1  f i  
2
2
   mi 4  2  6  2   4 mi 2  pi 2 1  1  2 





2
2
2
  i 4    pi 4  2  6  2   4 mi 3  pi 1  2  2 
  4   mi   mi  pi   pi  



  2 mi 2  pi 2 1  1  1  2   4 pi 3  mi 1  2  2  





4
4
2
2
 10   mi   pi   30 mi  pi 
  4  mi 4  2   mi 3  pi   mi  pi 3   3 mi 2  pi 2   pi 4
  i4  
  20   3    3  

mi
pi
pi
mi




  i 4 6   mi 4   pi 4   18 mi 2  pi 2  12   mi 3  pi   pi 3  mi 

2

(6)
With all delta’s equal to 1 this is 544. This says that 2 is calculated with an error of
(54/6) for each point. Or equivalently, each point predicts 2  3 2
K:\PUBLIC~1\Fittery\ERRORE~1>errest
THE CODE ERREST IS DESIGNED TO FIT THE ERROR AS
ERR=SQRT(A+B*ABS(F)+C*F**2)
OPTIONS ARE
1 FIT A ONLY, B AND C ARE ZERO
2 FIT A AND B ONLY C = 0
3 FIT A, B AND C
4 FIT B ONLY
5 FIT C ONLY
ENTER THE DESIRED OPTION
3
ENTER THE NAME OF THE DATA FILE
test.out
A =
1.7670219674060160
B =
-0.6021908354246110
C =
0.0842177393747761
ERR= -5.8207660913467410D-011
OUTPUT WILL BE IN test.EST




Figure 3 Black line connects individual estimates of 2. Blue line connects estimates based on
 2  a  bf  cf 2 .
The function f is f(x)+I and is shown in the figure above.
The curve should be at 13/(5000/3), where the last 3 reflects the fact that the
error estimates are not statistically independent.
K:\public_html\Fittery\ErrorEstimation>errest
THE CODE ERREST IS DESIGNED TO FIT THE ERROR AS
ERR=SQRT(A+B*ABS(F)+C*F**2)
OPTIONS ARE
1 FIT A ONLY, B AND C ARE ZERO
2 FIT A AND B ONLY C = 0
3 FIT A, B AND C
4 FIT B ONLY
5 FIT C ONLY
ENTER THE DESIRED OPTION
1
ENTER THE NAME OF THE DATA FILE
test.out
A =
0.9421409819383385 err = 0.058 compare 3/40 = 0.075
B =
0.0000000000000000
C =
0.0000000000000000
ERR=
0.0000000000000000
OUTPUT WILL BE IN test.EST
A simulation with a value of c =10-4 is in SIMEGRC.FOR
K:\PUBLIC~1\Fittery\ERRORE~1>errest
THE CODE ERREST IS DESIGNED TO FIT THE ERROR AS
ERR=SQRT(A+B*ABS(F)+C*F**2)
OPTIONS ARE
1 FIT A ONLY, B AND C ARE ZERO
2 FIT A AND B ONLY C = 0
3 FIT A, B AND C
4 FIT B ONLY
5 FIT C ONLY
ENTER THE DESIRED OPTION
5
ENTER THE NAME OF THE DATA FILE
test.out
A =
0.0000000000000000
B =
0.0000000000000000
C =
1.0816415929070580D-004
ERR=
0.0000000000000000
This illustrates that there is a problem as the error approaches or exceeds the size of the
function. At that point the value of the function is so poorly defined that it gets mixed
into the fit to the error.
K:\PUBLIC~1\Fittery\ERRORE~1>errest
THE CODE ERREST IS DESIGNED TO FIT THE ERROR AS
ERR=SQRT(A+B*ABS(F)+C*F**2)
OPTIONS ARE
1 FIT A ONLY, B AND C ARE ZERO
2 FIT A AND B ONLY C = 0
3 FIT A, B AND C
4 FIT B ONLY
5 FIT C ONLY
ENTER THE DESIRED OPTION
5
ENTER THE NAME OF THE DATA FILE
test.out
A =
0.0000000000000000
B =
0.0000000000000000
C =
0.0026304982451279  0.0016
ERR=
0.0000000000000000
OUTPUT WILL BE IN test.EST
K:\PUBLIC~1\Fittery\ERRORE~1>errest
THE CODE ERREST IS DESIGNED TO FIT THE ERROR AS
ERR=SQRT(A+B*ABS(F)+C*F**2)
OPTIONS ARE
1 FIT A ONLY, B AND C ARE ZERO
2 FIT A AND B ONLY C = 0
3 FIT A, B AND C
4 FIT B ONLY
5 FIT C ONLY
ENTER THE DESIRED OPTION
3
ENTER THE NAME OF THE DATA FILE
test.out
A = -2.0382455788840400D-004
B =
0.0005828916847791
C =
0.0024628186100235
ERR= -4.5474735088646410D-013
OUTPUT WILL BE IN test.EST