Error Analysis
by Monica Bobra
In this document, I derive analytical functions for the error in each active region parameter using
formal error propagation. It is important to note that the error formulae below assume that (i) all
the variables are linearly independent, and (ii) the relative errors are small. And neither of those
assumptions are true. But we are making them. Moving on to the derivations:
1
Φ=
X
|Bzi |dA
i
Φ is a sum of all pixels; therefore, the error in Φ is simply: (δΦ)2 =
X
(δBzi )2 dA2 .
i
2
γ=
1
N
X
i
Bhi
arctan
Bzi
The error in γ per pixel, (i, j), is as follows:
2 2
2
∂γ
∂γ
(δγ) = ∂Bh δBh + ∂Bz δBz .
The partial derivatives per pixel, (i, j), are as follows:
∂γ
∂γ
= B1 2 and ∂B
= − BBh2 .
∂Bh
z
h
Bz
1+
1+
Bz
Bz 2
h
Bz
Therefore, the error in γ per pixel, (i, j), is simply:
!2
!2
2
1
(δγ) =
2 δBh
B
1+ Bh
Bz
+
z
Bh δBz
2
B
1+ Bh
Bz 2
=
!2 1
2
B
1+ Bh
δBh
Bz
2
+
Bh δBz
Bz 2
2 .
z
z
We report
X the mean value of γ, which means we must also determine the error in γ:
1
γ=N
γi , where N represents the number of pixels. Therefore, the error in γ is:
i
2
(δγ) =
1
N2
X
(δγi )2 .
i
Additionally, the δBh array must be determined. We can calculate δBh per pixel, (i, j), as follows:
(δBh )2 =
3
(Bx δBx )2 +(By δBy )2
.
Bh 2
|∇Bt| =
1
N
s
X
i
∂Bti
∂x
2
+
∂Bti
∂y
2
There are two approaches one can take to calculate δ |∇Bt |: calculate the error terms assuming (i)
t
t
an analytical form for the derivative, e.g. ∂B
, or (ii) a numerical form, e.g. ∂B
= 0.5*(Bt[i,j+1]
∂y
∂y
1
- Bt[i,j-1]). I am going to go with the numerical form.
Using a simple finite difference method, the derivatives are defined as follows:
∂Bt
t
=0.5*(Bx[i,j+1] - Bx[i,j-1]) and ∂B
=0.5*(Bt[i+1,j] - Bt[i-1,j]).
∂y
∂x
Therefore, theqgradient at one position (i, j) is defined as follows:
|∇Bt[i, j]| = 0.25 ∗ (Bt[i + 1, j] − Bt[i − 1, j])2 + 0.25 ∗ (Bt[i, j + 1] − Bt[i, j − 1])2
The error in in |∇Bt | is as follows:
2 2
∂∇Bt
∂∇Bt
(δ |∇Bt[i, j]|) =
δBt[i, j + 1] +
δBt[i, j − 1] +
∂Bt[i, j + 1]
∂Bt[i, j − 1]
2 2
∂∇Bt
∂∇Bt
δBt[i + 1, j] +
δBt[i − 1, j] .
∂Bt[i + 1, j]
∂Bt[i − 1, j]
2
(1)
Let’s evaluate the first term of the equation above (we’ll define this as T ERM 1):
2
2 ∂|∇Bt |
Bt[i,j+1]−Bt[i,j−1]
δBx[i, j + 1] .
δBx[i, j + 1] = √
2
2
∂Bx[i,j+1]
4
0.25∗(Bt[i+1,j]−Bt[i−1,j]) +0.25∗(Bt[i,j+1]−Bt[i,j−1])
And here are the subsequent
terms. T ERM 2:
2 ∂|∇Bt |
−Bt[i,j+1]+Bt[i,j−1]
δBx[i, j − 1] = √
2
∂Bx[i,j−1]
4
T ERM 3:
∂|∇Bt |
δBx[i
∂Bx[i+1,j]
T ERM 4:
∂|∇Bt |
δBx[i
∂Bx[i−1,j]
+ 1, j]
2
− 1, j]
2
=
=
0.25∗(Bt[i+1,j]−Bt[i−1,j]) +0.25∗(Bt[i,j+1]−Bt[i,j−1])
2
√
Bt[i+1,j]−Bt[i−1,j]
√
−Bt[i+1,j]+Bt[i−1,j]
0.25∗(Bt[i+1,j]−Bt[i−1,j])2 +0.25∗(Bt[i,j+1]−Bt[i,j−1])2
4
2
δBx[i, j − 1] .
2
4
δBx[i + 1, j]
.
2
δBx[i − 1, j] .
2
0.25∗(Bt[i+1,j]−Bt[i−1,j])2 +0.25∗(Bt[i,j+1]−Bt[i,j−1])
Therefore, the error in |∇Bt | is as follows: (δ |∇Bt |)2 = T ERM 1 + T ERM 2 + T ERM 3 + T ERM 4.
We report the mean value of |∇Bt |, which means we must also determine the error in |∇Bt |:
X
|∇Bt | = N1
∇Bti , where N represents the number of pixels. Therefore, the error in |∇Bt | is:
2 i
δ|∇Bt | =
1
N2
X
(δ∇Bti )2 .
i
The δBt array must be determined. We can calculate δBt per pixel, (i, j), as follows:
(δBt )2 =
(Bx δBx )2 +(By δBy )+(Bz δBz )2
.
Bt 2
The errors in the quantities below follow similarly:
s
X ∂Bh 2 ∂Bh 2
i
i
|∇Bh | = N1
+
and |∇Bz | =
∂x
∂y
i
2
1
N
X
i
s
∂Bzi
∂x
2
+
∂Bzi
∂y
2
.
4
Jz =
X
1
N
Jzi
i
The error in Jz is as follows: δJz
2
=
1
N2
X
(δJzi )2 .
i
There are two approaches one can take to calculate δJz : calculate the error terms assuming (i)
y
x
− ∂B
, or (ii) a numerical form for the curl, e.g.
an analytical form for the curl, i.e. Jz = ∂B
∂y
∂x
0.5*(Bx[i,j+1] - Bx[i,j-1]) - 0.5*(By[i+1,j] - By[i-1,j]). I am going to go with the
numerical form.
Using a simple finite difference method, the derivatives are defined as follows:
y
∂Bx
=0.5*(Bx[i,j+1] - Bx[i,j-1]) and ∂B
=0.5*(By[i+1,j] - By[i-1,j]).
∂y
∂x
Thus, Jz[i,j] = 0.5*(Bx[i,j+1] - Bx[i,j-1] - By[i+1,j] + By[i-1,j]).
Therefore, the error in in Jz is as follows:
2
2
2
2
1
.
(δJz[i, j])2 = 2(∆x)
2 (δBx[i, j + 1]) + (δBx[i, j − 1]) + (δBx[i + 1, j]) + (δBx[i − 1, j])
5
Jztotal =
X
CJzi
i
The error in Jztotal is (δJztotal )2 =
X
C 2 (δJzi )2 .
i
X
6
αtotal
Jzi Bzi
i
=C X
Bz2i
i
The general formula for the twist parameter, α, is simply α = C BJzz , where C is a constant. We’ve
calculated a modified version from Hagino et al., which I’m explicitly calling αtotal because we are
dividing sums by sums; we are not strictly taking a mean, i.e. a moment of a distribution.
First, for purposes of simplicity, let’s assume that there are only two pixels and αtotal is reduced to:
αsimplif ied = C
Jz1 Bz1 +Jz2 Bz2
.
Bz21 +Bz22
The error in in αsimplif ied is as follows:
2 2 2 2
∂αsimplif ied
∂αsimplif ied
∂αsimplif ied
∂αsimplif ied
(δαsimplif ied )2 =
δJ
+
δJ
+
δB
+
δB
.
z1
z2
z1
z2
∂Jz
∂Jz
∂Bz
∂Bz
1
2
1
The partial derivatives are as follows:
Bz1
∂αsimplif ied
= C B 2 +B
2 , and similarly for Jz2 .
∂Jz
1
z1
z2
3
2
∂αsimplif ied
∂Bz1
=C
Jz1
Bz21 +Bz22
−
2Bz1 (Jz1 Bz1 +Jz2 Bz2 )
2
(Bz21 +Bz22 )
, and similarly for Bz2 .
Therefore, the error in αtotal is simply a generalization of the error in αsimplif ied :
2
 
!
2

X

 
2Bzi
Jzi Bzi 


 J
X
X

B




z
z
i
2
i
i


(δαtotal ) =
 δBzi  .
C  X 2 −
!2
C X 2 δJzi  +


 
X
Bzi
Bzi
i
i  


2
B
zi
i
i
i
7
Hc =
1
N
X
CBzi Jzi
i
The error in in Hc for one pixel, (i, j), is as follows:
2 2
∂Hc
c
δB
+
δJ
.
(δHc )2 = ∂H
z
z
∂Bz
∂Jz
(δHc )2 = (CJz δBz )2 + (CBz δJz )2 .
We report theX
mean value of Hc , which means we must also determine the error in Hc :
2
1
δHc = N 2
(δHci )2 .
i
X
X
We also compute two similar quantities: Hctotal =
|Bzi Jzi | and Hcabs = Bzi Jzi . The same
i
i
error formula applies to both of these quantities, namely:
X
(δHctotal )2 = (δHcabs )2 =
(δHci )2 .
i
8
Jzsummed
+
−
Bz
Bz
X X = C Jzi + C Jzi Because errors are always positive, the error in Jzsummed is the same as the error in
X
Therefore, the error in Jzsummed is simply: (Jzsummed )2 =
C 2 (δJzi )2 .
X
CJzi .
i
i
9
ρ=
1
N
X
C
~o
B
i
−
~p
B
i
2
i
Since the z-component of the observed field is used to generate the potential field, Bzo = Bzp and
2
ρ at one pixel, (i, j), is reduced to: ρ = C (Bxo − Bxp )2 + C Byo − Byp . We assume no error in
the potential components of the field; thus our value of δρ represents a lower bound. Another
4
~ o = δB
~ p , which means that our
approach, which we do not take, is to assume that the order of δ B
value of δρ would include some multiplicative factor. The error in ρ at one pixel, (i, j), is reduced to:
2 2
∂ρ
∂ρ
o
o
(δρ)2 = ∂B
δB
δB
+
.
o
x
y
∂B o
x
y
2
(δρ)2 = [2C (Bxo − Bxp ) δBxo ]2 + 2C Byo − Byp δByo .
We report the
of ρ, which means we must also determine the error in ρ:
X mean value
2
2
2
1
(δρ) = N 2
C (δρi ) .
i
10
ρtotal =
X
~ io − B
~p
B
i
2
dA2
i
Following from the section above, the error in ρtotal is:
X
2 X 2
(δρtotal )2 =
2 Bxoi − Bxpi δBxoi dA +
2 Byoi − Bypi δByoi dA .
i
11
χ=
i
1
N
X
arccos
i
~p
~o · B
B
i
i
o
|Bi | |Bip|
χ at one pixel, (i, j): χ = arccos
√
!
Bxo Bxp +Byo Byp +Bzo Bzp
Bxo 2 +Byo 2 +Bzo 2
√
2
2
.
p2
Bxp +Byp +Bz
Since δBxp = δByp = δBzp = 0, the error in χ is as follows:
2 2 2
∂ρ
∂ρ
∂ρ
o
o
o
(δχ)2 = ∂B
+
+
.
δB
δB
δB
o
x
y
z
∂B o
∂B o
x
y
z
Let’s compute each term:
T ERM 1 ≡
T ERM 2 ≡
T ERM 3 ≡
∂ρ
δBxo
∂Bxo
∂ρ
δByo
∂Byo
∂ρ
δBzo
∂Bxo
=
Bxo Byo Byp −Byo 2 Bxp +Bzo (Bxo Bzp −Bzo Bxp )
√ p2 p2 p2s
3/2
(B o B p +B o B p +B o B p )2
Bx +By +Bz
1− o 2 ox2 x o 2 y y p 2 z pz2
(Bxo 2 +Byo 2 +Bzo 2 )
(Bx +By +Bz )(Bx
(Bzo Byp −Byo Bzp )
s
+By +B p z 2
δBxo .
)
Bxo 2 Byp −Bxo Byo Bxp +Bzo
=−
(
Bxo 2 +Byo 2 +Bzo 2
3/2
)
√
2
2
2
Bxp +Byp +Bzp
1−
Bxo 2 Bzp −Bxo Bzo Bxp +Byo
=−
(
Bxo 2 +Byo 2 +Bzo 2
3/2
)
√
(Bxo 2 +Byo 2 +Bzo 2 )(Bxp 2 +Byp 2 +Bp z2 )
(Byo Bzp −Bzo Byp )
s
2
2
2
Bxp +Byp +Bzp
(Bxo Bxp +Byo Byp +Bzo Bzp )2
1−
(Bxo Bxp +Byo Byp +Bzo Bzp )2
δByo .
δBzo .
(Bxo 2 +Byo 2 +Bzo 2 )(Bxp 2 +Byp 2 +Bp z2 )
Therefore, (δχ)2 = T ERM 12 +T ERM 22 +T ERM 32 . We report the mean value of χ, which means
we must also determine the error in χ:
X
(δχ)2 = N12
(δχi )2 .
i
5