Christoph Florian Schaller - FU Berlin, FB MI

Christoph Schaller - STORMicroscopy 9
Once more we can limit the tting to the one-dimensional case as all occuring distributions are
rotationally symmetric. Hence we use C i = ∑ j S ij as the observation data and t with G i =
N ´ i+ 1 2
p
i− 1 1D (x)dx, where p 1D denotes the one dimensional Gaussian distribution centered in x 0 with
2
standard deviation σ, i.e. p 1D (x) = √ 1
2πσ
exp(− (x−x0)2
2σ
). 2
We repeat the least squares approach
0 = d
dx 0
( ∑ i
(C i − G i ) 2 )
⇐⇒ 0 = ∑ 2(C i − G i ) d
dx 0
G i ,
d
but now requiring to calculate
dx 0
G i = N d p
i− 1 1D (x)dx.
2
Luckily we can switch integral and dierentiation here as [i− 1 2 , i+ 1 2 ] is nite, p 1D(x) is continuous
and
d
dx 0
p 1D (x) exists and is continuous, too. Therefore we obtain
d
dx 0
G i = N ´ i+ 1 2 (x−x 0)
i− 1 σ
p 2 1D (x)dx
2
and thus 0 = ∑ (C i − G i ) ´ i+ 1 2
(x − x
i− 1 0 )p 1D (x)dx.
2
Knowing that e(x) = 1 x−x0
2erf( For simplicity we denote e i± = e(i ± 1 2
dx 0
´ i+ 1
2
σ √ ) satises 2 e′ (x) = p 1D (x), plugging in yields G i = N ( e(i + 1 2 ) − e(i − 1 2 )) .
). Furthermore we can integrate
ˆ i+ 1
2
i− 1 2
For our least squares problem follows
(x − x 0 )p 1D (x)dx = [ −σ 2 p 1D (x) ] i+ 1 2
.
i− 1 2
0 = ∑ (C i − Ne i+ + Ne i− )σ 2 [p 1D (i + 1 2 ) − p 1D(i − 1 2 )]
⇐⇒ 0 = ∑ (C i − Ne i+ + Ne i− )
(exp(− (i + 1 2 − x 0) 2
2σ 2 ) − exp(− (i − 1 2 − x 0) 2 )
2σ 2 ) .
} {{ }
=:f(x 0)
Now we just need to solve this nonlinear equation. If we do not want to approximate, an application
of Newton's method for f(x 0 ) started in the pixel center of the local maximum should suce. In order
to apply the iteration x n+1 = x n − f(xn) we need to know f ′ (x f ′ n)
(x 0 ), too. As e i± = e(i ± 1 2
) depends on
x 0 , the product rule yields
f ′ (x 0 ) = ∑ (Np 1D (i + 1 2 ) − Np 1D(i − 1 (
2 )) exp(− (i + 1 2 − x 0) 2
2σ 2 ) − exp(− (i − 1 2 − x 0) 2 )
2σ 2 )
+ ∑ ( i +
1
2
(C i − Ne i+ + Ne i− )
− x 0
σ 2 exp(− (i + 1 2 − x 0) 2
2σ 2 ) − i − 1 2 − x 0
σ 2 exp(− (i − 1 2 − x 0) 2 )
2σ 2 ) .
By denoting
and generalizing
p i± (x n ) := exp(− (i ± 1 2 − x n) 2
2σ 2 ) = √ 2πσp 1D (i ± 1 2 )
e i± (x n ) := 1 2 erf(i ± 1 2 − x n
σ √ )
2