Thanks Ian, that makes it clear indeed!
Pavel.
On 7/30/10 2:12 AM, Ian Tickle wrote:
Hi Pavel
The expressions Fo+(Fo-Fc) and Fc+2(Fo-Fc) are obviously equal so it
makes no difference whether you look at it as [half from Fo + half
from (Fo-Fc)] or [2 times half from (Fo-Fc)]. But that wasn't the
point I was making: I was saying that I thought the relationship
between the Fourier and difference map coefficients is clearer if you
write the former as Fc+2(Fo-Fc) (well it helped me anyway!).
Cheers
-- Ian
On Thu, Jul 29, 2010 at 11:55 PM, Pavel Afonine<pafon...@lbl.gov> wrote:
Hi Ian,
please correct me if I'm wrong in what I'm writing below...
My reasoning for writing it like this
2Fo-Fc = Fo + (Fo-Fc)
is:
1) the map (Fo, Pcalc) shows density for missing atoms at half size
(approximately)
2) the map (Fo-Fc, Pcalc) shows density for missing atoms at half size
(approximately)
3) then the map (2Fo-Fc, Pcalc) shows density for missing atoms at full size
(approximately), and this is why this map is preferred over (Fo, Pcalc).
And maximum-likelihood weighted map 2mFo-DFc is even better since in
addition it is expected to be less model biased.
This was my "rationale" to write 2Fo-Fc = Fo + (Fo-Fc) and not Fc + 2(Fo-Fc)
.
Pavel.
On 7/29/10 2:38 PM, Ian Tickle wrote:
On Thu, Jul 29, 2010 at 8:25 PM, Pavel Afonine<pafon...@lbl.gov> wrote:
Speaking of 3fo2fc or 5fo3fc, ... etc maps (see classic works on this
published 30+ years ago), I guess the main rationale for using them in
those
cases arises from the facts that
2Fo-Fc = Fo + (Fo-Fc),
3Fo-2Fc = Fo +2(Fo-Fc)
To be precise, it is actually
2mFo-DFc for acentric reflections
and mFo for centric reflections
I prefer to think of it rather as
2mFo - DFc = DFc + 2(mFo-DFc) for acentrics and
mFo = DFc + (mFo-DFc) for centrics.
Then it also becomes clear that to be consistent the corresponding
difference map coefficients should be 2(mFo-DFc) for acentrics and
(mFo-DFc) for centrics.
Cheers
-- Ian
