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 <[log in to unmask]> 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<[log in to unmask]> 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
>
>
|