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 <[email protected]> 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<[email protected]> 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 > >
