On Mon, Mar 23, 2015 at 9:45 PM, Scott Calvin
wrote:
> Hi Anatoly,
>
> The method Ifeffit uses to compute uncertainties in fitted parameters is
> independent of noise in the data because it, in essence, assumes the fit is
> statistically good and rescales accordingly. This means that the estimate
Hi Scott and Anatoly,
I didn't add any noise to the calculated data. So I can understand the
magnitude of chi-square and reduced chi-square are terribly large,
which is OK.
I see the small uncertainties in fitted parameters in the way that the
generated data is very consistent with the gu
Hi Anatoly,
The method Ifeffit uses to compute uncertainties in fitted parameters is
independent of noise in the data because it, in essence, assumes the fit is
statistically good and rescales accordingly. This means that the estimated
uncertainties really aren't dependable for fits that are kn
Hi Yuanyun and Bruce,
Your test is a very nice way to check the effect of the muffin tin radius on
the fit value of the amplitude fator.
Although I hadn't run your test myself I would have expected that the error
bars produced by the fit of a theory to "theoretically generated data" would
have
Hi Bruce,
Thanks for the comments.
Just to make an end to this topic so far: the samples which give large
amplitude are suspected to be more complicated than the normal
skutterudites I showned to you last fall. Normal skutterudites data
actually didn't cause me worried because its S02 is f
On 03/23/2015 04:10 PM, huyan...@physics.utoronto.ca wrote:
This is to follow up the *test* experiment you suggested. Attached are
three Artemis files.
I chose the Fe (im-3m) structure to do the test. The normal crystal
structure has its first-shell distance at 2.48 angstrom. A large
structure
Hi Anders,
Is there a reason you are trying to install from source? Have you tried to
install from macports?
George
On Fri, Mar 20, 2015 at 5:16 PM, Anders Glans wrote:
> Hi again,
>
> I have rebuilt things more carefully but I still have problems.
>
> - I built pgplot following the instructio
Hi Bruce,
Thanks for your comments and links of your lectures. I forgot to mention that I
use Artemis for fitting my data.
Best regards,
Jatin
-Original Message-
From: ifeffit-boun...@millenia.cars.aps.anl.gov
[mailto:ifeffit-boun...@millenia.cars.aps.anl.gov] On Behalf Of
ifeffit-req
Hi Scott,
Thanks for your explanation. It means the reverse can also be true, i.e., I can
guess N1 (nearest-neighbors in the first shell) and S02 by setting N2, N3 and
N4 to values known from other analysis. I did a quick check by fitting the data.
I conducted two fits:
1) setting S02 and guess
On 03/23/2015 08:54 AM, Rana, Jatinkumar Kantilal wrote:
The term N*S02 is fitted for each path of the FEFF calculation. So my
question is, even if we know N with a great certainty for some path,
how can we vary both N and S02 for other paths ? or Did I understand
it wrong ?
At no point in your
Hi Jatin,
The key is that S02 should be the same for all paths.
For example:
Suppose you are very confident path 1 has a coordination number of 6, because
of prior knowledge you have about the system. Paths 2, 3, and 4 have unknown
coordination numbers, however.
N1 (i.e. the degeneracy of pat
Hi Chris,
The term N*S02 is fitted for each path of the FEFF calculation. So my question
is, even if we know N with a great certainty for some path, how can we vary
both N and S02 for other paths ? or Did I understand it wrong ?
Best regards,
Jatin
-Original Message-
From: ifeffit-boun
I think Scott was pointing out that first neighbors may be known with high
certainty and therefore you can set this value thereby removing it and slightly
reducing the correlations.
Chris
Sent from my iPhone
> On Mar 23, 2015, at 7:00 AM, Rana, Jatinkumar Kantilal
> wrote:
>
> Hi Scott,
>
Hi Scott,
Thank you for your comments. Can you please elaborate a little bit more on this
"In cases like that, both N for all paths but one and S02 can be fit without
100% correlation."
Best regards,
Jatin
-Original Message-
From: ifeffit-boun...@millenia.cars.aps.anl.gov
[mailto:ifef
Hi Matt,
Thank you very much for your detailed explanation. As you pointed out that this
approach ignores the statistical significance of fits and assumes that all fits
are "good" fits. Also, the point that this approach yields a value of the
parameter which is only slightly less correlated wit
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