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| sean |
 Posted: Tue Feb 22, 2005 4:03 am |
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Could anyone here on this newsgroup help solve this disagreement
over what the results of these following calculations will be..
A) Where z=0 and s=1.1 in the formula w=s(1+z), does w = 1.21?
B) Where z=0 and s=1.1 in the formula w=s(1+z), does it follow
that s=1.21?
C) In this chi square fitting formula below ....
I(t) = Imax * ( fR((t-tmax)/s(1+z)) + b )
where z=0 and s is unknown, is it possible to calculate that s=1.21
just by looking at the formula above ? Or to calculate s does one have
to perform the chi squared fitting function?
thanks ahead |
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| Joseph Fagan |
| Posted: Tue Feb 22, 2005 4:03 am |
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"sean" <jaymoseley@hotmail.com> wrote in message
news:40f80aa1.0502220103.43b311af@posting.google.com...
[quote:a1ccb2b708]Could anyone here on this newsgroup help solve this disagreement
over what the results of these following calculations will be..
A) Where z=0 and s=1.1 in the formula w=s(1+z), does w = 1.21?
Typo? When z=0, w=s.[/quote:a1ccb2b708]
[quote:a1ccb2b708]
B) Where z=0 and s=1.1 in the formula w=s(1+z), does it follow
that s=1.21?
When z=0, w=s.
C) In this chi square fitting formula below ....
I(t) = Imax * ( fR((t-tmax)/s(1+z)) + b )
where z=0 and s is unknown, is it possible to calculate that s=1.21
just by looking at the formula above ? Or to calculate s does one have
to perform the chi squared fitting function?
Can't find s without knowing I(), Imax, t, tmax, f, R and b.[/quote:a1ccb2b708]
[quote:a1ccb2b708]
thanks ahead[/quote:a1ccb2b708] |
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| Bjoern Feuerbacher |
| Posted: Tue Feb 22, 2005 6:44 am |
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sean wrote:
[quote:1b346b111b]Could anyone here on this newsgroup help solve this disagreement
over what the results of these following calculations will be..
A) Where z=0 and s=1.1 in the formula w=s(1+z), does w = 1.21?
B) Where z=0 and s=1.1 in the formula w=s(1+z), does it follow
that s=1.21?
C) In this chi square fitting formula below ....
I(t) = Imax * ( fR((t-tmax)/s(1+z)) + b )
where z=0 and s is unknown, is it possible to calculate that s=1.21
just by looking at the formula above ? Or to calculate s does one have
to perform the chi squared fitting function?
[/quote:1b346b111b]
Oh my goodness. To all who want to reply to these questions: please note
that Sean has completely scrambled up the actual problem at hand
(for all interested in what it is about: see my discussions with sean
in sci.astro.research and sci.astro)
The *real* problem is the following:
Let's say we have a set of data points, x-axis is time (t), y-axis is
intensity (I). Let's also say we have a function fR(t) which should,
after some stretching and translating in x and y direction, describe
the data points. We hence try the following fit to the data:
I(t) = Imax * ( fR((t-tmax)/(s(1+z))) + b )
Imax, tmax, s and b are here four free fit parameters, z is a given
constant. Let's say z = 0.1, i.e. I do a fit to the formula:
I(t) = Imax * ( fR((t-tmax)/(0.1 s)) + b)
Now, let's assume that the chi squared fit tells us that for this data
set, s = 1.1.
So much for describing the preliminaries. Now on to the actual claim:
I say that if I *instead* had used a fit *of the same data* to the formula
I(t) = Imax * ( fR((t-tmax)/s) + b )
(i.e. without the factor 1+z stretching the time axis), the result would
have been s = 1.21.
Is there anywhere here who disagrees with that?
Bye,
Bjoern |
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| Bjoern Feuerbacher |
| Posted: Tue Feb 22, 2005 11:26 am |
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Bjoern Feuerbacher wrote:
[snip]
[quote:6ecaa38a53]I(t) = Imax * ( fR((t-tmax)/(s(1+z))) + b )
Imax, tmax, s and b are here four free fit parameters, z is a given
constant. Let's say z = 0.1, i.e. I do a fit to the formula:
I(t) = Imax * ( fR((t-tmax)/(0.1 s)) + b)
[/quote:6ecaa38a53]
Ooops. Obvious typo: I meant
[quote:6ecaa38a53]I(t) = Imax * ( fR((t-tmax)/(1.1 s)) + b)
[/quote:6ecaa38a53]
[snip]
Bye,
Bjoern |
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| sean |
| Posted: Thu Feb 24, 2005 9:50 am |
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Your logic in this argument does not make sense to me. For
instance I believe that you do agree that s is calculated
only by performing the chi squared fitting in maple or manually.
Ie in this formula
I(t) = Imax * ( fR((t-tmax)/w) + b
all the variables are known from the template and tables and
z is supplied as the redshift value for each SN. So s is
the unknown that is calculated by the chi fitting. In other
words you cant find out s until you have done the chi
squared fit.
Yet you contradict yourself and say that you dont have to do the
fit to find out what s is when z is either .49 or 0.
Dont you see how that does not make sense?
For instance here I will set you a test.
You claim that you can calculate s just by looking at the formula
.. So I will set you a test to derive s for a SN where z=0.49
I know what s is because I have taken it from another paper where
someone else has already calculated it so I can check if your
answer is correct.
Now according to your logic the formula `tells` you
what s will be. So please what does the formula tell you
what s is for that SN where z=0.49?
And then, what is s where z=0 for that SN?
Do you still `hear` the formula telling you the answer
or will you now admit that you have to do the chi
squared fit to derive s where z=.49 and z=0
And if you admit that you have to do the fit then that
is a direct contradiction of your earlier statement that
you can `hear `the answer because the formula is `telling`
you the answer. |
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| Bjoern Feuerbacher |
| Posted: Fri Feb 25, 2005 7:42 am |
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sean wrote:
[quote:5efa830629]Your logic in this argument does not make sense to me.
[/quote:5efa830629]
Thanks for including the "to me" here. The simple fact that
it makes sense for other people, like Steve Willner, should make
you thinking...
Unfortunately, no one else has answered my post here yet...
[quote:5efa830629]For
instance I believe that you do agree that s is calculated
only by performing the chi squared fitting in maple or manually.
[/quote:5efa830629]
Yes, or by other computer programs equipped to do that.
[quote:5efa830629]Ie in this formula
I(t) = Imax * ( fR((t-tmax)/w) + b
all the variables are known from the template and tables and
z is supplied as the redshift value for each SN.
[/quote:5efa830629]
Yes. But please notice that in the formula above, there actually is
no z. Probably you meant
I(t) = Imax * ( fR((t-tmax)/(s(1+z))) + b
?
[quote:5efa830629]So s is the unknown that is calculated by the chi fitting.
[/quote:5efa830629]
Yes. (or w in the formula you gave above)
[quote:5efa830629]In other
words you cant find out s until you have done the chi
squared fit.
[/quote:5efa830629]
Or unless I have information from another chi squared fit
which was already done to a similar formula. That's the part
of the argument which you apparently keep missing.
[quote:5efa830629]Yet you contradict yourself and say that you dont have to do the
fit to find out what s is when z is either .49 or 0.
Dont you see how that does not make sense?
[/quote:5efa830629]
It only makes no sense to you because you keep missing the part
of the argument where I use (hypothetical!) results obtained
from another fit to a similar formula.
[quote:5efa830629]For instance here I will set you a test.
You claim that you can calculate s just by looking at the formula.
[/quote:5efa830629]
No, I don't claim that. Read my argument again.
I'll repeat it yet again here, concentrating on the main point, in the
hope that you will see what I actually say better if I leave out all the
surrounding information.
What I am saying is:
*IF* a fit of the data to the formula
I(t) = Imax * ( fR((t-tmax)/(1.1 s)) + b )
gives s = 1.1, *THEN* a fit of the *same* data to the formula
I(t) = Imax * ( fR((t-tmax)/s) + b )
will give s = 1.21.
Or, more generally:
*IF* a fit of the data to the formula
I(t) = Imax * ( fR((t-tmax)/(s*(1+z))) + b )
gives s = s0, *THEN* a fit of the *same* data to the formula
I(t) = Imax * ( fR((t-tmax)/s) + b )
will give s = s0*(1+z).
[snip argument based on false premise]
Bye,
Bjoern |
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