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J.F.P.
Germany
7 Posts |
 Posted - 07/18/2012 : 04:14:18 AM
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Origin Ver. and Service Release (Select Help-->About Origin): 8.5.1G SR2
Operating System: winxp
Hallo,
usually I fit my data with the following function what works well:
y = (4/3)*(x/2*1E-9)^(1.5)*E*sqrt(R)*1E6
Now I fitted with the inverse function but don't get same value for fitting parameter E (R is fixed):
y = (3/4*x*1e-6/E/sqrt(R))^(2/3)*1e9*2
Shouldn't the two fits give equal values for E?
Thanks for any help!
Best regards |
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easwar
USA
1964 Posts |
Posted - 07/18/2012 : 1:16:03 PM
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Hi J.F.P.,
Is R supposed to be a constant? Or it is a parameter as well?
If R is also a parameter, then the function is over-parametrized.
The term
E*sqrt(R)
could be replaced with just one parameter, as this is just product of two parameters. In other words, the iterative procedure cannot find an optimal value for E and R, because if one changes E to any value, one can come up with a corresponding R value to give the same result for the product term, so there is no unique solution.
In the fitter, this can be seen from the Dependency value, which will be very close to, or equal to, 1.
This may be the reason your inverse function does not yield the same result.
Also, when fitting with either, are you assigning the "x" and the "y" alternately?
When I tried your functions with some test data, I got the same parameters, but the errors were very large (and in one case missing values), as the iterative procedure cannot converge due to the over-parameterization.
If R is not a parameter, but a constant, send your data and the value for R, to tech support, using the link on the top right of this page.
Easwar
OriginLab |
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easwar
USA
1964 Posts |
Posted - 07/18/2012 : 1:21:37 PM
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Hi J.F.P,
Sorry, you did mention that you had fixed R. When I do that with some test data, the value of E for the two cases is very close, and within the parameter error values.
If you see significant differences, send your OPJ to tech support so we can look.
Easwar
OriginLab |
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J.F.P.
Germany
7 Posts |
Posted - 07/20/2012 : 03:37:00 AM
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Hi Easwar,
thanks for your help.
I will send my data to tech support, E is not within the parameter values.
Best regards |
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J.F.P.
Germany
7 Posts |
Posted - 09/18/2012 : 06:53:04 AM
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Hallo,
as I understand this is inherent to least square fitting procedure. But since there is just one "real" value of E what is an appropriate fitting procedure to determine it? Does it make sens to fit data first by a polynomial and this polynomial with the actual fit function? I tested this procedure and obtained very similar values of E. But then the question arises if the x-y or y-x data is fitted with a polynomial...
Thanks for any help and comments!
Best regards
J.F.P. |
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Sam Fang
291 Posts |
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Fitting data with inverse function