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 Topic  |
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thitchen
44 Posts |
 Posted - 10/19/2012 : 06:35:27 AM
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Origin Ver. 8.6 and Service Release 0 (Select Help-->About Origin):
Operating System: win7
I have a set of data with
x:
0.03469
0.03476
0.03477
0.03483
0.0348
0.03485
0.03498
0.03503
0.03507
0.03511
0.03511
0.03512
0.03512
y:
0.56647
0.62063
0.67129
0.76376
0.84648
0.92132
1.0525
1.16489
1.26319
1.42917
1.56614
1.78425
1.95479
It is supposed to be fitted with function:
y=bx*(a-x)^(-0.5)+c
which seems like a very straight forward task.
I enter the expression into the fitting function builder, and fit the data, however, originlab keeps saying it's over-parameterized. I'm now totally lost and confused. Please any help is appreciated. |
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greg
USA
1378 Posts |
Posted - 10/19/2012 : 11:56:50 AM
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The correct definition should be
y = b*x*(a-x)^(-0.5)+c
with parameters a,b,c
Using that, and a simple initialization (all parameters equal to 1), I was able to get a fit of your data
Value Standard Error
a 0.03519 2.60275E-4
b 0.40068 0.40021
c 0.02726 0.54407
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thitchen
44 Posts |
Posted - 10/19/2012 : 1:53:21 PM
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Greg,
Thanks for the reply, I tried with your suggestion, now the problem I got it's that the fitting do not even try to iterate once with all initial parameters setting to 1.
the message I got is:
(1) Parameter Initialization was called.
(2) ----------Levenberg-Marquardt-----------
Reduced Chi-sqr = 188.65792818
Iterations Performed = 0
Total Iterations in Session = 0
(3) Fit did not converge - reason unknown.
Please help.
quote:
Originally posted by greg
The correct definition should be
y = b*x*(a-x)^(-0.5)+c
with parameters a,b,c
Using that, and a simple initialization (all parameters equal to 1), I was able to get a fit of your data
Value Standard Error
a 0.03519 2.60275E-4
b 0.40068 0.40021
c 0.02726 0.54407
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thitchen
44 Posts |
Posted - 10/19/2012 : 1:56:36 PM
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Problem solved. Thanks so much, Greg!
quote:
Originally posted by greg
The correct definition should be
y = b*x*(a-x)^(-0.5)+c
with parameters a,b,c
Using that, and a simple initialization (all parameters equal to 1), I was able to get a fit of your data
Value Standard Error
a 0.03519 2.60275E-4
b 0.40068 0.40021
c 0.02726 0.54407
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cc261
22 Posts |
Posted - 10/22/2012 : 05:01:47 AM
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The results from 1stOpt are more accurate:
a 0.035161773138033
b 0.287946048033968
c 0.221136990845496
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curve fitting please help