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KelvinS

Post subject: Can anyone find a formula to fit this data? Posted: Wed Jan 01, 2014 7:11 pm 

Joined: Mon Mar 30, 2009 5:13 pm

I have some data below, and am tring to find the best formula, y = f(x), with just 2 or 3 arbitrary constants, to give minimal discrepancy (residual) between calculated and observed y over the entire data set.
Would anyone like to have a crack at this:
x > y
0.9 > 0.292326 0.8 > 0.274275 0.7 > 0.260295 0.6 > 0.248933 0.5 > 0.239433 0.4 > 0.231333 0.3 > 0.224326 0.2 > 0.218196 0.1 > 0.212786 0.0 > 0.207978 0.1 > 0.203679 0.2 > 0.199818 0.3 > 0.196335 0.4 > 0.193183 0.5 > 0.190323 0.6 > 0.187722 0.7 > 0.185354 0.8 > 0.183193 0.9 > 0.181221 1.0 > 0.179432
_________________ If you want something you’ve never had, you’ve got to do something you’ve never done  Thomas Jefferson


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benpuzzles

Post subject: Re: Can anyone find a formula to fit this data? Posted: Wed Jan 01, 2014 7:23 pm 

Joined: Fri Dec 28, 2012 1:50 pm Location: Near Las Vegas, NV


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KelvinS

Post subject: Re: Can anyone find a formula to fit this data? Posted: Wed Jan 01, 2014 7:44 pm 

Joined: Mon Mar 30, 2009 5:13 pm

benpuzzles wrote: Fitting into a quadratic regression gives me this: y=0.0323x²0.056x+0.2075 It has very minimal error. Does this help? Thanks, but a quadratic fit seems to be quite poor (RSQ = 0.9906): Attachment:
Quadratic fit.png [ 36.43 KiB  Viewed 827 times ]
_________________ If you want something you’ve never had, you’ve got to do something you’ve never done  Thomas Jefferson


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Brandon Enright

Post subject: Re: Can anyone find a formula to fit this data? Posted: Wed Jan 01, 2014 7:46 pm 

Joined: Thu Dec 31, 2009 8:54 pm Location: Bay Area, California

Looks more like an exponential curve than an conic curve. Here is what I used: Code: g(x) = (a * exp((x + 1) * b)) + c
Final set of parameters Asymptotic Standard Error ======================= ==========================
a = 0.134809 +/ 0.001144 (0.849%) b = 1.34265 +/ 0.03847 (2.865%) c = 0.171799 +/ 0.001216 (0.7076%)
f(x) is Ben's proposal and g(x) is mine.
Attachments: 
kevin_fit.png [ 33.08 KiB  Viewed 823 times ]

_________________ Prior to using my real name I posted under the account named bmenrigh.


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KelvinS

Post subject: Re: Can anyone find a formula to fit this data? Posted: Wed Jan 01, 2014 7:54 pm 

Joined: Mon Mar 30, 2009 5:13 pm

Brandon Enright wrote: Looks more like an exponential curve than an conic curve. Here is what I used... That's much better (RSQ = 0.998787), thanks!
_________________ If you want something you’ve never had, you’ve got to do something you’ve never done  Thomas Jefferson
Last edited by KelvinS on Wed Jan 01, 2014 7:55 pm, edited 1 time in total.


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Brandon Enright

Post subject: Re: Can anyone find a formula to fit this data? Posted: Wed Jan 01, 2014 7:54 pm 

Joined: Thu Dec 31, 2009 8:54 pm Location: Bay Area, California

Also, you didn't specify any precision on your data. Is the y column really precise to 6 significant digits? How precise is the x column? Infinitely? One significant digit?
_________________ Prior to using my real name I posted under the account named bmenrigh.


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KelvinS

Post subject: Re: Can anyone find a formula to fit this data? Posted: Wed Jan 01, 2014 7:58 pm 

Joined: Mon Mar 30, 2009 5:13 pm

Brandon Enright wrote: Also, you didn't specify any precision on your data. Is the y column really precise to 6 significant digits? How precise is the x column? Infinitely? One significant digit? x column is indefinitely precise, y column is precise to 6 decimals (as given), I think.
_________________ If you want something you’ve never had, you’ve got to do something you’ve never done  Thomas Jefferson


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KelvinS

Post subject: Re: Can anyone find a formula to fit this data? Posted: Wed Jan 01, 2014 8:06 pm 

Joined: Mon Mar 30, 2009 5:13 pm

Brandon, what program are you using, and what do you get if you fit with 4 constants:
g(x) = (a * exp((x + d) * b)) + c
Where d may not be exactly 1 ?
_________________ If you want something you’ve never had, you’ve got to do something you’ve never done  Thomas Jefferson


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Brandon Enright

Post subject: Re: Can anyone find a formula to fit this data? Posted: Wed Jan 01, 2014 8:23 pm 

Joined: Thu Dec 31, 2009 8:54 pm Location: Bay Area, California

KelvinS wrote: Brandon, what program are you using, and what do you get if you fit with 4 constants:
g(x) = (a * exp((x + d) * b)) + c
Where d may not be exactly 1 ? I'm using gnuplot which is a pretty awesome tool. Code: gnuplot> g(x) = (a * exp((x + d) * b)) + c gnuplot> fit g(x) 'data.txt' via a,b,c,d gnuplot> plot 'data.txt' using 1:2 with lines, g(x) with lines
Code: Final set of parameters Asymptotic Standard Error ======================= ==========================
a = 0.302054 +/ 4.572e+09 (1.514e+12%) b = 1.34265 +/ 0.03977 (2.962%) c = 0.171799 +/ 0.001255 (0.7302%) d = 1.60086 +/ 1.128e+10 (7.049e+11%)
correlation matrix of the fit parameters:
a b c d a 1.000 b 0.068 1.000 c 0.047 0.926 1.000 d 1.000 0.068 0.047 1.000
Attachment:
kevin_fit_2.png [ 18.04 KiB  Viewed 790 times ]
_________________ Prior to using my real name I posted under the account named bmenrigh.


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KelvinS

Post subject: Re: Can anyone find a formula to fit this data? Posted: Wed Jan 01, 2014 8:33 pm 

Joined: Mon Mar 30, 2009 5:13 pm

Wow, that is a very cool program, thanks!
Interesting that fitting with 4 parameters does not improve the fit (no change in RSQ) even though d is changed from 1 to 1.6.
Doesn't seem right, but what do you think?
_________________ If you want something you’ve never had, you’ve got to do something you’ve never done  Thomas Jefferson


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Brandon Enright

Post subject: Re: Can anyone find a formula to fit this data? Posted: Wed Jan 01, 2014 8:36 pm 

Joined: Thu Dec 31, 2009 8:54 pm Location: Bay Area, California

Notice that d and a have perfect correlation in the correlation matrix? You can set d to anything and find an appropriate a to make it work. Here are some examples: Code: a = 6.531e08 +/ 1209 (1.851e+12%) b = 1.34266 +/ 0.04359 (3.247%) c = 0.171799 +/ 0.001369 (0.7967%) d = 9.82944 +/ 1.37e+10 (1.393e+11%)
a = 0.479834 +/ 6.738e+09 (1.404e+12%) b = 1.34264 +/ 0.03996 (2.976%) c = 0.171798 +/ 0.001267 (0.7372%) d = 1.94558 +/ 1.047e+10 (5.383e+11%)
a = 0.169406 +/ 1.778e+09 (1.05e+12%) b = 1.34266 +/ 0.0438 (3.262%) c = 0.171799 +/ 0.001366 (0.795%) d = 1.17014 +/ 7.824e+09 (6.687e+11%)
This is because the shape of an exp() curve is scale invariant modulo some appropriate perspective adjustments.
_________________ Prior to using my real name I posted under the account named bmenrigh.


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KelvinS

Post subject: Re: Can anyone find a formula to fit this data? Posted: Wed Jan 01, 2014 8:47 pm 

Joined: Mon Mar 30, 2009 5:13 pm

Ah yes, of course, thanks!
Then how/where would you add a fourth constant to get a much better fit?
What about:
g(x) = a * exp(x * b) + c + d * x
_________________ If you want something you’ve never had, you’ve got to do something you’ve never done  Thomas Jefferson
Last edited by KelvinS on Wed Jan 01, 2014 8:48 pm, edited 1 time in total.


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Brandon Enright

Post subject: Re: Can anyone find a formula to fit this data? Posted: Wed Jan 01, 2014 8:47 pm 

Joined: Thu Dec 31, 2009 8:54 pm Location: Bay Area, California

I probably shouldn't have shifted the curve at all. Using g(x) = (a * exp(x * b)) + c results in: Code: Final set of parameters Asymptotic Standard Error ======================= ==========================
a = 0.0352053 +/ 0.001413 (4.014%) b = 1.34266 +/ 0.03841 (2.861%) c = 0.171799 +/ 0.001216 (0.7076%)
As I said before, shifting the curve with d is completely redundant because of the use of a for scale.
_________________ Prior to using my real name I posted under the account named bmenrigh.


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KelvinS

Post subject: Re: Can anyone find a formula to fit this data? Posted: Wed Jan 01, 2014 8:49 pm 

Joined: Mon Mar 30, 2009 5:13 pm

What about:
g(x) = a * exp(b * x) + c + d * x
_________________ If you want something you’ve never had, you’ve got to do something you’ve never done  Thomas Jefferson


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Brandon Enright

Post subject: Re: Can anyone find a formula to fit this data? Posted: Wed Jan 01, 2014 8:57 pm 

Joined: Thu Dec 31, 2009 8:54 pm Location: Bay Area, California

KelvinS wrote: Ah yes, of course, thanks!
Then how/where would you add a fourth constant to get a much better fit?
What about:
g(x) = a * exp(x * b) + c + d * x Whoa I wouldn't have thought of adding a linear term but that's a way better fit (still with x although you could just do b instead): Code: Final set of parameters Asymptotic Standard Error ======================= ==========================
a = 0.0126512 +/ 0.001062 (8.395%) b = 2.05477 +/ 0.06713 (3.267%) c = 0.195112 +/ 0.001144 (0.5865%) d = 0.0178764 +/ 0.001092 (6.107%)
Attachment:
kevin_fit_3.png [ 16.84 KiB  Viewed 763 times ]
_________________ Prior to using my real name I posted under the account named bmenrigh.


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KelvinS

Post subject: Re: Can anyone find a formula to fit this data? Posted: Wed Jan 01, 2014 9:00 pm 

Joined: Mon Mar 30, 2009 5:13 pm

Superb, thank you!
I may need your help at some point to help me learn this great tool!
Cheers, K.
_________________ If you want something you’ve never had, you’ve got to do something you’ve never done  Thomas Jefferson


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Brandon Enright

Post subject: Re: Can anyone find a formula to fit this data? Posted: Wed Jan 01, 2014 9:02 pm 

Joined: Thu Dec 31, 2009 8:54 pm Location: Bay Area, California

What is this data from? What made you guess adding a linear term to the exponential term?
_________________ Prior to using my real name I posted under the account named bmenrigh.


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KelvinS

Post subject: Re: Can anyone find a formula to fit this data? Posted: Wed Jan 01, 2014 9:08 pm 

Joined: Mon Mar 30, 2009 5:13 pm

Brandon Enright wrote: What is this data from? What made you guess adding a linear term to the exponential term? The data is from a work project. I guessed adding the linear term purely by seeing that the right side of the curve tended towards a slope/gradient rather than a flat/horizontal line. So it seemed to be a composition of exponential and linear. Make sense?
_________________ If you want something you’ve never had, you’ve got to do something you’ve never done  Thomas Jefferson


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gingervergo

Post subject: Re: Can anyone find a formula to fit this data? Posted: Wed Jan 01, 2014 9:18 pm 

Joined: Fri Mar 06, 2009 9:23 pm

A quick minitab analysis yielded the following results, which looks like it conforms to what has already been posted.
an r^2 value of 99% is typically a lot higher than what I see whenever I'm running tests. Depending on the accuracy of your equipment/ repeatability of your study, I'd say this is a slam dunk, but I'm looking at this like an engineer.
The normality test of the residuals show that the residuals are not normally distributed, which might be indicative of the fact that the nature of the data is not actually quadratic. Another possible reason of the non normally distributed residuals could just be the fact that there is a relatively small sample size. Reading through the thread (I know I missed the boat a little bit)it looks like you are going through and adding terms to make the regression fit better. Knowing how the data is generated might lead to some theory that could drive the regression instead of just throwing stuff in and making it fit.
Attachments: 
Screenshot_5.png [ 39.79 KiB  Viewed 739 times ]

_________________ Eric Vergo
My Shapeways shop


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garathnor

Post subject: Re: Can anyone find a formula to fit this data? Posted: Wed Jan 01, 2014 9:26 pm 

Joined: Fri Dec 20, 2013 11:47 pm

I would have said 42, but I'm also bad at math


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idreamincubes

Post subject: Re: Can anyone find a formula to fit this data? Posted: Fri Jan 03, 2014 7:16 am 

Joined: Mon Dec 16, 2013 7:13 am

A neat tool I have used for stuff like this is the function finder at http://zunzun.com/. The top ranked result for your data was this: Code: y = 1.0 / exp(((xa)/b)^c) + Offset
Fitting target of lowest sum of squared absolute error = 2.2126873197635119E07
a = 1.3826027651964967E+00 b = 7.4496380258280037E02 c = 3.3644155009905519E01 Offset = 1.3882881100890193E01 which could be written as Code: y = exp(2.3958553356389 * (x + 1.38260276519649)^0.336441550099055) + 0.138828811008901
If you have an idea what type of equation you're looking for, there are many equally good matches returned by the function finder. // Per.


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KelvinS

Post subject: Re: Can anyone find a formula to fit this data? Posted: Fri Jan 03, 2014 8:17 am 

Joined: Mon Mar 30, 2009 5:13 pm

Very cool, thank you! In fact I see this tool can also fit surface functions to 2D data arrays (i.e., z = f(x,y)?), which is actually what I have been trying to do, by fitting curves to individual slices of data, then fitting curves to the parameters of those first curves. Wow, this is fantastic! AND, it's completely webbased! I wish I had known about zunzun earlier! And now I sound like one of those very annoying canned actors endorsing the product as a "real" customer in a commercial!
_________________ If you want something you’ve never had, you’ve got to do something you’ve never done  Thomas Jefferson


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idreamincubes

Post subject: Re: Can anyone find a formula to fit this data? Posted: Fri Jan 03, 2014 5:56 pm 

Joined: Mon Dec 16, 2013 7:13 am

Yes, it's cool, isn't it. I have also used this site when I wanted to fit a custom equation to data: http://statpages.org/nonlin.htmlI ended up making my own program based on the code from that site, though. // Per.


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Brandon Enright

Post subject: Re: Can anyone find a formula to fit this data? Posted: Fri Jan 03, 2014 8:03 pm 

Joined: Thu Dec 31, 2009 8:54 pm Location: Bay Area, California

Kevin, can you get more data points? And can you get them to a higher precision? I'd like to see how various functions stack up and the more data the better.
_________________ Prior to using my real name I posted under the account named bmenrigh.


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KelvinS

Post subject: Re: Can anyone find a formula to fit this data? Posted: Fri Jan 03, 2014 9:02 pm 

Joined: Mon Mar 30, 2009 5:13 pm

Yes, in fact what I've been trying to do is find a formula to solve the following function for p, since it does not have a closed solution, to avoid calculating it numerically: Code: g = (p ^k  (1  p) ^k) / (p ^k + (1  p) ^ k  2 / (k + 1)), where:
0 <= p <= 1, 0.9 <= k <=1, and k <> 0
Basically, I fit curves to individual slices of data at different values of k, fit another curve to the parameters of the first set of curves, and then substitute in the formulae for the individual parameters to get the surface function, p = f(g,k). So far, the best fit (RSQ = 0.999992) I have found is: Code: p = 0.5 + x * Arctan(g * Tan((0.5  (k + 1) ^ (1 / k)) / x)), where:
x = 0.0127 * Exp(2.0856 * k)  0.0168 * k + 0.1941
But I haven't tried zunzun yet...
_________________ If you want something you’ve never had, you’ve got to do something you’ve never done  Thomas Jefferson


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