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 November 21st, 2014, 06:48 AM #1 Newbie   Joined: Aug 2012 Posts: 17 Thanks: 0 Is this the best hyperbolic fit to this data? I tried to fit a hyperbolic trendline to my data using the inxtructions here: How can I add in a hyperbolic regression curve in Excel 2010? - Microsoft Community Since this way excel wont give me R^2 I used this to calculate it https://www.khanacademy.org/math/pro...ting-r-squared and the result was: R^2 = 0,437225524 This was a bit less than I was anticipating so I would like to know if there is a better way to fit a hyperbolic curve to my data. I think my R^2 calculation is ok since it works well with real excel trendlines. Data and all calculations can be found here: https://dl.dropboxusercontent.com/u/54506834/test.xlsx 70k, 500 rows (It's a bit much to copypaste) Thank you for the help
November 21st, 2014, 10:23 AM   #2
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From: Erewhon

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Quote:
 Originally Posted by nekdolan I tried to fit a hyperbolic trendline to my data using the inxtructions here: How can I add in a hyperbolic regression curve in Excel 2010? - Microsoft Community Since this way excel wont give me R^2 I used this to calculate it https://www.khanacademy.org/math/pro...ting-r-squared and the result was: R^2 = 0,437225524 This was a bit less than I was anticipating so I would like to know if there is a better way to fit a hyperbolic curve to my data. I think my R^2 calculation is ok since it works well with real excel trendlines. Data and all calculations can be found here: https://dl.dropboxusercontent.com/u/54506834/test.xlsx 70k, 500 rows (It's a bit much to copypaste) Thank you for the help
That is not how I would do it but it gives a pretty good result for a fit of the type you are attempting.

Do you have a theoretical reason for seeking a hyperbolic fit? There are other models that will give better fits but they are of no use if your are forced to a hyperbolic.

A model of the form:
$$y=\frac{m(x+a)}{(x+b)}$$
will give a very much better fit

The attachment shows plots of your data, the hyperbolic fit and the least squares fit of the above model ($r^2=0.966$).

The fit is obtained using the non-linear solver in Gnumeric, there is one for Excel also but it is not always installed.

CB
Attached Images
 fit plots.jpg (12.2 KB, 9 views)

Last edited by CaptainBlack; November 21st, 2014 at 10:28 AM.

 November 21st, 2014, 11:20 AM #3 Newbie   Joined: Aug 2012 Posts: 17 Thanks: 0 Wow that's awesome. I do have a reason to look for the best possible hyperbolic fit, but I am also interested in the best possible fit of any kind of simple function. Is it reasonable to believe that a better hyperbolic curve could have a significantly better r2? Like 0.6 or 0.8? I tried a power law function and that had a r2 = 0.95 so yours is indeed better. I will try out the non-linear solver, although I'm not really sure how I can get m, a and b using it. EDIT: Ok I managed to do it. My values are: m = 0,077276363 a = 58163,47999 b = 63221,98938 I wonder does this function have a limit x->infinity? Last edited by nekdolan; November 21st, 2014 at 11:52 AM.
November 21st, 2014, 11:39 AM   #4
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Quote:
 Originally Posted by nekdolan Wow that's awesome. I do have a reason to look for the best possible hyperbolic fit, but I am also interested in the best possible fit of any kind of simple function. I tried a power law function and that had a r2 = 0.95 so yours is indeed better. I will try out the non-linear solver, although I'm not really sure how I can get m, a and b using it. (EDIT: seems I will be able to do that) Is it reasonable to believe that a better hyperbolic curve could have a significantly better r2? Like 0.6 or 0.8?
No the improvement in $r^2$ for a non-linear fit hyperbolic model is negligible.

CB

 November 21st, 2014, 12:41 PM #5 Newbie   Joined: Aug 2012 Posts: 17 Thanks: 0 Thanks than that's that, but I still would like to know (see my second edit) the limit of the function you have found. It seems to be around 0.0.07727? That's a very rough guess on my part. Not sure how you can do this...
November 21st, 2014, 08:57 PM   #6
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Quote:
 Originally Posted by nekdolan Thanks than that's that, but I still would like to know (see my second edit) the limit of the function you have found. It seems to be around 0.0.07727? That's a very rough guess on my part. Not sure how you can do this...
The limit as $x \to \infty$ is $m$.

CB

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