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I have the following data available. 

1572415440  1.110919
1572415500  1.110919
1572415560  1.110915
1572415620  1.110908
1572415680  1.110892

It forms a curve which is somewhat concave.
You can see that here:
https://mycurvefit.com/share/b7de79ee-dafc-42f8-9e2b-8c7140b16b87

I was tying to find the equation of the curve and the degree of the curvature without using the radius of curvature. I thought the site might be helpful in finding the equation which it does but I am still not able to find a general equation for such curves.

The equation I got is: 

y = 1.110889 + (1.110911 - 1.110889)/(1 + (x/2222773000)^25236300)

But I cannot understand how I can find the equation of such a curve and calculate the degree of the curve without using the radius of the curve.
If anyone has any suggestion please let me know.

Jyrki Lahtonen
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Jaffer Wilson
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    I hope you understand that there are infinitely many functions that would pass through these points. Do you have a reason to believe that the one you're looking for is of the form $$ y = A + \frac{B}{1+(x/C)^{\alpha}} $$ or could it as well be a polynomial? – Matti P. Oct 30 '19 at 12:13
  • The above equation is given by the site. I thought it is correct. But I doubt it. I guess it might be a polynomial and not what is shown there. – Jaffer Wilson Oct 30 '19 at 12:17
  • @Moo I do not have idea of this mathematica. I will try it – Jaffer Wilson Oct 30 '19 at 12:23
  • The site you linked to is still in beta and does not appear to be fully fleshed out yet. It looks like there should have been ways in which you can choose a different best fit model to use, but at a cursory glance It does not appear to be working. I get a console error "Fitter and Model use different best fit parameters." That said, using the method of least squares one can find a "best fitting" polynomial of a desired degree that fits a given set of data points and a measurement of how well it fit the data. – JMoravitz Oct 30 '19 at 12:26
  • Sure I will try it out – Jaffer Wilson Oct 30 '19 at 12:27
  • @JMoravitz BUt I am unable to understand how least square method would be helpful. I have given a try to it. but did not find how it is helpful. – Jaffer Wilson Oct 30 '19 at 12:29
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    Least square regression provides a way to find the curve of a specified type whose error is least. "How is it helpful"? It is always helpful. It is the standard approach. That said, five data points is a horrendously small sample size. In any respectable experiment it should have been thousands or millions or more data points. You can through lagrange interpolation find an exact fit polynomial of degree n-1 passing through n points (who all have distinct x values). It often doesn't make sense to have such a high degree polynomial though. Exponentiatial and parabolas are more common in nature – JMoravitz Oct 30 '19 at 12:41

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