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So I'm working on a small project that will use a calculation in excel, I'm nearly there and the only piece missing is being able to accurately estimate a data point from the following set of data points. I have tried using online calculators to generate formulas, but they are accurate to 3d.p. at best, which unfortunately isn't any good. I have tried the FORECAST tool in Excel which again, estimates some values below 0.75, when that's clearly where they converge too.

Any ideas? Or even if this is possible. Data points are:

0 0.7894735

0.1 0.7857143

0.3 0.78

0.5 0.77586205

0.75 0.7720588

1 0.7692306

1.5 0.7653061

2 0.762712

3 0.759494

4 0.757575757

5 0.7563025

6 0.7553956

7 0.754717

8 0.75419

9 0.753769

11 0.753138

13 0.752688

15 0.752351

19 0.7518795

29 0.7512519

39 0.7509388

49 0.7507508

99 0.75037517

249 0.75015001

999 0.7500375

Many Thanks

Robert Howard
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user994319
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  • It will be helpful if you add some context. Where does the data come from? what are you trying to predict? It is pretty un-reasonable to think that from 20-30 data points you could predict more data points to 3 digits of accuracy (beyond the decimal)... unless you know an awful lot about what is generating the data. – TravisJ Jun 23 '15 at 00:09
  • Of course, I have obtained the data manually by inputting into my Excel spreadsheet in the hope to obtain a formula that can be used instead. The spreadsheet is quite complex, but it is to do with betting, pulling data from an xml and then running formulas on it to find anomalies. In reality the only important data points are 0 through to about 7-8. I was just hoping there might be something that could calculate it. – user994319 Jun 23 '15 at 00:21

1 Answers1

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On the off chance that this is still useful, I've found a formula that describes that data set very accurately, to at least six decimal places: $$f(x)=\frac{0.0375003}{x+0.950011}+0.75$$ I used Desmos to find that formula, and you can check out what I did here. In cell 8, you can see that the greatest amount by which this formula's prediction differs from the data is about $3\times10^{-7}$.

Robert Howard
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