How to extrapolate with these values

Question

I'm reading a paper by Yallop and in it he gives the following table of values: $$\begin{array}{|c|c|c|c|c|c|c|}\hline x &0.3 &0.5 &0.7 &1 &2 &3\\ \hline y &? & 8.4 &7.5 &6.4 &4.7 &4.3 \\\hline\end{array}$$

How do I extrapolate to find the value of $y$ when $x=0.3$?

I tried several different ways and all of them give values of $9.3$ or $ 9.4$. However the extrapolated value given by Yallop is $10.0$.

Lagrange interpolation gives $9.43$. Could you provide more information? — Michael Galuza, Jun 28 '15 at 07:31
Could you give a link tot the paper or send it to me by e-mail (my address is in my profile). — Claude Leibovici, Jun 28 '15 at 07:48
I know. Lagrange interpolation gives $9.4$, linear extrapolation gives $9.3$. Michael Galuza what extra information do you need? — Nabigh, Jun 28 '15 at 07:53
Here is the link <astro.ukho.gov.uk/download/NAOTN69.pdf> @ClaudeLeibovici. See 3. (c), Table 3. — Nabigh, Jun 28 '15 at 08:00

Claude Leibovici · Answer 1 · 2015-06-28T08:17:57.743

1

Plotting the data, it looks that we face an exponential function. So, using the available data points and model $$y=a\,e^{bx}+c$$ a nonlinear regression leads to $$y=8.14843\, e^{-1.25978 x}+4.08646 $$ with $R^2=0.999977$ which is very good.

Comparing to the data points, the corresponding computed values are $8.427$, $7.460$, $6.398$, $4.742$ and $4.273$.

Extrapolating the model at $x=0.3$ gives $y=9.670$.

Edit

Using a fit with a cubis polynomial as it seems to be done in the paper, for $x=0.3$, $y=9.502$.

edited Jun 28 '15 at 08:17

answered Jun 28 '15 at 07:37

Claude Leibovici

260,315

That is a good fit, but to 1 d.p. it's still $9.7$, not $10.0$. The Lagrange interpolation formula give $10.0$ for $ x=2 $. – Nabigh Jun 28 '15 at 08:09

How to extrapolate with these values

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