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I've stumbled upon a method of extrapolation that I haven't seen before.

We are trying to approximate $f(0)$ for a certain function $f$, which we have only measured at points $x_0, \ldots, x_N$ in an interval $[a,b]$ that does not contain $0$. We have reason to believe that $f$ is analytic in a neighbourhood of a region of $\mathbb C$ (containing both $0$ and $[a,b]$) bounded by a simple positively oriented closed contour $\Gamma$. Suppose we can approximate $1/z$ on $\Gamma$ by a linear combination of $1/(z-x_j)$, say $$ \left| \frac{1}{z} - \sum_{j=0}^N \frac{a_j}{z-x_j} \right| \le \varepsilon \ \text{for}\ z \in \Gamma $$ Then using Cauchy's formula, $$ \left|f(0) - \sum_{j=0}^N a_j f(x_j) \right| \le \frac{M\; \text{length}(\Gamma) \varepsilon}{2\pi}$$

Rather than uniform approximation, it is more convenient to use an $L^2$ approximation. This will let us find the $a_j$ by minimizing a quadratic form. In the case where $\Gamma$ is a circle of radius $r$ centred at $0$, I get a nice closed form: $$ a_j = \frac{r^2 - x_j^2}{r^{2N+2}} \prod_{k \ne j} \frac{x_k (r^2 - x_j x_k)}{x_k - x_j}$$

I can't believe I'm the first to think of this idea. Has anyone seen something like this?

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1 Answer 1

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There are extrapolation schemes in use that take essential recourse to analyticity, for example, the "z-expansion fits" described in arXiv:1008.4619 (it's sufficient to read the first two pages to see what the method is). In the details, however, these fits differ substantially from the scheme you are proposing. I have not seen anything that is very similar to your scheme.

Having said that, I worry about the stability of your scheme. You say that you "measure" the function $f$, which I assume means that, generically, you don't have exact data on $f$, but they come with statistical or systematic error. In that case, it is important that the scheme be robust against these errors. Let's look at an example that, I think, by no stretch of the imagination can be dismissed as contrived:

10 equally spaced data points at $x$ ranging from $x=0.1$ to $x=1$, with $r=1.5$, i.e., in your notation, $N=9$, $x_j = 0.1(j+1)$, $r=1.5$. Then, your coefficients $a_j $ are:

$a[0]=7.80096236306711$

$a[1]=-27.161848927028$

$a[2]=55.5523167543375$

$a[3]=-73.8548460044706$

$a[4]=66.6361882416331$

$a[5]=-41.2843129034312$

$a[6]=17.3237609884424$

$a[7]=-4.70612791676158$

$a[8]=0.746272144521954$

$a[9]=-0.0523647403501327$

The coefficients are large and alternating! This means they are prone to magnify fluctuations of the data. Let's take the simplest possible true functional dependence, $f(x)=1$. As long as you have no errors in the data, $f(x_j )=1$, you predict $f(0)=1$ with high accuracy. However, as soon as I put small random fluctuations on the data $f(x_j )$, the prediction fluctuates wildly. Here is a sequence of 10 predictions for $f(0)$ when I put random fluctuations on the $f(x_j )$ that are smaller than 1% (i.e., I'm adding $0.01 (2 \cdot rand()-1)$ in perl):

$1.978$

$1.114$

$0.529$

$1.286$

$1.032$

$-0.179$

$1.384$

$0.102$

$0.371$

$1.656$

Thus, 1% errors in the data are amplified to 100% errors in the predictions of $f(0)$, which is of course not at all surprising in view of the $a_j $. So, unfortunately, I'm not sure whether the fact that we're having a hard time finding someone who has seen a scheme of the type you describe used means that no one has thought of it before ...

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  • $\begingroup$ Numerical stability is indeed an issue, which I think is inherent to all extrapolation methods. I think you can do quite a bit better than with equally spaced points: you might try an optimization to minimize the sum of squares of coefficients. $\endgroup$ Commented Sep 3, 2019 at 16:10
  • $\begingroup$ Yes, extrapolation is always dangerous, however you do it. I'm not an expert on the theory, I just have the experience of frequently grappling with data. Indeed, equal spacing is not the best-behaved thing, and if you have the luxury of choosing where to measure, you'd bunch at the edges, like at Chebyshev points. Still, I gave you a decent lever arm in my example ... it's worth exploring how much one can smooth the behavior of the coefficients. There's also the dependence on the contour - if you reduce r to be close to minimal, e.g., r=1.05 in the example, fluctuations are only 1/2 as large. $\endgroup$ Commented Sep 3, 2019 at 19:48
  • $\begingroup$ Another thought, which may lead absolutely nowhere: Given the ubiquitous Cauchy/Weierstrass dichotomy in analytic function theory, I wonder whether your Cauchy-style formulation doesn't in the end have a Weierstrass equivalent, i.e., a realization in terms of an expansion in powers, polynomials or some such, more akin to standard schemes. Then one could compare and contrast ... $\endgroup$ Commented Sep 3, 2019 at 20:02
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    $\begingroup$ @RobertIsrael: I think the instability is a property of the analytic continuation and not the point selection, though point selection could ostensibly make the situation worse. See arxiv.org/abs/1908.11097 $\endgroup$
    – user14717
    Commented Sep 4, 2019 at 21:31

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