MMP.17:5.1 - Refine an interpolation only until it settles the comparison

A repeated calculation needs the response (r(x)) of a costly reference model for (0\leq x\leq1). The present question is whether (r(0.4)\leq0.3). Available analysis of the reference model gives (|r’’(x)|\leq2). Its accurate case values are (r(0)=0) and (r(1)=1).

The first surrogate is the line (\widehat r(x)=x). For linear interpolation on an interval of width (h), the curvature bound gives an error at most (2h^2/8). With (h=1), the response at (0.4) is therefore enclosed by (0.4\pm0.25). This interval crosses (0.3); the replacement has not answered the question.

Query the reference model at (x=0.5), obtaining (r(0.5)=0.25), and use the two half-intervals. In the first half, (\widehat r(x)=0.5x), so (\widehat r(0.4)=0.2). The bound is now (2(0.5)^2/8=0.0625), giving

[ r(0.4)\in[0.1375,,0.2625]. ]

The upper endpoint is below (0.3). One added case and a specified interpolation rule settle the comparison under the curvature premise. The three values alone would not justify the bound.

Now the receiving limit changes to (0.18). The same enclosure crosses that limit. For this one query, the practitioner returns to the reference model at (0.4), which gives (0.16), and can answer the stricter comparison. If many similar queries are expected, further subdivision may instead be worthwhile. There is no need to improve the surrogate over the entire interval to finish the single question.

These calculations establish agreement with the reference model under its stated smoothness and case-accuracy conditions. Whether that model’s response represents the subject phenomenon remains a separate modeling question.