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I was reading a passage from an old essay by Martin Gardner on the calculus of finite differences, and it seems to me that there is more that can and should be said about seemingly anomalous values of the function that counts the number of pieces into which some shape can be divided by $n$ straight cuts. Here's the passage:

"In applying Newton's formula to empirically obtained data, one sometimes comes up against an anomaly for the zero case. For instance, The Scientific American Book of Mathematical Puzzles & Diversions, page 149, gives the formula for the maximum number of pieces that can be produced by n simultaneous plane cuts through a doughnut. The formula is a cubic that can be obtained by applying Newton's formula to results obtained empirically, but it does not seem to apply to the zero case. When a doughnut is not cut at all, clearly there is one piece, whereas the formula says there should be no pieces. To make the formula applicable, we must define "piece" as part of a doughnut produced by cutting. Where there is ambiguity about the zero case, one must extrapolate backward in the chart of differences and assume for the zero case a value that produces the desired first number in the last row of differences."

(The formula Gardner is referring to here is the cubic function $(n^3+3n^2+8n)/6$.)

My first thought was that a doughnut-shaped piece should count as "zero pieces" because its Euler measure is zero. However, if you think about slicing an annulus in 2D, you'll find that there's more to justifying anomalous cases than just taking Euler measure into account.

Has anyone come up with the "right" way to look at these functions of $n$? My instinct is that with properly chosen definition of "pieces" and "counting", the anomalies go away.

It would be especially nice if there were some combinatorial significance for values of these functions obtained by plugging in negative values of $n$ (cf. Ehrhart reciprocity and Stanley's result about evaluating the chromatic polynomial at $-1$).

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  • $\begingroup$ What does happen when slicing an annulus? $\endgroup$ Commented Sep 2, 2022 at 0:55
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    $\begingroup$ Oops! Gerry’s question prompted me to think harder, and I found a mistake in my analysis. So I don’t know what happens with the annulus. One might indeed get a quadratic function of $n$ for all $n$ sufficiently large, and plugging $n=0$ into this quadratic might indeed give 0 (as an Euler measure viewpoint would suggest) — I don’t know yet. $\endgroup$ Commented Sep 2, 2022 at 3:26
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    $\begingroup$ Update: According to A000096 in the OEIS the answer is $n(n+3)/2$, which indeed takes the value $0$ at $n=0$. So maybe counting pieces by Euler measure works after all. $\endgroup$ Commented Sep 2, 2022 at 16:08

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