The “method of finite differences” as used in the Difference Engine is closely related to, but slightly different from, Newton’s method of divided differences (used, for example, for polynomial interpolation or for boundary-value problems in ordinary differential equations) and the finite difference method used in the solution of ODEs and PDEs.
The Wikipedia page on the Difference Engine explains how to calculate the initial values, but I am skeptical of its explanation. However, it cites Nathan Myhrvold’s instructions for setting up the machine, which we can suppose have been tested on the one he had built.
Reading F. Baily’s 1823 comments on the Difference Engine we find:
2º. Tables of Square Numbers. The squares of all numbers, as far as 1000, were a long time ago published … In computing a table of this kind by the machine, even if extended to the most remote point that could be desired, the whole of the mental labour would be saved: and when the numbers 1, 1, 2 are once placed in it, it will continue to produce all the square numbers in succession without interruption. This is, in fact, one of those tables which the engine already made is capable of computing, as far as its limited number of wheels will admit.
3º. Tables of Cube Numbers. Tables of this kind have likewise been already computed… In computing such a table by the machine, the whole of the mental labour would be in this case also saved: since it would be merely necessary to place in the machine the numbers 1, 7, 6, 6; and it would then produce in succession all the cube numbers.
The table of cubes begins 1, 8, 27, 64, 125; its first differences are 7, 19, 37, 61…, and its second differences are thus 12, 18, 24…, its third differences 6, 6…, and its fourth differences merely a sequence of zeroes, since the third-order approximation is actually precisely correct. The first item of each of these sequences gives the (1, 7, 6, 6) setup described in Baily’s article.
XXX no it doesn’t; 6 is the difference between 1 and 7, so I think Baily actually took into account the half-cycle offset I claim below that he didn’t take into account; his numbers give the correct answer as shown at the end.
However, to update this state (1, 7, 6, 6) to its successor state (8, 13, 12, 6) without any extra storage, we must work strictly from left to right, because the previous value of each number (except the 1, in the lowest-order register, T) is needed to update the next-lower-order term, before being itself updated (except the second 6, in the highest-order register, Δ³). So we must proceed as follows:
| T | Δ¹ | Δ² | Δ³ |
| 1 | 7 | 6 | 6 |
| 8 | 7 | 6 | 6 |
| 8 | 13 | 6 | 6 |
| 8 | 13 | 12 | 6 |
XXX note that 13 is wrong because it should be 19
However, at least in the “Difference Engine No. 2” design Babbage developed between 1846 and 1849 (23–26 years after Baily’s letter), the calculations are not performed in this serial fashion. As explained in Swade’s analysis, instead a set of four parallel additions is carried out, then a set of three parallel additions:
However, the sequence of additions as executed by the engine does not proceed in a stepwise way from right to left as one might expect from the manual method. One complete calculating cycle consists of two symmetrical half-cycles. During the first half-cycle the number values on the odd-numbered axes are simultaneously added to those of the even-numbered axes to the immediate left i.e. axes 1 [Δ⁷], 3 [Δ⁵], 5 [Δ³], 7 [Δ¹] are added to 2 [Δ⁶], 4 [Δ⁴], 6 [Δ²],and 8 [T] respectively. Similarly, during the second half-cycle all the even-numbered axes are simultaneously added to the odd-numbered axes again to the immediate left. Provided this is taken account of when setting up at the start of a run by offsetting the initial values on alternate axes by a half-cycle, the end result is the same i.e. each full calculating cycle results in a new tabular value which is the cumulative sum of the number of active differences.
So, in fact, if we want to compute a table of cubes, to get 8 as our second cube, we do indeed need to set Δ¹ (column 7 on the machine) to 7, as Baily says. And then, having been used, it is incremented by Δ² (column 6 on the machine) in the second half-cycle — but that Δ² had previously been incremented by Δ³ in the first half-cycle. So the correct initial state we would need is not Baily’s (1, 7, 6, 6) but (1, 7, 0, 6), proceeding as follows:
| T | Δ¹ | Δ² | Δ³ |
| 1 | 7 | 0 | 6 |
| 8 | 7 | 6 | 6 |
| 8 | 13 | 6 | 6 |
XXX this is the wrong answer. If we start from Baily’s setup we get the right answer:
| T | Δ¹ | Δ² | Δ³ |
| 1 | 7 | 6 | 6 |
| 8 | 7 | 12 | 6 |
| 8 | 19 | 12 | 6 |
| 27 | 19 | 18 | 6 |
| 27 | 37 | 18 | 6 |
| 64 | 37 | 24 | 6 |
| 64 | 61 | 24 | 6 |
| 125 | 61 | 30 | 6 |