| (θ + 1) θ (θ − 1) | δ2u1 + | (θ + 2) (θ + 1) θ (θ − 1) (θ − 2) | δ4u1 + ... |
| 3! | 5! |
(10)
by θV1, we have, for interpolation between u0 and u1,
uθ = u0 + θΔu0 + θV1 + 1−θV0
(11),
the successive values of θ being 1/n, 2/n, ... (n − 1)/n. For interpolation between u1 and u2 we have, with the same succession of values of θ,
u1+θ = u1 + θV1, V2 + 1−θV1
(12).
The values of 1−θV1 in (12) are exactly the same as those of θV1 in (11), but in the reverse order. The process is therefore that (i.) we find the successive values of u0 + θΔu0, &c., i.e. we construct a table, with the required intervals of x, as if we had only to take first differences into account; (ii.) we construct, in a parallel column, a table giving the values of θV1, &c.; (iii.) we repeat these latter values, placing the set belonging to each interval h in the interval next following it, and writing the values in the reverse order; and (iv.) by adding horizontally we get the final values for the new table.
As an example, take the values of tan x by intervals of ½° in x, as found above (Ex. 5). The first diagram below is a portion of this table, with the differences, and the second shows the calculation of the terms of (11) so as to get a table in which the intervals are 0.1 of 1°. The last column but one in the second diagram is introduced for convenience of calculation.