6. When the values of u have been tabulated for values of x proceeding
by a difference h, it is often desirable to deduce a table in which
the differences of x are h/n, where n is an integer.
If n is even it may be advisable to form an intermediate table in
which the intervals are (1/2)h. For this purpose we have
u_(1/2) = (1/2)(U0 + U1) (7),
where
U = u - (1/8)[delta]^2u + (3/128)[delta]^4u - (5/1024)[delta]^6u + ...
= u - (1/8)[[delta]^2u - (3/16){[delta]^4u - (5/24)([delta]^6u - ...)}] (8).
The following is an example; the data are the values of tan x to five
places of decimals, the interval in x being 1 deg. The differences of
odd order are omitted for convenience of printing.
_Example 5._
+--------+---------+-----------+-----------+-----------+------------+------------+-------------+
| | u [eq] | | | | |u = mean of | |
| x. | tan x. |[delta]^2u.|[delta]^4u.|[delta]^6u.| U. |values of U.| x. |
+--------+---------+-----------+-----------+-----------+------------+------------+-------------+
| | | + | + | + | | | |
| 73 deg.| 3.27085 | 2339 | 100 | 5 | 3.26794 95 | | |
| | | | | | | 3.37594 | 73(1/2) deg.|
| 74 deg.| 3.48741 | 2808 | 132 | 23 | 3.48392 98 | | |
| | | | | | | 3.60588 | 74(1/2) deg.|
| 75 deg.| 3.73205 | 3409 | 187 | 18 | 3.72783 17 | | |
| | | | | | | 3.86671 | 75(1/2) deg.|
| 76 deg.| 4.01078 | 4197 | 260 | 51 | 4.00559 22 | | |
| | | | | | | 4.16530 | 76(1/2) deg.|
| 77 deg.| 4.33148 | 5245 | 384 | 64 | 4.32501 07 | | |
+--------+---------+-----------+-----------+-----------+------------+------------+-------------+
If a new table is formed from these values, the intervals being 1/2
deg., it will be found that differences beyond the fourth are
negligible.
To subdivide h into smaller intervals than (1/2)h, various methods may
be used. One is to calculate the sets of quantities which in the new
table will be the successive differences, corresponding to u0, u1, ...
and to find the intermediate terms by successive additions. A better
method is to use a formula due to J. D. Everett. If we write [phi] = 1
- [theta], Everett's formula is, in its most symmetrical form,
([theta] + 1)[theta]([theta] - 1)
u_([theta]) = [theta]u1 + ---------------------------------[delta]^2u1
3!
([theta] + 2)([theta] + 1)[theta]([theta] - 1)([theta] - 2)
+ -----------------------------------------------------------[delta]^4u1 + ...
([phi] + 1)[phi]([phi] - 1)
+ [phi]u0 + ---------------------------[delta]^2u0
3!
([phi] + 2)([phi] + 1)[phi]([phi] - 1)([phi] - 2)
+ -------------------------------------------------[delta]^4u0 + ... (9).
5!