(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°. The differences of odd order are omitted for convenience of printing.
Example 5.
| x. | u ≡ tan x. | δ2u. | δ4u. | δ6u. | U. | u = mean of values of U. | x. |
| + | + | + | |||||
| 73° | 3.27085 | 2339 | 100 | 5 | 3.26794 95 | ||
| 3.37594 | 73½° | ||||||
| 74° | 3.48741 | 2808 | 132 | 23 | 3.48392 98 | ||
| 3.60588 | 74½° | ||||||
| 75° | 3.73205 | 3409 | 187 | 18 | 3.72783 17 | ||
| 3.86671 | 75½° | ||||||
| 76° | 4.01078 | 4197 | 260 | 51 | 4.00559 22 | ||
| 4.16530 | 76½° | ||||||
| 77° | 4.33148 | 5245 | 384 | 64 | 4.32501 07 |
If a new table is formed from these values, the intervals being ½°, it will be found that differences beyond the fourth are negligible.
To subdivide h into smaller intervals than ½h, 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 φ = 1 − θ, Everett’s formula is, in its most symmetrical form,
| uθ = θu1 + | (θ + 1) θ (θ − 1) | δ2u1 + | (θ + 2) (θ + 1) θ (θ − 1) (θ − 2) | δ4u1 + ... |
| 3! | 5! |
| + φu0 + | (φ + 1) φ (φ − 1) | δ2u0 + | (φ + 2) (φ + 1) φ (φ − 1) (φ − 2) | δ4u0 + ... |
| 3! | 5! |
(9).
For actual calculations a less symmetrical form may be used. Denoting