102 
Miscellanea 
and it follows that if we use a prime number {p) as our interval, we get with each interpolation 
(p-\\(p-l 
line ' ^2^ cases ; but as we require 
2 
" + 1 
cases in all, the number of interpolation-lines required is wf^—r— -^+1 
This aives the 
following table : 
Interval 
No. of lines 
11 
13 
4 
17 
5 
19 
5 
Turning from prime numbers to those containing factors we find at ouce that many more 
lines ai'e required. Thus we have already seen that an interval of 10 calls for no fewer than 
six interpolation-lines, and the intervals of 6 and 8 will both be found to require four. The 
interval 9 requires three lines. 
Let us now compare the 1 1 interval with the 8 and 10 intervals and see which is the best 
grouping. 
Interval 11. 
Ordinary interpolation Line 1, 1 
gives 
0, 1 
0, 2 
0, 3 
0, 4 
0, 5 
gives 
1, 1 
2, 2 
3, 3 
4,4 
5, 5 
Line 1, 2 
gives 
1, 2 
2, 4 
3, 6 = 3, 5 
4, 8 = 4,3 
5, 10 = 5, 1 
Line 1, 3 
gives 
1, 3 
2, 6 = 2, 5 
3, 9 = 3, 2 
4, 12 = 4, 1 
5, 15 = 5, 4 
Average interval in interpolation in x 
)) )) )) 5) 11 y 
Maximum interval in ... x 
y 
-■ 8-25 
= 15-4 
= 11 
= 33 
Interval 10. 
Ordinary 
interpolation 
Line 1, 1 
Line 1, 2 Line 1, 3 
Line 1, 4*' 
Line 1, 5 
Line 2, 5 
gives 
gives 
gives gives 
gives 
gives 
gives 
0, 1 
1, 1 
1,2 1,3 
1, 4 
1, 5 
2,5 
0, 2 
2, 2 
2,4 2,6 = 2,4| 
also by 1,2 ^ 
given above 
0, 3 
3, 3 
3,6 = 3,4 3,9 = 3,1 
3, 12 = 3,2 
3,5 
6,5 = 4,5 
0,4 
4, 4 
4, 8 = 4, 2 
0,5 
5,5 
which has already 
been obtained 
Average interpolation interval in x = 8-5 or 9'5 
„ =23-5 or 22-5 
Maximum interval in ... x =20 
y =50 
2, 3 would give the same results and the averages resulting in each case are given. 
