Page 21 - Foundations Of Differential Calculus

4 1. On Finite Differences
Fifth Differences:

5
∆ y, ...
Each of these series comes from the preceding series by subtracting each
term from its successor. Hence, no matter what function of x we substitute
I
for y, it is easy to find each of the series of differences, since the values y ,
II
III
y , y ,... are easily found from the definition of the function.
I
I
8. Let y = x, so that y = x = x + ω, y II = x II = x +2ω, and so
II
I
forth. When we take the differences, ∆x = ω,∆x = ω,∆x = ω, ... , the
result is that all of the first differences of x are constant, so that all of the
second differences vanish, as do the third differences and all those of higher
orders. Since ∆x = ω, it is convenient to use the notation ∆x instead of ω.
I II III
Since we are assuming that the successive values x, x , x , x , ... form
I II
an arithmetic progression, the differences ∆x,∆x ,∆x , ... are constants
3 4
and mutually equal. It follows that ∆∆x =0, ∆ x =0, ∆ x = 0, and so
forth.
9. We have assumed that the successive values of x are terms of an arith-
metic progression, so that the values of its first differences are constant and
its second and succeeding differences vanish. Although the choice is freely
ours to make among all possible progressions, still we usually choose the
progression to be arithmetic, since it is both the simplest and easiest to
understand, and also it has the greatest versatility, in that x can assume
absolutely any value. Indeed, if we give ω either negative or positive values
in this series, the values of x will always be real numbers. On the other
hand, if the series we have chosen is geometric, there is no place for neg-
ative values. For this reason the nature of functions y is best determined
from the values of x chosen from an arithmetic progression.
I
10. Just as ∆y = y − y, so all the higher differences can also be defined
I
III
II
from the terms of the first series: y, y , y , y , ... . Since
I II I
∆y = y − y ,
we have
II I
∆∆y = y − 2y + y
and
I
II
I
∆∆y = y III − 2y + y .
Furthermore,
3 I III II I
∆ y =∆∆y − ∆∆y = y − 3y +3y − y;