An introduction to the study of the elements of the differential and integral calculus. From the German of the late Axel Harnack, With the permission of the author.

58

Successive differentiation of explicit functions.

Bk. ]. ch. VII.

We perceive as before, that h may be assumed = Ax and then both
made simultaneously converge to zero.
There is no difficulty in establishing the general equation of this
kind, by showing that when it holds for n it is also true for n + 1.
33. These new expressions are no doubt less suited for calculating
the higher differential quotients than those first formed; but they exhibit
them to us as limiting values of higher quotients of differences, which
is of importance for the theory. Euler (1707-1783) in his work:
Institutiones calculi differentialis, Petrop. 1755, gave the following
convenient exposition of this. If we denote the values of the function
y = f(x) which belong to the arguments
x, x + Ax, x + 2Anx,.. x + nAx, respectively by
Y, Y, Y 2 * Yn,
we get the series of first differences:
Y1 - Y = Ay, Y2 - Y = AY1, Y3 - 2 = AY2,.. n - y-1 = AYn-1.
From these we form the series of second differences:
A y1 - Ay = A'Y, A y2 -- Al - a, A3 -- A A22, ~= ~.
A Yn- - A yn-2 = A2Yn-_
and so of third differences:
A2y1 - A2y == A3y, A2y2 - A2Y1 = A3y1, A2 3 A- 2 y2==...
A2Yn-2 -- A2n-3 = A3Yn_3,
on to the nth difference:
An-ly1 - An-1 Y = An Y.
If we propose to express the higher differences by the original values
of the function we find:
Ay ==y - y,
A2 = (Y2 - 1) - (Y - Y) = 2 - 2y1 +- =-f(x+2Ax)-2f(x+Ax)+f(x),
A3y= { (Y3-Y2)-(Y2 - Y1) - { (Y2 — Y)-(1 - Y)} I = Y-33Y2+ 3y-1-y
=f(x+3Ax)-3f(x+2 [ Ax) +3f(x+A x)-f(x),..
whence results, that f"(x) = Lim Xy, fY() = Lim A3 etc.. This
accounts for the notation for the higher differential quotients
d2y d'y _. c _y im Ay
d~z' d~3' etc.; in general: - Limr " n
dx^l dX3 5 0 dxn A Xn
We can also form higher differences of the values of the independent
variable, but they are found to vanish. For, from the values
x1 = x - Ax, x 2= x + 2Ax, x3 = x + 3Ax,... x = x + nAx
we find: