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.

~ 188. 189.

The implicit algebraic function.

351

is not a branching point for this first leaf; if it do, let us establish by
repeated expansion, employing the newly found value for the point
z = a, how many leaves are connected with the first in the point.
Let these be called 1, 2,... p, then for each of these leaves the
significance of any other critical point can be determined by means
of the loops. In this way too we shall become acquainted in another
branching point with the value for - = a belonging to a new leaf
that did not occur among the values 1, 2,... p.
If the case were to occur, that the circuit of all the loops does
not lead beyond the leaves 1 to p, or, in general, that all the n values
which belong to z =- a are not obtained in this process, this signifies,
as we shall subsequently prove (~ 199), that the algebraic function
f(zm, wtv) may be resolved into factors which are rational in s and tw.
In the case of an irreducible function this does not occur, the theorem
holds good for it as above enunciated.
189. Having established that analytic functions can be expressed
by series of ascending positive integer powers, it is still necessary that
we should discuss their singular points, in order that we may ascertain
how analytic functions admit of expansion in the neighbourhood of
any singular point according to its kind.
We must first of all premise that in a domain wherein a function
is generally a unique analytic function, it can have no other singularities
than the non-essential and essential points in which it becomes infinite.
For, if an analytic function f(z) have finite discontinuities in separate
points a of a domain, the product ( - a)f(z) is an analytic function
in the neighbourhood of and including the point a at which it is
zero, and its first derived function is equal to Lim f(z) for S =a.
Accordingly there is an expansion of the form:
(z - a)f(s) = Lim f(z) ~ (z - a) + a (z - a)2 + ca (z - c)3 -+. etc..
Z=a
Therefore the value of the function f(z) at the point a must be
Lim f(z), thus this point is regular.
z=-a
It is likewise impossible (see ~ 181. 3. c) that an analytic function
while itself finite and continuous in a domain should become singular
in separate points or along an entire curve in consequence of any
af af
violation of the equation f + i - = 0 between its partial derivates.
The integral ff(z)dz even up to any point a of the irregular curve
would then be an analytic function; for the integral is continuous
and its derived function is Lim f(z) for 2 = —. But an analytic
function, as was proved, has arbitrarily many successive derivates all of
which are continuous functions. Therefore f(S) must have a determinate