5.5 The Process of Continuation.

Near every point \(P\), \(z_0\), in the neighbourhood of which a function \(f(z)\) is analytic, we have seen that an expansion exists for the function as a series of ascending positive integral powers of \((z - z_0)\), the coefficients in which involve the successive derivates of the function at \(z_0\).

Now let \(A\) be the singularity of \(f(z)\) which is nearest to \(P\). Then the circle within which this expansion is valid has \(P\) for centre and \(PA\) for radius.

Suppose that we are merely given the values of a function at all points of the circumference of a circle slightly smaller than the circle of convergence and concentric with it together with the condition that the function is to be analytic throughout the interior of the larger circle. Then the preceding theorems enable us to find its value at all points within the smaller circle and to determine the coefficients in the Taylor series proceeding in powers of \(z - z_0\). The question arises, Is it possible to define the function at points outside the circle in such a way that the function is analytic throughout a larger domain than the interior of the circle?

In other words, given a power series which converges and represents a function only at points within a circle, to define by means of it the values of the function at points outside the circle.

For this purpose choose any point \(P_1\) within the circle, not on the line \(PA\). We know the value of the function and all its derivates at \(P_1\), from the series, and so we can form the Taylor series (for the same function) with \(P_1\) as origin, which will define a function analytic throughout some circle of centre \(P_1\). Now this circle will extend as far as the singularity[1] which is nearest to \(P_1\), which may or may not be \(A\); but in either case, this new circle will usually[2] lie partly outside the old circle of convergence, and for points in the region which is included in the new circle but not in the old circle, the new series may be used to define the values of the function, although the old series failed to do so.

[1]Of the function defined by the new series.  ↩
[2]The word ‘usually’ must be taken as referring to the cases which are likely to come under the reader’s notice while studying the less advanced parts of the subject. ↩

Similarly we can take any other point \(P_2\), in the region for which the values of the function are now known, and form the Taylor series with \(P_2\) as origin, which will in general enable us to define the function at other points, at which its values were not previously known; and so on.

[3]French, prolongement; German, Fortsetzung.

Editor’s Note: In modern terminology, analytic continuation. ↩

This process is called continuation[3]. By means of it, starting from a representation of a function by any one power series we can find any number of other power series, which between them define the value of the function at all points of a domain, any point of which can be reached from \(P\) without passing through a singularity of the function;[4] and the aggregate of all the power series thus obtained[5] constitutes the analytical expression of the function.

[4]Editor’s Note: Any point of an open, path-connected set can be reached in this way. Whittaker and Watson refer to such a set as a continuum (§3.21). ↩
[5]Such an aggregate of power series has been obtained for various functions by M. J. M. Hill, by purely algebraical processes, Proc. London Math. Soc. xxxv. (1903), pp. 388–416. ↩

It is important to know whether continuation by two different paths \(PBQ\), \(PB'\!Q\) will give the same final power series; it will be seen that this is the case, if the function have no singularity inside the closed curve \(PBQB'\!P\), in the following way: Let \(P_1\) be any point on \(PBQ\), inside the circle \(C\) with centre \(P\); obtain the continuation of the function with \(P_1\) as origin, and let it converge inside a circle \(C_1\); let \(P_1'\) be any point inside both circles and also inside the curve \(PBQB'\!P\); let \(S\), \(S_1\), \(S_1'\) be the power series with \(P\), \(P_1\), \(P_1'\) as origins; then \(S_1 \equiv S_1'\)[6] over a certain domain which will contain \(P_1\), if \(P_1'\) be taken sufficiently near \(P_1\); and hence \(S_1\) will be the continuation of \(S_1'\); for if \(T_1\) were the continuation of \(S_1'\), we have \(T_1 \equiv S_1\) over a domain containing \(P_1\), and so (§3.73) corresponding coefficients in \(S_1\) and \(T_1\) are the same. By carrying out such a process a sufficient number of times, we deform the path \(PBQ\) into the path \(PB'\!Q\) if no singular point is inside \(PBQB'\!P\). The reader will convince himself by drawing a figure that the process can be carried out in a finite number of steps.

[6]Since each is equal to \(S\). ↩
Example. The series \[\frac{1}{a}+\frac{z}{a^2}+\frac{z^2}{a^3}+\frac{z^3}{a^4}+ \cdots\] represents the function \[f(z) = \frac{1}{a - z}\] only for points \(z\) within the circle \(\left|\, z \,\right| = \left|\, a\vphantom{z} \,\right|\).

But any number of other power series exist, of the type \[\frac{1}{a-b}+ \frac{z-b}{(a-b)^2}+\frac{(z-b)^2}{(a-b)^3}+\frac{(z-b)^3}{(a-b)^4} + \cdots ;\] if \(\left. b\middle/ a\right.\) is not real and positive these converge at points inside a circle which is partly inside and partly outside \(\left|\, z \,\right| = \left|\, a\vphantom{z} \,\right|\); these series represent this same function at points outside this circle.

5.501 On functions to which the continuation-process cannot be applied.

It is not always possible to carry out the process of continuation. Take as an example the function \(f(z)\) defined by the power series \[f(z) = 1+z^2 + z^4 + z^8 + z^{16} +...+ z^{2^n} + \cdots , \] which clearly converges in the interior of a circle whose radius is unity and whose centre is at the origin.

Now it is obvious that, as \(z \rightarrow 1-0\), \(f(z) \rightarrow +\infty\); the point \(+1\) is therefore a singularity of \(f(z)\).

But \[f(z) = z^2 +f(z^2),\] and if \(z^2 \rightarrow 1-0\), \(f(z^2) \rightarrow \infty\) and so \(f(z) \rightarrow \infty\), and hence the points for which \(z^2 =1\) are singularities of \(f(z)\); the point \(z= - 1\) is therefore also a singularity of \(f(z)\).

Similarly since \[f(z)=z^2 +z^4+f(z^4),\] we see that if \(z\) is such that \(z^4 = 1\), then \(z\) is a singularity of \(f(z)\); and, in general, any root of any of the equations \[z^2 = 1, \;z^4 = 1,\; z^8 = 1,\; z^{16} = 1, \;\dots ,\] is a singularity of \(f(z)\). But these points all lie on the circle \(\left|\, z \,\right| = 1\); and in any arc of this circle, however small, there are an unlimited number of them. The attempt to carry out the process of continuation will therefore be frustrated by the existence of this unbroken front of singularities, beyond which it is impossible to pass.

[7]Editor’s Note: The term lacunary refers to the increasingly large gaps in the power series, which give rise to the limiting circle. (See Mandelbrojt, Researches on the Singularities of Functions Defined by Taylor’s Series (1927), Chs. v., xii.) ↩

In such a case the function \(f(z)\) cannot be continued at all to points \(z\) situated outside the circle \(\left|\, z \,\right| = 1\); such a function is called a lacunary function, and the circle is said to be a limiting circle for the function.[7]

5.51 The identity of two functions.

The two series \[1 + z + z^2 + z^3 + \cdots\] and \[ -1 +(z - 2) - (z - 2)^2 + (z - 2)^3 - (z - 2)^4 + \cdots\] do not both converge for any value of \(z\), and are distinct expansions. Nevertheless, we generally say that they represent the same function, on the strength of the fact that they can both be represented by the same rational expression \(\dfrac{1}{1-z}\).

This raises the question of the identity of two functions. When can two different expansions be said to represent the same function?

We might define a function (after Weierstrass), by means of the last article, as consisting of one power series together with all the other power series which can be derived from it by the process of continuation. Two different analytical expressions will then define the same function, if they represent power series derivable from each other by continuation.

Since if a function is analytic (in the sense of Cauchy, §5.12) at and near a point it can be expanded into a Taylor’s series, and since a convergent power series has a unique differential coefficient (§5.3), it follows that the definition of Weierstrass is really equivalent to that of Cauchy.[8]

[8]Editor’s Note: The definitions are equivalent provided, in Cauchy’s definition, we call \(f(z)\) analytic at \(a\) if it is differentiable in an open set containing \(a\).  ↩

It is important to observe that the limit of a combination of analytic functions can represent different analytic functions in different parts of the plane. This can be seen by considering the series \[\frac{1}{2}\left(z+\frac{1}{z}\right) + \sum_{n-1}^\infty \left(z-\frac{1}{z}\right)\left(\frac{1}{1+z^n}-\frac{1}{1+z^{n-1}}\right).\]

The sum of the first \(n + 1\) terms of this series is \[\frac{1}{z}+\left(z-\frac{1}{z}\right)\frac{1}{1+z^n}.\] The series therefore converges for all values of \(z\) (zero excepted) not on the circle \(\left|\, z\, \right| = 1\). Put, as \(n \rightarrow \infty\), \(\left|\, z^n\, \right| \rightarrow 0\) or \(\left|\, z^n\, \right| \rightarrow \infty\) according as \(\left|\, z\, \right|\) is less or greater than unity; hence we see that the sum to infinity of the series is \(z\) when \(\left|\, z\, \right|< 1\), and \(\dfrac{1}{z}\) when \(\left|\, z\, \right| > 1\). This series therefore represents one function at points in the interior of the circle \(\left|\, z\, \right| = 1\), and an entirely different function at points outside the same circle. The reader will see from §5.3 that this result is connected with the non-uniformity of the convergence of the series near \(\left|\, z\, \right|= 1\).

[9]Proc. Math. Congress, Cambridge (1912), i. pp. 137–138. Leçons sur les fonctions monogènes (1917). The functions are not monogenic strictly in the sense of §5.12 because, in the example quoted, in working out \(\left.\{f(z + h)-f(z)\}\middle/ h\right.\), it must be supposed that \(\mathfrak{Re}(z + h)\) and \(\mathfrak{Im}(z + h)\) are not both rational fractions.  ↩

It has been shewn by Borel[9] that if a region \(C\) is taken and a set of points \(S\) such that points of the set \(S\) are arbitrarily near every point of \(C\), it may be possible to define a function which has a unique differential coefficient (i.e. is monogenic) at all points of \(C\) which do not belong to \(S\); but the function is not analytic in \(C\) in the sense of Weierstrass.

Such a function is \[f(z) \equiv \sum_{n=1}^\infty \sum_{p=0}^n \sum_{q=0}^n \frac{\exp(-\exp n^4)}{z-\left.(p+qi)\middle/n\right.}.\]