Page:EB1911 - Volume 11.djvu/329

Rh if n1, n2, ... nr, be positive integers, the rational function

is finite at = 1, and has a pole of order n1 at  = c1; the rational function

is thus finite except for = c2, where it has a pole of order n1n2; finally, writing

the rational function

has a pole only at = 1 + c, of order n1n2 ... nr.

The difference (1 − )−1 − U is of the form (1 − )−1P, where P, of the form

1 − (1 − 1) (1 − 2)...(1 − k),

in which there are equalities among 1, 2, ... k, is of the form

1 − 12 + 123 − ...;

therefore, if |ri&#8202;| = |i&#8202;|, we have

now, so long as is without the closed curve above described round = 1, = 1 + c, we have

and hence

Take an arbitrary real positive, and , a positive number, so that mu − 1 < a, then a value of n1 such that n1 < /(1 + ) and therefore n1 /(1 − n1 <, and values for n2, n3 ... such that  n2 < 1/n1  2n1 , n3 < 1/n1n2  3n1 , ... nr < 1/(n1 ... n r−1 ) nr n1 ; then, as 1 + x < ex, we have

and therefore less than

a−1 {exp ( n1 + 2n1 + ... +  nr n1 ) − 1},

which is less than

and therefore less than.

The rational function U, with a pole at = 1 + c, differs therefore from (1 − )−1, for all points outside the closed region put about = 1, = l + c, by a quantity numerically less than. So long as a remains the same, r and will remain the same, and a less value of will require at most an increase of the numbers n1, n2, ... nr; but if a be taken smaller it may be necessary to increase r, and with this the complexity of the function U.

Now put

thereby the points = 0, 1, 1 + c become the points z = 0, 1, ∞, the function (1 − z)−1 being given by (1 − z)−1 = c(c + 1)−1 (1 − )−1 + (c + 1)−1; the function U becomes a rational function of z with a pole only at z = ∞, that is, it becomes a polynomial in z, say [(c + 1)/c] H − 1/c, where H is also a polynomial in z, and

the lines = ±a become the two circles expressed, if z = x + iy, by

the points ( = 0, = 1 − a), ( = 0,  = 1 + c + a) become respectively the points (y = 0, x = c(1 − a)/(c + a), (y = 0, x = −c(l + c + a)/a), whose limiting positions for a = 0 are respectively (y = 0, x = 1), (y = 0, x = −∞). The circle (x + c)2 + y2 = c(c + 1)y/a can be written

where = c(c + 1)/a; its ordinate y, for a given value of x, can therefore be supposed arbitrarily small by taking a sufficiently small.

We have thus proved the following result; taking in the plane of z any finite region of which every interior and boundary point is at a finite distance, however short, from the points of the real axis for which 1 &#8924; x &#8924; ∞, we can take a quantity a, and hence, with an arbitrary c, determine a number r; then corresponding to an arbitrary s, we can determine a polynomial Ps, such that, for all points interior to the region, we have

|(1 − z−1) − Ps| < s;

thus the series of polynomials

P1 + (P2 − P1) + (P3 − P2) + ...,

constructed with an arbitrary aggregate of real positive numbers 1, 2, 3, ... with zero as their limit, converges uniformly and represents (1 − z)−1 for the whole region considered.

§ 12. Expansion of a Monogenic Function in Polynomials, over a Star Region.—Now consider any monogenic function ƒ(z) of which the origin is not a singular point; joining the origin to any singular point by a straight line, let the part of this straight line, produced beyond the singular point, lying between the singular point and z = ∞, be regarded as a barrier in the plane, the portion of this straight line from the origin to the singular point being erased. Consider next any finite region of the plane, whose boundary points constitute a path of integration, in a sense previously explained, of which every point is at a finite distance greater than zero from each of the barriers before explained; we suppose this region to be such that any line joining the origin to a boundary point, when produced, does not meet the boundary again. For every point x in this region R we can then write

where ƒ(x) represents a monogenic branch of the function, in case it be not everywhere single valued, and t is on the boundary of the region. Describe now another region R0 lying entirely within R, and let x be restricted to be within R0 or upon its boundary; then for any point t on the boundary of R, the points z of the plane for which zt − 1 is real and positive and equal to or greater than 1, being points for which |z| = |t| or |z| > |t|, are without the region R0, and not infinitely near to its boundary points. Taking then an arbitrary real positive we can determine a polynomial in xt − 1, say P(xt−1), such that for all points x in R0 we have

|(1 − xt−1)−1 − P(xt−1)| < ;

the form of this polynomial may be taken the same for all points t on the boundary of R, and hence, if E be a proper variable quantity of modulus not greater than ,

where L is the length of the path of integration, the boundary of R, and M is a real positive quantity such that upon this boundary
 * t−1 ƒ(t)| < M. If now

P (xt−1) = c0 + c1xt−1 + ... + c m xm t −m ,

and

this gives

|ƒ(x) − {c00 + c11x + ... + c m m xm}| &#8924; LM/2,

where the quantities 0, 1, 2, ... are the coefficients in the expansion of ƒ(x) about the origin.

If then an arbitrary finite region be constructed of the kind explained, excluding the barriers joining the singular points of ƒ(x) to x = ∞, it is possible, corresponding to an arbitrary real positive number, to determine a number m, and a polynomial Q(x), of order m, such that for all interior points of this region

|ƒ(x) − Q(x)| <.

Hence as before, within this region ƒ(x) can be represented by a series of polynomials, converging uniformly; when ƒ(x) is not a single valued function the series represents one branch of the function.

The same result can be obtained without the use of Cauchy’s integral. We explain briefly the character of the proof. If a monogenic function of t, (t) be capable of expression as a power series in t − x about a point x, for |t − x| &#8924;, and for all points of this circle |(t)| < g, we know that |(n)(x)| < g−n(n!). Hence, taking for n > m we have ( + n) < n, we have
 * z| <, and, for any assigned positive integer , taking m so that

and therefore

where

Now draw barriers as before, directed from the origin, joining the singular point of (z) to z = ∞, take a finite region excluding all these barriers, let be a quantity less than the radii of convergence of all the power series developments of (z) about interior points of this region, so chosen moreover that no circle of radius with centre at an interior point of the region includes any singular point of (z), let g be such that |(z)| < g for all circles of radius whose centres are interior points of the region, and, x being any interior point of the region, choose the positive integer n so that 1/n |x| < ; then take the points a1 = x/n, a2 = 2x/n, a3 = 3x/n, ... an = x; it is supposed that the region is so taken that, whatever x may be, all these are interior points of the region. Then by what has been said, replacing x, z respectively by 0 and x/n, we have