7.4 The expansion of a class of functions in rational fractions.[1]
Consider a function \(f(z)\), whose only singularities in the finite part of the plane are simple poles \(a_{1}\), \(a_{2}\), \(a_{3},\ldots\), where \(\left|\, a_{1} \,\right| \leq \left|\, a_{2} \,\right| \leq \left|\, a_{3} \,\right| \leq \ldots \:\!\); let \(b_{1}\), \(b_{2}\), \(b_{3},\ldots\), be the residues at these poles, and let it be possible to choose a sequence of circles \(C_{m}\) (the radius of \(C_{m}\) being \(R_{m}\)) with centre at \(O\), not passing through any poles, such that \(\left|\, f(z) \,\right|\) is bounded on \(C_{m}\). (The function \(\mathrm{cosec}\;\! z\) may be cited as an example of the class of functions considered, and we take \(R_{m} = (m + \frac{1}{2})\pi\).) Suppose further that \(R_{m} \rightarrow \infty\) as \(m \rightarrow \infty\) and that the upper bound[2] of \(\left|\, f(z) \,\right|\) on \(C_{m}\) is itself bounded as \(m\rightarrow\infty\);[3] so that, for all points on the circle \(C_{m}\), \(\left|\, f(z) \,\right| < M\), where \(M\) is independent of \(m\).
Then, if \(x\) be not a pole of \(f(z)\), since the only poles of the integrand are the poles of \(f(z)\) and the point \(z = x\), we have, by §6.1, \[ \frac{1}{2 \pi i} \!\int_{C_{m}}\! \frac{f(z)}{z-x} \, d z = f(x) + \sum_{r} \frac{b_{r}}{a_{r}-x}. \] where the summation extends over all poles in the interior of \(C_m\).
But \[\begin{align*} \frac{1}{2 \pi i} \!\int_{C_{m}}\! \frac{f(z)}{z-x} \, d z &= \frac{1}{2 \pi i} \!\int_{C_{m}}\! \frac{f(z)}{z} \, d z + \frac{x}{2 \pi i} \!\int_{C_{m}}\! \frac{f(x)}{z(z-x)} \, d z \\ \\ &= f(0) + \sum_{r} \frac{b_{r}}{a_{r}} + \frac{x}{2 \pi i} \!\int_{C_{m}}\! \frac{f(z)}{z(z-x)} \, d z, \end{align*}\] if we suppose the function \(f(z)\) to be analytic at the origin.
Now as \(m \rightarrow \infty\), \(\displaystyle \int_{C_{m}}\! \frac{f(z)}{z(z-x)} \, d z\) is \(O(R_{m}^{-1})\), and so tends to zero as \(m\) tends to infinity.
Therefore, making \(m \rightarrow \infty\), we have \[ 0 = f(x) - f(0) + \sum_{n=1}^{\infty} b_{n} \left( \frac{1}{a_{n}-x} - \frac{1}{a_{n}} \right) - \lim_{m\rightarrow\infty} \frac{x}{2 \pi i} \!\int_{C_{m}}\! \frac{f(z)}{z(z-x)} \, d x, \] i.e. \[ f(x) = f(0) + \sum_{n=1}^{\infty} b_{n} \left\{ \frac{1}{x-a_{n}} + \frac{1}{a_{n}} \right\} \] which is an expansion of \(f(x)\) in rational fractions of \(x\); and the summation extends over all the poles of \(f(x)\).
If \(\left|\, a_{n} \,\right| < \left|\, a_{n+1} \,\right|\), this series converges uniformly throughout the region given by \(\left|\, x \,\right| < a\), where \(a\) is any constant (except near the points \(a_{n}\)). For if \(R_{m}\) be the radius of the circle which encloses the points \(\left|\, a_{1} \,\right|, \ldots, \left|\, a_{n} \,\right|\), the modulus of the remainder of the terms of the series after the first \(n\) is \[ \left|\, \frac{x}{2 \pi i} \!\int_{C_{m}}\! \frac{f(z)}{z(z-x)} \, d z \,\right| < \frac{Ma}{R_{m}-a}, \] by §4.62; and, given \(\epsilon\), we can choose \(n\) independent of \(x\) such that \(\displaystyle \frac{Ma}{R_{m}-a} < \epsilon\).
The convergence is obviously still uniform even if \(\left|\, a_{n} \,\right| \leq \left|\, a_{n+1} \,\right|\) provided the terms of the series are grouped so as to combine the terms corresponding to poles of equal moduli.
If, instead of the condition \(\left|\, f(z) \,\right| < M\), we have the condition \(\left|\, z^{-p} f(z) \,\right| < M\), where \(M\) is independent of \(m\) when \(z\) is on \(C_{m}\), and \(p\) is a positive integer, then we should have to expand \(\displaystyle \int_{C} \frac{f(z)}{z-x} \, d z\) by writing \[ \frac{1}{z-x} = \frac{1}{z} + \frac{x}{z^{2}} + \cdots + \frac{x^{p+1}}{z^{p+1}(z-x)}, \] and should obtain a similar but somewhat more complicated expansion.
Example 1. Prove that \[ \mathrm{cosec}\;\! z = \frac{1}{z} + \sum (-1)^{n} \left(\frac{1}{z-n\pi} + \frac{1}{n\pi}\right) \] the summation extending to all positive and negative values of \(n\).
To obtain this result, let \(\mathrm{cosec}\;\! z - \dfrac{1}{z} = f(z)\). The singularities of this function are at the points \(z=n\pi\), where \(n\) is any positive or negative integer.
The residue of \(f(z)\) at the singularity \(n\pi\) is therefore \((-1)^{n}\), and the reader will easily see that \(\left|\, f(z) \,\right|\) is bounded on the circle \(\left|\, z \,\right| = (n + \frac{1}{2}\:\!) \pi\) as \(n \rightarrow \infty\).
Applying now the general theorem \[ f(z) = f(0) + \sum c_{n} \left[ \frac{1}{z-a_{n}} + \frac{1}{a_{n}} \right], \] where \(c_{n}\) is the residue at the singularity \(a_{n}\), we have \[ f(z) = f(0) + \sum (-1)^{n} \left\{ \frac{1}{z-n\pi} + \frac{1}{n\pi} \right\}. \]
But \[ f(0) = \lim_{z \rightarrow 0} \frac{z - \sin z}{z \sin z} = 0. \]
Therefore \[ \mathrm{cosec}\;\! z = \frac{1}{z} + \sum (-1)^{n} \left[ \frac{1}{z-n\pi} + \frac{1}{n\pi} \right], \] which is the required result.
Example 2. If \(0 < a < 1\), shew that \[ \frac{e^{az}}{e^{z}-1} = \frac{1}{z} + \sum_{n=1}^{\infty} \frac{2z \cos 2na\pi - 4n\pi \sin 2na\pi}{z^{2}+4n^{2}\pi^{2}}. \]
Example 3. Prove that \[\begin{align*} \frac{1}{2 \pi x^{2} (\cosh x - \cos x)} =& \frac{1}{2 \pi x^{4}} - \frac{1}{e^{\pi}-e^{-\pi}} \frac{1}{\pi^{4} + \frac{1}{4} x^{4}}\\ &+ \frac{2}{e^{2 \pi}-e^{-2 \pi}} \frac{1}{(2 \pi)^{4} + \frac{1}{4} x^{4}} - \frac{3}{e^{3 \pi}-e^{-3 \pi}} \frac{1}{(3 \pi)^{4} + \frac{1}{4} x^{4}} + \cdots. \end{align*}\]
The general term of the series on the right is \[ \frac{(-1)^{r} r}{(e^{r \pi}-e^{-r \pi}) \left\{(r \pi)^{4} + \frac{1}{4} x^{4}\right\}}, \] which is the residue at each of the four singularities \(r\), \(-r\), \(ri\), \(-ri\) of the function \[ \frac{\pi z}{(\pi^{4}z^{4} + \frac{1}{4} x^{4}) (e^{\pi z} - e^{-\pi z}) \sin \pi z}. \]
The singularities of this latter function which are not of the type \(r\), \(-r\), \(ri\), \(-ri\) are at the five points \[ 0, \frac{(\pm 1 \pm i\:\!) x}{2 \pi}. \] At \(z=0\) the residue is \[ \frac{2}{\pi x^{4}}; \] at each of the four points \(\displaystyle \frac{(\pm 1 \pm i\:\!) x}{2 \pi}\), the residue is \[ \left\{ 2 \pi x^{2} (\cos x - \cosh x) \right\}^{-1} \]
Therefore \[\begin{align*} 4 \sum_{r=1}^{\infty} \frac{(-1)^{r} r}{e^{r\pi} - e^{-r\pi}}& \frac{1}{(r\pi)^{4} + \frac{1}{4} x^{4}} + \frac{2}{\pi x^{4}} - \frac{2}{ \pi x^{2} (\cosh x - \cos x)} \\ & = \frac{1}{2 \pi i} \lim_{n \rightarrow \infty} \!\int_{C}\! \frac{\pi z}{ (\pi^{4}z^{4} + \frac{1}{x^{4}}) (e^{\pi z}-e^{-\pi z}) \sin \pi z } \, d z, \end{align*}\] where \(C\) is the circle whose radius is \(n + \frac{1}{2}\), (\(n\) an integer), and whose centre is the origin. But, at points on \(C\), this integrand is \(O( \left|\, z \,\right|^{-3} )\); the limit of the integral round \(C\) is therefore zero. From the last equation the required result is now obvious.
Example 4. Prove that \[ \sec x = 4 \pi \left( \frac{1}{\pi^{2} - 4 x^{2}} - \frac{3}{9 \pi^{2} - 4 x^{2}} + \frac{5}{25 \pi^{2} - 4 x^{2}} - \cdots \right). \]
Example 5. Prove that \[ \mathrm{cosech}\;\! x = \frac{1}{x} - 2x \left( \frac{1}{\pi^{2} + x^{2}} - \frac{1}{4\pi^{2} + x^{2}} + \frac{1}{9 \pi^{2} + x^{2}} - \cdots \right). \]
Example 6. Prove that \[ \mathrm{sech}\;\! x = 4 \pi \left( \frac{1}{\pi^{2} + 4 x^{2}} - \frac{3}{9\pi^{2} + 4 x^{2}} + \frac{5}{25 \pi^{2} + 4 x^{2}} - \cdots \right). \]
Example 7. Prove that \[ \coth x = \frac{1}{x} + 2x \left( \frac{1}{\pi^{2} + x^{2}} + \frac{1}{4 \pi^{2} + x^{2}} + \frac{1}{9 \pi^{2} + x^{2}} + \cdots \right). \]
Example 8. Prove that \[ \sum_{m=-\infty}^{\infty} \sum_{n=-\infty}^{\infty} \frac{1}{ (m^{2}+a^{2}) (n^{2}+b^{2}) } = \frac{\pi^{2}}{ab} \coth \pi a \coth \pi b. \] (Math. Trip. 1899.)
7.5 The expansion of a class of functions as infinite products.
The theorem of the last article can be applied to the expansion of a certain class of functions as infinite products.
For let \(f(z)\) be a function which has simple zeros at the points \(a_{1}\), \(a_{2}\), \(a_{3}, \ldots\),[4] where \(\lim_{n \rightarrow \infty} \left|\, a_{n} \,\right|\) is infinite; and let \(f(z)\) be analytic for all values of \(z\).
Then \(f'(z)\) is analytic for all values of \(z\) (§5.22), and so \(\frac{f'(z)}{f(z)}\) can have singularities only at the points \(a_{1}\), \(a_{2}\), \(a_{3}, \ldots\).
Consequently, by Taylor’s theorem, \[ f(z) = (z-a_{r})\:\! f'(a_{r}) + \frac{ (z-a_{r})^{2} }{2} f''(a_{r}) + \cdots \] and \[ f'(z) = f'(a_{r}) + (z-a_{r}) \:\! f''(a_{r}) + \cdots. \]
It follows immediately that at each of the points \(a_{r}\), the function \(\frac{f'(z)}{f(z)}\) has a simple pole, with residue \(+1\).
If then we can find a sequence of circles \(C_{m}\) of the nature described in §7.4, such that \(\frac{f'(z)}{f(z)}\) is bounded on \(C_{m}\) as \(m \rightarrow \infty\), it follows, from the expansion given in §7.4, that \[ \frac{f'(z)}{f(z)} = \frac{f'(0)}{f(0)} + \sum_{n=1}^{\infty} \left\{ \frac{1}{z-a_{n}} - \frac{1}{a_{n}} \right\}. \]
Since this series converges uniformly when the terms are suitably grouped (§7.4), we may integrate term-by-term (§4.7}. Doing so, and taking the exponential of each side, we get \[ f(z) = c e^{ z \:\!\frac{f'(0)}{f(0)} } \prod_{n=1}^{\infty} \left\{ \left( 1 - \frac{z}{a_{n}} \right) e^{ \frac{z}{a_{n}} } \right\}, \] where \(c\) is independent of \(z\).
Putting \(z = 0\), we see that \(f(0) = c\), and thus the general result becomes \[ f(z) = f(0) e^{ z \:\!\frac{f'(0)}{f(0)} } \prod_{n=1}^{\infty} \left\{ \left( 1 - \frac{z}{a_{n}} \right) e^{ \frac{z}{a_{n}} } \right\}. \]
This furnishes the expansion, in the form of an infinite product, of any function \(f(z)\) which fulfils the conditions stated.
Example 1. Consider the function \(f(z) = \dfrac{\sin z}{z}\), which has simple zeros at the points \(r \pi\), where \(r\) is any positive or negative integer.
In this case we have \[ f(0) = 1, \quad f'(0) = 0, \] and so the theorem gives immediately \[ \frac{\sin z}{z} = \prod_{n=1}^{\infty} \left\{ \left(1 - \frac{z}{n \pi}\right) e^{ \frac{z}{n \pi} } \right\} \left\{ \left(1 + \frac{z}{n \pi}\right) e^{ -\frac{z}{n \pi} } \right\}; \] for it is easily seen that the condition concerning the behaviour of \(\frac{f'(z)}{f(z)}\) as \(\left|\, z \,\right| \rightarrow \infty\) is fulfilled.
Example 2. Prove that \[\begin{align*} \left\{ 1 + \left( \frac{k}{x} \right)^{2} \right\} & \left\{ 1 + \left( \frac{k}{2\pi - x} \right)^{2} \right\} \left\{ 1 + \left( \frac{k}{2\pi + x} \right)^{2} \right\} \\ \times & \left\{ 1 + \left( \frac{k}{4\pi - x} \right)^{2} \right\} \left\{ 1 + \left( \frac{k}{4\pi + x} \right)^{2} \right\} \times \cdots \\ \\ & \quad = \frac{\cosh x - \cos x}{1-\cos x}. \end{align*}\] (Trinity, 1899.)
7.6 The factor theorem of Weierstrass.[5]
The theorem of §7.5 is very similar to a more general theorem in which the character of the function \(f(z)\), as \(\left|\, z \,\right| \rightarrow \infty\), is not so narrowly restricted.
Let \(f(z)\) be a function of \(z\) with no essential singularities (except at ‘the point infinity’); and let the zeros and poles of \(f(z)\) be at \(a_{1}\), \(a_{2}\), \(a_{3}, \ldots\), where \(0 < \left|\, a_{1} \,\right| \leq \left|\, a_{2} \,\right| \leq \left|\, a_{3} \,\right| \ldots\). Let the zero at \(a_{n}\) be of (integer) order \(m_{n}\).[6]
If the number of zeros and poles is unlimited, it is necessary that \(\left|\, a_{n} \,\right| \rightarrow \infty\), as \(n \rightarrow \infty\); for, if not, the points \(a_{n}\) would have a limit point,[7] which would be an essential singularity of \(f(z)\).
We proceed to shew first of all that it is possible to find polynomials \(g(z)\) such that \[ \prod_{n=1}^{\infty} \left\{ \left( 1 - \frac{z}{a_{n}} \right) e^{ g_{n}(z)} \right\}^{m_{n}} \] converges for all finite values of \(z\).[8]
Let \(K\) be any constant, and let \(\left|\, z \,\right| < K\); then, since \(\left|\, a_{n} \,\right| \rightarrow \infty\), we can find \(N\) such that, when \(n > N\), \(\left|\, a_{n} \,\right| > 2K\).
The first \(N\) factors of the product do not affect its convergence;[8] consider any value of \(n\) greater than \(N\), and let \[ g_{n}(z) = \frac{z}{a_{n}} + \frac{1}{2} \left( \frac{z}{a_{n}} \right)^{2} + \cdots + \frac{1}{k_{n}-1} \left( \frac{z}{a_{n}} \right)^{k_{n} - 1}. \] Then \[\begin{align*} \left|\, - \sum_{m=1}^{\infty} \frac{1}{m} \left( \frac{z}{a_{n}} \right)^{m} + g_{n} \,\right| &= \left|\, \sum_{m=k_{n}}^{\infty} \frac{1}{m} \left( \frac{z}{a_{n}} \right)^{m} \,\right| \\ &< \left|\, \frac{z}{a_{n}} \,\right|^{k_{n}} \sum_{m=0}^{\infty} \left|\, \frac{z}{a_{n}} \,\right|^{m} \\ &< 2 \left|\, (K a_{n}^{-1})^{k_{n}} \,\right|, \end{align*}\] since \( \left|\, z_{n} a_{n}^{-1} \,\right| < \frac{1}{2}\).
Hence \[ \left\{ \left( 1 - \frac{z}{a_{n}} \right) e^{g_{n}(z)} \right\}^{m_{n}} = e^{u_{n}(z)}, \] where \[ \left|\, u_{n}(z) \,\right| \leq 2 \left|\, m_{n} (K a_{n}^{-1})^{k_{n}} \,\right|. \]
Now \(m_{n}\) and \(a_{n}\) are given, but \(k_{n}\) is at our disposal; since \(K a_{n}^{-1} < \frac{1}{2}\), we choose \(k_{n}\) to be the smallest number such that \(2 \left|\, m_{n} (K a_{n}^{-1})^{k_{n}} \,\right| < b_{n}\), where \(\sum_{n=1}^{\infty} b_{n}\) is any convergent series of positive terms.[10]
Hence \[ \prod_{n = N+1}^{\infty} \left\{ \left( 1 - \frac{z}{a_{n}} \right) e^{g_{n}(z)} \right\}^{m_{n}} = \prod_{n = N+1}^{\infty} e^{u_{n}(z)}, \] where \( \left|\, u_{n}(z) \,\right| < b_{n} \); and therefore, since \(b_{n}\) is independent of \(z\), the product converges absolutely and uniformly when \(\left|\, z \,\right| < K\), except near the points \(a_{n}\).
Now let \[ F(z) = \prod_{n=1}^{\infty} \left[ \left\{ \left( 1 - \frac{z}{a_{n}} \right) e^{g_{n}(z)} \right\}^{m_{n}} \right]. \]
Then, if \(f(z) \div F(z) = G_{1}(z)\), \(G_{1}(z)\) is an integral function (§5.64) of \(z\) and has no zeros.
It follows that \(\dfrac{1}{G_{1}(z)} \dfrac{d}{d z} G_{1}(z)\) is analytic for all finite values of \(z\); and so, by Taylor’s theorem, this function can be expressed as a series \(\sum_{n=1}^{\infty} n b_{n} z^{n-1}\) converging everywhere; integrating, it follows that \[ G_{1}(z) = c e^{G(z)}, \] where \(G(z) = \sum_{n=1}^{\infty} b_{n} z^{n}\) and \(c\) is a constant; this series converges everywhere, and so \(G(z)\) is an integral function.
Therefore, finally, \[ f(z) = f(0) e^{G(z)} \prod_{n=1}^{\infty} \left\{ \left( 1 - \frac{z}{a_{n}} \right) e^{g_{n}(z)} \right\}^{m_{n}}, \] where \(G(z)\) is some integral function such that \(G(0) = 0\).
[Note. The presence of the arbitrary element \(G(z)\) which occurs in this formula for \(f(z)\) is due to the lack of conditions as to the behaviour of \(f(z)\) as \(\left|\, z \,\right| \rightarrow \infty\).]
Corollary. If \(m_{n} =1\), it is sufficient to take \(k_{n} = n\), by §2.36.