A COURSE OF MODERN ANALYSIS

7.3 Bürmann’s theorem[1]

We shall next consider several theorems which have for their object the expansion of one function in powers of another function.

Let \(\phi(z)\) be a function of \(z\) which is analytic in a closed region \(S\) of which \(a\) is an interior point; and let \[ \phi(a) = b. \]

Suppose also that \(\phi'(a) \neq 0\). Then Taylor’s theorem furnishes the expansion \[ \phi(z) - b = \phi'(a) (z-a) + \frac{1}{2} \phi''(a) (z-a)^{2} + \cdots, \] and if it is legitimate to revert this series we obtain \[ z - a = \frac{1}{\phi'(a)} \left\{ \phi(z) - b \right\}- \frac{1}{2} \frac{\phi''(a)}{\left\{\phi'(a)\right\}^{3}} \left\{ \phi(z) - b \right\}^{2} + \cdots, \] which expresses \(z\) as an analytic function of the variable \(\left\{ \phi(z) - b \right\}\), for sufficiently small values of \(\left|\,z-a\,\right|\). If then \(f(z)\) be analytic near \(z = a\), it follows that \(f(z)\) is an analytic function of \(\left\{ \phi(z) - b \right\}\) when \(\left|\,z - a\,\right|\) is sufficiently small, and so there will be an expansion of the form \[ f(z) = f(a) + a_{1} \left\{ \phi(z) - b \right\} + \frac{a_{2}}{2!} \left\{ \phi(z) - b \right\}^{2} + \frac{a_{3}}{3!} \left\{ \phi(z) - b \right\}^{3} + \cdots. \]

The actual coefficients in the expansion are given by the following theorem, which is generally known as Bürmann’s theorem.

Let \(\psi(z)\) be a function of \(z\) defined by the equation \[ \psi(z) = \frac{z-a}{ \phi(z) - b }; \] then an analytic function \(f(z)\) can, in a certain domain of values of \(z\), be expanded in the form \[ f(z) = f(a) + \sum_{m=1}^{n-1} \frac{ \left\{\phi(z)-b\right\}^{m} }{m!} \frac{d^{m-1}}{d a^{m-1}}\! \left[ \:\!f'(a) \left\{\psi(a)\right\}^{m} \right] + R_{n}, \] where \[ R_{n} = \frac{1}{2 \pi i} \!\int_{a}^{z}\!\! \int_{\gamma} \left[ \frac{\phi(\zeta) - b}{\phi(t) - b} \right]^{n-1} \!\! \frac{ f'(t) \;\!\phi'(\zeta) }{\phi(t) - \phi(\zeta)} \, d t \, d \zeta, \] and \(\gamma\) is a contour in the \(t\)-plane, enclosing the points \(a\) and \(z\) and such that, if \(\zeta\) be any point inside it, the equation \(\phi(t) = \phi(\zeta)\) has no roots on or inside the contour except a simple root \(t=\zeta\).[2]

To prove this, we have[3] \[\begin{align*} f(z) - f(a) = & \int_{a}^{z}\! f'(\zeta) \, d \zeta = \frac{1}{2 \pi i} \!\int_{a}^{z}\!\! \int_{\gamma}\! \frac{f'(t) \;\!\phi'(\zeta)}{\phi(t)-\phi(\zeta)}dt \,d\zeta \\ \\ = & \frac{1}{2 \pi i} \!\int_{a}^{z}\!\! \int_{\gamma} \frac{f'(t) \;\!\phi'(\zeta)}{\phi(t)-b}dt \,d\zeta \left[\sum_{m=0}^{n-2}\left\{\frac{\phi(\zeta)-b}{\phi(t)-b}\right\}^m\right. \\ & \quad + \left. \frac{\left\{\phi(\zeta)-b\right\}^{n-1}}{\left\{\phi(t)-b\right\}^{n-2}\left\{\phi(t)-\phi(\zeta)\right\}} \right] . \end{align*}\]

But, by §4.3, \[\begin{align*} \frac{1}{2 \pi i} \!\int_{a}^{z}\!\! \int_{\gamma}\;\! & \left[ \frac{\phi(\zeta) - b}{\phi(t) - b} \right]^{m} \frac{f'(t)\:\!\phi'(\zeta)}{\phi(t) - b} \, d t \, d \zeta \\ \\ &= \frac{ \left\{\phi(z) - b\right\}^{m+1} }{2 \pi i (m+1)} \!\int_{\gamma}\! \frac{f'(t)}{[\phi(t) - b]^{m+1}} \, d t \\ \\ &= \frac{ \left\{\phi(z) - b\right\}^{m+1} }{2 \pi i (m+1)} \!\int_{\gamma}\;\! \frac{ f'(t) \left\{\psi(t)\right\}^{m+1} }{ (t-a)^{m+1} } \, d t \\ \\ &= \frac{ \left\{\phi(z) - b \right\}^{m+1} }{ (m+1)! } \frac{ d^{m} }{ d a^{m} }\! \left[\;\! f'(a) \left\{\psi(a)\right\}^{m+1} \right]. \end{align*}\] Therefore, writing \(m - 1\) for \(m\), \[\begin{align*} f(z) = f(a) + & \sum_{m=1}^{n-1} \frac{ [\phi(z) - b]^{m} }{m!} \frac{d^{m-1}}{d a^{m-1}}\! \left[\;\! f'(a) \left\{ \psi(a) \right\}^{m} \right] \\ & +\frac{1}{2 \pi i} \!\int_{a}^{z}\!\! \int_{\gamma} \left[ \frac{\phi(\zeta) - b}{\phi(t) - b} \right]^{n-1}\!\! \frac{f'(t) \:\!\phi'(\zeta)}{\phi(t) - \phi(\zeta)} \, d t \, d \zeta. \end{align*}\]

If the last integral tends to zero as \(n \rightarrow \infty\), we may write the right-hand side of this equation as an infinite series.

Example 1. Prove that \[ z = a + \sum_{n=1}^{\infty} \frac{(-1)^{n-1} C_{n} (z-a)^{n} e^{n\:\! (z^{2} - a^{2})}}{n!}, \] where \[\begin{align*} C_{n} =& (2na)^{n-1} - \frac{n(n-1)(n-2)}{1!} (2na)^{n-3} \\ &\quad + \frac{n^{2}(n-1)(n-2)(n-3)(n-4)}{2!} (2na)^{n-5} - \cdots. \end{align*}\] To obtain this expansion, write \[ f(z) = z, \quad \phi(z) - b = (z-a) e^{z^{2} - a^{2}}, \quad \psi(z) = e^{a^{2} - z^{2}} \] in the above expression of Bürmann’s theorem; we thus have \[ z = a + \sum_{n=1}^{\infty} \frac{1}{n!} (z-a)^{n} e^{n\:\! (z^{2} - a^{2})} \left\{ \frac{d^{n-1}}{d z^{n-1}} e^{n\:\! (a^{2} - z^{2})} \right\}_{z=a}. \]

But, putting \(z = a + t\), \[\begin{align*} &\left\{ \frac{d^{n-1}}{d z^{n-1}} e^{n\:\! (a^{2} - z^{2})} \right\}_{z=a} \!\! = \left\{ \frac{d^{n-1}}{d t^{n-1}} e^{-n\:\! (2at + t^{2})} \right\}_{t=0} \\ \\ &\qquad = (n-1)! \times \text{ the coefficient of } t^{n-1} \text{ in the expansion of } e^{-n\:\!t\:\!(2a+t)} \\ &\qquad = (n-1)! \times \text{ the coefficient of } t^{n-1} \text{ in } \sum_{r=0}^\infty \frac{(-1)^r n^r t^r (2a+t)^r}{r!} \\ &\qquad = (n-1)! \times \sum_{r=\left\lceil\frac{n-1}{2}\right\rceil}^{n-1} \frac{(-1)^r n^r (2a)^{2r-n+1}}{ (n-1-r)! \;\! (2r-n+1)! } . \end{align*}\] The highest value of \(r\) which gives a term in the summation is \(r = n-1\). Arranging therefore the summation in descending indices \(r\), beginning with \(r = n-1\), we have \[\begin{align*} \left\{ \frac{d^{n-1}}{d z^{n-1}} e^{n(a^{2} - z^{2})} \right\}_{z=a}\!\! =& (-1)^{n-1}\!\left\{\!(2na)^{n-1}-\frac{n(n-1)(n-2)}{1!}(2na)^{n-3}+\cdots \!\right\} \\ =& (-1)^{n-1} C_{n}, \end{align*}\] which gives the required result.

Example 2. Obtain the expansion \[ z^{2} = \sin^{2} z + \frac{2}{3} \frac{1}{2} \sin^{4} z + \frac{2 \cdot 4}{3 \cdot 5} \frac{1}{3} \sin^{6} z + \frac{2 \cdot 4 \cdot 6}{3 \cdot 5 \cdot 7} \frac{1}{4} \sin^{8} z + \cdots. \]

Example 3. Let a line \(p\) be drawn through the origin in the \(z\)-plane, perpendicular to the line which joins the origin to any point \(a\). If \(z\) be any point on the \(z\)-plane which is on the same side of the line \(p\) as the point \(a\) is, shew that \[ \log z = \log a + 2 \sum_{m=1}^{\infty} \frac{1}{2m+1} \left(\frac{z-a}{z+a}\right)^{2m+1}. \]

7.31 Teixeira’s extended form of Bürmann’s theorem.

In the last section we have not investigated closely the conditions of convergence of Bürmann’s series, for the reason that a much more general form of the theorem will next be stated; this generalisation bears the same relation to the theorem just given that Laurent’s theorem bears to Taylor’s theorem: viz., in the last paragraph we were concerned only with the expansion of a function in positive powers of another function, whereas we shall now discuss the expansion of a function in positive and negative powers of the second function.

The general statement of the theorem is due to Teixeira,[4] whose exposition we shall follow in this section.

Suppose:

The equation \[ \theta(z) - \theta(x) = 0 \] has, in this case, a single root \(z = x\) in the interior of \(C\), as is seen from the equation[5] \[\begin{align*} \frac{1}{2 \pi i} \!\int_{C} \frac{\theta'(z)}{\theta(z) - \theta(x)} d z =& \frac{1}{2 \pi i} \left[ \int_{C}\! \frac{\theta'(z)}{\theta(z)} \, d z + \theta(x) \!\int_{C}\! \frac{\theta'(z)}{ \left\{\theta(z)\right\}^{2}} \, d z + \cdots \right] \\ =& \frac{1}{2 \pi i}\! \int_{C}\! \frac{\theta'(z)}{\theta(z)} d z \end{align*}\] of which the left-hand and right-hand members represent respectively the number of roots of the equation considered (§6.31) and the number of the roots of the equation \(\theta(z) = 0\) contained within \(C\).

Cauchy’s theorem therefore gives \[ f(x) = \frac{1}{2 \pi i}\! \left[ \int_{C}\! \frac{f(z) \;\!\theta'(z)}{\theta(z) - \theta(x)} d z - \int_{c}\! \frac{f(z) \;\!\theta'(z)}{\theta(z) - \theta(x)} d z \right]. \]

The integrals in this formula can be expanded, as in Laurent’s theorem, in powers of \(\theta(x)\), by the formulae \[\begin{align*} \int_C \! \frac{f(z)\;\!\theta'(z)}{\theta(z)-\theta(x)}dz &= \sum_{n=0}^\infty \left\{\theta(x)\right\}^n \! \int_C \! \frac{f(z)\;\!\theta'(z)}{\{\theta(z)\}^{n+1}} dz ,\\ \int_c \! \frac{f(z)\;\!\theta'(z)}{\theta(z)-\theta(x)}dz &= -\sum_{n=1}^\infty \frac{1}{\left\{\theta(x)\right\}^n} \! \int_c \! f(z) \{\theta(z)\}^{n-1}\theta'(z)\, dz. \end{align*}\]

We thus have the formula \[ f(x) = \sum_{n=0}^\infty A_n \left\{\theta(x)\right\}^n + \sum_{n=1}^\infty \frac{B_n}{ \left\{\theta(x)\right\}^n} , \] where \[ A_n = \frac{1}{2\pi \:\! i} \! \int_C \! \frac{f(z)\;\!\theta'(z)}{\{\theta(z)\}^{n+1}} dz, \quad B_n = \frac{1}{2\pi\:\! i} \! \int_c \! f(z) \{\theta(z)\}^{n-1}\theta'(z)\, dz. \]

Integrating by parts, we get, if \(n \neq 0\), \[ A_n = \frac{1}{2\pi \:\! in} \! \int_C \! \frac{f'(z)}{\{\theta(z)\}^{n}} dz, \quad B_n = \frac{-1}{2\pi \:\! in} \! \int_c \! \{\theta(z)\}^{n} \:\! f'(z) \, dz. \]

This gives a development of \(f(x)\) in positive and negative powers of \(\theta(x)\), valid for all points \(x\); within the ring-shaped space \(A\).

If the zeros and poles of \(f(z)\) and \(\theta(z)\) inside \(C\) are known, \(A_{n}\) and \(B_{n}\) can be evaluated by §5.22 or by §6.1.

Example 1. Shew that, if \(\left|\,x\,\right| < 1\), then \[ x = \frac{1}{2} \left( \frac{2x}{1+x^{2}} \right) + \frac{1}{2 \cdot 4} \left( \frac{2x}{1+x^{2}} \right)^{3} + \frac{1\cdot 3}{2 \cdot 4 \cdot 6} \left( \frac{2x}{1+x^{2}} \right)^{5} + \cdots. \]

Shew that, when \(\left|\,x\,\right| > 1\), the second member (right-hand side) represents \(x^{-1}\).

Example 2. If \(S^{\:\!(m)}_{2n}\) denote the sum of all combinations of the numbers \[ 2^{2},\, 4^{2},\, 6^{2}, \ldots, (2n-2)^{2}, \] taken \(m\) at a time, shew that \[ \frac{1}{z} = \frac{1}{\sin z} + \sum_{n=0}^{\infty} \frac{(-1)^{n+1}}{ (2n+2)! } \left( \frac{1}{2n+3} - \frac{S^{\:\!(1)}_{2(n+1)}}{2n+1} + \cdots + \frac{ (-1)^{n} S^{\:\!(n)}_{2(n+1)}}{3} \right) (\sin z)^{2n+1} \] the expansion being valid for all values of \(z\) represented by points within the oval whose equation is \(\left|\,\sin z\,\right| = 1\) and which contains the point \(z = 0\). \(\vphantom{\\ 3\\}\)
(Teixeira.)

7.32 Lagrange’s theorem.

Suppose now that the function \(f(z)\) of §7.31 is analytic at all points in the interior of \(C\), and let \(\theta(x) = (x - a) \theta_{1}(x)\). Then \(\theta_{1}(x)\) is analytic and not zero on or inside \(C\) and the contour \(c\) can be dispensed with; therefore the formulae which give \(A_{n}\) and \(B_{n}\) now become, by §5.22 and §6.1, \[\begin{align*} A_{n} =& \frac{1}{2\pi \:\! in} \!\int_{C}\! \frac{f'(z)}{(z-a)^{n} \theta_{1}^{n}(z)} \, d z = \frac{1}{n!} \frac{d^{n-1}}{d a^{n-1}}\!\! \left\{ \frac{f'(a)}{\theta_{1}^{n}(a) } \right\} \quad (n \geq 1), \\ A_{0} =& \frac{1}{2\pi\:\! i} \!\int_{C}\! \frac{f(z) \theta'(z)}{\theta_{1}(z)} \frac{\, d z}{z-a} = f(a), \\ \\ B_{n} =& \, 0. \end{align*}\]

The theorem of the last section accordingly takes the following form, if we write \(\theta_{1}(z) = 1 / \phi(z)\):

Let \(f(z)\) and \(\phi(z)\) be functions of \(z\) analytic on and inside a contour \(C\) surrounding a point \(a\), and let \(t\) be such that the inequality \[ \left|\,t \phi(z) \,\right| < \left|\,z - a\,\right| \] is satisfied at all points \(z\) on the perimeter of \(C\); then the equation \[ \zeta = a + t \phi(\zeta), \] regarded as an equation in \(\zeta\), has one root in the interior of \(C\); and further any function of \(\zeta\) analytic on and inside \(C\) can be expanded as a power series in \(t\) by the formula \[ f(\zeta) = f(a) + \sum_{n=1}^{\infty} \frac{t^{n}}{n!} \frac{d^{n-1}}{d a^{n-1}}\!\! \left[\:\! f'(a) \phi^{n}(a) \right]. \]

This result was published by Lagrange in 1770.[6]

Example 1. Within the contour surrounding \(a\) defined by the inequality \(\left|\,z (z - a)\,\right| > \left|\,\vphantom{z} \boldsymbol{\alpha}\,\right|\), where \(\left|\,\vphantom{z} \boldsymbol{\alpha}\,\right| < \frac{1}{2} \left|\,\vphantom{z} a \,\right|\),[7] the equation \[ z - a - \frac{\boldsymbol{\alpha}}{z} = 0 \] has one root \(\zeta\), the expansion of which is given by Lagrange’s theorem in the form \[ \zeta = a + \sum_{n=1}^{\infty} \frac{(-1)^{n-1} (2n-2)!}{n! (n-1)! a^{2n-1}} \boldsymbol{\alpha}^{n}. \]

Now, from the elementary theory of quadratic equations, we know that the equation \[ z - a - \frac{\boldsymbol{\alpha}}{z} = 0 \] has two roots, namely \[\frac{a}{2}\left\{1+\sqrt{1+\frac{4\boldsymbol{\alpha}}{a^2}}\right\}\,\text{ and }\, \frac{a}{2}\left\{1-\sqrt{1+\frac{4\boldsymbol{\alpha}}{a^2}}\right\};\] and our expansion represents the former[8] of these only — an example of the need for care in the discussion of these series.

Example 2. If \(y\) be that one of the roots of the equation \[ y=1+zy^2 \] which tends to \(1\) when \(z \rightarrow 0\), shew that \[\begin{align*} y^n= 1 &+ nz + \frac{n(n+3)}{2!}z^2 + \frac{n(n+4)(n+5)}{3!}z^3 \\ &+ \frac{n(n+5)(n+6)(n+7)}{4!}z^4 + \frac{n(n+6)(n+7)(n+8)(n+9)}{5!}z^5 + \cdots \end{align*}\] so long as \(\left|\,z\,\right| < \frac{1}{4}\).

Example 3. If \(x\) be that one of the roots of the equation \[ x = 1 + y x^{a} \] which tends to \(1\) when \(y \rightarrow 0\), shew that \[ \log x = y + \frac{2a-1}{2}y^2 + \frac{(3a-1)(3a-2)}{2 \cdot 3}y^3 + \cdots, \] the expansion being valid so long as \[ \left|\,y\,\right| < \left|\, (a-1)^{a-1} a^{-a} \,\right|. \] (McClintock.)