The Expansion of Functions in Infinite Series

7.1 A formula due to Darboux.[1]

[1]Journal de Math. (3), ii. (1876), p. 291. ↩

Let \(f(z)\) be analytic at all points of the straight line joining \(a\) to \(z\), and let \(\phi(t)\) be any polynomial of degree \(n\) in \(t\).

Then if \(0 \leq t \leq 1\), we have by differentiation \[\begin{align*} &\frac{d }{d t} \sum_{m=1}^{n} (-1)^{m} (z-a)^{m} \phi^{(n-m)}(t) f^{(m)}(a + t(z-a)) \\ &\quad =-(z-a) \phi^{(n)}(t) f'(a + t(z-a)) + (-1)^{n} (z-a)^{n+1} \phi(t) f^{(n+1)}(a + t (z-a)). \end{align*}\]

Noting that \(\phi^{(n)}(t)\) is constant \(= \phi^{(n)}(0)\), and integrating between the limits \(0\) and \(1\) of \(t\), we get \[\begin{align*} &\phi^{(n)}(0) \left\{ f(z) - f(a) \right\} \\ &\quad= \sum_{m=1}^{n} (-1)^{m-1} (z-a)^{m} \left\{ \phi^{(n-m)}(1) f^{(m)}(z) - \phi^{(n-m)}(0) f^{(m)}(a) \right\} \\ &\qquad + (-1)^{n} (z-a)^{n+1} \!\int_{0}^{1}\! \phi(t) f^{(n+1)}(a + t(z-a)) \, dt, \end{align*}\] which is the formula in question.

Taylor’s series may be obtained as a special case of this by writing \(\phi(t) = (t-1)^{n}\) and making \(n\rightarrow\infty\).

Example. By substituting \(2n\) for \(n\) in the formula of Darboux, and taking \(\phi(t) = t^{n} (t-1)^{n}\), obtain the expansion (supposed convergent) \[ f(z) - f(a) = \sum_{n=1}^{\infty} \frac{ (-1)^{n-1} (z-a)^{n} }{2^{n} n!} \left\{ f^{(n)}(z) + (-1)^{n-1} f^{(n)}(a) \right\}, \] and find the expression for the remainder after \(n\) terms in this series.

7.2 The Bernoullian numbers and the Bernoullian polynomials.

The function \(\frac{1}{2} z \cot \frac{1}{2} z\) is analytic when \(\left|\, z\,\right| < 2\pi\), and, since it is an even function of \(z\), it can be expanded into a Maclaurin series, thus \[ \frac{1}{2} z \cot \frac{1}{2} z = 1 - B_{1} \frac{z^{2}}{2!} - B_{2} \frac{z^{4}}{4!} - B_{3} \frac{z^{6}}{6!} \cdots ; \] then \(B_{n}\) is called the \(n\)th Bernoullian number.[2] It is found that[3] [4] \[ B_{1} = \frac{1}{6}, \quad B_{2} = \frac{1}{30}, \quad B_{3} = \frac{1}{42}, \quad B_{4} = \frac{1}{30}, \quad B_{5} = \frac{5}{66}, \quad \ldots. \]

[2]These numbers were introduced by Jakob Bernoulli in his Ars Conjectandi, p. 97 (published posthumously, 1713). (See also, a German translation by Haussner, Wahrscheinlichkeitsrechnung, Ars conjectandi (1899), p. 99 – Editor.) ↩
[3]Tables of the first sixty-two Bernoullian numbers have been given by Adams, Brit. Ass. Reports, 1877, p. 8. ↩
[4]Editor’s Note: Modern practice defines the Bernoulli numbers somewhat differently (see “Bernoulli Number.” at MathWorld). In terms of the original Bernoullian numbers, \(B_n\), defined here, the modern Bernoulli numbers, \(\mathbb{B}_n\), are\[\mathbb{B}_n = \left\{\begin{array}{lcl} 1 && \text{ for } n=1 \\ -\tfrac{1}{2} && \text{ for } n=2 \\ (-1)^{\left. n\middle/ 2\right.-1} B_{\left. n\middle/ 2\right.} && \text{ for } n \text{ even } \\ 0 && \text{ for } n \text{ odd. }\end{array} \right.\] ↩

These numbers can be expressed as definite integrals as follows:

We have, by miscellaneous example 2 of Chapter vi, \[\begin{align*} \int_{0}^{\infty}\! \frac{\sin px \, dx}{e^{\pi x} - 1} =& -\frac{1}{2p} + \frac{i}{2} \cot ip \\ =& -\frac{1}{2p} + \frac{1}{2p} \left\{ 1 + B_{1} \frac{(2p)^{2}}{2!} - B_{2} \frac{(2p)^{4}}{4!} + \cdots \right\}. \end{align*}\]

Since \[ \int_{0}^{\infty}\! \frac{x^{n} \sin \left(px + \frac{1}{2} n \pi \right)}{e^{\pi x} - 1} dx \] converges uniformly (by de la Vallée Poussin’s test) near \(p=0\) we may, by §4.44 corollary, differentiate both sides of this equation any number of times and then put \(p = 0\); doing so and writing \(2t\) for \(x\), we obtain \[ B_{n} = 4n \!\int_{0}^{\infty}\! \frac{t^{2n-1} \, dt}{e^{2\pi\:\! t} - 1}. \]

Note: A proof of this result, depending on contour integration, is given by Carda, Monatshefte für Math. und Phys. v. (1894), pp. 321–4.

Example 1. Shew that \[ B_{n} = \frac{2n}{\pi^{2n} (2^{2n}-1)} \!\int_{0}^{\infty}\! \frac{x^{2n-1}dx}{\sinh x} > 0. \]

Now consider the function \(t \dfrac{e^{z\:\!t}-1}{e^{t}-1}\), which may be expanded into a Maclaurin series in powers of \(t\) valid when \(\left|\, t\,\right| < 2\pi\).

[5]The name was given by Raabe, Journal für Math. xlii. (1851), p. 348. ↩

The Bernoullian polynomial[5] of order \(n\) is defined to be the coefficient of \(\left.\vphantom{g_n}t^{n}\!\middle/n!\right.\) in this expansion. It is denoted by \(\phi_{n}(z)\), so that \[ t \frac{e^{z\:\!t}-1}{e^{t}-1} = \sum_{n=1}^{\infty} \frac{\phi_{n}(z) t^{n}}{n!}. \]

This polynomial possesses several important properties. Writing \(z+1\) for \(z\) in the preceding equation and subtracting, we find that \[ t e^{z\:\! t} = \sum_{n=1}^{\infty} \left\{ \phi_{n}(z+1) - \phi_{n}(z) \right\} \frac{t^{n}}{n!}. \]

On equating coefficients of \(t^{n}\) on both sides of this equation we obtain \[ n z^{n-1} = \phi_{n}(z+1) - \phi_{n}(z), \] which is a difference-equation satisfied by the function \(\phi_{n}(z)\).

An explicit expression for the Bernoullian polynomials can be obtained as follows. We have \[ e^{zt} - 1 = zt + \frac{z^{2}t^{2}}{2!} + \frac{z^{3}t^{3}}{3!} + \cdots, \] and \[ \frac{t}{e^{t}-1} = \frac{t}{2i} \cot \frac{t}{2i} - \frac{t}{2} = 1 - \frac{t}{2} + \frac{B_{1} t^{2}}{2!} - \frac{B_{2} t^{4}}{4!} + \cdots. \]

Hence \[ \sum_{n=1}^{\infty} \frac{\phi_{n}(z) t^{n}}{n!} = \!\! \left\{ zt + \frac{z^{2} t^{2}}{2!} + \frac{z^{3} t^{3}}{3!} + \cdots \right\} \!\! \left\{ 1 - \frac{t}{2} + \frac{B_{1} t^{2}}{2!} - \frac{B_{2} t^{4}}{4!} + \cdots \right\}\! . \]

From this, by equating coefficients of \(t^{n}\) (§3.73), we have \[ \phi_{n}(z) = z^{n} - \frac{1}{2} n z^{n-1} + {n \choose 2} B_{1} z^{n-2} - {n \choose 4} B_{2} z^{n-4} + {n \choose 6} B_{3} z^{n-6} - \cdots, \] the last term being that in \(z\) or \(z^{2}\) and \({n \choose 2}, {n \choose 4},\ldots\) being the binomial coefficients; this is the Maclaurin series for the \(n\)th Bernoullian polynomial.

When \(z\) is an integer, it may be seen from the difference-equation that \[ \left.\phi_{n}(z)\middle/n \right. = 1^{n-1} + 2^{n-1} + \cdots + (z-1)^{n-1}. \]

The Maclaurin series for the expression on the right was given by Bernoulli.

Example 2. Shew that, when \(n > 1\), \[ \phi_{n}(z) = (-1)^{n} \phi_{n}(1-z). \]

7.21 The Euler-Maclaurin expansion.

In the formula of Darboux (§7.1) write \(\phi_{n}(t)\) for \(\phi(t)\), where \(\phi_{n}(t)\) is the \(n\)th Bernoullian polynomial.

Differentiating the equation \[ \phi_{n}(t+1) - \phi_{n}(t) = n t^{n-1} \] \(n - k\) times, we have \[ \phi_{n}^{(n-k)}(t+1) - \phi_{n}^{(n-k)}(t) = n (n-1) \cdots (k+1)k t^{k-1} \] Putting \(t=0\) in this, we have \(\phi_{n}^{(n-k)}(1) = \phi_{n}^{(n-k)}(0).\)

Now, from the Maclaurin series for \(\phi_{n}(z)\), we have if \(k > 0\) \[\begin{align*} &\phi_{n}^{(n-2k-1)}(0) = 0,& \; &\phi_{n}^{(n-2k)}(0) = \frac{n!}{(2k)!} (-1)^{k-1} B_{k}, \\ &\phi_{n}^{(n-1)}(0) = -\frac{1}{2} n!,& \; &\phi_{n}^{(n)}(0) = n!. \end{align*}\]

Substituting these values of \(\phi_{n}^{(n-k)}(1)\) and \(\phi_{n}^{(n-k)}(0)\) in Darboux’s result, we obtain the Euler-Maclaurin sum formula,[6] \[\begin{align*} (z-a) f'(a) =& f(z) - f(a) - \frac{z-a}{2} \left\{ f'(z) - f'(a) \right\} \\ & + \sum_{m=1}^{n-1} \frac{ (-1)^{m-1} B_{m} (z-a)^{2m}}{(2m)!} \left\{ f^{(2m)}(z) - f^{(2m)}(a) \right\} \\ & -\frac{(z-a)^{2n+1}}{(2n)!} \!\int_{0}^{1}\! \phi_{2n}(t) \, f^{(2n+1)}(a + (z-a) t ) \, dt. \end{align*}\]

[6]A history of the formula is given by Barnes, Proc. London Math. Soc. (2), iii. (1905), p. 253. It was discovered by Euler (1732), but was not published at the time. Euler communicated it (June 9, 1736) to Stirling who replied (April 16, 1738) that it included his own theorem (see §12.33) as a particular case, and also that the more general theorem had been discovered by Maclaurin; and Euler, in a lengthy reply, waived his claims to priority. The theorem was published by Euler, Comm. Acad. Imp. Petrop. vi. (1732), [Published 1738], pp. 68–97, and by Maclaurin in 1742, Treatise on Fluxions, p. 672. For information concerning the correspondence between Euler and Stirling, we are indebted to Mr C. Tweedie (see his James Stirling; a sketch of his life and works (1922) – Editor). ↩

In certain cases the last term tends to zero as \(n \rightarrow \infty\), and we can thus obtain an infinite series for \(f(z) - f(a)\).

If we write \(\omega\) for \(z - a\) and \(F(x)\) for \(f'(x)\), the last formula becomes \[\begin{align*} \!\int_{a}^{a+\omega}\! F(x) \, dx = & \frac{1}{2} \omega \left\{ F(a) + F(a + \omega) \right\} \\ & + \sum_{m=1}^{n-1} \frac{(-1)^{m} B_{m} \omega^{2m}}{(2m)!} \left\{ F^{(2m-1)}(a+\omega) - F^{(2m-1)}(a) \right\} \\ & + \frac{\omega^{2n+1}}{(2n)!} \!\int_{0}^{1}\! \phi_{2n}(t) F^{(2n)}(a + \omega t) \, dt. \end{align*}\]

Writing \(a + \omega\), \(a + 2\omega, \ldots, a + (r-1) \omega\) for \(a\) in this result and adding up, we get \[\begin{align*} \int_{a}^{a + r\omega}\! \!\! F(x) \, dx & = \omega \left\{ \frac{1}{2} F(a) + F(a+\omega) + F(a+2\omega) + \cdots + \frac{1}{2} F(a + r\omega) \right\} \\ & + \sum_{m=1}^{n-1} \frac{(-1)^{m} B_{m} \omega^{2m}}{(2m)!} \left\{ F^{(2m-1)}(a + r\omega) - F^{(2m-1)}(a) \right\} + R_{n} , \end{align*}\] where \[ R_{n} = \frac{\omega^{2n+1}}{(2n)!} \!\int_{0}^{1}\! \phi_{2n}(t) \left\{ \sum_{m=0}^{r-1} F^{(2n)}(a + m\omega + \omega t)\right\} dt. \]

This last formula is of the utmost importance in connexion with the numerical evaluation of definite integrals. It is valid if \(F(x)\) is analytic at all points of the straight line joining \(a\) to \(a + r \omega\).

Example 1. If \(f(z)\) be an odd function of \(z\), shew that \[\begin{align*} z f'(z) =& f(z) + \sum_{m=2}^{n} (-1)^{m} \frac{B_{m-1} (2z)^{2m-2}}{(2m-2)!} f^{(2m-2)}(z) \\ &- \frac{2^{2n} z^{2n+1}}{(2n)!} \!\int_{0}^{1}\!\! \phi_{2n}(t)\, f^{(2n+1)}(-z + 2\:\! z\:\! t) \, dt. \end{align*}\]

Example 2. Shew, by integrating by parts, that the remainder after \(n\) terms of the expansion of \(\frac{1}{2} z \cot \frac{1}{2} z\) may be written in the form \[ \frac{ (-1)^{n+1} z^{2n+1} }{ (2n)! \sin z } \!\int_{0}^{1} \phi_{2n}(t) \cos (z\:\! t) \, dt. \] Math. Trip. 1904.