234. The logarithmic series.

We found in § 213 that \[\begin{equation*} \log(1 + z) = z – \tfrac{1}{2} z^{2} + \tfrac{1}{3} z^{3} – \dots \tag{1} \end{equation*}\] when \(z\) is real and numerically less than unity. The series on the right-hand side is convergent, indeed absolutely convergent, when \(z\) has any complex value whose modulus is less than unity. It is naturally suggested that the equation (1) remains true for such complex values of \(z\). That this is true may be proved by a modification of the argument of § 213. We shall in fact prove rather more than this, viz. that (1) is true for all values of \(z\) such that \(|z| \leq 1\), with the exception of the value \(-1\).

It will be remembered that \(\log(1 + z)\) is the principal value of \(\log(1 + z)\), and that \[\log(1 + z) = \int_{C} \frac{du}{u},\] where \(C\) is the straight line joining the points \(1\) and \(1 + z\) in the plane of the complex variable \(u\). We may suppose that \(z\) is not real, as the formula (1) has been proved already for real values of \(z\).

If we put \[z = r(\cos\theta + i\sin\theta) = \zeta r,\] so that \(|r| \leq 1\), and \[u = 1 + \zeta t,\] then \(u\) will describe \(C\) as \(t\) increases from \(0\) to \(r\). And \[\begin{aligned} \int_{C} \frac{du}{u} &= \int_{0}^{r} \frac{\zeta\, dt}{1 + \zeta t} \\ &= \int_{0}^{r} \left\{\zeta – \zeta^{2} t + \zeta^{3} t^{2} – \dots + (-1)^{m-1} \zeta^{m} t^{m-1} + \frac{(-1)^{m} \zeta^{m+1} t^{m}}{1 + \zeta t}\right\} dt \\ &= \zeta r – \frac{(\zeta r)^{2}}{2} + \frac{(\zeta r)^{3}}{3} – \dots + (-1)^{m-1} \frac{(\zeta r)^{m}}{m} + R_{m} \\ &= z – \frac{z^{2}}{2} + \frac{z^{3}}{3} – \dots + (-1)^{m-1} \frac{z^{m}}{m} + R_{m}, \begin{equation*} \tag{2} \end{equation*}\end{aligned}\] where \[\begin{equation*} R_{m} = (-1)^{m} \zeta^{m+1} \int_{0}^{r} \frac{t^{m}\, dt}{1 + \zeta t}. \tag{3} \end{equation*}\]

It follows from (1) of § 164 that \[\begin{equation*} |R_{m}| \leq \int_{0}^{r} \frac{t^{m}\, dt}{|1 + \zeta t|}. \tag{4} \end{equation*}\] Now \(|1 + \zeta t|\) or \(|u|\) is never less than \(\varpi\), the perpendicular from \(O\) on to the line \(C\).1 Hence \[|R_{m}| \leq \frac{1}{\varpi} \int_{0}^{r} t^{m}\, dt = \frac{r^{m+1}}{(m + 1) \varpi} \leq \frac{1}{(m + 1) \varpi},\] and so \(R_{m} \to 0\) as \(m \to \infty\). It follows from that \[\begin{equation*} \log(1 + z) = z – \tfrac{1}{2} z^{2} + \tfrac{1}{3} z^{3} – \dots. \tag{5} \end{equation*}\]

We have of course shown in the course of our proof that the series is convergent: this however has been proved already (Ex. LXXX. 4). The series is in fact absolutely convergent when \({|z|} < 1\) and conditionally convergent when \(|z| = 1\).

Changing \(z\) into \(-z\) we obtain \[\begin{equation*} \log \left(\frac{1}{1 – z}\right) = -\log(1 – z) = z + \tfrac{1}{2} z^{2} + \tfrac{1}{3} z^{3} + \dots. \tag{6} \end{equation*}\]

 

235.

Now \[\begin{aligned} \log(1 + z) &= \log\{(1 + r \cos\theta) + ir\sin\theta\} \\ &= \tfrac{1}{2} \log(1 + 2r\cos\theta + r^{2}) + i\arctan \left(\frac{r\sin\theta}{1 + r\cos\theta}\right).\end{aligned}\] That value of the inverse tangent must be taken which lies between \(-\frac{1}{2}\pi\) and \(\frac{1}{2}\pi\). For, since \(1 + z\) is the vector represented by the line from \(-1\) to \(z\), the principal value of \(\operatorname{am}(1 + z)\) always lies between these limits when \(z\) lies within the circle \(|z| = 1\).2

Since \(z^{m} = r^{m}(\cos m\theta + i\sin m\theta)\), we obtain, on equating the real and imaginary parts in equation (5) of § 234, \[\begin{aligned} \tfrac{1}{2} \log(1 + 2r\cos\theta + r^{2}) &= r\cos\theta – \tfrac{1}{2}r^{2} \cos 2\theta + \tfrac{1}{3}r^{3} \cos 3\theta – \dots, \\ \arctan \left(\frac{r\sin\theta}{1 + r\cos\theta}\right) &= r\sin\theta – \tfrac{1}{2}r^{2} \sin 2\theta + \tfrac{1}{3}r^{3} \sin 3\theta – \dots.\end{aligned}\] These equations hold when \(0 \leq r \leq 1\), and for all values of \(\theta\), except that, when \(r = 1\), \(\theta\) must not be equal to an odd multiple of \(\pi\). It is easy to see that they also hold when \(-1 \leq r \leq 0\), except that, when \(r = -1\), \(\theta\) must not be equal to an even multiple of \(\pi\).

A particularly interesting case is that in which \(r = 1\). In this case we have \[\begin{aligned} \log(1 + z) = \log(1 + \operatorname{Cis}\theta) &= \tfrac{1}{2} \log(2 + 2\cos\theta) + i\arctan\left(\frac{\sin\theta}{1 + \cos\theta}\right) \\ &= \tfrac{1}{2} \log(4\cos^{2} \tfrac{1}{2}\theta) + \tfrac{1}{2}i\theta,\end{aligned}\] if \(-\pi < \theta < \pi\), and so \[\begin{aligned} {4} \cos\theta &- \tfrac{1}{2} \cos 2\theta &&+ \tfrac{1}{3} \cos 3\theta &&- \dots &&= \tfrac{1}{2} \log(4\cos^{2} \tfrac{1}{2}\theta), \\ \sin\theta &- \tfrac{1}{2} \sin 2\theta &&+ \tfrac{1}{3} \sin 3\theta &&- \dots &&= \tfrac{1}{2} \theta.\end{aligned}\] The sums of the series, for other values of \(\theta\), are easily found from the consideration that they are periodic functions of \(\theta\) with the period \(2\pi\). Thus the sum of the cosine series is \(\frac{1}{2} \log(4\cos^{2} \frac{1}{2}\theta)\) for all values of \(\theta\) save odd multiples of \(\pi\) (for which values the series is divergent), while the sum of the sine series is \(\frac{1}{2} (\theta – 2k\pi)\) if \((2k – 1)\pi < \theta < (2k + 1)\pi\), and zero if \(\theta\) is an odd multiple of \(\pi\). The graph of the function represented by the sine series is shown in Fig. 58. The function is discontinuous for \(\theta = (2k + 1)\pi\).

If we write \(iz\) and \(-iz\) for \(z\) in , and subtract, we obtain \[\frac{1}{2i} \log\left(\frac{1 + iz}{1 – iz}\right) = z – \tfrac{1}{3}z^{3} + \tfrac{1}{5}z^{5} – \dots.\] If \(z\) is real and numerically less than unity, we are led, by the results of § 231, to the formula \[\arctan z = z – \tfrac{1}{3}z^{3} + \tfrac{1}{5}z^{5} – \dots,\] already proved in a different manner in § 214.

Example XCVII
1. Prove that, in any triangle in which \(a > b\), \[\log c = \log a – \frac{b}{a} \cos C – \frac{b^{2}}{2a^{2}} \cos 2C – \dots.\]

[Use the formula \(\log c = \frac{1}{2} \log(a^{2} + b^{2} – 2ab\cos C )\).]

2. Prove that if \(-1 < r < 1\) and \(-\frac{1}{2}\pi < \theta < \frac{1}{2}\pi\) then \[r\sin 2\theta – \tfrac{1}{2}r^{2} \sin 4\theta + \tfrac{1}{3}r^{3} \sin 6\theta – \dots = \theta – \arctan \left\{\left(\frac{1 – r}{1 + r}\right) \tan\theta\right\},\] the inverse tangent lying between \(-\frac{1}{2}\pi\) and \(\frac{1}{2}\pi\). Determine the sum of the series for all other values of \(\theta\).

3. Prove, by considering the expansions of \(\log(1 + iz)\) and \(\log(1 – iz)\) in powers of \(z\), that if \(-1 < r < 1\) then \[\begin{gathered} \begin{alignedat}{4} r\sin\theta &+ \tfrac{1}{2}r^{2} \cos 2\theta &&- \tfrac{1}{3}r^{3} \sin 3\theta &&- \tfrac{1}{4}r^{4} \cos 4\theta + \dots &&= \tfrac{1}{2} \log(1 + 2r \sin\theta + r^{2}),\\ r\cos\theta &+ \tfrac{1}{2}r^{2} \sin 2\theta &&- \tfrac{1}{3}r^{3} \cos 3\theta &&- \tfrac{1}{4}r^{4} \sin 4\theta + \dots &&= \arctan \left(\frac{r\cos\theta}{1 – r\sin\theta}\right), \end{alignedat} \\ \begin{alignedat}{2} r\sin\theta &- \tfrac{1}{3}r^{3} \sin 3\theta + \dots &&= \tfrac{1}{4} \log\left(\frac{1 + 2r \sin\theta + r^{2}} {1 – 2r \sin\theta + r^{2}}\right),\\ r\cos\theta &- \tfrac{1}{3}r^{3} \cos 3\theta + \dots &&= \tfrac{1}{2} \arctan \left(\frac{2r\cos\theta}{1 – r^{2}}\right), \end{alignedat}\end{gathered}\] the inverse tangents lying between \(-\frac{1}{2}\pi\) and \(\frac{1}{2}\pi\).

4. Prove that \[\begin{aligned} {3} \cos\theta \cos\theta &- \tfrac{1}{2} \cos 2\theta \cos^{2}\theta &&+ \tfrac{1}{3} \cos 3\theta \cos^{3} \theta – \dots &&= \tfrac{1}{2} \log(1 + 3\cos^{2} \theta),\\ \sin\theta \sin\theta &- \tfrac{1}{2} \sin 2\theta \sin^{2}\theta &&+ \tfrac{1}{3} \sin 3\theta \sin^{3} \theta – \dots &&= \operatorname{arccot} (1 + \cot\theta + \cot^{2}\theta),\end{aligned}\] the inverse cotangent lying between \(-\frac{1}{2}\pi\) and \(\frac{1}{2}\pi\); and find similar expressions for the sums of the series \[\cos\theta \sin\theta – \tfrac{1}{2} \cos 2\theta \sin^{2}\theta + \dots,\quad \sin\theta \cos\theta – \tfrac{1}{2} \sin 2\theta \cos^{2}\theta + \dots.\]


  1. Since \(z\) is not real, \(C\) cannot pass through \(O\) when produced. The reader is recommended to draw a figure to illustrate the argument.↩︎
  2. See the preceding footnote.↩︎

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