237. The binomial series

We have proved already (§ 215) that the sum of the series \[1 + \binom{m}{1} z + \binom{m}{2} z^{2} + \dots\] is \((1 + z)^{m} = \exp\{m\log(1 + z)\}\), for all real values of \(m\) and all real values of \(z\) between \(-1\) and \(1\). If \(a_{n}\) is the coefficient of \(z^{n}\) then \[\left|\frac{a_{n+1}}{a_{n}}\right| = \left|\frac{m – n}{n + 1}\right| \to 1,\] whether \(m\) is real or complex. Hence (Ex. LXXX. 3) the series is always convergent if the modulus of \(z\) is less than unity, and we shall now prove that its sum is still \(\exp\{m\log(1 + z)\}\), the principal value of \((1 + z)^{m}\).

It follows from § 236 that if \(t\) is real then \[\frac{d}{dt}(1 + tz)^{m} = mz(1 + tz)^{m-1},\] \(z\) and \(m\) having any real or complex values and each side having its principal value. Hence, if \(\phi(t) = (1 + tz)^{m}\), we have \[\phi^{(n)}(t) = m(m – 1) \dots (m – n + 1)z^{n} (1 + tz)^{m-n}.\] This formula still holds if \(t = 0\), so that \[\frac{\phi^{n}(0)}{n!} = \binom{m}{n} z^{n}.\]

Now, in virtue of the remark made at the end of § 164, we have \[\phi(1) = \phi(0) + \phi'(0) + \frac{\phi”(0)}{2!} + \dots + \frac{\phi^{(n-1)}(0)}{(n – 1)!} + R_{n},\] where \[R_{n} = \frac{1}{(n – 1)!}\int_{0}^{1} (1 – t)^{n-1} \phi^{(n)}(t)\, dt.\] But if \(z = r(\cos\theta + i\sin\theta)\) then \[|1 + tz| = \sqrt{1 + 2tr\cos\theta + t^{2}r^{2}} \geq 1 – tr,\] and therefore \[\begin{aligned} |R_{n}| &< \frac{|m(m – 1) \dots (m – n + 1)|}{(n – 1)!}\, r^{n} \int_{0}^{1} \frac{(1 – t)^{n-1}}{(1 – tr)^{n-m}}\, dt\\ &< \frac{|m(m – 1) \dots (m – n + 1)|}{(n – 1)!}\, \frac{(1 – \theta)^{n-1} r^{n}}{(1 – \theta r)^{n-m}},\end{aligned}\] where \(0 < \theta < 1\); so that (cf. § 163) \[|R_{n}| < K\frac{|m(m – 1) \dots (m – n + 1)|}{(n – 1)!}\, r^{n} = \rho_{n},\] say. But \[\frac{\rho_{n+1}}{\rho_{n}} = \frac{|m – n|}{n}r \to r,\] and so (Ex. XXVII. 6) \(\rho_{n} \to 0\), and therefore \(R_{n} \to 0\), as \(n \to \infty\). Hence we arrive at the following theorem.

Theorem. The sum of the binomial series \[1 + \binom{m}{1} z + \binom{m}{2} z^{2} + \dots\] is \(\exp\{m\log(1 + z)\}\), where the logarithm has its principal value, for all values of \(m\), real or complex, and all values of \(z\) such that \({|z|} < 1\).

A more complete discussion of the binomial series, taking account of the more difficult case in which \(|z| = 1\), will be found on pp. 225 et seq. of Bromwich’s Infinite Series.

Example XCVIII
1. Suppose \(m\) real. Then since \[\log(1 + z) = \tfrac{1}{2} \log(1 + 2r\cos\theta + r^{2}) + i\arctan\left(\frac{r\sin\theta}{1 + r\cos\theta}\right),\] we obtain \[\begin{aligned} \sum_{0}^{\infty} \binom{m}{n} z^{n} &= \exp\{\tfrac{1}{2}m \log(1 + 2r\cos\theta + r^{2})\} \operatorname{Cis} \left\{m\arctan\left(\frac{r\sin\theta}{1 + r\cos\theta}\right)\right\} \\ &= (1 + 2r\cos\theta + r^{2})^{\frac{1}{2}m} \operatorname{Cis} \left\{m\arctan\left(\frac{r\sin\theta}{1 + r\cos\theta}\right)\right\},\end{aligned}\] all the inverse tangents lying between \(-\frac{1}{2}\pi\) and \(\frac{1}{2}\pi\). In particular, if we suppose \(\theta = \frac{1}{2}\pi\), \(z = ir\), and equate the real and imaginary parts, we obtain \[\begin{aligned} 1 – \binom{m}{2} r^{2} + \binom{m}{4} r^{4} – \dots &= (1 + r^{2})^{\frac{1}{2}m} \cos(m\arctan r), \\ \binom{m}{1} r – \binom{m}{3} r^{3} + \binom{m}{5} r^{5} – \dots &= (1 + r^{2})^{\frac{1}{2}m} \sin(m\arctan r).\end{aligned}\]

2. Verify the formulae of Ex. 1 when \(m = 1\), \(2\), \(3\). [Of course when \(m\) is a positive integer the series is finite.]

3. Prove that if \(0 \leq r < 1\) then \[\begin{aligned} 1 – \frac{1\cdot3}{2\cdot4} r^{2} + \frac{1\cdot3\cdot5\cdot7}{2\cdot4\cdot6\cdot8} r^{4} – \dots &= \sqrt{\frac{\sqrt{1 + r^{2}} + 1}{2(1 + r^{2})}}, \\ \frac{1}{2} r – \frac{1\cdot3\cdot5}{2\cdot4\cdot6} r^{3} + \frac{1\cdot3\cdot5\cdot7\cdot9}{2\cdot4\cdot6\cdot8\cdot10} r^{5} – \dots &= \sqrt{\frac{\sqrt{1 + r^{2}} – 1}{2(1 + r^{2})}}.\end{aligned}\]

[Take \(m = -\frac{1}{2}\) in the last two formulae of Ex. 1.]

4. Prove that if \(-\frac{1}{4}\pi < \theta < \frac{1}{4}\pi\) then \[\begin{aligned} \cos m\theta &= \cos^{m} \theta \left\{1 – \binom{m}{2} \tan^{2} \theta + \binom{m}{4} \tan^{4} \theta – \dots\right\}, \\ \sin m\theta &= \cos^{m} \theta \left\{\binom{m}{1} \tan\theta – \binom{m}{3} \tan^{3} \theta + \dots\right\},\end{aligned}\] for all real values of \(m\). [These results follow at once from the equations \[\cos m\theta + i\sin m\theta = (\cos\theta + i\sin\theta )^{m} = \cos^{m} \theta(1 + i\tan\theta)^{m}.]\]

5. We proved (Ex. LXXXI. 6), by direct multiplication of series, that \(f(m, z) = \sum\dbinom{m}{n} z^{n}\), where \(|z| < 1\), satisfies the functional equation \[f(m, z) f(m’, z) = f(m + m’, z).\] Deduce, by an argument similar to that of § 216, and without assuming the general result above, that if \(m\) is real and rational then \[f(m, z) = \exp\{m\log(1 + z)\}.\]

6. If \(z\) and \(\mu\) are real, and \(-1 < z < 1\), then \[\sum \binom{i\mu}{n} z^{n} = \cos\{\mu\log(1 + z)\} + i\sin\{\mu\log(1 + z)\}.\]

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