If \(f(x) = (1 + x)^{m}\), where \(m\) is not a positive integer, then Cauchy’s form of the remainder is \[R_{n} = \frac{m(m – 1)\dots (m – n + 1)}{1\cdot2\dots (n – 1)}\, \frac{(1 – \theta )^{n-1} x^{n}}{(1 + \theta x)^{n-m}}.\]
Now \((1 – \theta)/(1 + \theta x)\) is less than unity, so long as \(-1 < x < 1\), whether \(x\) is positive or negative; and \((1 + \theta x)^{m-1}\) is less than a constant \(K\) for all values of \(n\), being in fact less than \((1 + |x|)^{m-1}\) if \(m > 1\) and than \((1 – |x|)^{m-1}\) if \(m < 1\) Hence \[|R_{n}| < K |m| \left|\binom{m – 1}{n – 1}\right| |x^{n}| = \rho_{n},\] say. But \(\rho_{n} \to 0\) as \(n \to \infty\), by Ex. XXVII. 13, and so \(R_{n} \to 0\). The truth of the Binomial Theorem is thus established for all rational values of \(m\) and all values of \(x\) between \(-1\) and \(1\). It will be remembered that the difficulty in using Lagrange’s form, in Ex. LVI. 2, arose in connection with negative values of \(x\).
$\leftarrow$ 162. Alternative proof of Taylor’s Theorem | Main Page | 164. Integrals of complex functions $\rightarrow$ |