Suppose that \(f(x)\) is a function all of whose differential coefficients are continuous in an interval \({[a – \eta, a + \eta]}\) surrounding the point \(x = a\). Then, if \(h\) is numerically less than \(\eta\), we have \[f(a + h) = f(a) + hf'(a) + \dots + \frac{h^{n-1}}{(n – 1)!} f^{(n-1)}(a) + \frac{h^{n}}{n!} f^{(n)}(a + \theta_{n} h),\] where \(0 < \theta_{n} < 1\), for all values of \(n\). Or, if \[S_{n} = \sum_{0}^{n-1} \frac{h^{\nu}}{\nu!} f^{(\nu)}(a),\quad R_{n} = \frac{h^{n}}{n!} f^{(n)}(a + \theta_{n} h),\] we have \[f(a + h) – S_{n} = R_{n}.\]

Now let us suppose, in addition, that we can prove that \(R_{n} \to 0\) as \(n \to \infty\). Then \[f(a + h) = \lim_{n\to\infty} S_{n} = f(a) + hf'(a) + \frac{h^{2}}{2!} f”(a) + \dots.\]

This expansion of \(f(a + h)\) is known as Maclaurin’s Series. When \(a = 0\) the formula reduces to \[f(h) = f(0) + hf'(0) + \frac{h^{2}}{2!} f”(0) + \dots,\] which is known as . The function \(R_{n}\) is known as Lagrange’s form of the remainder.

The reader should be careful to guard himself against supposing that the continuity of all the derivatives of \(f(x)\) is a sufficient condition for the validity of Taylor’s series. A direct discussion of the behaviour of \(R_{n}\) is always essential.

Example LVI

1. Let \(f(x) = \sin x\). Then all the derivatives of \(f(x)\) are continuous for all values of \(x\). Also \(|f^{n}(x)| \leq 1\) for all values of \(x\) and \(n\). Hence in this case \(|R_{n}| \leq h^{n}/n!\), which tends to zero as \(n \to \infty\) (Ex. XXVII 12) whatever value \(h\) may have. It follows that \[\sin(x + h) = \sin x + h\cos x – \frac{h^{2}}{2!}\sin x – \frac{h^{3}}{3!}\cos x + \frac{h^{4}}{4!}\sin x + \dots,\] for all values of \(x\) and \(h\). In particular \[\sin h = h – \frac{h^{3}}{3!} + \frac{h^{5}}{5!} – \dots,\] for all values of \(h\). Similarly we can prove that \[\cos(x + h) = \cos x – h\sin x – \frac{h^{2}}{2!}\cos x + \frac{h^{3}}{3!} \sin x + \dots,\quad \cos h = 1 – \frac{h^{2}}{2!} + \frac{h^{4}}{4!} – \dots.\]

2. The Binomial Series. Let \(f(x) = (1 + x)^{m}\), where \(m\) is any rational number, positive or negative. Then \(f^{(n)}(x) = m(m – 1) \dots (m – n + 1) (1 + x)^{m-n}\) and Maclaurin’s Series takes the form \[(1 + x)^{m} = 1 + \binom{m}{1}x + \binom{m}{2}x^{2} + \dots.\]

When \(m\) is a positive integer the series terminates, and we obtain the ordinary formula for the Binomial Theorem with a positive integral exponent. In the general case \[R_{n} = \frac{x^{n}}{n!} f^{(n)}(\theta_{n}x) = \binom{m}{n}x^{n}(1 + \theta_{n}x)^{m-n},\] and in order to show that Maclaurin’s Series really represents \((1 + x)^{m}\) for any range of values of \(x\) when \(m\) is not a positive integer, we must show that \(R_{n} \to 0\) for every value of \(x\) in that range. This is so in fact if \(-1 < x < 1\), and may be proved, when \(0\leq x < 1\), by means of the expression given above for \(R_{n}\), since \((1 + \theta_{n}x)^{m-n} < 1\) if \(n > m\), and \(\dbinom{m}{n} x^{n} \to 0\) as \(n \to \infty\) (Ex. XXVII. 13). But a difficulty arises if \(-1 < x < 0\), since \(1 + \theta_{n}x < 1\) and \((1 + \theta_{n}x)^{m-n} > 1\) if \(n > m\); knowing only that \(0 < \theta_{n} < 1\), we cannot be assured that \(1 + \theta_{n}x\) is not quite small and \((1 + \theta _{n}x)^{m-n}\) quite large.

In fact, in order to prove the Binomial Theorem by means of Taylor’s Theorem, we need some different form for \(R_{n}\), such as will be given later (§ 162).


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