148. Taylor’s Series

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).

$\leftarrow$ 147. Taylor’s Theorem

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149. Applications of Taylor’s Theorem to maxima and minima $\rightarrow$