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Formal Statement of Taylor's Theorem

Let $ f(x)$ be continuous on a real interval $ I$ containing $ x_0$ (and $ x$ ), and let $ f^{(n)}(x)$ exist at $ x$ and $ f^{(n+1)}(\xi)$ be continuous for all $ \xi\in I$ . Then we have the following Taylor series expansion:

\begin{eqnarray*}
f(x) = f(x_0) &+& \frac{1}{1}f^\prime(x_0)(x-x_0) \\ [10pt]
&+& \frac{1}{1\cdot 2}f^{\prime\prime}(x_0)(x-x_0)^2 \\ [10pt]
&+& \frac{1}{1\cdot 2\cdot 3}f^{\prime\prime\prime}(x_0)(x-x_0)^3\\ [10pt]
&+& \cdots\\ [10pt]
&+& \frac{1}{n!}f^{(n+1)}(x_0)(x-x_0)^n\\ [10pt]
&+& R_{n+1}(x)
\end{eqnarray*}

where $ R_{n+1}(x)$ is called the remainder term. Then Taylor's theorem [65, pp. 95-96] provides that there exists some $ \xi$ between $ x$ and $ x_0$ such that

$\displaystyle R_{n+1}(x) = \frac{f^{(n+1)}(\xi)}{(n+1)!}(x-x_0)^{n+1}.
$

In particular, if $ \vert f^{(n+1)}\vert\leq M$ in $ I$ , then

$\displaystyle R_{n+1}(x) \leq \frac{M \vert x-x_0\vert^{n+1}}{(n+1)!}
$

which is normally small when $ x$ is close to $ x_0$ .

When $ x_0=0$ , the Taylor series reduces to what is called a Maclaurin series [58, p. 96].


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``Mathematics of the Discrete Fourier Transform (DFT), with Audio Applications --- Second Edition'', by Julius O. Smith III, W3K Publishing, 2007, ISBN 978-0-9745607-4-8.
Copyright © 2014-04-06 by Julius O. Smith III
Center for Computer Research in Music and Acoustics (CCRMA),   Stanford University
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