Taylor's Theorem/One Variable/Proof by Rolle's Theorem

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Theorem

Let $f$ be a real function which is continuous on the closed interval $\left[{a \,.\,.\, b}\right]$ and $n + 1$ times differentiable on the open interval $\left({a \,.\,.\, b}\right)$.

Let $\xi \in \left({a \,.\,.\, b}\right)$.


Then, given any $x \in \left({a \,.\,.\, b}\right)$, there exists some $\eta \in \R: x \le \eta \le \xi$ or $\xi \le \eta \le x$ such that:

\(\displaystyle f \left({x}\right)\) \(=\) \(\displaystyle \frac 1 {0!} f \left({\xi}\right)\)
\(\displaystyle \) \(\) \(\, \displaystyle + \, \) \(\displaystyle \frac 1 {1!} \left({x - \xi}\right) f^{\prime} \left({\xi}\right)\)
\(\displaystyle \) \(\) \(\, \displaystyle + \, \) \(\displaystyle \frac 1 {2!} \left({x - \xi}\right)^2 f^{\prime \prime} \left({\xi}\right)\)
\(\displaystyle \) \(\) \(\, \displaystyle + \, \) \(\displaystyle \cdots\)
\(\displaystyle \) \(\) \(\, \displaystyle + \, \) \(\displaystyle \frac 1 {n!} \left({x - \xi}\right)^n f^{\left({n}\right)} \left({\xi}\right)\)
\(\displaystyle \) \(\) \(\, \displaystyle + \, \) \(\displaystyle R_n\)

where $R_n$ (sometimes denoted $E_n$) is known as the error term, and satisfies:

$R_n = \dfrac 1 {\left({n + 1}\right)!} \left({x - \xi}\right)^{n + 1} f^{\left({n + 1}\right)} \left({\eta}\right)$


Note that when $n = 0$ Taylor's Theorem reduces to the Mean Value Theorem.


The expression:

$\displaystyle f \left({x}\right) = \sum_{n \mathop = 0}^\infty \frac {\left({x - \xi}\right)^n} {n!} f^{\left({n}\right)} \left({\xi}\right)$

where $n$ is taken to the limit, is known as the Taylor series expansion of $f$ about $\xi$.


Proof

Let the function $g$ be defined as:

$g \left({t}\right) = R_n \left({t}\right) - \dfrac {\left({t - a}\right)^{n + 1} } {\left({x - a}\right)^{n + 1} } R_n \left({x}\right)$

Then:

$g^{\left({k}\right)} \left({a}\right) = 0$

for $k = 0, \dotsc, n$, and $g \left({x}\right) = 0$.


Apply Rolle's Theorem successively to $g, g', \dotsc, g^{\left({n}\right)}$.

Then there exist:

$\xi_1, \ldots, \xi_{n + 1}$

between $a$ and $x$ such that:

$g' \left({\xi_1}\right) = 0, g'' \left({\xi_2}\right) = 0, \ldots, g^{\left({n + 1}\right)} \left({\xi_{n + 1} }\right) = 0$


Let $\xi = \xi_{n + 1}$.

Then:

$0 = g^{\left({n + 1}\right)} \left({\xi}\right) = f^{\left({n + 1}\right)} \left({\xi}\right) - \dfrac {\left({n + 1}\right)!} {\left({x - a}\right)^{n + 1} } R_n \left({x}\right)$

and the formula for $R_n \left({x}\right)$ follows.

$\blacksquare$


Source of Name

This entry was named for Brook Taylor.


Sources