Taylor's Theorem/One Variable/Proof by Cauchy Mean Value 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 $G$ be a real-valued function continuous on $\left[{a \,.\,.\, x}\right]$ and differentiable with non-vanishing derivative on $\left({a \,.\,.\, x}\right)$.

Let:

$F \left({t}\right) = f \left({t}\right) + \dfrac {f' \left({t}\right)} {1!} \left({x - t}\right) + \dotsb + \dfrac {f^{\left({n}\right)} \left({t}\right)} {n!} \left({x - t}\right)^n$

By the Cauchy Mean Value Theorem:

$(1): \quad \dfrac {F' \left({\xi}\right)} {G' \left({\xi}\right)} = \dfrac {F \left({x}\right) - F \left({a}\right)} {G \left({x}\right) - G \left({a}\right)}$

for some $\xi \in \left({a \,.\,.\, x}\right)$.

Note that the numerator:

$F \left({x}\right) - F \left({a}\right) = R_n$

is the remainder of the Taylor polynomial for $f \left({x}\right)$.

On the other hand, computing $F' \left({\xi}\right)$:

$F' \left({\xi}\right) = f' \left({\xi}\right) - f' \left({\xi}\right) + \dfrac {f'' \left({\xi}\right)} {1!} \left({x - \xi}\right) - \dfrac {f'' \left({\xi}\right)} {1!} \left({x - \xi}\right) + \dotsb + \dfrac {f^{\left({n + 1}\right)} \left({t}\right)} {n!} \left({x - \xi}\right)^n = \dfrac {f^{\left({n + 1}\right)} \left({\xi}\right)} {n!} \left({x - \xi}\right)^n$

Putting these two facts together and rearranging the terms of $(1)$ yields:

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

which was to be shown.


Note that the Lagrange Form of the Remainder comes from taking $G \left({t}\right) = \left({x - t}\right)^{n + 1}$ and the given Cauchy Form of the Remainder comes from taking $G \left({t}\right) = t - a$.

$\blacksquare$


Source of Name

This entry was named for Brook Taylor.