Taylor's Theorem/One Variable/Proof by Cauchy Mean Value Theorem
Theorem
Let $f$ be a real function which is:
- of differentiability class $C^n$ on the closed interval $\closedint a x$
and:
- at least $n + 1$ times differentiable on the open interval $\openint a x$.
Then:
\(\ds \map f x\) | \(=\) | \(\ds \frac 1 {0!} \map f a\) | ||||||||||||
\(\ds \) | \(\) | \(\, \ds + \, \) | \(\ds \frac 1 {1!} \paren {x - a} \map {f'} a\) | |||||||||||
\(\ds \) | \(\) | \(\, \ds + \, \) | \(\ds \frac 1 {2!} \paren {x - a}^2 \map {f} a\) | |||||||||||
\(\ds \) | \(\) | \(\, \ds + \, \) | \(\ds \cdots\) | |||||||||||
\(\ds \) | \(\) | \(\, \ds + \, \) | \(\ds \frac 1 {n!} \paren {x - a}^n \map {f^{\paren n} } a\) | |||||||||||
\(\ds \) | \(\) | \(\, \ds + \, \) | \(\ds R_n\) |
where $R_n$ (sometimes denoted $E_n$) is known as the error term or remainder, and can be presented in one of $2$ forms:
- Lagrange Form
- $R_n = \dfrac {\map {f^{\paren {n + 1} } } \xi} {\paren {n + 1}!} \paren {x - a}^{n + 1}$
for some $\xi \in \openint a x$.
- Cauchy Form
- $R_n = \dfrac {\map {f^{\paren {n + 1} } } \xi \paren {x - \xi}^n} {n!} \paren {x - a}$
for some $\xi \in \openint a x$.
Proof
Let $G$ be a real-valued function continuous on $\closedint a x$ and differentiable with non-vanishing derivative on $\openint a x$.
Let:
- $\map F t = \map f t + \dfrac {\map {f'} t} {1!} \paren {x - t} + \dotsb + \dfrac {\map {f^{\paren n} } t} {n!} \paren {x - t}^n$
By the Cauchy Mean Value Theorem:
- $(1): \quad \dfrac {\map {F'} x} {\map {G'} \xi} = \dfrac {\map F x - \map F a} {\map G x - \map G a}$
for some $\xi \in \openint a x$.
Note that the numerator:
- $\map F x - \map F a = R_n$
is the remainder of the Taylor polynomial for $\map f x$.
On the other hand, computing $\map {F'} \xi$:
- $\map {F'} \xi = \map {f'} \xi - \map {f'} \xi + \dfrac {\map {f} \xi} {1!} \paren {x - \xi} - \dfrac {\map {f} \xi} {1!} \paren {x - \xi} + \dotsb + \dfrac {\map {f^{\paren {n + 1} } } t} {n!} \paren {x - \xi}^n = \dfrac {\map {f^{\paren {n + 1} } } \xi} {n!} \paren {x - \xi}^n$
Putting these two facts together and rearranging the terms of $(1)$ yields:
- $R_n = \dfrac {\map {f^{\paren {n + 1} } } \xi} {n!} \paren {x - \xi}^n \dfrac {\map G x - \map G a} {\map {G'} \xi}$
which was to be shown.
Although this article appears correct, it's inelegant. There has to be a better way of doing it. In particular: Find a way of neatly handling the two different remainder forms You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by redesigning it. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Improve}} from the code.If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page. |
Note that the Lagrange Form of the Remainder comes from taking $\map G t = \paren {x - t}^{n + 1}$ and the given Cauchy Form of the Remainder comes from taking $\map G t = t - a$.
$\blacksquare$
Source of Name
This entry was named for Brook Taylor.