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.