# 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.Find a way of neatly handling the two different remainder formsYou can help Proof Wiki 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.