# Taylor's Theorem/One Variable/Proof by Rolle's 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 the function $g$ be defined as:

$\map g t = \map {R_n} t - \dfrac {\paren {t - a}^{n + 1} } {\paren {x - a}^{n + 1} } \map {R_n} x$

Then:

$\map {g^{\paren k} } a = 0$

for $k = 0, \dotsc, n$, and $\map g x = 0$.

Apply Rolle's Theorem successively to $g, g', \dotsc, g^{\paren n}$.

Then there exist:

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

between $a$ and $x$ such that:

$\map {g'} {\xi_1} = 0, \map {g''} {\xi_2} = 0, \ldots, \map {g^{\paren {n + 1} } } {\xi_{n + 1} } = 0$

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

Then:

$0 = \map {g^{\paren {n + 1} } } \xi = \map {f^{\paren {n + 1} } } \xi - \dfrac {\paren {n + 1}!} {\paren {x - a}^{n + 1} } \map {R_n} x$

and the formula for $\map {R_n} x$ follows.

$\blacksquare$

## Source of Name

This entry was named for Brook Taylor.