# 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.

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.