# Taylor's Theorem/One Variable with Two Functions

## Theorem

Let $f$ and $g$ be real functions satisfying following conditions:

$(1): \quad f$ is $n + 1$ times differentiable on the open interval $\openint a x$
$(2): \quad f$ is of differentiability class $C^n$ on the closed interval $\closedint a x$
$(3): \quad g$ is $k + 1$ times differentiable on the open interval $\openint a x$
$(4): \quad g$ is of differentiability class $C^k$ on the closed interval $\closedint a x$
$(5): \quad \map {g^{\paren {k + 1}}} t \ne 0$ for any $t \in \openint a x$

Then the following equation holds for some real number $\xi \in \openint a x$:

$\dfrac {\map {f^{\paren {n + 1} } } \xi /n!} {\map {g^{\paren {k + 1} } } \xi /k!} \paren {x - \xi}^{n - k} = \dfrac {\map f x - \map f a - \map {f'} a \paren {x - a} - \dfrac {\map {f^{\prime\prime} } a} {2!} \paren {x - a}^2 - \dotsb - \dfrac {\map {f^{\paren n} } a} {n!} \paren {x - a}^n} {\map g x - \map g a - \map {g'} a \paren {x - a} - \dfrac {\map {g^{\prime\prime} } a} {2!} \paren {x - a}^2 - \dotsb - \dfrac {\map {g^{\paren k} } a} {k!} \paren {x - a}^k}$

or equivalently:

 $\ds \map f x$ $=$ $\ds \map f a + \map {f'} a \paren {x - a} + \dfrac {\map {f^{\prime\prime} } a} {2!} \paren {x - a}^2 + \dotsb + \dfrac {\map {f^{\paren n} } a} {n!} \paren {x - a}^n + R_n$ $\ds R_n$ $=$ $\ds \dfrac {\map {f^{\paren {n + 1} } } \xi / n!} {\map {g^{\paren {k + 1} } } \xi / k!} \paren {x - \xi}^{n - k} \paren {\map g x - \map g a - \map {g'} a \paren {x - a} - \dfrac {\map {g^{\prime\prime} } a} {2!} \paren {x - a}^2 - \dotsb - \dfrac {\map {g^{\paren k} } a} {k!} \paren {x - a}^k}$

## Proof

We define $F$ and $G$ as follows:

 $\ds \map F t$ $=$ $\ds \map f t + \map {f'} t \paren {x - t} + \dfrac {\map {f^{\prime\prime} } t} {2!} \paren {x - t}^2 + \dotsb + \dfrac {\map {f^{\paren n} } t} {n!} \paren {x - t}^n$ $\ds \map G t$ $=$ $\ds \map g t + \map {g'} t \paren {x - t} + \dfrac {\map {g^{\prime\prime} } t} {2!} \paren {x - t}^2 + \dotsb + \dfrac {\map {g^{\paren k} } t} {k!} \paren {x - t}^k$

Then $F$ and $G$ are continuous on $\closedint a x$ and differentiable on $\openint a x$.

Differentiating with respect to $t$:

 $\ds \map {F'} t$ $=$ $\ds \map {f'} t + \paren {\map {f{\prime\prime} } t \paren {x - t} - \map {f'} t} + \paren {\frac {\map {f^{\paren 3} } t} {2!} \paren {x - t}^2 - \frac {\map {f^{\paren 2} } t} {1!} \paren {x - t} } + \dotsb + \paren {\frac {\map {f^{\paren {n + 1} } } t} {n!} \paren {x - t}^n - \frac {\map {f^{\paren n} } t} {\paren {n - 1}!} \paren {x - t}^{n - 1} }$ $\ds$ $=$ $\ds \frac {\map {f^{\paren {n + 1} } } t} {n!} \paren {x - t}^n$ $\ds \map {G'} t$ $=$ $\ds \frac {\map {g^{\paren {k + 1} } } t} {k!} \paren {x - t}^k$

By condition $(5)$ of the statement of the theorem, it follows that $G$ does not vanish on $\openint a x$.

By Cauchy Mean Value Theorem, there exists a real number $\xi \in \openint a x$ such that:

$\dfrac {\map {F'} \xi} {\map {G'} \xi} = \dfrac {\map F x - \map F a} {\map G x - \map G a}$

That is:

$\dfrac {\map {f^{\paren {n + 1} } } \xi / n!} {\map {g^{\paren {k + 1} } } \xi /k!} \paren {x - \xi}^{n - k} = \dfrac {\map f x - \map f a - \map {f'} a \paren {x - a} - \dfrac {\map {f^{\prime\prime} } a} {2!} \paren {x - a}^2 - \dotsb - \dfrac {\map {f^{\paren n} } a} {n!} \paren {x - a}^n} {\map g x - \map g a - \map {g'} a \paren {x - a} - \dfrac {\map {g^{\prime\prime} } a} {2!} \paren {x - a}^2 - \dotsb - \dfrac {\map {g^{\paren k}} a} {k!} \paren {x - a}^k}$

$\blacksquare$

## Source of Name

This entry was named for Brook Taylor.