Taylor's Theorem/One Variable with Two Functions
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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:
- $\dfrac {\map {f^{\paren {n + 1} } } \xi / n!} {\map {g^{\paren {k + 1} } } \xi /k!} \paren {x - \xi}^{n - k} = \dfrac {\map f x - \map f a - \map {f'} a \paren {x - a} - \dfrac {\map {f^{\prime\prime} } a} {2!} \paren {x - a}^2 - \dotsb - \dfrac {\map {f^{\paren n} } a} {n!} \paren {x - a}^n} {\map g x - \map g a - \map {g'} a \paren {x - a} - \dfrac {\map {g^{\prime\prime} } a} {2!} \paren {x - a}^2 - \dotsb - \dfrac {\map {g^{\paren k}} a} {k!} \paren {x - a}^k}$
$\blacksquare$
Also see
- Definition:Lagrange Form of Remainder of Taylor Series: taking $k = n$ and $\map g x = \paren {x - a}^{n + 1}$.
- Definition:Cauchy Form of Remainder of Taylor Series: taking $k = 0$ and $\map g x = x - a$.
Source of Name
This entry was named for Brook Taylor.