Proof of Taylor's Theorem 📂Calculus

Proof of Taylor's Theorem

Theorem¹

If a function $f(x)$ is continuous at $[a,b]$ and differentiable up to $n$ times at $(a,b)$, then there exists $\xi \in (a,b)$ that satisfies

$$ \begin{align*} f(b) =& \sum_{k=0}^{n-1} {{(b-a)^{k}\over{k!}}{f^{(k)}( a )}} + {(b-a)^{n}\over{n!}}{f^{(n)}(\xi)} \\ =& {f(a)} + {(b-a)f ' (a)} + \cdots + {(b-a)^{n-1}\over{(n-1)!}}{f^{(n-1)}(a)} + {(b-a)^{n}\over{(n)!}}{f^{(n)}(\xi)} \end{align*} $$

Explanation

This theorem, which is widely used throughout mathematics, has lent its name to the Taylor series. In terms of being differentiated $n$ times, it can be considered a generalization of the mean value theorem.

Customarily, when using the Taylor theorem, $c$ is not used, but $\xi$ is used instead.

Proof

Let’s consider

$$ \begin{align*} f(b) :=& {(b-a)^0\over{0!}}{f(a)} + {(b-a)^1\over{1!}}{f ' (a)} + {(b-a)^2\over{2!}}{f '' (a)} \\ &+ \cdots + {(b-a)^{n-1}\over{(n-1)!}}{f^{(n-1)}(a)} + {(b-a)^{n}\over{(n)!}}c \end{align*} $$

Showing $c={f^{(n)}(\xi)}$ concludes the proof. Let’s define function $g$ as follows.

$$ \begin{align*} g(x):=& -f(b) + f(x) + {(b-x)^1\over{1!}}{f ' (x)} + {(b-x)^2\over{2!}}{f '' (x)} \\ & + \cdots + {(b-x)^{n-1}\over{(n-1)!}}{f^{(n-1)}(x)} + {(b-x)^{n}\over{(n)!}}c \\ =& -f(b) + \sum_{k=0}^{n-1}{(b-x)^{k}\over{(k)!}}{f^{(k)}(x)} + {(b-x)^{n}\over{(n)!}}c \end{align*} $$

$g$ is continuous at $[a,b]$, differentiable at $(a,b)$, and by definition of $c$, it is $g(b)=g(a)=0$.

Rolle’s Theorem: If function $f(x)$ is continuous at $[a,b]$, differentiable at $(a,b)$, and $f(a)=f(b)$, then there exists at least one $\xi$ in $(a,b)$ that satisfies $f ' (\xi)=0$.

If $h(x)$ is set as

$$ \begin{align*} h(x):=& \left[ \sum_{k=0}^{n-1}{(b-x)^{k}\over{(k)!}}{f^{(k)}(x)} \right] ' \\ =& \left[ {(b-x)^{0}\over{(0)!}}{f^{(0)}(x)} + {(b-x)^{1}\over{(1)!}}{f^{(1)}(x)} + \cdots + {(b-x)^{n-1}\over{(n-1)!}}{f^{(n-1)}(x)} \right] ' \\ =& \left[ f (x) + {(b-x)^{1}\over{(1)!}}{f^{(1)}(x)} + \cdots + {(b-x)^{n-1}\over{(n-1)!}}{f^{(n-1)}(x)} \right] ' \\ =& f^{(1)} (x) - \left[ f^{(1)} (x) + {(b-x)^{1}\over{(1)!}}{f^{(2)}(x)} \right] \\ & + \left[ - {(b-x)^{1}\over{(1)!}} f^{(2)} (x) + {(b-x)^{2}\over{(2)!}}{f^{(3)}(x)} \right] \\ & \vdots \\ & + \left[ - {(b-x)^{n-3}\over{(n-3)!}} f^{(n-2)} (x) + {(b-x)^{n-2}\over{(n-2)!}}{f^{(n-1)}(x)} \right] \\ & + \left[ - {(b-x)^{n-2}\over{(n-2)!}} f^{(n-1)} (x) + {(b-x)^{n-1}\over{(n-1)!}}{f^{(n)}(x)} \right] \\ =& {(b-x)^{n-1}\over{(n-1)!}}{f^{(n)}(x)} \end{align*} $$

and since $\displaystyle g ' (x) = 0 + h(x) - {(b-x)^{n-1}\over{(n-1)!}}c$, by Rolle’s theorem,

$$ \begin{align*} g ' (\xi) =& h(\xi) - {(b-\xi)^{n-1}\over{(n-1)!}}c \\ =& {(b-\xi)^{n-1}\over{(n-1)!}}{f^{(n)}(\xi)} - {(b-\xi)^{n-1}\over{(n-1)!}}c \\ =& 0 \end{align*} $$

there exists at least one $\xi$ in $(a,b)$ that satisfies it. Thus,

$$ \begin{align*} && {(b-\xi)^{n-1}\over{(n-1)!}}{f^{(n)}(\xi)} - {(b-\xi)^{n-1}\over{(n-1)!}}c =& 0 \\ \implies && {(b-\xi)^{n-1}\over{(n-1)!}}{f^{(n)}(\xi)} =& {(b-\xi)^{n-1}\over{(n-1)!}}c \\ \implies && {f^{(n)}(\xi)} =& c \end{align*} $$

Since $c={f^{(n)}(\xi)}$ is shown, the proof is concluded.

■

The proof is as shown above; however, a more commonly used form is as follows. Of course, as $x \in [a,b]$ and set as $x_{0} \in (a,b)$, it essentially becomes $[x_{0} , x] \subset [a,b]$.

Taylor’s Theorem

If a function $f(x)$ is continuous at $[a,b]$ and differentiable up to $n$ times at $(a,b)$, then for $x_{0} \in (a,b)$,

$$ f(x) = \sum_{k=0}^{n-1} {{( x - x_{0} )^{k}\over{ k! }}{f^{(k)}( x_{0} )}} + {(x - x_{0} )^{n}\over{ n! }}{f^{(n)}(\xi)} $$

there exists $\xi \in (a,b)$ that satisfies it.