
# 7.6: Taylor's Theorem Revisited


The following is a version of Taylor's Theorem with an alternative form of the remainder term.

## Theorem $$\PageIndex{1}$$

(Taylor's Theorem)

Suppose $$f \in C^{(n+1)}(a, b), \alpha \in(a, b),$$ and

$P_{n}(x)=\sum_{k=0}^{n} \frac{f^{(k)}(\alpha)}{k !}(x-\alpha)^{k}.$

Then, for any $$x \in(a, b)$$,

$f(x)=P_{n}(x)+\int_{\alpha}^{x} \frac{f^{(n+1)}(t)}{n !}(x-t)^{n} d t.$

Proof

By the Fundamental Theorem of Calculus, we have

$\int_{\alpha}^{x} f^{\prime}(t) d t=f(x)-f(\alpha),$

which implies that

$f(x)=f(\alpha)+\int_{\alpha}^{x} f^{\prime}(t) d t.$

Hence the theorem holds for $$n=0 .$$ Suppose the result holds for $$n=k-1,$$ that is,

$f(x)=P_{k-1}(x)+\int_{\alpha}^{x} \frac{f^{(k)}(t)}{(k-1) !}(x-t)^{k-1} d t.$

Let

$F(t)=f^{(k)}(t),$

$g(t)=\frac{(x-t)^{k-1}}{(k-1) !},$

and

$G(t)=-\frac{(x-t)^{k}}{k !}.$

Then

\begin{aligned} \int_{\alpha}^{x} \frac{f^{(k)}(t)}{(k-1) !}(x-t)^{k-1} d t &=\int_{\alpha}^{x} F(t) g(t) d t \\ &=F(x) G(x)-F(\alpha) G(\alpha)-\int_{\alpha}^{x} F^{\prime}(t) G(t) d t \\ &=\frac{f^{(k)}(\alpha)(x-\alpha)^{k}}{k !}+\int_{\alpha}^{x} \frac{f^{(k+1)}(t)}{k !}(x-t)^{k} d t, \end{aligned}

Hence

$f(x)=P_{k}(x)+\int_{\alpha}^{x} \frac{f^{(k+1)}(t)}{k !}(x-t)^{k} d t,$

and so the theorem holds for $$n=k$$. $$\quad$$ Q.E.D.

## Exercise $$\PageIndex{1}$$

(Cauchy form of the remainder)

Under the conditions of Taylor's Theorem as just stated, show that

$\int_{\alpha}^{x} \frac{f^{(n+1)}(t)}{n !}(x-t)^{n} d t=\frac{f^{(n+1)}(\gamma)}{n !}(x-\gamma)^{n}(x-\alpha)$

for some $$\gamma$$ between $$\alpha$$ and $$x .$$

## Exercise $$\PageIndex{2}$$

(Lagrange form of the remainder)

Under the conditions of Taylor's Theorem as just stated, show that

$\int_{\alpha}^{x} \frac{f^{(n+1)}(t)}{n !}(x-t)^{n} d t=\frac{f^{(n+1)}(\gamma)}{(n+1) !}(x-\alpha)^{n+1}$

for some $$\gamma$$ between $$\alpha$$ and $$x .$$ Note that this is the form of the remainder in Theorem $$6.6 .1,$$ although under slightly more restrictive assumptions.