7.6: Taylor's Theorem Revisited
- Page ID
- 22683
The following is a version of Taylor's Theorem with an alternative form of the remainder term.
(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
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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.
(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 \(x .\)
(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.