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7.6: Taylor's Theorem Revisited

  • Page ID
    22683
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    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.


    This page titled 7.6: Taylor's Theorem Revisited is shared under a CC BY-NC-SA 1.0 license and was authored, remixed, and/or curated by Dan Sloughter via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.