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In multiple places, the requirements for Taylor's Theorem with integral form of remainder state that the assumption is slightly stronger then the usual form of Taylor's theorem, since as opposed to assuming only that the (n+1)th derivative exists, we now assume that the (n+1)th derivative is continuous

(e.g. https://en.wikipedia.org/wiki/Taylor%27s_theorem#Taylor's_theorem_in_one_real_variable and https://brilliant.org/wiki/taylors-theorem-with-lagrange-remainder/)

My question is, why is assuming the (n+1)th derivative is continuous necessary, and assuming that is is merely integrable not sufficient? I am familiar with the typical proof using induction coupled with integration by parts, and I don't see any point at which continuity seems necessary, but it was so consistently stated as a condition in online sources I worried I might have missed something important.

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The different requirements can already be seen in the case $n=0$: The mean-value theorem $$ f(x) = f(a) + (x-a) f'(\xi) $$ holds whenever $f$ is differentiable on $[a, x]$, but the Fundamental theorem of calculus $$ f(x) = f(a) + \int_a^x f'(t) \, dt $$ requires that $f$ is absolutely continuous. That is in particular the case if $f'$ is continuous.

The derivation for the integral form of the remainder uses the Fundamental theorem of calculus and then integration by parts on the terms $$ \int_a^x \frac{f^{(k)}(t)}{(k-1)!} (x-t)^{k-1} \, dt $$ for $k=1, \ldots, n$. Again these steps are valid if the $k$-th derivative of $f$ is absolutely continuous (see for example Weakest hypothesis for integration by parts).

Therefore Taylor's theorem with the integral remainder is valid if $f^{(n)}$ is absolutely continuous.

If $f^{(n+1)}$ is continuous then $f^{(n)}$ is absolutely continuous, so that is a stronger (but perhaps easier to verify) requirement.

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  • $\begingroup$ Thank you for the detailed response! I wasn't familiar yet with the concept of absolute continuity; had only been introduced to continuous and uniformly continuous as concepts up to this point. In the course that i'm taking, we are taught that that fundamental theorem of calculus requires g to be differentiable, and g' to be integrable in order to have g(x) - g(a) = integral from a to b of g', instead of using this notion of absolutely continuous. $\endgroup$ Mar 16, 2021 at 5:56
  • $\begingroup$ @ChairmanMeow: Actually your version seems to be correct as well: Theorem 8.21 in Rudin's “Real and Complex Analysis” states that $f(x) = f(a) + \int_a^x f'(t) \, dt$ also holds if $f$ is differentiable everywhere in the interval, and the derivative is Lebesgue integrable. $\endgroup$
    – Martin R
    Mar 16, 2021 at 17:17

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