Taylor's Theorem with Peano's Form of Remainder

The following form of Taylor's Theorem with minimal hypotheses is not widely popular and goes by the name of Taylor's Theorem with Peano's Form of Remainder:

Taylor's Theorem with Peano's Form of Remainder: If $f$ is a function such that its $n^{\text{th}}$ derivative at $a$ (i.e. $f^{(n)}(a)$) exists then $$f(a + h) = f(a) + hf'(a) + \frac{h^{2}}{2!}f''(a) + \cdots + \frac{h^{n}}{n!}f^{(n)}(a) + o(h^{n})$$ where $o(h^{n})$ represents a function $g(h)$ with $g(h)/h^{n} \to 0$ as $h \to 0$.

One of the proofs (search "Proof of Taylor's Theorem" in this blog post) of this theorem uses repeated application of L'Hospital's Rule. And it appears that proofs of the above theorem apart from the one via L'Hospital's Rule are not well known. I have asked this question to get other proofs of this theorem which do not rely on L'Hospital's Rule and instead use simpler ideas.

BTW I am also posting one proof of my own as a community wiki.

• +1. Just one observation: nowhere did I say that "I think L'Hopital Rule is the only way to prove it". I said that I didn't know another way, and even asked for one (thus implying heavily that I actually didn't think it was the only one). – Aloizio Macedo Jun 2 '16 at 15:27
• @AloizioMacedo: I am extremely sorry if I misinterpreted your comment and if I offended you even slightly. My main objection however was that you thought that the use of Taylor was somewhat circular. I will change the wording of my question to remove "only way to prove it." – Paramanand Singh Jun 2 '16 at 15:29
• Paramanand, this should be a protected question inasmuch as a proof of Taylor with the Peano form of the remainder is not trivial to find. And a proof that does not appeal to LHR is even more difficult to find. – Mark Viola Jul 12 '16 at 13:48
• @Dr.MV: How to protect questions? I see only close and delete links. – Paramanand Singh Jul 12 '16 at 14:45
• Paramanand, I just learned about protected questions HERE. So, I might have incorrectly understood the intent behind protection. One would also need an answer with 10 up votes or more. – Mark Viola Jul 12 '16 at 16:38

We will prove the result for $h \to 0^{+}$ and the argument for $h \to 0^{-}$ is similar. The proof is taken from my favorite book A Course of Pure Mathematics by G. H. Hardy.

Since $f^{(n)}(a)$ exists it follows that $f^{(n - 1)}(x)$ exists in some neighborhood of $a$ and $f^{(n - 2)}(x)$ is continuous in that neighborhood of $a$. Let $h \geq 0$ and we define another function $$F_{n}(h) = f(a + h) - \left\{f(a) + hf'(a) + \frac{h^{2}}{2!}f''(a) + \cdots + \frac{h^{n - 1}}{(n - 1)!}f^{(n - 1)}(a)\right\}\tag{1}$$ Then $F_{n}(h)$ and its first $(n - 1)$ derivatives vanish at $h = 0$ and $F_{n}^{(n)}(0) = f^{(n)}(a)$. Hence if we write $$G(h) = F_{n}(h) - \frac{h^{n}}{n!}\{f^{(n)}(a) - \epsilon\}\tag{2}$$ where $\epsilon > 0$, then we have $$G(0) = 0, G'(0) = 0, \ldots, G^{(n - 1)}(0) = 0, G^{(n)}(0) = \epsilon > 0\tag{3}$$ Since $G^{(n)}(0) > 0$ it follows that there is a number $\delta_{1} > 0$ such that $G^{(n - 1)}(h) > 0$ for all values of $h$ with $0 < h < \delta_{1}$. Using mean value theorem and noting that $G^{(n - 1)}(0) = 0$ we can see that $G^{(n - 2)}(h) > 0$ for all $h$ with $0 < h < \delta_{1}$. Applying the same argument repeatedly we can see that $G(h) > 0$ for all $h$ wih $0 < h < \delta_{1}$. Thus $$F_{n}(h) > \frac{h^{n}}{n!}\{f^{(n)}(a) - \epsilon\}\tag{4}$$ for $0 < h < \delta_{1}$. Similarly we can prove that $$F_{n}(h) < \frac{h^{n}}{n!}\{f^{(n)}(a) + \epsilon\}\tag{5}$$ for all $h$ with $0 < h < \delta_{2}$.

Thus for every $\epsilon > 0$ there is a $\delta = \min(\delta_{1}, \delta_{2}) > 0$ such that $$\frac{h^{n}}{n!}\{f^{(n)}(a) - \epsilon\} < F_{n}(h) < \frac{h^{n}}{n!}\{f^{(n)}(a) + \epsilon\}\tag{6}$$ for all values of $h$ with $0 < h < \delta$. This proves the theorem for $h \to 0^{+}$.

Slight care has to be taken when dealing with negative values of $h$ for the case $h \to 0^{-}$ because here the nature of inequalities will depend on whether $n$ is odd or even and thus we need to handle both the cases of even $n$ and odd $n$ separately.

• This is a very useful proof! Well done. +1 – Mark Viola Jul 12 '16 at 13:44
• Paramanand, I have a question. In the first sentence following the separation line, why did you write the seemingly obvious "$f^{(n-2)}(x)$ is continuous in that neighborhood of $a$?" – Mark Viola Jul 12 '16 at 14:00
• @Dr.MV: Yeah its obvious, but I wanted to highlight the consequences of existence of $f^{(n)}(a)$ explicitly. – Paramanand Singh Jul 12 '16 at 14:41
• @MathematicsStudent1122: Thanks! It is a bit unfortunate that Hardy's book is not so popular nowadays and such proofs are virtually unknown. Only Michael Spivak's calculus comes nearest to Hardy's book and Spivak mentions Hardy's book in his bibliography. – Paramanand Singh Sep 4 '16 at 5:01
• @PedroGomes: see the definition of $G$. It has two terms. The first term is $F_n$ whose nth derivative is $f^{(n)} (a)$ and the second term is $h^n/n! (f^{(n)} (a) - \epsilon)$ whose $n$th derivative is $f^{(n)} (a) - \epsilon$. Combining these we get $G^{(n)} (0)=\epsilon$. – Paramanand Singh Jul 7 '18 at 0:49

The following argument is for $n=2$, but it can be extended to higher derivatives without much pain. Suppose $f:(\mathbb R\supseteq )D\to\mathbb R$ is twice differentiable at $a\in D$ (open), then $f'$ exists in a neighborhood (without loss of generality call it $D$), it is thus gauge (or Kurzweil-Henstock) integrable (see Lamoreaux & Armstrong (1998) for an undergraduate level discussion) and satisfies \begin{equation} f(a+h)=f(a)+\int_0^1 f'(a+th) \operatorname dt\,h. \end{equation} Differentiability of $f'$ at $a$ is equivalent to say for all $k:a+k\in D$ we have \begin{equation} f'(a+k)=f'(a)+f''(a)k+\hat g(k) \text{ where } \frac{\hat g(k)}k=:\bar g(k)\to0 \text{ as }k\to0. \end{equation} Note that since $f'(a+k)$, $f'(a)$ and $f''(a)k$ are gauge integrable in $k$ (the first by the cited article and the latter two are Riemann integrable) also $\hat g(k)$ is gauge integrable in $k$. Taking $k=ht$ in the first equation we see \begin{equation} \begin{split} f(a+h)-f(a) &= \int_0^1f'(a)+f''(a)th+\hat g(th)\operatorname dt\,h \\ &= f'(a)h+(\smallint_0^1t\operatorname dt)f''(a)h^2+\int_0^1\hat g(th)\operatorname dt\,h \\ &= f'(a)h+\frac12f''(a)h^2+g(h) \end{split} \end{equation} where \begin{equation} g(h):= h\int_0^1\hat g(th)\operatorname dt =\int_0^h\hat g(k)\operatorname dk =\int_0^h\bar g(k)k\operatorname dk \end{equation} and $\bar g(k)\to0$ as $k\to0$. Taylor's theorem (as stated in the question) will follow from \begin{equation} \frac{g(h)}{h^2}\to0 \text{ as }h\to0. \end{equation} To see this, suppose $\epsilon>0$ and let $\delta>0$ such that $|\bar g(k)|<2\epsilon$ if $|k|<\delta$, and use monotonicity for the gauge integral (summarised in Heikkilä (2011)) to get \begin{equation} g(h)\leq\frac{h^2}\epsilon \text{ if } |h|<\delta. \end{equation}

In the general case (including this one) all we need is the gauge integrability of $f^{(n-1)}$ in a neighborhood of $a$, but this is guaranteed by the fact that $f^{(n-1)}$ is the derivative of $f^{(n-2)}$ in a whole such neighborhood.

The concept of gauge integral may seem sophisticated, but in fact it is quite elementary and many people teach it to second and even first year undergrads. A letter underwritten by many analysis researchers and teachers, calling for a review of the "standard" calculus curriculum, has been around for some years.

• All this proof is very standard and routinely presented in most analysis textbooks. This approach requires the integrability of the derivatives involved and thus requires more assumptions. The Peano's form has very minimal assumptions and the approach in your answer can't really be used to prove it. – Paramanand Singh Jun 7 '18 at 0:31
• Sure. I wouldn't know how to extend Hardy's argument to more dimensions, unless you assume something more about the $n$-th derivative than just existing at $a$. Was your question restricted to one dimension? – Oskar Limka Jun 8 '18 at 21:47
• Yeah my question was for one dimensional case, although this was not mentioned explicitly. – Paramanand Singh Jun 8 '18 at 23:32
• Apologies, I should've realised that from your own answer to your question. I've rewritten completely my answer, focusing on the one dimensional case but insisting with using integrals. Note the use of the somewhat less known gauge (Kurzweil-Henstock) integral. I don't need to assume more than what you assume. – Oskar Limka Jun 9 '18 at 22:20