5. Taylor’s approximation

In what follows, fix some $x,\delta\in\mathbb{U}$ with $x>\mathbb{R}$ and $x+\delta>\mathbb{R}$ . We shall abbreviate $\frac{f^{\prime}}{f}$ with $f^{\dagger}$ , provided $f\neq 0$ . Note that $f^{\dagger}=(\log|f|)^{\prime}$ .

Proposition 5.1.

For all non-zero $f\in\mathbb{R}\langle\!\langle T\rangle\!\rangle$ , if $f\not\asymp 1$ , $x+\delta\asymp x$ , and $(f^{\dagger}\circ x)\delta\preceq 1$ , then

f\circ(x+\delta)\asymp f\circ x\quad\text{and}\quad f\circ(x+\delta)-f\circ x% \asymp(f^{\prime}\circ x)\delta.

In particular, if $(f^{\dagger}\circ x)\delta\prec 1$ , then $f\circ(x+\delta)\sim f\circ x$ .

Proof.

Note that $x+\delta\asymp x$ implies $\delta\preceq x$ . We work by induction on $\mathrm{ER}(f)$ . For the base case, suppose that $f=\log^{\circ n}(T)$ for some $n\in\mathbb{N}$ . Both conclusions are trivial for $n=0$ . For $n=1$ , we know that

\log(x+\delta)-\log(x)=\log\left(1+\frac{\delta}{x}\right)\asymp\frac{\delta}{% x}=(f^{\prime}\circ x)\delta.

Note that $\frac{\delta}{x}\preceq 1\prec\log(x)$ , so in particular $\log(x+\delta)\sim\log(x)$ . For $n>1$ , recall that the Taylor series of $\log$ implies that $\log(y+\varepsilon)-\log(y)\sim\frac{\varepsilon}{y}$ when $\varepsilon\prec y$ . Using $y=\log^{\circ n-1}(x)$ , $\varepsilon=\log^{\circ n-1}(x+\delta)-y$ , we can verify inductively that $\varepsilon\prec y$ , or in other words $\log^{\circ n-1}(x+\delta)\sim\log^{\circ n-1}(x)$ , and

\log(\log^{\circ n-1}(x+\delta))-\log(\log^{\circ n-1}(x))\sim\frac{\log^{% \circ n-1}(x+\delta)-\log^{\circ n-1}(x)}{\log^{\circ n-1}(x)}\\ \sim\frac{\log(x+\delta)-\log(x)}{\log(x)\cdots\log^{\circ n-1}(x)}\asymp\frac% {\delta}{x\log(x)\cdots\log^{\circ n-1}(x)}=(f^{\prime}\circ x)\delta.

For $\mathrm{ER}(f)>0$ , we first prove the conclusions for $f=e^{\gamma}$ with $\gamma\in\mathbb{J}^{\neq 0}$ . Suppose that $(f^{\dagger}\circ x)\delta\preceq 1$ ; since $f^{\dagger}=\frac{e^{\gamma}\gamma^{\prime}}{e^{\gamma}}=\gamma^{\prime}$ , we have $(\gamma^{\prime}\circ x)\delta\preceq 1$ , and in particular $(\gamma^{\dagger}\circ x)\delta\prec 1$ , as $\gamma\succ 1$ . By inductive assumption, $\gamma\circ(x+\delta)-\gamma\circ x\asymp(\gamma^{\prime}\circ x)\delta\preceq 1$ . Since $e^{y}-1\asymp y$ for $y\preceq 1$ , we get

e^{\gamma}\circ(x+\delta)-e^{\gamma}\circ x=e^{\gamma\circ x}\left(e^{\gamma% \circ(x+\delta)-\gamma\circ x}-1\right)\asymp e^{\gamma\circ x}(\gamma^{\prime% }\circ x)\delta=((e^{\gamma})^{\prime}\circ x)\delta.

Since $e^{y}\asymp 1$ for $y\preceq 1$ , we similarly find that

e^{\gamma}\circ(x+\delta)=e^{\gamma\circ x}e^{\gamma\circ(x+\delta)-\gamma% \circ x}\asymp e^{\gamma\circ x}=e^{\gamma}\circ x.

For general $f$ , let $re^{\gamma}$ be the leading term of $f$ . Note that by assumption $\gamma\neq 0$ . Since $f\sim re^{\gamma}$ , we have

f\circ(x+\delta)\sim re^{\gamma}\circ(x+\delta)\asymp re^{\gamma}\circ x\sim f% \circ x.

Since $f\not\asymp 1$ , we also have $f^{\prime}\sim(re^{\gamma})^{\prime}$ and $f^{\dagger}\sim{(re^{\gamma})}^{\dagger}=\gamma^{\prime}$ , hence by Corollary 4.4,

f\circ(x+\delta)-f\circ x\sim re^{\gamma}\circ(x+\delta)-re^{\gamma}\circ x% \asymp r((e^{\gamma})^{\prime}\circ x)\delta\sim(f^{\prime}\circ x)\delta.

When $(f^{\dagger}\circ x)\delta\prec 1$ , or in other words $(f^{\prime}\circ x)\delta\prec f\circ x$ , we find that indeed $f\circ(x+\delta)\sim f\circ x$ . ∎

To prove Taylor’s theorem, we now want to iterate the above approximation. To do that, we need to control the assumption $(f^{\dagger}\circ x)\delta\preceq 1$ when we replace $f$ with its derivatives.

Lemma 5.2.

Let $f\in\mathbb{R}\langle\!\langle T\rangle\!\rangle$ be non-zero such that $f\not\asymp T^{k}$ for all $k\in\mathbb{N}$ .

(1)

If $f^{\dagger}\succ\frac{1}{T}$ , then ${(f^{(n)})}^{\dagger}\sim f^{\dagger}$ for all $n$ ; in particular, $f^{(n)}\sim{(f^{\dagger})}^{n}f$ .
(2)

Otherwise, for some $\ell\in\mathbb{N}$ , ${(f^{(n)})}^{\dagger}\asymp\frac{1}{T}$ for all $n\neq\ell-1$ , and if $\ell>0$ , then ${(f^{(\ell-1)})}^{\dagger}\prec\frac{1}{T}$ ; in particular, $f^{(n)}\preceq\frac{f}{T^{n}}$ , $f^{(\ell+n)}\asymp\frac{f^{(\ell)}}{T^{n}}$ for all $n$ .

Proof.

(1) We have $\frac{1}{f^{\dagger}}=\frac{f}{f^{\prime}}\prec T$ , hence

T^{\prime}=1\succ\left(\frac{f}{f^{\prime}}\right)^{\prime}=1-\frac{ff^{\prime% \prime}}{{(f^{\prime})}^{2}}=1-\frac{{(f^{\prime})}^{\dagger}}{f^{\dagger}}.

This says that ${(f^{\prime})}^{\dagger}\sim f^{\dagger}$ , and in particular also that ${(f^{\prime})}^{\dagger}\succ\frac{1}{T}$ . By induction, ${(f^{(n)})}^{\dagger}\sim f^{\dagger}$ for all $n\in\mathbb{N}$ . Since $f^{(n+1)}={(f^{(n)})}^{\dagger}f^{(n)}\sim f^{\dagger}f^{(n)}$ , we also find $f^{(n)}\sim{(f^{\dagger})}^{n}f$ .

(2) Let $r$ be the unique real number such that $f^{\dagger}-\frac{r}{T}\prec\frac{1}{T}$ . Then $f^{\prime}T-rf\prec f$ . Since $f\not\asymp 1$ , we find

f^{\prime\prime}T+f^{\prime}-rf^{\prime}\prec f^{\prime},\quad\text{or in % other words}\quad{(f^{\prime})}^{\dagger}-\frac{r-1}{T}\prec\frac{1}{T}.

In particular, ${(f^{\prime})}^{\dagger}\preceq\frac{1}{T}$ . Since $f\not\asymp T^{k}$ for all $k$ , we have $f^{(k)}\not\asymp 1$ for all $k$ , so we deduce by induction that

{\left(f^{(n)}\right)}^{\dagger}-\frac{r-n}{T}\prec\frac{1}{T}.

In turn, ${(f^{(n)})}^{\dagger}\asymp\frac{1}{T}$ for $n\neq r$ , and ${(f^{(n)})}^{\dagger}\prec\frac{1}{T}$ if $n=r$ , which can only happen if $r\in\mathbb{N}$ . Let $\ell=r+1$ if $r\in\mathbb{N}$ and $\ell=0$ otherwise to recover the desired conclusion. For $n\in\mathbb{N}$ we have $f^{(\ell+n+1)}={(f^{(\ell+n)})}^{\dagger}f^{(\ell+n)}\asymp\frac{f^{(\ell+n)}}% {T}$ , so by induction $f^{(\ell+n)}\asymp\frac{f^{(\ell)}}{T^{n}}$ . Similarly, $f^{(n+1)}={(f^{(n)})}^{\dagger}f^{(n)}\preceq\frac{f^{(n)}}{T}$ , hence $f^{(n)}\preceq\frac{f}{T^{n}}$ . ∎

Two illustrative examples are the following. Take $f=e^{e^{T}}$ . In this case, $f^{\dagger}=e^{T}\succ\frac{1}{T}$ , and Lemma 5.2 predicts that $f^{(n)}\sim{(f^{\dagger})}^{n}f$ . Indeed,

f^{\prime}=e^{T}e^{e^{T}}=f^{\dagger}f,\quad f^{\prime\prime}=e^{T}e^{e^{T}}+e% ^{2T}e^{e^{T}}\sim{\left(f^{\dagger}\right)}^{2}f,\quad f^{\prime\prime\prime}% \sim e^{3T}e^{e^{T}}={\left(f^{\dagger}\right)}^{3}f\quad\dotsc

Now take $f=T\log(T)$ . Here $f^{\dagger}=\frac{\log(T)+1}{T\log(T)}\sim\frac{1}{T}$ , so we expect to see $f^{(n+1)}\asymp\frac{f^{(n)}}{T}$ , with at most one exceptional $f^{(n+1)}\prec\frac{f^{(n)}}{T}$ . Indeed,

f^{\prime}=\log(T)+1\sim\frac{f}{T},\quad f^{\prime\prime}=\frac{1}{T}\prec% \frac{f^{\prime}}{T},\quad f^{\prime\prime\prime}=-\frac{1}{T^{2}}\asymp\frac{% f^{\prime\prime}}{T},\quad f^{\prime\prime\prime\prime}=\frac{2}{T^{3}}\asymp% \frac{f^{\prime\prime\prime}}{T}\quad\dotsc

Corollary 5.3.

Let $f\in\mathbb{R}\langle\!\langle T\rangle\!\rangle$ be non-zero such that $f\not\asymp T^{k}$ for all $k\in\mathbb{N}$ . Suppose that $\delta\neq 0$ . Then the sequence $(f^{(n)}\circ x)\delta^{n}$ is strictly $\prec$ -decreasing when $\delta\prec x$ , and $(f^{\dagger}\circ x)\delta\prec 1$ , weakly $\prec$ -increasing when $\delta\preceq x$ and $(f^{\dagger}\circ x)\delta\preceq 1$ , and eventually strictly $\prec$ -increasing otherwise.

Proof.

We apply Lemma 5.2 to the ratio between two successive elements of the sequence:

\frac{(f^{(n+1)}\circ x)\delta^{n+1}}{(f^{(n)}\circ x)\delta^{n}}=\left(\left(% \frac{f^{(n+1)}}{f^{(n)}}\right)\circ x\right)\delta=\left({\left(f^{(n)}% \right)}^{\dagger}\circ x\right)\delta.

When $f^{\dagger}\succ\frac{1}{T}$ , then ${(f^{(n)})}^{\dagger}\sim f^{\dagger}$ (Lemma 5.2(1)), so

\left({\left(f^{(n)}\right)}^{\dagger}\circ x\right)\delta\sim\left(f^{\dagger% }\circ x\right)\delta.

The sequence is then strictly $\prec$ -decreasing when $(f^{\dagger}\circ x)\delta\prec 1$ , which implies $\delta\prec x$ since $f^{\dagger}\circ x\succ\frac{1}{x}$ , weakly $\prec$ -increasing when $(f^{\dagger}\circ x)\delta\preceq 1$ , which also implies $\delta\prec x$ , and it is strictly $\prec$ -increasing otherwise.

When $f^{\dagger}\preceq\frac{1}{T}$ , then ${(f^{(n)})}^{\dagger}\preceq\frac{1}{T}$ , with ${(f^{(n)})}^{\dagger}\asymp\frac{1}{T}$ for all but possibly one value of $n$ (Lemma 5.2(2)), hence

\left({\left(f^{(n)}\right)}^{\dagger}\circ x\right)\delta\preceq\frac{\delta}% {x},

with equivalence for all but possibly one $n$ . The sequence is then strictly $\prec$ -decreasing when $\delta\prec x$ , which implies $(f^{\dagger}\circ x)\delta\prec 1$ since $f^{\dagger}\circ x\preceq\frac{1}{x}$ , weakly $\prec$ -decreasing when $\delta\preceq x$ , which implies $(f^{\dagger}\circ x)\delta\preceq 1$ , and it is eventually strictly $\prec$ -increasing otherwise. ∎

Corollary 5.4.

Let $f\in\mathbb{R}\langle\!\langle T\rangle\!\rangle$ be non-zero such that $f\not\asymp T^{k}$ for all $k\in\mathbb{N}$ . If $x+\delta\asymp x$ and $(f^{\dagger}\circ x)\delta\preceq 1$ , then $f^{(n)}\circ(x+\delta)\asymp f^{(n)}\circ x$ for all $n\in\mathbb{N}$ .

Proof.

Note that $\delta\preceq x$ . By Corollary 5.3, for every $n\in\mathbb{N}$ we have

{\left({\left(f^{(n)}\right)}^{\dagger}\circ x\right)}\delta=\frac{\left(f^{(n% +1)}\circ x\right)\delta^{n+1}}{\left(f^{(n)}\circ x\right)\delta^{n}}\preceq 1,

thus the conclusion follows from Proposition 5.1 applied to each $f^{(n)}$ . ∎

Proof of Theorem C.

When $\mathrm{ER}(f)=0$ , $f=\log^{\circ k}(T)$ , and the conclusion follows directly from the Taylor expansion of $\log$ . For $\mathrm{ER}(f)>0$ , fix some $x$ , $\delta$ as in the assumptions. Note that $\delta\preceq x$ . Write $f=\sum_{\gamma}r_{\gamma}e^{\gamma}$ for $\gamma$ ranging in $\mathbb{J}$ , where by construction $r_{\gamma}\neq 0$ implies $\mathrm{ER}(\gamma)<\mathrm{ER}(f)$ . Split $f$ as follows:

f_{0}=\sum_{(\gamma^{\prime}\circ x)\delta\preceq 1}r_{\gamma}e^{\gamma},\quad f% _{1}=f-f_{0}=\sum_{(\gamma^{\prime}\circ x)\delta\succ 1}r_{\gamma}e^{\gamma}.

We first show that $f_{1}$ can be ignored. Suppose that $f_{1}\neq 0$ . On the one hand, by construction $(f_{1}^{\dagger}\circ x)\delta\sim(\gamma^{\prime}\circ x)\delta\succ 1$ , where $e^{\gamma}$ is the leading monomial of $f_{1}$ , hence $f_{1}^{\dagger}\succ f^{\dagger}$ , thus necessarily $f_{1}\prec f$ and $f_{1}\prec 1$ , whence $f_{1}\not\asymp T^{k}$ for all $k\in\mathbb{N}$ ; since $f\not\asymp T^{k}$ for all $k$ , $f^{(n)}\succ f_{1}^{(n)}$ . On the other, $(f_{1}^{\dagger}\circ x)\succ\frac{1}{\delta}\succeq\frac{1}{x}$ , so we cannot have $f_{1}^{\dagger}\preceq\frac{1}{T}$ , hence $f_{1}^{\dagger}\succ\frac{1}{T}$ , and so $f_{1}^{(n)}\sim{(f_{1}^{\dagger})}^{n}f_{1}$ by Lemma 5.2(1). It follows that

f_{1}\circ x\sim\frac{f_{1}^{(n)}\circ x}{{(f_{1}^{\dagger}\circ x)}^{n}}\prec% \frac{f^{(n)}\circ x}{{(f_{1}^{\dagger}\circ x)}^{n}}\prec(f^{(n)}\circ x)% \delta^{n}.

Similarly, consider $f_{1}\circ(x+\delta)$ . Note that $(f_{1}^{\dagger}\circ(x+\delta))\delta\succ 1$ : if not, by Corollary 5.4 applied to $f_{1}$ at $x+\delta$ , with $-\delta$ in place of $\delta$ , we would get

1\prec(f_{1}^{\dagger}\circ x)\delta=\frac{f_{1}^{\prime}\circ x}{f_{1}\circ x% }\delta\asymp\frac{f_{1}^{\prime}\circ(x+\delta)}{f_{1}\circ(x+\delta)}\delta=% (f_{1}^{\dagger}\circ(x+\delta))\delta\preceq 1,

a contradiction. Therefore, just as in the previous argument, we find

f_{1}\circ(x+\delta)\sim\frac{f^{(n)}\circ(x+\delta)}{{(f_{1}^{\dagger}\circ(x% +\delta))}^{n}}\prec(f^{(n)}\circ(x+\delta))\delta^{n}.

Since $f^{(n)}\circ(x+\delta)\asymp f^{(n)}\circ x$ by Corollary 5.4,

f_{1}\circ(x+\delta)\prec(f^{(n)}\circ x)\delta^{n}.

Therefore,

f\circ(x+\delta)-f\circ x=f_{0}\circ(x+\delta)-f_{0}\circ x+o\left((f^{(n)}% \circ x)\delta^{n}\right)

Since $f\asymp f_{0}$ , so $f^{(n)}\asymp f_{0}^{(n)}$ , it is now enough to prove the conclusion with $f_{0}$ in place of $f$ . Suppose that $e^{\gamma}$ is a monomial in the support of $f_{0}$ , thus $\gamma\in\mathbb{J}$ , $(\gamma^{\prime}\circ x)\delta\preceq 1$ , and $\mathrm{ER}(\gamma)<\mathrm{ER}(f_{0})\leq\mathrm{ER}(f)$ . We distinguish two cases.

If $\gamma\asymp T^{k}$ for some $k\in\mathbb{N}$ , then $k>0$ , hence $\gamma^{\dagger}\asymp\frac{1}{T}$ . For each monomial $\mathfrak{m}$ in the support of $\gamma$ , we have $\gamma\succeq\mathfrak{m}\succ 1$ , so in particular $\mathfrak{m}^{\dagger}\preceq\gamma^{\dagger}\asymp\frac{1}{T}$ , hence $(\mathfrak{m}^{\dagger}\circ x)\delta\preceq\frac{\delta}{x}\preceq 1$ . When $\mathfrak{m}\not\asymp T^{d}$ for all $d\in\mathbb{N}$ , since $\mathrm{ER}(\mathfrak{m})\leq\mathrm{ER}(\gamma)<\mathrm{ER}(f)$ , we may apply the inductive hypothesis to deduce

\mathfrak{m}\circ(x+\delta)=\sum_{i=0}^{n-1}\frac{\mathfrak{m}^{(i)}\circ x}{i% !}\delta^{i}+O\left(\mathfrak{m}^{(n)}\delta^{n}\right).

When $\mathfrak{m}\asymp T^{d}$ for some $d\in\mathbb{N}$ , then in fact $\mathfrak{m}=T^{d}$ , hence the above approximation still holds by the binomial theorem and the fact that $\delta\preceq x$ . Moreover, in either case $\mathfrak{m}^{(i)}\preceq\frac{\mathfrak{m}}{T^{i}}\preceq T^{k-i}$ (by Lemma 5.2(2) in the former case, trivially in the latter), so $(\mathfrak{m}^{(i)}\circ x)\delta^{i}\preceq 1$ for all $i>0$ . By strong linearity of composition and derivation, we can sum all the terms of $\gamma$ to deduce that $(\gamma^{(i)}\circ x)\delta^{i}\preceq 1$ for all $i>0$ , and that

\gamma\circ(x+\delta)=\sum_{i=0}^{n-1}\frac{\gamma^{(i)}\circ x}{i!}\delta^{i}% +O\left(\gamma^{(n)}\delta^{n}\right).

If $\gamma\not\asymp T^{k}$ for all $k\in\mathbb{N}$ , then we simply observe that $(\gamma^{\dagger}\circ x)\delta\prec 1$ , since $\gamma\succ 1$ , so the above equality holds by inductive hypothesis, and $(\gamma^{(i)}\circ x)\delta^{i}\preceq 1$ for $i>0$ by Corollary 5.3.

Therefore,

\begin{split}f_{0}\circ(x+\delta)&=\sum_{(\gamma^{\prime}\circ x)\delta\preceq 1% }r_{\gamma}e^{\gamma\circ(x+\delta)}=\sum_{(\gamma^{\prime}\circ x)\delta% \preceq 1}r_{\gamma}e^{\sum_{i=0}^{n-1}\frac{\gamma^{(i)}\circ x}{i!}\delta^{i% }+O\left((\gamma^{(n)}\circ x)\delta^{n}\right)}\\ &=\sum_{(\gamma^{\prime}\circ x)\delta\preceq 1}r_{\gamma}e^{\gamma\circ x}% \exp\left(\sum_{i=1}^{n-1}\frac{\gamma^{(i)}\circ x}{i!}\delta^{i}+O\left((% \gamma^{(n)}\circ x)\delta^{n}\right)\right)\\ &=\sum_{(\gamma^{\prime}\circ x)\delta\preceq 1}\left(\sum_{i=0}^{n-1}\frac{{(% r_{\gamma}e^{\gamma})}^{(i)}\circ x}{i!}\delta^{i}+O\left(({(r_{\gamma}e^{% \gamma})}^{(n)}\circ x)\delta^{n}\right)\right)\\ &=\sum_{i=0}^{n-1}\frac{f_{0}^{(i)}\circ x}{i!}\delta^{i}+O\left((f_{0}^{(n)}% \circ x)\delta^{n}\right),\end{split}

where on the second line, the argument of $\exp$ is $\preceq 1$ , since $(\gamma^{(i)}\circ x)\delta^{i}\preceq 1$ for $i>0$ , and so we may use the fact that $e^{y}=1+y+\dotsb+\frac{y^{n-1}}{(n-1)!}+O(y^{n})$ for any $y\preceq 1$ to proceed to the following step. ∎

Remark 5.5.

The conclusion of Theorem C loses its significance at the boundary: if $f^{\dagger}\succ\frac{1}{T}$ and $(f^{\dagger}\circ x)\delta\asymp 1$ , then the error terms all have the same size by Lemma 5.2, and the conclusion collapses to $f\circ(x+\delta)=O(f\circ x)$ ; a comparable remark can be made for $f^{\dagger}\preceq\frac{1}{T}$ and $\delta\succeq x$ , where the error terms can get smaller at most once.

Error terms even increase in size if $(f^{\dagger}\circ x)\delta\succ 1$ or $\delta\succ x$ . In those cases, the conclusion of Theorem C may or may not be valid depending on $f$ . Consider the ‘first-order approximation’

f\circ(x+\delta)-f\circ x=O((f^{\prime}\circ x)\delta).

When $(f^{\dagger}\circ x)\delta\succ 1$ , consider $f=e^{T}$ , thus assume $\delta\succ 1$ . Then the first-order approximation is valid for $f=e^{T}$ if and only if $\delta<0$ . More generally, the approximation remains valid if $f\succ 1$ and $\delta<0$ , or if $f\prec 1$ and $\delta>0$ , since Theorem A implies $f\circ(x+\delta)\preceq f\circ x$ .

For $\delta\succ x$ , the first-order approximation is valid for $f=\log(T)$ and it fails for $f=\sqrt{T^{3}}$ . Analogy with real functions suggests that the approximation is valid for $\delta\succ x$ exactly when $f\preceq T$ . This is related to whether $f^{\prime\prime}\geq 0$ implies that $f$ is convex, namely $f\circ(x+\delta)-f\circ x\geq(f^{\prime}\circ x)\delta$ for $\delta\geq 0$ . As alluded to in the introduction, this does not seem to follow in a direct way from Theorem A.

The boundary $x+\delta\prec x$ is more subtle: the error terms do not increase in size, but the Taylor approximation may still hold or fail. For example, the first-order approximation fails for $f=\frac{1}{T}$ and $f=\log(T)$ (note that $\delta\asymp x$ , so $(f^{\dagger}\circ x)\delta\preceq 1$ in both examples):

\frac{1}{x+\delta}-\frac{1}{x}=-\frac{\delta}{(x+\delta)x}\succ-\frac{\delta}{% x^{2}},\quad\log(x+\delta)-\log(x)=\log\left(\frac{x+\delta}{x}\right)\succ 1% \asymp\frac{\delta}{x}.

On the other hand, the approximation is valid for $f=\sqrt{T}$ :

\sqrt{x+\delta}-\sqrt{x}=\frac{\delta}{\sqrt{x+\delta}+\sqrt{x}}\asymp-\frac{% \delta}{2\sqrt{x}}.

We can in fact give a full classification. Assume $x+\delta\prec x$ and $f^{\dagger}\preceq\frac{1}{T}$ (we ignore $f^{\dagger}\succ\frac{1}{T}$ , as in that case $(f^{\dagger}\circ x)\delta\preceq 1$ implies $\delta\prec x$ ). Then the first-order approximation is valid if and only if $f^{\dagger}\asymp\frac{1}{T}$ and $f\succ 1$ .

Indeed, suppose that $f^{\dagger}\asymp\frac{1}{T}$ . Note that the first-order approximation collapses to $f\circ(x+\delta)=O(f\circ x)$ . We have $\log(f)\sim r\log(T)$ for some non-zero $r\in\mathbb{R}$ , so

\log(f\circ(x+\delta))-\log(f\circ x)\sim r(\log(x+\delta)-\log(x))\succ 1

by Corollary 4.4. Note that $\log(x+\delta)-\log(x)$ is negative infinite. When $f\prec 1$ , we have $r<0$ , so $f\circ(x+\delta)\succ f\circ x$ , hence the approximation fails. When $f\succ 1$ , then $r>0$ , so $f\circ(x+\delta)\prec f\circ x$ , hence the approximation is valid.

Now suppose that $f^{\dagger}\prec\frac{1}{T}$ . Let $\varepsilon=\frac{f\circ(x+\delta)-f\circ x}{f\circ x}$ . We claim that $f$ satisfies the first-order approximation if and only if $\log|f|$ does. Since $(f^{\prime}\circ x)\delta\asymp(f^{\prime}\circ x)x\prec f\circ x$ , the approximation for $f$ implies $\varepsilon\prec 1$ , while the one for $\log|f|$ implies $\log(1+\varepsilon)\prec 1$ . Crucially, in either case $\varepsilon\sim\log(1+\varepsilon)$ . The claim follows at once from $(\log|f|)^{\prime}=\frac{f^{\prime}}{f}$ . Moreover, ${(\log|f|)}^{\dagger}=\frac{f^{\dagger}}{\log|f|}\prec f^{\dagger}\prec\frac{1% }{T}$ , since $f\not\asymp 1$ . Therefore, to check whether $f$ satisfies the first-order approximation, we may replace $f$ with $\log|f|$ until $f\sim\log^{\circ k}(T)$ . Since $f^{\prime}\sim(\log^{\circ k}(T))^{\prime}$ , Corollary 4.4 implies that we may further replace $f$ with $\log^{\circ k}(T)$ . Applying the same argument in reverse, we may replace $\log^{\circ k}(T)$ with $\log(T)$ . The approximation fails for $\log(T)$ , hence it fails for the starting $f$ .

Corollary 5.6.

Let $f\in\mathbb{R}\langle\!\langle T\rangle\!\rangle$ and $x,\delta\in\mathbb{U}$ with $x>\mathbb{R}$ , $x+\delta>\mathbb{R}$ , $x+\delta\asymp x$ . If $f\asymp T^{k}$ for some $k\in\mathbb{N}$ and $f^{(k+1)}\neq 0$ , suppose that $({(f^{(k+1)})}^{\dagger}\circ x)\delta\preceq 1$ ; otherwise, if $f\neq 0$ , suppose that $(f^{\dagger}\circ x)\delta\preceq 1$ . Then for all $n\geq 0$ ,

f\circ(x+\delta)=\sum_{i=0}^{n-1}\frac{f^{(i)}\circ x}{i!}\delta^{i}+O\left((f% ^{(n)}\circ x)\delta^{n}\right).

Proof.

If $f\not\asymp T^{k}$ for all $k\in\mathbb{N}$ and $f\neq 0$ , this is just Theorem C. If $f=0$ , the conclusion is trivial.

Now suppose that $f\asymp T^{k}$ for some $k\in\mathbb{N}$ . Let $p$ be the sum of all the terms of $f$ that are $\asymp T^{d}$ for some $d\in\mathbb{N}$ , where necessarily $d\leq k$ . Note that $p\asymp f\asymp T^{k}$ . Then $p$ is a polynomial in $T$ of degree $k$ , and by construction $g=f-p$ satisfies $g\not\asymp T^{d}$ for all $d\in\mathbb{N}$ . In particular, $p^{(k+1)}=0$ , so $f^{(k+1)}=g^{(k+1)}$ .

The conclusion of Theorem C is valid for $p$ by the binomial theorem and the fact that $\delta\preceq x$ (and for $n>k$ , it is true even for $\delta\succ x$ , as the error term becomes $0$ ). If $g=0$ , we are done. If $g\neq 0$ , we distinguish two cases. If $g^{\dagger}\preceq\frac{1}{T}$ , then $\delta\preceq x$ implies $(g^{\dagger}\circ x)\delta\preceq\frac{\delta}{x}\preceq 1$ . If $g^{\dagger}\succ\frac{1}{T}$ , then ${(f^{(k+1)})}^{\dagger}={(g^{(k+1)})}^{\dagger}\sim g^{\dagger}$ by Lemma 5.2(1), so the assumptions guarantee that $(g^{\dagger}\circ x)\delta\preceq 1$ . In either case, we can apply Theorem C to $g$ . The conclusion now follows immediately from $f=p+g$ and the observations $f^{(n)}\asymp p^{(n)}\succ g^{(n)}$ for $n\leq k$ , $f^{(n)}=g^{(n)}$ for $n>k$ . ∎