CommeUnJeu · L1 PCSI

Asymptotic analysis

⌚ ~163 min ▢ 20 blocks ✓ 70 exercises ➣ Prerequisites : Differentiability, Standard functions

Asymptotic analysis is the toolbox of local approximation in this course. Faced with a complicated expression involving a function near a point or a sequence at $+\infty$, the goal is to compare it to simpler reference objects --- powers, logarithms, exponentials --- and to extract enough information (limit, equivalent, position relative to a tangent or to an asymptote) without computing it exactly. The chapter introduces three comparison relations $f = o(g)$, $f = O(g)$, $f \sim g$, the algebraic theory of développements limités (DL), the central Taylor-Young formula, and the applications to local study (limits, equivalents, position, asymptotes, extrema).
The chapter has six sections. The first section introduces the three comparison relations for functions, the « croissances comparées » table at $+\infty$ and at $0^+$, and the operations + forbidden operations on equivalents. The second section defines a DL, proves the uniqueness of its coefficients, links DL at order $1$ with differentiability, proves the primitivation lemma, and culminates in Taylor-Young ; it closes with the boxed table of usual DLs at $0$ that the student must know by heart. The third section treats the operations on DLs (linear combination, product, inverse, quotient, primitivation, composition by example), with explicit warnings on the operations that fail (differentiation, composition without an inner limit at $0$). The fourth section adapts the comparison relations to sequences --- we keep this as a rapid adaptation, so the section is deliberately short. The fifth section applies DLs to four local-study problems : computing a limit, extracting an equivalent, determining the position of a curve relative to its tangent (with a standalone Proposition for CN/CS at order $2$ for an extremum), determining an asymptote at $\pm \infty$. The final section sketches four flavors of asymptotic problems --- inverse function, parametric equation, recurrent sequence, sequence of integrals --- states the harmonic asymptotic ($H_n = \ln n + \gamma + o(1)$) and the Stirling formula (both admitted), and derives as a corollary the central binomial $\binom{2n}{n} \sim 4^n / \sqrt{\pi n}$.
Four reflexes the reader should leave with :

the two equivalent forms of $\sim$ : $f \sim g \iff f/g \to 1 \iff f = g + o(g)$ --- the second form is the workhorse of every DL manipulation ;
the forbidden operations on equivalents (no sum, no composition à gauche) and on DLs (no differentiation without $\mathcal C^n$) ;
« le premier terme non nul d'un DL tient lieu d'équivalent » --- the bridge from the DL section to the applications ;
the « factoriser par le terme prépondérant » reflex to anticipate the order at which a DL must be computed.

I Comparison relations for functions

I.1 Negligibility $o$

We open with the negligibility symbol $o$, the most operationally useful of the three. To say $f$ is negligible compared to $g$ at a point $a$ means that, near $a$, $f$ is much smaller than $g$ : the ratio $f/g$ tends to $0$. The advantage of writing this as $f = o(g)$ rather than $f/g \to 0$ is twofold : (i) it does not require $g$ to be nonzero locally (we can work directly with the factorisation $f = \varepsilon g$ where $\varepsilon \to 0$), and (ii) it composes algebraically (sum, product, transitivity), giving a calculus. We immediately state the croissances comparées table --- powers vs.\ logarithms vs.\ exponentials --- as a single named Theorem ; it is the most-used quantitative content of the chapter.

Definition — Negligibility $f \equal o(g)$

Let $a \in \overline{\mathbb R}$ and let $f, g$ be two real-valued functions defined on a neighbourhood of $a$ (possibly punctured at $a$). We say that $f$ is negligible compared to $g$ at $a$, written $f(x) = o(g(x))$ when $x \to a$ (or simply $f = o(g)$ at $a$), if there exists a function $\varepsilon$ defined on a neighbourhood of $a$ such that $$ f(x) \ = \ \varepsilon(x) \, g(x) \qquad \text{near } a, \qquad \text{with} \qquad \varepsilon(x) \ \underset{x \to a}{\longrightarrow} \ 0. $$ When $g$ does not vanish on a punctured neighbourhood of $a$, this is equivalent to $f(x)/g(x) \underset{x \to a}{\longrightarrow} 0$. The notation $o(1)$ stands for any function tending to $0$ at $a$.

Theorem — Croissances comparées usuelles

The two following groups of negligibility relations hold for functions.

At $+\infty$. For all $\alpha, \beta, \gamma > 0$ : $$ (\ln x)^\beta \ = \ o(x^\alpha) \quad \text{at } +\infty, \qquad x^\alpha \ = \ o(e^{\gamma x}) \quad \text{at } +\infty. $$
At $0^+$. For all $\alpha, \beta > 0$ : $$ |\ln x|^\beta \ = \ o(x^{-\alpha}) \quad \text{at } 0^+, \qquad \text{equivalently} \qquad x^\alpha \, |\ln x|^\beta \ \underset{x \to 0^+}{\longrightarrow} \ 0. $$

The analogous discrete table for sequences (powers vs.\ exponentials vs.\ factorials) is stated in the subsection on discrete croissances comparées.

Status of the proof

The proofs reduce to limits already established in chapter Dérivabilité (via l'Hôpital or monotonicity of $x \mapsto x e^{-x}$, etc.). They are admitted here ; the sequence-side discrete table is stated separately in the subsection on discrete croissances comparées.

Method — Reading and manipulating an $o$

The symbol $o(g)$ is a placeholder, not a function : « $f = o(g)$ » means « there exists a function $\varepsilon \to 0$ such that $f = \varepsilon g$ ». Two different occurrences of $o(g)$ in the same line denote two different placeholders.

Sum and external multiplication. $o(g_1) + o(g_2)$ stays distinct unless one dominates : $o(g_1) + o(g_2) = o(g_1)$ when $g_2 = o(g_1)$. Multiplication by an unrelated function : $f \cdot o(g) = o(fg)$.
Transitivity. If $f = o(g)$ at $a$ and $g = o(h)$ at $a$, then $f = o(h)$ at $a$ (product of two functions tending to $0$).
$o(o(g)) = o(g)$. If $f = o(\eta)$ and $\eta = o(g)$, then $f = o(g)$. In particular, $o(1) \cdot g = o(g)$.
Reflex. To verify $f = o(g)$ at $a$ when $g$ is nonzero locally, compute $\lim_{x \to a} f(x)/g(x)$ : if it is $0$, the relation holds.

Example — Polynomial growths

At $0$, $x^3 = o(x^2)$ because $x^3/x^2 = x \to 0$. At $+\infty$, the opposite holds : $x^2 = o(x^3)$ because $x^2/x^3 = 1/x \to 0$. The roles of « small » and « big » swap depending on whether the base point is finite or infinite.

Example — Usual croissances comparées

At $+\infty$, $\ln x = o(x^{1/100})$ (logarithms are negligible compared with every positive power) and $x^{100} = o(e^x)$ (exponentials dominate every power). At $0^+$, $x \ln x \to 0$ because $x \ln x = - x \, |\ln x|$ and $x \, |\ln x|^1 \to 0$ (croissance comparée case $\alpha = 1, \beta = 1$). These three patterns recur on every page of the chapter ; one should be able to read them off the table instantly.

Skills to practice

Computing $o$-relations and using usual comparisons

I.2 Domination $O$

The domination symbol $O$ is a weaker cousin of $o$. To say $f = O(g)$ at $a$ means $f$ is at most of the size of $g$ near $a$ : the ratio $f/g$ is bounded near $a$. Every $o$ is a $O$ (a function tending to $0$ is in particular bounded), but the converse fails : $\cos x = O(1)$ on $\mathbb R$ yet $\cos x \ne o(1)$ at $+\infty$. We will use $O$ mainly to write « error of order $g$ » without committing to negligibility.

Definition — Domination $f \equal O(g)$

Let $a \in \overline{\mathbb R}$ and $f, g$ defined on a neighbourhood of $a$. We say that $f$ is dominated by $g$ at $a$, written $f(x) = O(g(x))$ when $x \to a$ (or $f = O(g)$ at $a$), if there exists a function $\lambda$ defined and bounded on a neighbourhood of $a$ such that $$ f(x) \ = \ \lambda(x) \, g(x) \qquad \text{near } a. $$ When $g$ does not vanish on a punctured neighbourhood of $a$, this is equivalent to $f(x)/g(x)$ bounded near $a$. The notation $O(1)$ stands for any function bounded near $a$.

Proposition — Links between $o$ and $O$

For two functions $f, g$ defined on a neighbourhood of $a \in \overline{\mathbb R}$ :

If $f = o(g)$ at $a$, then $f = O(g)$ at $a$ ;
$f = O(1)$ at $a$ if and only if $f$ is bounded on a neighbourhood of $a$ ;
Operations on $O$ mirror those on $o$ : $O(g_1) + O(g_2) \subset O(|g_1| + |g_2|)$ ; if moreover $g_2 = O(g_1)$, then $O(g_1) + O(g_2) = O(g_1)$ ; $f \cdot O(g) = O(fg)$ ; $O(o(g)) = o(g)$ ; $o(O(g)) = o(g)$ ; transitivity.

Example — $O$ at $0$ and at $+\infty$

At $0$, $\sin x = O(x)$ because $|\sin x| \le |x|$ on $\mathbb R$, so $\sin x / x$ is bounded by $1$ near $0$ ; in fact more is true, $\sin x = o(1)$ at $0$ since $\sin x \to 0$. At $+\infty$, $\cos x = O(1)$ on $\mathbb R$ since $|\cos| \le 1$, but $\cos x$ has no limit at $+\infty$ : it is bounded without being equivalent to any constant. This is the genuine difference between $O$ and $\sim$ : a $O$ controls size, not asymptotic behaviour.

Skills to practice

Computing $O$-relations

I.3 Equivalence $\sim$$\virgule$ operations$\virgule$ forbidden operations

The equivalence symbol $\sim$ is the central relation of the chapter. We say $f \sim g$ at $a$ when $f$ and $g$ have the same dominant behaviour near $a$ : the ratio tends to $1$. We teach both characterisations side-by-side --- $f \sim g \iff f/g \to 1 \iff f = g + o(g)$ --- because the second form is the one that drives every DL manipulation in the Operations on DLs and Applications of DLs sections. We list the legal operations on $\sim$ (product, quotient, power) and we display the forbidden operations as boxed warnings : equivalents do not sum, and they do not pass through a left-composition (e.g.\ $e^{f}$ when $f \sim g$).

Definition — Equivalence $f \sim g$

Let $a \in \overline{\mathbb R}$ and $f, g$ defined on a neighbourhood of $a$. We say that $f$ is equivalent to $g$ at $a$, written $f(x) \underset{x \to a}{\sim} g(x)$ (or $f \sim g$ at $a$), if there exists a function $\eta$ defined on a neighbourhood of $a$ such that $$ f(x) \ = \ \eta(x) \, g(x) \qquad \text{near } a, \qquad \text{with} \qquad \eta(x) \ \underset{x \to a}{\longrightarrow} \ 1. $$ In practice, we use this relation when $g$ does not vanish on a punctured neighbourhood of $a$, so that the equivalent form $f / g \to 1$ is available and the sign and limit statements below are unambiguous.

Proposition — Two-form characterisation of $\sim$

Let $f, g$ be defined on a neighbourhood of $a \in \overline{\mathbb R}$ with $g$ not vanishing on a punctured neighbourhood of $a$. The three statements are equivalent :

$f \sim g$ at $a$ ;
$f(x)/g(x) \underset{x \to a}{\longrightarrow} 1$ ;
$f(x) = g(x) + o(g(x))$ at $a$.

Proof

By definition, $f \sim g$ iff $f = \eta g$ near $a$ with $\eta \to 1$. Since $g$ does not vanish locally, this is equivalent to $f/g = \eta \to 1$, which is (ii). To get (iii), write $\eta = 1 + (\eta - 1)$ and note $\eta - 1 \to 0$ ; then $f = g + (\eta - 1) g = g + o(g)$ since $(\eta - 1) g = o(g)$. Conversely, $f = g + o(g) = g(1 + o(1))$ with $1 + o(1) \to 1$, so $f \sim g$ by the definition.

Proposition — Equivalence is a relation$\virgule$ preserves sign and limit

Let $a \in \overline{\mathbb R}$. The relation $\sim$ at $a$ is an equivalence relation on the set of functions defined and nonzero on a punctured neighbourhood of $a$. Moreover, if $f \sim g$ at $a$, then :

$f$ and $g$ have the same sign on a punctured neighbourhood of $a$ ;
$f$ and $g$ have the same limit at $a$ (whether finite, $+\infty$, or $-\infty$).

Proof

Equivalence relation. Reflexivity : $f = 1 \cdot f$ with $1 \to 1$. Symmetry : if $f = \eta g$ with $\eta \to 1$, then $1/\eta \to 1$ and $g = (1/\eta) f$ on a neighbourhood where $\eta$ stays close to $1$ (hence nonzero). Transitivity : if $f = \eta_1 g$ with $\eta_1 \to 1$ and $g = \eta_2 h$ with $\eta_2 \to 1$, then $f = \eta_1 \eta_2 h$ with $\eta_1 \eta_2 \to 1$.
Same sign. Since $f = \eta g$ with $\eta \to 1$, there is a neighbourhood of $a$ on which $\eta > 1/2 > 0$ ; on this neighbourhood, $f$ and $g$ have the same sign.
Same limit. If $g(x) \to \ell \in \overline{\mathbb R}$, then $f(x) = \eta(x) g(x) \to 1 \cdot \ell = \ell$ by the product rule for limits.

Method — Operations on equivalents (and forbidden operations)

« Il ne doit en rester qu'un » : in a multiplicative chain only one equivalent appears at the end. The legal operations on $\sim$ at $a$ are :

Product. If $f_1 \sim g_1$ and $f_2 \sim g_2$, then $f_1 f_2 \sim g_1 g_2$.
Quotient. If $f_1 \sim g_1$ and $f_2 \sim g_2$ with $g_2$ nonzero near $a$, then $f_1/f_2 \sim g_1/g_2$.
Integer (or real) power. If $f \sim g$ with $g > 0$ near $a$, then $f^\alpha \sim g^\alpha$ for every $\alpha \in \mathbb R$.

Two forbidden operations --- never apply them :

[(F1)] Sum of equivalents. $f_1 \sim g_1$ and $f_2 \sim g_2$ do not imply $f_1 + f_2 \sim g_1 + g_2$. Counter-example : $f_1 = n+1 \sim n = g_1$, $f_2 = -n \sim -n = g_2$ ; the target sum $g_1 + g_2 = 0$ vanishes identically, so it cannot serve as an equivalent in the usual nonzero-denominator sense, while $f_1 + f_2 = 1 \ne 0$. The operation is meaningless on this pair.
[(F2)] Left-composition. $f \sim g$ does not imply $\varphi(f) \sim \varphi(g)$ for a general function $\varphi$. Counter-example : $n + \ln n \sim n$ at $+\infty$, but $e^{n + \ln n} = n e^n \not\sim e^n$ at $+\infty$.

Proposition — Squeeze for equivalents

Let $a \in \overline{\mathbb R}$ and $f, g, h$ defined on a neighbourhood of $a$ such that :

$f(x) \le g(x) \le h(x)$ on a neighbourhood of $a$ ;
$f \sim h$ at $a$ ;
$f$ is eventually of constant nonzero sign on a punctured neighbourhood of $a$.

Then $g \sim f$ at $a$ (and equivalently $g \sim h$).

Proof

Since $f$ has constant nonzero sign on a punctured neighbourhood of $a$, we distinguish two cases.
Case $f > 0$ near $a$. Divide the inequality $f \le g \le h$ by $f > 0$ : $1 \le g/f \le h/f$. By hypothesis $f \sim h$, so $h/f \to 1$ at $a$. By the squeeze theorem for limits, $g/f \to 1$ at $a$, hence $g \sim f$.
Case $f < 0$ near $a$. Dividing $f \le g \le h$ by $f < 0$ reverses the inequalities : $1 \ge g/f \ge h/f$. Again $h/f \to 1$, so by squeeze $g/f \to 1$, hence $g \sim f$.

Example — Limit via equivalent at $0$

At $0$, $\sin x \sim x$ (since $\sin x / x \to 1$, classical limit). Hence $$ \lim_{x \to 0} \frac{\sin x}{x} \ = \ 1, \qquad \lim_{x \to 0} \frac{\sin(2x)}{x} \ = \ \lim_{x \to 0} \frac{2x}{x} \ = \ 2 $$ where in the last step we replaced $\sin(2x)$ by its equivalent $2x$ at $0$ (since $2x \to 0$ when $x \to 0$, the substitution into $\sin$ is legitimate via composition with an inner limit at $0$, see the Composition subsection below).

Example — Forbidden sum --- counterexample

At $+\infty$, $n + 1 \sim n$ and $-n \sim -n$. If summing equivalents were legal, we would conclude that the sum $(n+1) + (-n) = 1$ is equivalent to $n + (-n) = 0$ --- but the target $0$ vanishes identically, so the relation is meaningless. The operational lesson : never substitute one summand by its equivalent inside a sum --- always reduce to a single multiplicative chain first, e.g.\ by factoring the dominant term.

Skills to practice

Manipulating equivalents and avoiding forbidden operations

II Limited expansions$\virgule$ generalities and Taylor-Young

II.1 Definition$\virgule$ uniqueness$\virgule$ truncation$\virgule$ parity

A développement limité (DL) at order $n$ at $a$ is a polynomial approximation of $f$ near $a$ up to a remainder $o((x-a)^n)$. It is the workhorse data structure of asymptotic analysis : it bundles, in a finite list of coefficients, all the local information needed for limits, equivalents, position relative to the tangent, and asymptotes. We define a DL, prove the uniqueness of its coefficients (the structural fact that lets us do arithmetic on DLs), state the truncation Proposition (a DL at order $n$ gives a DL at every lower order), and finish with the parity rule on DLs at $0$.

Definition — Limited expansion at order $n$

Let $a \in \mathbb R$, $n \in \mathbb N$, and $f$ a real-valued function defined on a neighbourhood of $a$ (possibly punctured at $a$). We say that $f$ admits a limited expansion (DL) at order $n$ at $a$ if there exist real numbers $a_0, a_1, \ldots, a_n$ such that, near $a$, $$ f(x) \ = \ a_0 + a_1 (x - a) + \ldots + a_n (x - a)^n + o\bigl((x - a)^n\bigr). $$ The polynomial $P(x) = \sum_{k=0}^n a_k (x - a)^k$ is the regular part of the DL, and the term $o((x-a)^n)$ is the remainder. When $a = 0$ we simply say « at $0$ ».

Theorem — Uniqueness of a DL

If $f$ admits a DL at order $n$ at $a$, the coefficients $a_0, \ldots, a_n$ of its regular part are uniquely determined by $f, a, n$.

Proof

Suppose $f$ admits two DLs at order $n$ at $a$ : $$ f(x) \ = \ \sum_{k=0}^n a_k (x-a)^k + o((x-a)^n) \ = \ \sum_{k=0}^n b_k (x-a)^k + o((x-a)^n). $$ Subtracting, $\sum_{k=0}^n (a_k - b_k) (x-a)^k = o((x-a)^n)$. Suppose for contradiction that not all coefficients agree ; let $p$ be the smallest index with $a_p \ne b_p$. Then $$ (a_p - b_p) (x-a)^p + \sum_{k=p+1}^n (a_k - b_k) (x-a)^k \ = \ o((x-a)^n). $$ Divide by $(x-a)^p$ (valid for $x \ne a$ near $a$) : $a_p - b_p + \sum_{k=p+1}^n (a_k - b_k) (x-a)^{k-p} = o((x-a)^{n-p})$. As $x \to a$, the left side tends to $a_p - b_p \ne 0$, while the right side tends to $0$ since $n - p \ge 0$ and $o((x-a)^{n-p})$ is either an explicit power tending to $0$ or, when $n = p$, an $o(1)$ that still tends to $0$. Contradiction.

Proposition — Truncation of a DL

If $f$ admits a DL at order $n$ at $a$ with regular part $\sum_{k=0}^n a_k (x-a)^k$, then for every $m \in \{0, 1, \ldots, n\}$, $f$ admits a DL at order $m$ at $a$, of regular part $\sum_{k=0}^m a_k (x-a)^k$.

Proposition — Parity of a DL at $0$

Let $f$ be defined on a symmetric neighbourhood $]-r, r[$ of $0$ and admit a DL at order $n$ at $0$ of regular part $\sum_{k=0}^n a_k x^k$. Then :

if $f$ is even, the odd-degree coefficients vanish : $a_1 = a_3 = a_5 = \ldots = 0$ ;
if $f$ is odd, the even-degree coefficients vanish : $a_0 = a_2 = a_4 = \ldots = 0$.

Example — DL of $1/(1-x)$ at order $n$ at $0$

For $x \ne 1$, the geometric sum identity gives $$ \frac{1}{1 - x} \ = \ 1 + x + x^2 + \ldots + x^n + \frac{x^{n+1}}{1 - x}. $$ Near $0$, $x^{n+1}/(1-x) = x^{n+1} \cdot \frac{1}{1-x}$ where $1/(1-x) \to 1$, so $x^{n+1}/(1-x) = o(x^n)$ (since $x^{n+1} = x \cdot x^n$ and $x \to 0$). Hence $$ \frac{1}{1 - x} \ = \ 1 + x + x^2 + \ldots + x^n + o(x^n) \qquad \text{at } 0. $$ This is the first « usual DL » of the boxed table in the subsection Table of usual DLs at $0$.

Skills to practice

Computing a DL by definition or by Taylor-Young

II.2 DL and differentiability$\virgule$ primitivation lemma

A DL at order $0$ is the statement that $f$ has a limit at $a$ ; a DL at order $1$ is precisely the statement that $f$ is differentiable at $a$ (with the natural identification between the coefficient $a_1$ and the derivative $f'(a)$). Beyond order $1$, differentiability ceases to be equivalent to a DL : a function may admit a DL at order $n$ without being differentiable $n$ times. We then prove a small but central technical lemma --- the primitivation lemma : a primitive of a function with a DL at order $n$ has a DL at order $n + 1$. This is the engine that drives Taylor-Young in the next subsection and the DL of $\ln(1+x)$ from that of $1/(1-x)$ in the subsection on the table of usual DLs at $0$.

Proposition — Characterisation of differentiability by a DL at order $1$

Let $f$ be defined on a neighbourhood of $a \in \mathbb R$. The two statements are equivalent :

$f$ is differentiable at $a$ ;
$f$ admits a DL at order $1$ at $a$.

In this case the DL reads $f(x) = f(a) + f'(a) (x - a) + o(x - a)$.

Proof

$f$ is differentiable at $a$ with derivative $\ell$ iff $\frac{f(x) - f(a)}{x - a} \to \ell$ as $x \to a$, iff $f(x) - f(a) = \ell (x - a) + (x - a) \varepsilon(x)$ with $\varepsilon \to 0$, iff $f(x) = f(a) + \ell (x - a) + o(x - a)$. By uniqueness of the DL, $\ell$ equals the coefficient of $(x - a)$ in the DL ; we write $\ell = f'(a)$.

Proposition — Primitivation lemma

Let $g$ be a real-valued function admitting a DL at order $n$ at $0$, $g(x) = \sum_{k=0}^n b_k x^k + o(x^n)$. Let $G$ be any primitive of $g$ on a neighbourhood of $0$ (assuming $g$ continuous near $0$). Then $G$ admits a DL at order $n + 1$ at $0$, given by $$ G(x) \ = \ G(0) + \sum_{k=0}^n \frac{b_k}{k + 1} \, x^{k+1} + o(x^{n+1}). $$

Proof

Write $g(x) = P(x) + x^n \varepsilon(x)$ where $P(x) = \sum_{k=0}^n b_k x^k$ and $\varepsilon \to 0$ at $0$. Let $H(x) = G(x) - G(0) - \sum_{k=0}^n \frac{b_k}{k+1} x^{k+1}$. Then $H(0) = 0$ and, by differentiating term by term (using that $G' = g$ and that the polynomial part is differentiated to $P$), $$ H'(x) \ = \ g(x) - P(x) \ = \ x^n \varepsilon(x). $$ Fix $\eta > 0$. Since $\varepsilon \to 0$, there is $\delta > 0$ such that $|\varepsilon(t)| \le \eta$ for $|t| \le \delta$. For $|x| \le \delta$, $$ |H(x)| \ = \ \left| \int_0^x t^n \varepsilon(t) \, dt \right| \ \le \ \int_{\min(0,x)}^{\max(0,x)} |t|^n \, |\varepsilon(t)| \, dt \ \le \ \eta \cdot \frac{|x|^{n+1}}{n+1}, $$ hence $|H(x)/x^{n+1}| \le \eta/(n+1)$. Therefore $H(x) = o(x^{n+1})$, which is the claimed DL at order $n+1$.

Skills to practice

Translating between differentiability and DL at order $1$$\virgule$ primitivating DLs

II.3 Taylor-Young formula

We arrive at the central named theorem of the chapter : the Taylor-Young formula. It says that a $\mathcal C^n$ function on a neighbourhood of $a$ admits a DL at order $n$ at $a$, with explicit coefficients $f^{(k)}(a)/k!$. The proof is a clean induction on $n$ using the primitivation lemma just proved. Combined with the table of usual derivatives (chapter FonctionsUsuelles), Taylor-Young gives a direct way to compute the DL of every classical function at $0$ --- and that is exactly what we tabulate in the next subsection.

Theorem — Taylor-Young formula

Let $a \in \mathbb R$, $n \in \mathbb N$, and $f$ a real-valued function of class $\mathcal C^n$ on a neighbourhood of $a$. Then $f$ admits a DL at order $n$ at $a$, given by $$ f(a + h) \ = \ \sum_{k=0}^n \frac{f^{(k)}(a)}{k!} \, h^k + o(h^n) \qquad \text{as } h \to 0. $$

Proof

Set $g(h) = f(a + h)$ ; then $g$ is $\mathcal C^n$ on a neighbourhood of $0$ with $g^{(k)}(0) = f^{(k)}(a)$. It suffices to prove the formula for $g$ at $0$. We induct on $n$.
Base case $n = 0$. $g$ is continuous at $0$, so $g(h) = g(0) + o(1)$, which is the DL at order $0$.
Induction step. Assume the formula holds at order $n - 1$ for every $\mathcal C^{n-1}$ function. Let $g$ be $\mathcal C^n$ on a neighbourhood of $0$. Then $g'$ is $\mathcal C^{n-1}$ on this neighbourhood, so by the induction hypothesis applied to $g'$, $$ g'(h) \ = \ \sum_{k=0}^{n-1} \frac{(g')^{(k)}(0)}{k!} \, h^k + o(h^{n-1}) \ = \ \sum_{k=0}^{n-1} \frac{g^{(k+1)}(0)}{k!} \, h^k + o(h^{n-1}). $$ Apply the primitivation lemma to this DL of $g'$ at order $n - 1$. The primitive $g$ has a DL at order $n$ at $0$ : $$ g(h) \ = \ g(0) + \sum_{k=0}^{n-1} \frac{g^{(k+1)}(0)}{k! (k+1)} \, h^{k+1} + o(h^n) \ = \ g(0) + \sum_{j=1}^{n} \frac{g^{(j)}(0)}{j!} \, h^j + o(h^n), $$ (where we substituted $j = k+1$ and used $k! (k+1) = (k+1)! = j!$). Including the $j = 0$ term $g(0)$ gives the Taylor-Young formula at order $n$, completing the induction.

Method — Computing a DL via Taylor-Young

To compute the DL at order $n$ at $a$ of a $\mathcal C^n$ function $f$ :

Compute $f(a), f'(a), \ldots, f^{(n)}(a)$ (close-form derivatives).
Assemble the Taylor polynomial $\sum_{k=0}^n \frac{f^{(k)}(a)}{k!} (x - a)^k$.
The DL at order $n$ at $a$ is this polynomial plus $o((x-a)^n)$.

Example — Taylor-Young at $0$ for $\exp$

The exponential is $\mathcal C^\infty$ on $\mathbb R$ with $\exp^{(k)} = \exp$ for all $k$, so $\exp^{(k)}(0) = 1$. By Taylor-Young at order $n$ at $0$, $$ e^x \ = \ 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \ldots + \frac{x^n}{n!} + o(x^n) \qquad \text{at } 0. $$ This is the first row of the encadré in the next subsection and the most ubiquitous DL in all of this course.

Skills to practice

Applying Taylor-Young to a $\mathcal C^n$ function

II.4 Table of usual DLs at $0$ (par cœur)

This subsection is a single boxed theorem : the table of usual DLs at $0$ that every student must know by heart. Each row follows from Taylor-Young applied to the function (when its higher derivatives have a simple closed form, as for $\exp$, $\sin$, $\cos$, $\sinh$, $\cosh$) or from the primitivation lemma applied to a simpler DL (as for $\ln(1+x)$ from $1/(1-x)$, or for $\operatorname{Arctan}$ from $1/(1+x^2)$). The chapter's entire computational part rests on this table : the Operations on DLs section combines these DLs (linear combination, product, inverse, composition) and the Applications of DLs section applies them to local-study problems.

Theorem — Usual DLs at $0$ (par cœur)

The following standard DLs hold at $0$. In the first five rows, $n \ge 0$, except for $\ln(1+x)$ where $n \ge 1$. In the trigonometric and hyperbolic rows, $p \in \mathbb N$. The last row is valid only at order $3$. \begin{align*} \frac{1}{1 - x} \ &= \ \sum_{k=0}^n x^k + o(x^n),
\frac{1}{1 + x} \ &= \ \sum_{k=0}^n (-1)^k x^k + o(x^n),
\ln(1 + x) \ &= \ \sum_{k=1}^n (-1)^{k-1} \frac{x^k}{k} + o(x^n),
e^x \ &= \ \sum_{k=0}^n \frac{x^k}{k!} + o(x^n),
(1 + x)^\alpha \ &= \ \sum_{k=0}^n \binom{\alpha}{k} x^k + o(x^n) \quad (\alpha \in \mathbb R), \quad \binom{\alpha}{k} = \frac{\alpha (\alpha - 1) \cdots (\alpha - k + 1)}{k!},
\cos x \ &= \ \sum_{k=0}^p (-1)^k \frac{x^{2k}}{(2k)!} + o(x^{2p+1}),
\sin x \ &= \ \sum_{k=0}^p (-1)^k \frac{x^{2k+1}}{(2k+1)!} + o(x^{2p+2}),
\cosh x \ &= \ \sum_{k=0}^p \frac{x^{2k}}{(2k)!} + o(x^{2p+1}),
\sinh x \ &= \ \sum_{k=0}^p \frac{x^{2k+1}}{(2k+1)!} + o(x^{2p+2}),
\operatorname{Arctan} x \ &= \ \sum_{k=0}^p (-1)^k \frac{x^{2k+1}}{2k + 1} + o(x^{2p+2}),
\tan x \ &= \ x + \frac{x^3}{3} + o(x^3) \quad \text{(order $3$ only)}. \end{align*}

Proof

$1/(1-x)$ and $1/(1+x)$. Geometric sum (see the Example in the subsection on Definition, uniqueness).
$\ln(1+x)$. Primitivation lemma applied to the DL of $1/(1+x) = 1 - x + x^2 - \ldots + (-1)^n x^n + o(x^n)$ ; the primitive that vanishes at $0$ is $\ln(1+x)$.
$e^x$. Taylor-Young, see the Example in the subsection on DL and differentiability.
$(1+x)^\alpha$. Taylor-Young : $f(x) = (1+x)^\alpha$ has $f^{(k)}(x) = \alpha (\alpha - 1) \cdots (\alpha - k + 1) (1+x)^{\alpha - k}$, hence $f^{(k)}(0) = \alpha (\alpha - 1) \cdots (\alpha - k + 1)$.
$\cos x, \sin x, \cosh x, \sinh x$. Taylor-Young : derivatives cycle through $\pm \sin, \pm \cos$ (resp.\ $\pm \sinh, \pm \cosh$) ; the even/odd parity rule (above) kills the missing parity of coefficients.
$\operatorname{Arctan}$. Primitivation of $1/(1+x^2) = 1 - x^2 + x^4 - \ldots + (-1)^p x^{2p} + o(x^{2p+1})$ (DL of $1/(1+u)$ with $u = x^2$ ; rigorously justified by the geometric-sum identity as in the subsection on Definition, uniqueness).
$\tan x = \sin x / \cos x$. Direct quotient at order $3$ (see Example in the Inverse and quotient subsection).

Method — Learn the table by heart

The 11 lines of the previous theorem are the operational alphabet of the chapter. Every problem in the Operations, Applications and Asymptotic problems sections starts by recalling at least one of them. Memorise them. Memorise also the parity (which lines have only even powers, which only odd), the sign pattern (alternating vs.\ all positive : $\ln, \sin, \cos, \operatorname{Arctan}$ alternate ; $\exp, \sinh, \cosh, 1/(1-x)$ do not), and the special cases $\alpha = -1, 1/2, -1/2$ of $(1+x)^\alpha$ which recover $1/(1+x)$, $\sqrt{1+x}$, $1/\sqrt{1+x}$.

Skills to practice

Recalling the usual DLs

III Operations on DLs

III.1 Linear combination$\virgule$ product$\virgule$ truncation

The arithmetic operations on DLs --- linear combination and product --- preserve the « DL at order $n$ » property : the result has a DL at order $n$ whose regular part is computed from the regular parts of the inputs, truncated at order $n$. The « truncation » step is the only subtlety : multiplying two regular parts of degree $n$ gives a polynomial of degree $2n$, but only the part up to degree $n$ is reliable ; the rest is absorbed into the $o(x^n)$ remainder. We codify this with a Method : « factor by the dominant term to anticipate the order ».

Proposition — Linear combination of DLs

Let $f, g$ be defined on a neighbourhood of $0$ and admit DLs at order $n$ at $0$ : $$ f(x) = P(x) + o(x^n), \qquad g(x) = Q(x) + o(x^n). $$ Then for every $\lambda, \mu \in \mathbb R$, the function $\lambda f + \mu g$ admits a DL at order $n$ at $0$ of regular part $\lambda P(x) + \mu Q(x)$.

Proposition — Product of DLs

With the notations of the previous Proposition, the product $fg$ admits a DL at order $n$ at $0$ of regular part $T_n(P \cdot Q)$, where $T_n$ denotes truncation at order $n$ (keep only the monomials of degree $\le n$).

Method — Factor by the dominant term to anticipate the order

When computing a DL of $f g$ or $f / g$ at order $n$ at $0$, write each factor as $x^p \, P(x) + o(x^{p + \deg \text{(remaining factor)}})$ to identify the dominant term. If $f(x) = x^p (a_0 + a_1 x + \ldots) + o(x^{p + n})$ with $a_0 \ne 0$, then in any product $f g$ the order is governed by $p + $ (order of $g$). The reflex « factoriser par le terme prépondérant » avoids computing too many --- or too few --- terms.

Example — Product $e^x \cos x$ at order $3$ at $0$

We use the table : $$ e^x \ = \ 1 + x + \tfrac{x^2}{2} + \tfrac{x^3}{6} + o(x^3), \qquad \cos x \ = \ 1 - \tfrac{x^2}{2} + o(x^3). $$ Multiplying and keeping terms up to degree $3$ : $$ e^x \cos x \ = \ \bigl(1 + x + \tfrac{x^2}{2} + \tfrac{x^3}{6}\bigr) \cdot \bigl(1 - \tfrac{x^2}{2}\bigr) + o(x^3) \ = \ 1 + x + \bigl(\tfrac{1}{2} - \tfrac{1}{2}\bigr) x^2 + \bigl(\tfrac{1}{6} - \tfrac{1}{2}\bigr) x^3 + o(x^3), $$ giving $$ e^x \cos x \ = \ 1 + x - \tfrac{x^3}{3} + o(x^3) \qquad \text{at } 0. $$ The $x^2$ coefficient vanishes because $\cos$ has no degree-$1$ term and the $\cos$'s $-x^2/2$ cancels the $e^x$'s $+x^2/2$.

Skills to practice

Computing DLs via product$\virgule$ inverse and quotient

III.2 Inverse and quotient

To compute $1/f$ when $f$ has a DL, the trick is to factor $f$ as $a_0 (1 + u)$ where $a_0 = f(0) \ne 0$ and $u \to 0$ at $0$, then use the DL of $1/(1 + u)$. The quotient $f/g$ is treated by writing $f/g = (1/g) \cdot f$. We also derive the DL of $\tan x = \sin x / \cos x$ at order $5$ as a consistency check with the entry of the usual-DLs table at order $3$.

Proposition — Inverse via composition with $1/(1+u)$

Let $f$ admit a DL at order $n$ at $0$ of the form $f(x) = a_0 + a_1 x + \ldots + a_n x^n + o(x^n)$ with $a_0 \ne 0$. Set $u(x) = (f(x) - a_0)/a_0$, so $u(0) = 0$ and $u$ has a DL at order $n$ at $0$. Then $1/f$ admits a DL at order $n$ at $0$ : write $$ \frac{1}{f(x)} \ = \ \frac{1}{a_0} \cdot \frac{1}{1 + u(x)} \ = \ \frac{1}{a_0} \bigl(1 - u(x) + u(x)^2 - \ldots + (-1)^n u(x)^n\bigr) + o(x^n), $$ the polynomial $1 - u + u^2 - \ldots + (-1)^n u^n$ being truncated at order $n$ in $x$.

Method — Quotient

To compute the DL at order $n$ at $0$ of $f / g$ when $g(0) \ne 0$ :

Compute the DL at order $n$ at $0$ of $1/g$ by the inverse-via-$1/(1+u)$ trick above.
Multiply by the DL of $f$ at order $n$ and truncate at order $n$.

Example — DL of $\tan x$ at order $5$ at $0$

Start from the usual-DLs table at order $5$ : $$ \sin x \ = \ x - \tfrac{x^3}{6} + \tfrac{x^5}{120} + o(x^5), \qquad \cos x \ = \ 1 - \tfrac{x^2}{2} + \tfrac{x^4}{24} + o(x^5). $$ Set $u(x) = -x^2/2 + x^4/24 + o(x^5)$, so $\cos x = 1 + u$. Then $$ \frac{1}{\cos x} \ = \ 1 - u + u^2 + o(x^5) \ = \ 1 - \bigl(-\tfrac{x^2}{2} + \tfrac{x^4}{24}\bigr) + \bigl(-\tfrac{x^2}{2}\bigr)^2 + o(x^5) \ = \ 1 + \tfrac{x^2}{2} + \tfrac{5 x^4}{24} + o(x^5), $$ (using $u^2 = x^4/4 + o(x^5)$ and $u^3, u^4, u^5 = o(x^5)$). Multiplying by $\sin x$ and truncating at order $5$ : $$ \tan x \ = \ \bigl(x - \tfrac{x^3}{6} + \tfrac{x^5}{120}\bigr) \bigl(1 + \tfrac{x^2}{2} + \tfrac{5 x^4}{24}\bigr) + o(x^5) \ = \ x + \tfrac{x^3}{3} + \tfrac{2 x^5}{15} + o(x^5). $$ At order $3$ this reads $\tan x = x + x^3/3 + o(x^3)$, consistent with the usual-DLs table.

Skills to practice

Computing DLs of an inverse and of a quotient

III.3 Primitivation and differentiation (warning)

The primitivation lemma of the previous section already gives the rule for primitivating a DL : add one to the order, divide each coefficient by its new exponent. We restate it here for the operations toolbox, then issue an emphatic warning : the dual operation --- differentiating a DL --- is not legal in general. A function may admit a DL at some order while its derivative admits no DL at the next-lower order. We display this with the standard counter-example $f(x) = x^3 \sin(1/x^2)$ : $f = o(x^2)$ at $0$, but $f'$ has no limit at $0$.

Proposition — Primitivation of a DL (restated)

If $g$ admits a DL at order $n$ at $0$, $g(x) = \sum_{k=0}^n b_k x^k + o(x^n)$, and $G$ is a primitive of $g$ on a neighbourhood of $0$, then $G$ admits a DL at order $n + 1$ at $0$ given by $G(x) = G(0) + \sum_{k=0}^n \frac{b_k}{k+1} x^{k+1} + o(x^{n+1})$. (Restatement of the primitivation lemma above, for the operations toolbox.)

Differentiation of a DL is NOT automatic

If $f$ admits a DL at order $n$ at $0$, it does not follow that $f'$ admits a DL at order $n - 1$ at $0$. Counter-example : let $f : \mathbb R \to \mathbb R$ be defined by $f(x) = x^3 \sin(1/x^2)$ for $x \ne 0$ and $f(0) = 0$. Since $|x^3 \sin(1/x^2)| \le |x|^3 = o(x^2)$ at $0$, $f$ admits the DL at order $2$ at $0$ given by the zero polynomial : $f(x) = 0 + 0 \cdot x + 0 \cdot x^2 + o(x^2)$. However, for $x \ne 0$, $$ f'(x) \ = \ 3 x^2 \sin(1/x^2) - 2 \cos(1/x^2), $$ which oscillates and has no limit at $0$ (the term $-2 \cos(1/x^2)$ alone oscillates between $\pm 2$). So $f'$ admits no DL at order $0$ at $0$ --- not even a limit. Never differentiate a DL without a $\mathcal C^n$ hypothesis on $f$.

Skills to practice

Primitivating a DL and identifying the differentiation pitfall

III.4 Composition (by example only)

No general result is required for the composition of DLs at this level. We give no general theorem and instead a single Method block + three worked examples illustrating the standard tricks. The recipe is uniform : write $u = f(x)$, where $u \to 0$ at $0$, then plug $u$ into the DL of $g$ ; truncate at the right order. The condition « $u \to 0$ at $0$ » is essential : if $f(0) \ne 0$, the composition $g \circ f$ near $0$ is anchored at $g(f(0)) \ne 0$, not at $g(0)$, and one must do a preliminary change of variables.

Method — Composing two DLs (computational recipe$\virgule$ not a theorem to quote)

On simple examples, the following procedure is used (no general theorem on composition of DLs is required at this level). To compute the DL of $g \circ f$ at order $n$ at $0$ when $f(0) = 0$ :

Write the DL of $f$ at order $n$ at $0$ : $f(x) = a_1 x + a_2 x^2 + \ldots + a_n x^n + o(x^n)$ (no constant term since $f(0) = 0$).
Write the DL of $g$ at order $n$ at $0$ : $g(u) = b_0 + b_1 u + b_2 u^2 + \ldots + b_n u^n + o(u^n)$.
Substitute $u = f(x)$ and expand the powers $u, u^2, \ldots, u^n$ ; truncate at order $n$. Each power $u^k$ contributes at least at order $k$, so terms beyond degree $n$ can be dropped early.
Caveat. If $f(0) \ne 0$, the formula above is wrong : in that case $g \circ f$ at $0$ is anchored at $u = f(0) \ne 0$, where one must use the DL of $g$ at $f(0)$ (Taylor-Young around $f(0)$).

Example — DL of $\exp(\sin x)$ at order $3$ at $0$

Set $u = \sin x = x - x^3/6 + o(x^3)$, so $u \to 0$ at $0$. Then $$ e^{u} \ = \ 1 + u + \tfrac{u^2}{2} + \tfrac{u^3}{6} + o(u^3). $$ Compute $u^2 = (x - x^3/6 + o(x^3))^2 = x^2 + o(x^3)$ and $u^3 = x^3 + o(x^3)$. Substituting and truncating at order $3$ : $$ e^{\sin x} \ = \ 1 + (x - \tfrac{x^3}{6}) + \tfrac{1}{2} x^2 + \tfrac{1}{6} x^3 + o(x^3) \ = \ 1 + x + \tfrac{x^2}{2} + o(x^3). $$ The $x^3$ terms $-1/6$ from $u$ and $+1/6$ from $u^3/6$ cancel.

Example — DL of $\sqrt{1 + \sin x}$ at order $3$ at $0$

Set $u = \sin x = x - x^3/6 + o(x^3)$. The function is $(1 + u)^{1/2}$, with usual DL $$ (1 + u)^{1/2} \ = \ 1 + \tfrac{u}{2} - \tfrac{u^2}{8} + \tfrac{u^3}{16} + o(u^3). $$ Compute $u^2 = x^2 + o(x^3)$, $u^3 = x^3 + o(x^3)$. Substituting and truncating : $$ \sqrt{1 + \sin x} \ = \ 1 + \tfrac{1}{2}(x - \tfrac{x^3}{6}) - \tfrac{1}{8} x^2 + \tfrac{1}{16} x^3 + o(x^3) \ = \ 1 + \tfrac{x}{2} - \tfrac{x^2}{8} - \tfrac{x^3}{48} + o(x^3), $$ where the $x^3$ coefficient is $-1/12 + 1/16 = -4/48 + 3/48 = -1/48$.

Example — DL of $\ln(1 + \sin^2 x)$ at order $4$ at $0$ --- factor first

The function $\sin^2 x$ vanishes at order $2$ at $0$, so we first compute $\sin^2 x$ at order $4$ then compose with $\ln(1 + u)$ at order $2$ in $u$ (since $u = \sin^2 x = O(x^2)$, $u^2 = O(x^4)$ is the last term that matters at order $4$ in $x$).
Step 1. $\sin x = x - x^3/6 + o(x^4)$, so $$ \sin^2 x \ = \ \bigl(x - \tfrac{x^3}{6}\bigr)^2 + o(x^4) \ = \ x^2 - \tfrac{x^4}{3} + o(x^4). $$ Step 2. Set $u = \sin^2 x = x^2 - x^4/3 + o(x^4)$. Then $u^2 = x^4 + o(x^4)$ and $\ln(1 + u) = u - u^2/2 + o(u^2)$, so $$ \ln(1 + \sin^2 x) \ = \ (x^2 - \tfrac{x^4}{3}) - \tfrac{1}{2} x^4 + o(x^4) \ = \ x^2 - \tfrac{5 x^4}{6} + o(x^4). $$ The « factor by the dominant term » reflex (Method above, in the Linear combination, product, truncation subsection) is what told us to take only order $2$ of $\ln(1 + u)$ : higher orders contribute beyond $x^4$ and can be dropped.

Skills to practice

Composing DLs on simple examples

IV Comparison relations for sequences

IV.1 Definitions and transfer rules

We give a rapid adaptation to sequences of the function relations. Concretely, we replace « $x \to a$ » by « $n \to +\infty$ » in the definitions of $o$, $O$, $\sim$ ; everything else transfers verbatim, including the operations (product, quotient, integer power) and the forbidden operations (sum, left-composition). This section is deliberately short --- it codifies what was already implicit in the croissances comparées table for sequences.

Definition — Comparison relations for sequences

Let $(u_n)_{n \ge n_0}$ and $(v_n)_{n \ge n_0}$ be two real sequences with $v_n \ne 0$ for $n$ large enough.

$(u_n)$ is negligible compared to $(v_n)$ at $+\infty$, written $u_n = o(v_n)$, if $u_n / v_n \to 0$ as $n \to +\infty$ ; equivalently, $u_n = \varepsilon_n v_n$ for $n$ large enough with $\varepsilon_n \to 0$.
$(u_n)$ is dominated by $(v_n)$, written $u_n = O(v_n)$, if $(u_n / v_n)$ is bounded for $n$ large enough.
$(u_n)$ is equivalent to $(v_n)$, written $u_n \sim v_n$, if $u_n / v_n \to 1$, equivalently $u_n = v_n + o(v_n)$.

Method — Operations and forbidden operations on $u_n \sim v_n$

All rules of the Equivalence subsection (functions) transfer verbatim to sequences.

Legal : product, quotient (when the denominator is nonzero for $n$ large), integer (or real) power if $v_n > 0$ for $n$ large.
Forbidden : sum of equivalents (e.g.\ $n + 1 \sim n$, $-n \sim -n$, but $(n+1) + (-n) = 1 \not\sim 0$) ; left-composition (e.g.\ $n + \ln n \sim n$ but $e^{n + \ln n} \not\sim e^n$).
Preserved : sign (for $n$ large), limit (as $n \to +\infty$).

Example — Equivalents at $+\infty$

(i) $n + \ln n \sim n$ at $+\infty$ because $(n + \ln n)/n = 1 + (\ln n)/n \to 1$ (using $\ln n = o(n)$).
(ii) $\sqrt{n + 1} - \sqrt n \sim 1/(2\sqrt n)$ at $+\infty$. Indeed, $\sqrt{n+1} - \sqrt n = (\sqrt{n+1} - \sqrt n)(\sqrt{n+1} + \sqrt n)/(\sqrt{n+1} + \sqrt n) = 1/(\sqrt{n+1} + \sqrt n)$, and $\sqrt{n+1} + \sqrt n \sim 2 \sqrt n$ (factor $\sqrt n$ : $\sqrt n (\sqrt{1 + 1/n} + 1) \to \sqrt n \cdot 2$). Hence $\sqrt{n+1} - \sqrt n \sim 1/(2 \sqrt n)$.

Skills to practice

Comparing sequences

IV.2 Discrete croissances comparées

The discrete version of the croissances comparées table is the natural sequence-side counterpart of the function table from the Negligibility subsection ; we state it here as a Proposition for ease of reference, and we illustrate it with a few classical examples that come up on every exam.

Proposition — Discrete croissances comparées

For all $a > 0$ and $b > 1$ : at $+\infty$, $$ n^a \ = \ o(b^n), \qquad b^n \ = \ o(n!), \qquad n! \ = \ o(n^n). $$ For all $a, b > 0$ : at $+\infty$, $(\ln n)^a = o(n^b)$.

Example — Standard discrete croissances comparées

$n^{10} = o(2^n)$ at $+\infty$ (every power, however large, loses against any base-$b$ exponential with $b > 1$). $2^n = o(n!)$ at $+\infty$ (factorial beats every exponential). $\ln n = o(\sqrt n)$ at $+\infty$ (log loses against every power). These three facts will be invoked dozens of times in the series chapter (`Series`) when testing convergence by equivalent.

Skills to practice

Applying discrete usual comparisons

V Applications of DLs to local study

V.1 Computing a limit via DL

The first application : turning an indeterminate-form limit into a routine computation by replacing each piece of the expression by its DL at a well-chosen order. The whole game is to choose the right order : large enough so the leading non-zero term appears, small enough to keep the algebra manageable. The Method below codifies this and is reused in every subsequent application.

Method — Computing an indeterminate-form limit via DL

Faced with a limit of indeterminate form ($0/0$, $\infty - \infty$, $1^\infty$, $0 \cdot \infty$) at $a \in \overline{\mathbb R}$ :

Identify the form. At $a = \pm \infty$, change variables $u = 1/x$ to bring the problem to $0$.
For each piece of the expression, choose the order of the DL : minimal order such that the first non-zero term of the numerator and of the denominator both appear (a $0/0$ form usually requires order $1$ or $2$ on each side ; check by trial).
Replace each piece by its DL, simplify the leading terms, conclude using $o$-arithmetic.

Example — $\lim_{x \to 0} (\sin x - x)/x^3$

DL of $\sin x$ at order $3$ at $0$ : $\sin x = x - x^3/6 + o(x^3)$. Then $$ \sin x - x \ = \ -\frac{x^3}{6} + o(x^3), \qquad \frac{\sin x - x}{x^3} \ = \ -\frac{1}{6} + o(1) \ \underset{x \to 0}{\longrightarrow} \ -\frac{1}{6}. $$

Example — $\lim_{x \to 0} ((1+x)^{1/x} - e)/x$

The form is $0/0$ ($1^\infty - e \to 0$). Write $(1+x)^{1/x} = \exp\bigl(\tfrac{\ln(1+x)}{x}\bigr)$. By the table, $\ln(1 + x) = x - x^2/2 + o(x^2)$, so $\ln(1+x)/x = 1 - x/2 + o(x)$. Then $$ \exp\bigl(1 - x/2 + o(x)\bigr) \ = \ e \cdot \exp(-x/2 + o(x)) \ = \ e \bigl(1 + (-x/2) + o(x)\bigr) \ = \ e - \frac{e}{2} x + o(x). $$ Hence $((1+x)^{1/x} - e)/x = -e/2 + o(1) \to -e/2$.

Skills to practice

Computing a limit via DL

V.2 Extracting an equivalent from a DL

The bridge from DLs (generalities and operations) to equivalents : the first non-zero term of a DL provides an equivalent. This is the most-used recipe of the chapter --- it converts every DL computation into an asymptotic statement on the function near the base point.

Proposition — Equivalent from a DL

Let $a \in \mathbb R$ and $f$ a function with DL at order $n$ at $a$ whose regular part starts with $a_p (x - a)^p$ ($p \le n$, $a_p \ne 0$, and $a_k = 0$ for $k < p$). Then $$ f(x) \ \underset{x \to a}{\sim} \ a_p (x - a)^p. $$

Proof

By hypothesis, $f(x) = a_p (x - a)^p + a_{p+1} (x - a)^{p+1} + \ldots + a_n (x - a)^n + o((x - a)^n)$. Factor $a_p (x - a)^p$ : $$ f(x) \ = \ a_p (x - a)^p \left( 1 + \frac{a_{p+1}}{a_p} (x - a) + \ldots + \frac{a_n}{a_p} (x - a)^{n - p} + \frac{o((x-a)^n)}{a_p (x-a)^p} \right). $$ The bracket tends to $1$ as $x \to a$ (every term except $1$ is a positive power of $(x-a)$ or an $o((x-a)^{n-p}) \to 0$). Hence $f(x) \sim a_p (x - a)^p$.

Method — « Le premier terme non nul d'un DL tient lieu d'équivalent »

To find an equivalent of $f$ at $a$ : compute a DL of $f$ at $a$ at a sufficient order ; the first non-zero monomial $a_p (x-a)^p$ is an equivalent of $f$ at $a$. The order $p$ is the order of vanishing of $f$ at $a$.

Example — Equivalents from the table

$\cos x - 1 \underset{x \to 0}{\sim} -x^2/2$ (from $\cos x = 1 - x^2/2 + o(x^2)$). $e^x - 1 - x \underset{x \to 0}{\sim} x^2/2$ (from $e^x = 1 + x + x^2/2 + o(x^2)$). $\ln(1 + x) - x \underset{x \to 0}{\sim} -x^2/2$ (from $\ln(1+x) = x - x^2/2 + o(x^2)$). These three equivalents are the workhorses of every limit calculation in the chapter on series.

Skills to practice

Extracting an equivalent from a DL

V.3 Position relative to the tangent$\virgule$ local extrema

A DL at order $\ge 2$ contains more local information than the tangent : the first non-zero coefficient after the linear term tells whether the curve lies above or below its tangent (when its order is even) or crosses it (when its order is odd, an inflection point). The same parity argument, applied at $p = 2$ to a function with $f'(a) = 0$, gives the standard CN and CS at order $2$ for an extremum --- which we state as a standalone Proposition because it is a named required result.

Proposition — Position of a curve relative to its tangent

Let $a \in \mathbb R$ and $f$ a $\mathcal C^p$ function on a neighbourhood of $a$ with DL at order $p$ at $a$ of the form $$ f(x) \ = \ f(a) + f'(a) (x - a) + a_p (x - a)^p + o((x - a)^p), $$ where $p \ge 2$, $a_p \ne 0$, and the coefficients of $(x-a)^2, \ldots, (x-a)^{p-1}$ vanish. Set $\Delta(x) = f(x) - f(a) - f'(a)(x-a)$ (signed distance from the curve to its tangent at $a$). Then near $a$ :

if $p$ is even, $\Delta(x)$ has the constant sign of $a_p$ : the curve lies above its tangent (if $a_p > 0$) or below it (if $a_p < 0$) ;
if $p$ is odd, $\Delta(x)$ changes sign at $a$ : the tangent crosses the curve, which in the usual local-study vocabulary corresponds to an inflection-type behaviour at $a$ (with non-horizontal tangent if $f'(a) \ne 0$).

Proof

By the equivalent-from-DL Proposition, $\Delta(x) \sim a_p (x - a)^p$ at $a$. Equivalence preserves sign : $\Delta(x)$ has the sign of $a_p (x - a)^p$ on a punctured neighbourhood of $a$. If $p$ is even, $(x-a)^p > 0$ for $x \ne a$, so $\Delta$ has the constant sign of $a_p$. If $p$ is odd, $(x-a)^p$ changes sign at $a$, so $\Delta$ does too.

Proposition — CN and CS at order $2$ for a local extremum

Let $a$ be an interior point of an interval $I$ and $f$ a $\mathcal C^2$ function on a neighbourhood of $a$.

Necessary condition at order $1$. If $f$ has a local extremum at $a$, then $f'(a) = 0$.
Sufficient condition at order $2$ (strict). If $f'(a) = 0$ and $f''(a) > 0$ (resp.\ $f''(a) < 0$), then $f$ has a strict local minimum (resp.\ maximum) at $a$.
Necessary condition at order $2$. If $f$ has a local minimum (resp.\ maximum) at $a$, then $f'(a) = 0$ and $f''(a) \ge 0$ (resp.\ $f''(a) \le 0$).

Proof

(1) A standard chapter-Dérivabilité result : at an interior local extremum of a differentiable function, the derivative vanishes (Fermat's lemma).
(2) By Taylor-Young at order $2$ at $a$, $f(x) = f(a) + 0 \cdot (x-a) + \tfrac{f''(a)}{2} (x-a)^2 + o((x-a)^2)$. The previous Proposition applied with $p = 2$ and $a_2 = f''(a)/2$ : if $f''(a) > 0$, then $\Delta(x) = f(x) - f(a) > 0$ on a punctured neighbourhood of $a$, hence $f(a)$ is a strict local minimum ; similarly $f''(a) < 0$ gives a strict local maximum.
(3) If $f$ has a local minimum at $a$, then $f'(a) = 0$ by (1). Suppose for contradiction $f''(a) < 0$. Then by (2), $a$ would be a strict local maximum, contradicting the minimum assumption. Hence $f''(a) \ge 0$. The argument for the maximum case is symmetric.

Method — Local study via DL

To determine, at a point $a$, the position of the curve of $f$ relative to its tangent and the nature of the critical point (extremum / inflection) :

Compute the DL of $f$ at $a$ at a sufficient order so that the first non-zero coefficient after the linear term appears (typically order $2$ if $f''(a) \ne 0$ ; otherwise go higher).
Read off the parity of the order $p$ of this first non-zero coefficient :
- $p$ even $\Rightarrow$ curve on one side of tangent (sign of $a_p$) ; if moreover $f'(a) = 0$, it is a local extremum ;
- $p$ odd $\Rightarrow$ inflection point with the tangent crossing the curve.

Example — $f(x) \equal \cos x$ at $0$

$f(x) = \cos x = 1 - x^2/2 + o(x^2)$. So $f(0) = 1$, $f'(0) = 0$, and the first non-zero coefficient after the linear term is $a_2 = -1/2$ at order $p = 2$ (even). By the position Proposition, $f(x) - f(0) - 0 \cdot x = -x^2/2 + o(x^2)$ has the constant sign $-1/2 < 0$ : the curve lies below the tangent $y = 1$. By the CN/CS Proposition (case $f''(0) = -1 < 0$), $f(0) = 1$ is a strict local maximum.

Skills to practice

Studying position and extrema locally

V.4 Asymptotes at $\pm \infty$

At $\pm \infty$, the analogous local study is the search for an asymptote : an affine line $y = a x + b$ such that the graph of $f$ comes arbitrarily close to it. The technique is to change variables $u = 1/x$ to reduce the question to a DL at $0$, then read off the asymptote from the first two terms of the expansion. The sign of the remainder gives the position of the curve relative to its asymptote.

Definition — Asymptote at $\pm \infty$

Let $f$ be defined on a neighbourhood of $+\infty$. We say that the affine line $y = a x + b$ (with $a, b \in \mathbb R$) is an asymptote of $f$ at $+\infty$ if $$ f(x) - (a x + b) \ \underset{x \to +\infty}{\longrightarrow} \ 0. $$ The analogous definition holds at $-\infty$.

Method — Finding an affine asymptote at $+\infty$

To search for an affine asymptote of $f$ at $+\infty$ :

Compute $a = \lim_{x \to +\infty} f(x)/x$. If this limit is finite, continue ; otherwise no affine asymptote.
Compute $b = \lim_{x \to +\infty} (f(x) - a x)$. If this limit is finite, the asymptote is $y = a x + b$ ; otherwise no affine asymptote.
Position. The sign of $f(x) - (a x + b)$ near $+\infty$ tells whether the curve is above ($> 0$) or below ($< 0$) its asymptote. This sign is read off the next term of the DL after the constant.

Example — Asymptote of $f(x) \equal x \sqrt{1 + 1/x^2}$ at $+\infty$

Set $u = 1/x \to 0^+$. Then $f(x) = (1/u) \sqrt{1 + u^2} = (1/u)(1 + u^2/2 - u^4/8 + o(u^4))$. Multiplying : $$ f(x) \ = \ \frac{1}{u} + \frac{u}{2} - \frac{u^3}{8} + o(u^3) \ = \ x + \frac{1}{2 x} - \frac{1}{8 x^3} + o(1/x^3) \qquad \text{at } +\infty. $$ So $a = 1$, $b = 0$ : the asymptote is $y = x$. The position is given by $f(x) - x = 1/(2x) + o(1/x)$, which is $> 0$ near $+\infty$ : the curve is above its asymptote.

Skills to practice

Finding asymptotes at $\pm \infty$

VI Asymptotic problems

The chapter closes with a final paragraph on Problèmes d'analyse asymptotique, with the standing warning : « La notion d'échelle de comparaison est hors programme. » We honor this strictly : we do not formalise the notion of an ordered family of reference functions $\varphi_0, \varphi_1, \ldots$ with $\varphi_{i+1} = o(\varphi_i)$. Instead, each « développement asymptotique » in this section is treated as its own example : an expression giving several successive terms of the function (or sequence) under study with a remainder $o$ of the last term retained. We give two named theorems (Euler's harmonic asymptotic and Stirling's formula, both admitted) and three short example sketches of the three flavors (inverse function, parametric equation, recurrent sequence). The exo file carries the full computations.

VI.1 Euler's constant$\virgule$ $H_n \equal \ln n + \gamma + o(1)$

The harmonic sum $H_n = \sum_{k=1}^n 1/k$ is one of the first sequences whose asymptotic behaviour is non-trivial : it grows to $+\infty$, but slowly. The exact rate is captured by a two-term development $H_n = \ln n + \gamma + o(1)$, where $\gamma \approx 0{,}577$ is a famous mathematical constant introduced by Euler. The full proof (by comparison of the sum with the integral $\int_1^n dt/t$) was done in chapter `SuitesReelles` ; we restate the result here for use as a reference asymptotic.

Theorem — Asymptotic development of the harmonic sum

The harmonic sum $H_n = \sum_{k=1}^n 1/k$ admits the asymptotic development $$ H_n \ = \ \ln n + \gamma + o(1) \qquad \text{at } n \to +\infty, $$ where $\gamma = \lim_{n \to +\infty} (H_n - \ln n) \approx 0{,}5772156\ldots$ is the Euler-Mascheroni constant.

Status of the proof

The proof uses an integral-comparison argument $\int_k^{k+1} dt/t \le 1/k \le \int_{k-1}^{k} dt/t$ to show that the sequence $u_n = H_n - \ln n$ is decreasing and bounded below ; it therefore converges, and $\gamma$ denotes its limit. The proof is fully written in chapter `SuitesReelles` ; we admit it here.

Example — Numerical estimate of $H_{1000}$

At $n = 1000$, $\ln(1000) \approx 6{,}908$ and $\gamma \approx 0{,}577$, so $H_{1000} \approx 7{,}485$. The exact value is $H_{1000} \approx 7{,}4854709\ldots$ : the two-term asymptotic captures the result to four decimal places.

Skills to practice

Using the asymptotic development of $H_n$

VI.2 Stirling's formula

The second named asymptotic of the chapter is the Stirling formula, giving the asymptotic of $n!$ at $n \to +\infty$. Like Euler's harmonic asymptotic, it is admitted at this level --- the proof is not required. A direct corollary is the asymptotic of the central binomial coefficient $\binom{2n}{n}$, which controls every centered random walk in probability.

Theorem — Stirling's formula

The factorial $n!$ admits the equivalent $$ n! \ \underset{n \to +\infty}{\sim} \ \sqrt{2 \pi n} \, \left( \frac{n}{e} \right)^n. $$ Equivalently, taking $\ln$ : $$ \ln(n!) \ = \ n \ln n - n + \tfrac{1}{2} \ln n + \tfrac{1}{2} \ln(2 \pi) + o(1) \qquad \text{at } n \to +\infty. $$

Status of the proof

The proof is not required at this level: « La démonstration n'est pas exigible. » The proof is admitted here but given in full in chapter Series, following the standard demonstration : study the sequence $u_n = \ln(n!) - (n + \tfrac{1}{2}) \ln n + n$ via its telescoping series (lien suite-série), show $u_{n+1} - u_n = O(1/n^2)$ by DL, deduce convergence of $(u_n)$ to a limit $C$ ; identify $e^C = \sqrt{2 \pi}$ via the Wallis integrals $W_n = \int_0^{\pi/2} \sin^n t \, dt$ (treated in chapter IntegrationSegment).

Example — Central binomial coefficient

A direct consequence of Stirling : $$ \binom{2n}{n} \ = \ \frac{(2n)!}{(n!)^2} \ \underset{n \to +\infty}{\sim} \ \frac{\sqrt{4 \pi n} \, (2n / e)^{2n}}{2 \pi n \, (n/e)^{2n}} \ = \ \frac{4^n}{\sqrt{\pi n}}. $$ This asymptotic is the gold-standard tool for centered random walks (it counts the number of returns to $0$ after $2n$ steps of a simple symmetric walk on $\mathbb Z$).

Skills to practice

Applying Stirling's formula

VI.3 Four flavors of asymptotic study (sketches)

The four flavors of « problèmes d'analyse asymptotique » are : (i) the asymptotic of a functional inverse $f^{-1}$ at $+\infty$ when $f$ is a $\mathcal C^\infty$ bijection ; (ii) the asymptotic of the unique solution $x_n$ of an equation depending on a parameter $n$ as $n \to +\infty$ ; (iii) the asymptotic of a recurrent sequence $u_{n+1} = \varphi(u_n)$ ; (iv) the asymptotic of a sequence of integrals $I_n = \int g_n$. Each is illustrated below by a one-Example sketch ; the full computations belong to the exo file.

Method — Conducting an asymptotic study

General recipe for a problem in this section :

Identify the asymptotic regime. What variable tends where ? What is the dominant behaviour ?
Start from a known equivalent or DL. The first term of the asymptotic comes from a known reference (usual DL, croissance comparée, Stirling).
Iterate. Once the first term is known, substitute and refine to get the second term, then the third, etc. This is the bootstrapping mechanism : each step uses the previous as input.

Example — Inverse function --- $f(x) \equal x + e^x$ at $+\infty$

$f(x) = x + e^x$ is $\mathcal C^\infty$ on $\mathbb R$ with $f'(x) = 1 + e^x > 0$, so $f$ is a $\mathcal C^\infty$-bijection from $\mathbb R$ to $\mathbb R$. As $x \to +\infty$, $e^x$ dominates $x$, so $f(x) \sim e^x$. Inverting, $y \to +\infty$ means $e^x \to +\infty$, i.e.\ $x \to +\infty$ ; and $y \sim e^x$ gives $\ln y \sim x$. The first-order asymptotic is $f^{-1}(y) \sim \ln y$ at $+\infty$. Further terms are obtained by iteration (left for the exo).

Example — Parametric equation --- $x e^x \equal n$

For each $n \ge 1$, the function $x \mapsto x e^x$ is a $\mathcal C^\infty$-bijection from $[0, +\infty[$ to $[0, +\infty[$ (strictly increasing, going from $0$ to $+\infty$). It admits a unique solution $x_n > 0$ of $x_n e^{x_n} = n$. Taking $\ln$ : $\ln(x_n) + x_n = \ln n$, so $x_n \le \ln n$ ; conversely $x_n \to +\infty$ since $x_n e^{x_n} = n \to +\infty$. Since $x_n + \ln x_n = \ln n$ and $x_n \le \ln n$, we also have $x_n = \ln n - \ln x_n \ge \ln n - \ln(\ln n)$, hence $x_n / \ln n \to 1$, that is $x_n \sim \ln n$. Consequently $\ln x_n \sim \ln(\ln n)$, and one final substitution gives $$ x_n \ = \ \ln n - \ln(\ln n) + o(1) \qquad \text{at } n \to +\infty. $$

Example — Recurrent sequence --- $u_{n+1} \equal \sin u_n$

Let $u_0 \in \;]0, \pi/2]$ and $u_{n+1} = \sin u_n$. Since $0 < \sin t < t$ for $t \in \;]0, \pi/2]$, the sequence $(u_n)$ is positive and strictly decreasing, hence converges to some $\ell \ge 0$ with $\ell = \sin \ell$ ; the only solution in $[0, \pi/2]$ is $\ell = 0$. So $u_n \to 0$. From the DL of $\sin$ : $\sin u = u - u^3/6 + o(u^3)$, hence $u_{n+1} = u_n - u_n^3 / 6 + o(u_n^3)$. The trick (Cesàro on $v_n = 1/u_n^2$) : $$ \frac{1}{u_{n+1}^2} - \frac{1}{u_n^2} \ = \ \frac{u_n^2 - u_{n+1}^2}{u_n^2 u_{n+1}^2}, \qquad u_n^2 - u_{n+1}^2 \sim \frac{u_n^4}{3}, \qquad u_n^2 u_{n+1}^2 \sim u_n^4, $$ so $1/u_{n+1}^2 - 1/u_n^2 \to 1/3$. By Cesàro, $v_n / n \to 1/3$, i.e.\ $u_n^2 \sim 3/n$, i.e.\ $u_n \sim \sqrt{3/n}$ at $+\infty$. (Full proof in the exo.)

Example — Sequence of integrals --- endpoint domination

Let $f \in C^0([0, 1], \mathbb R)$ with $f(1) \ne 0$, and consider $I_n = \int_0^1 x^n f(x) \, dx$. The weight $x^n$ concentrates near $1$ as $n \to +\infty$, so heuristically only the value $f(1)$ matters. Precisely : $$ I_n \ - \ f(1) \int_0^1 x^n \, dx \ = \ \int_0^1 x^n \bigl(f(x) - f(1)\bigr) \, dx. $$ Fix $\varepsilon > 0$. By continuity of $f$ at $1$, choose $\eta > 0$ such that $|f(x) - f(1)| \le \varepsilon$ for $x \in [1 - \eta, 1]$. Split the integral and bound : $$ \left| \int_0^1 x^n (f(x) - f(1)) \, dx \right| \ \le \ 2 \|f\|_\infty \int_0^{1 - \eta} x^n \, dx \ + \ \varepsilon \int_{1 - \eta}^1 x^n \, dx \ \le \ \frac{2 \|f\|_\infty (1 - \eta)^{n+1}}{n+1} \ + \ \frac{\varepsilon}{n+1}. $$ The first term is $o(1/(n+1))$ (geometric decay), so dividing by $\int_0^1 x^n \, dx = 1/(n+1)$, $$ \frac{I_n}{1/(n+1)} \ - \ f(1) \ \underset{n \to +\infty}{\longrightarrow} \ 0 \quad \text{(up to } \varepsilon\text{)}, \qquad I_n \ \underset{n \to +\infty}{\sim} \ \frac{f(1)}{n + 1}. $$

Skills to practice

Carrying out an asymptotic study

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