---
title: "Taylor Formulas and Asymptotic Expansions"
book: "University Mathematics — Year 1"
subject: math
language: en
chapter: 16
exercises: 12
source: https://one-course.com/books/math/3/en/chapter/16-taylor-formulas-and-asymptotic-expansions
---

# Chapter 16 — Taylor Formulas and Asymptotic Expansions

Near a point, a smooth function is as good as a [polynomial](https://one-course.com/books/math/3/en/chapter/8-polynomials#def-b1-poly-def) — with a controllable error. The Taylor formulas make this exact in three flavors ([integral](https://one-course.com/books/math/3/en/chapter/15-integration-on-a-segment#thm-b1-integration-def) remainder, Lagrange remainder, Young remainder), and the resulting *asymptotic expansions*, manipulated algebraically, become the sharpest tool of elementary analysis: limits, equivalents, local behavior, asymptotes.

## 16.1 Comparison notation

**Definition 16.1 (Landau notation).**

Let $f, g$ be defined near $x_0$ ($x_0 \in \R$ or $\pm\infty$). One writes, as $x \to x_0$:

- $f = o(g)$ (“little-o”) when $f = \varepsilon g$ with $\varepsilon(x) \to 0$ ;
- $f = O(g)$ (“big-O”) when $f = u g$ with $u$ bounded near $x_0$ ;
- $f \sim g$ (“equivalent”) when $f = (1 + \varepsilon) g$ with $\varepsilon \to 0$ — equivalently $f - g = o(g)$ .

The same notation applies to sequences ($n \to \infty$).

**Proposition 16.2 (Rules).**

As $x \to x_0$:

1. $\sim$ is an [equivalence relation](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-equiv) ; $f \sim g$ implies that $f$ and $g$ share limits, signs (near $x_0$ ), and zeros’ absence;
2. equivalents *multiply and divide* : $f_1 \sim g_1$ , $f_2  \sim g_2$ imply $f_1 f_2 \sim g_1 g_2$ and $\frac{f_1}{f_2}  \sim \frac{g_1}{g_2}$ ;
3. equivalents do *not* add: $x + 1 \sim x$ and $-x \sim  -x + 2$ at $+\infty$ , yet the sums $1$ and $2$ are not equivalent. To add, return to expansions with explicit $o(\cdot)$ terms;
4. $o(g) + o(g) = o(g)$ , $\;u \cdot o(g) = o(ug)$ , $\;o(o(g)) =  o(g)$ , and $f \sim g \iff f = g + o(g)$ .

**Proof.** Each is a short manipulation of the definitions; for instance $f_1 f_2 = (1+\varepsilon_1)(1+\varepsilon_2) g_1 g_2$ and $(1 +
\varepsilon_1)(1+\varepsilon_2) \to 1$. Two items of (4) deserve their line. $u\cdot o(g) = o(ug)$: if $f = \varepsilon g$ with $\varepsilon \to 0$, then $uf = \varepsilon\,(ug)$ with the same $\varepsilon$. $o(o(g)) = o(g)$: if $f = \varepsilon_1 h$ and $h = \varepsilon_2 g$ with both $\varepsilon_i \to 0$, then $f =
(\varepsilon_1\varepsilon_2) g$ and the product of the two infinitesimals is one. The equivalence $f \sim g \iff f = g +
o(g)$ is the definition read twice: $f - g = \varepsilon g$. The counterexample in (3) is the proof of (3). ∎

**Example 16.3 (The comparison scale).**

As $x \to +\infty$, the standard scale reads, in increasing order of strength:

$$
1 \;=\; o(\ln x), \quad
\ln x = o(x^{0.01}), \quad
x^{0.01} = o(\sqrt x), \quad
\sqrt x = o(x^{10}), \quad
x^{10} = o(\eu^{x}), \quad
\eu^x = o(\eu^{2x}) ,
$$

every step being an instance of the [growth comparisons](https://one-course.com/books/math/3/en/chapter/4-standard-functions#prop-b1-functions-powerrules) of [Proposition 4.6](https://one-course.com/books/math/3/en/chapter/4-standard-functions#prop-b1-functions-powerrules) (powers beat logarithms, exponentials beat powers, and within one family the exponent decides). Two habits worth forming: first, an $O(\cdot)$ landing in a smaller class upgrades silently ($O(\ln x)$ is also $o(x^{0.01})$); second, at $x \to 0^+$ the whole ladder reverses through the substitution $x \mapsto \frac1x$ — $\ln x = o(x^{-0.01})$ there, so “$x^\alpha \ln x \to 0$” holds for every $\alpha > 0$. Keeping the scale straight is half of every asymptotic argument in [Chapter 17](https://one-course.com/books/math/3/en/chapter/17-numerical-series#ch-b1-series).

**Example 16.4 (Uniqueness of expansions, and a parity dividend).**

If a function admits two expansions at $0$ to the same order,

$$
a_0 + a_1 x + \dots + a_n x^n + o(x^n)
= b_0 + b_1 x + \dots + b_n x^n + o(x^n),
$$

then $a_k = b_k$ for every $k$: subtracting and setting $c_k =
a_k - b_k$, evaluate the identity $c_0 + c_1 x + \dots + c_n x^n
= o(x^n)$ at $x \to 0$ to get $c_0 = 0$; divide by $x$ and repeat — each division is legitimate because the remaining expression is again $o(x^{n-k})$. Coefficients are therefore intrinsic, and one may compute them by *any* route (Taylor [derivatives](https://one-course.com/books/math/3/en/chapter/14-differentiation#def-b1-derivative-def), algebra on known expansions, integration): all routes must agree. Dividend: an *even* function has only even powers in its expansion — replace $x$ by $-x$ and invoke uniqueness; likewise odd functions have odd powers. This is why $\cos$ carries $o(x^{2p+1})$ rather than $o(x^{2p})$ in the table below: the absent odd term is free information, one order of precision for nothing.

## 16.2 The three Taylor formulas

**Theorem 16.5 (Taylor with integral remainder).**

Let $f$ be of class $C^{n+1}$ on an [interval](https://one-course.com/books/math/3/en/chapter/10-real-numbers#prop-b1-reals-intervals) containing $a$ and $x$. Then

$$
f(x) = \sum_{k=0}^{n} \frac{f^{(k)}(a)}{k!}\,(x - a)^k
+ \int_a^x \frac{(x - t)^n}{n!}\, f^{(n+1)}(t)\, \dd t .
$$

**Proof.** Induction on $n$. For $n = 0$: $f(x) = f(a) + \int_a^x f'(t)\dd t$ is the fundamental theorem ([Theorem 15.9](https://one-course.com/books/math/3/en/chapter/15-integration-on-a-segment#thm-b1-integration-ftc)). Step: integrate the remainder by parts,

$$
\int_a^x \frac{(x-t)^n}{n!} f^{(n+1)}(t)\,\dd t
= \Bigl[-\frac{(x-t)^{n+1}}{(n+1)!} f^{(n+1)}(t)\Bigr]_a^x
+ \int_a^x \frac{(x-t)^{n+1}}{(n+1)!} f^{(n+2)}(t)\,\dd t ,
$$

the bracket contributing the term $\frac{f^{(n+1)}(a)}{(n+1)!}(x -
a)^{n+1}$. ∎

**Example 16.6 (An exact expansion with its remainder).**

For $\ln(1 + x)$ the [integral](https://one-course.com/books/math/3/en/chapter/15-integration-on-a-segment#thm-b1-integration-def) remainder can be made completely explicit without differentiating anything $n$ times: integrate the finite geometric identity $\frac{1}{1+t} =
\sum_{k=0}^{n-1}(-t)^k + \frac{(-t)^n}{1+t}$ from $0$ to $x$:

$$
\ln(1 + x) = \sum_{k=1}^{n} \frac{(-1)^{k-1}x^k}{k}
+ (-1)^n \int_0^x \frac{t^n}{1 + t}\,\dd t ,
$$

and for $0 \leq x \leq 1$ the remainder is bounded by $\int_0^x
t^n\,\dd t = \frac{x^{n+1}}{n+1}$. This is stronger than Taylor–Young in two ways: it is an *identity* valid for a fixed $x$ (not only $x \to 0$), and the error bound is numerical. The weekend problem ([Problem 16.1](#pb-b1-taylor-1)) lives on such exact forms; Taylor–Young below is the lighter tool for limits, where only the shape of the error matters.

**Theorem 16.7 (Taylor–Lagrange inequality).**

Let $f$ be $C^{n+1}$ with $\abs{f^{(n+1)}} \leq M$ between $a$ and $x$. Then

$$
\Bigl| f(x) - \sum_{k=0}^{n} \frac{f^{(k)}(a)}{k!}(x-a)^k \Bigr|
\leq M\, \frac{\abs{x - a}^{n+1}}{(n+1)!} .
$$

**Proof.** Bound the [integral](https://one-course.com/books/math/3/en/chapter/15-integration-on-a-segment#thm-b1-integration-def) remainder: $\bigl|\int_a^x \frac{(x-t)^n}{n!}
f^{(n+1)}(t)\,\dd t\bigr| \leq M \bigl|\int_a^x
\frac{\abs{x-t}^n}{n!}\dd t\bigr| = M\frac{\abs{x-a}^{n+1}}{(n+1)!}$. ∎

**Example 16.8 (Certified numerics).**

What is $\sqrt{1.02}$? Apply Taylor–Lagrange to $f(t) =
\sqrt{1 + t}$ at $a = 0$, order $2$, $x = 0.02$:

$$
\sqrt{1.02} \approx 1 + \frac{0.02}{2} - \frac{0.02^2}{8}
= 1.00995 ,
\qquad
\abs{f'''(t)} = \frac{3}{8}(1+t)^{-5/2} \leq \frac 38 ,
$$

so the error is at most $\frac38 \cdot \frac{0.02^3}{6} =
5\cdot10^{-7}$: $\sqrt{1.02} = 1.00995$ with six certified decimals (true value $1.0099504938\dots$ — the bound is nearly sharp). The closing insight: Taylor–Young says only *how fast* the error vanishes; Taylor–Lagrange converts the same [polynomial](https://one-course.com/books/math/3/en/chapter/8-polynomials#def-b1-poly-def) into a *certificate*, a number plus a proven error bar. Whenever a decimal claim is made in this book, a Lagrange-type bound is standing behind it; the weekend problem ([Problem 16.1](#pb-b1-taylor-1)) industrializes the idea.

**Theorem 16.9 (Taylor–Young).**

Let $f$ be $n$ times [differentiable](https://one-course.com/books/math/3/en/chapter/14-differentiation#def-b1-derivative-def) at $a$. Then, as $x \to a$:

$$
f(x) = \sum_{k=0}^{n} \frac{f^{(k)}(a)}{k!}\,(x-a)^k
+ o\bigl((x-a)^n\bigr) .
$$

**Proof.** Induction on $n$. For $n = 1$ this is the definition of the [derivative](https://one-course.com/books/math/3/en/chapter/14-differentiation#def-b1-derivative-def) ([Definition 14.1](https://one-course.com/books/math/3/en/chapter/14-differentiation#def-b1-derivative-def)). Assume the [statement](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-statement) at order $n - 1$, and let $f$ be $n$ times [differentiable](https://one-course.com/books/math/3/en/chapter/14-differentiation#def-b1-derivative-def) at $a$. Apply the induction hypothesis to $f'$ (which is $n-1$ times [differentiable](https://one-course.com/books/math/3/en/chapter/14-differentiation#def-b1-derivative-def) at $a$):

$$
f'(t) = \sum_{k=0}^{n-1} \frac{f^{(k+1)}(a)}{k!}(t-a)^k + r(t),
\qquad r(t) = o\bigl((t-a)^{n-1}\bigr).
$$

Let $g(x) = f(x) - \sum_{k=0}^{n} \frac{f^{(k)}(a)}{k!}(x-a)^k$; then $g' = r$ and $g(a) = 0$. Given $\varepsilon > 0$, choose $\delta$ with $\abs{r(t)} \leq \varepsilon\abs{t - a}^{n-1}$ for $\abs{t-a}
\leq \delta$; the mean value inequality ([Theorem 14.9](https://one-course.com/books/math/3/en/chapter/14-differentiation#thm-b1-derivative-mvt)) applied on the segment from $a$ to $x$ (where $\abs{g'} \leq \varepsilon\abs{x-a}^{n-1}$) yields $\abs{g(x)} \leq \varepsilon\abs{x - a}^n$: exactly $g(x) =
o((x-a)^n)$. ∎

**Remark 16.10 (Three formulas, three prices, three products).**

The hypotheses grade exactly with the conclusions. Taylor–Young asks the least ($n$ [derivatives](https://one-course.com/books/math/3/en/chapter/14-differentiation#def-b1-derivative-def) *at the point* only) and yields the least: a qualitative $o((x-a)^n)$, perfect for limits, useless for certified digits. The Lagrange inequality asks for $C^{n+1}$ *on the [interval](https://one-course.com/books/math/3/en/chapter/10-real-numbers#prop-b1-reals-intervals)* and a bound $M$ there, and returns a numerical error bar. The [integral](https://one-course.com/books/math/3/en/chapter/15-integration-on-a-segment#thm-b1-integration-def) form asks the same regularity and returns the most: the error as an explicit object one can transform (integrate by parts, bound piecewise, change variables) — it is the form that powered the irrationality machine of [Problem 15.1](https://one-course.com/books/math/3/en/chapter/15-integration-on-a-segment#pb-b1-integration-1). Choosing the weakest formula that supports the goal is not pedantry: the flat function of [Problem 16.1](#pb-b1-taylor-1) satisfies Taylor–Young at every order while every stronger conclusion about it is false away from $0$.

**Proposition 16.11 (Standard expansions at 000).**

As $x \to 0$, for every fixed order $n$:

$$
\begin{align*}
\eu^x &= 1 + x + \frac{x^2}{2!} + \dots + \frac{x^n}{n!} + o(x^n),\\
\cos x &= 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \dots
+ \frac{(-1)^p x^{2p}}{(2p)!} + o(x^{2p+1}),\\
\sin x &= x - \frac{x^3}{3!} + \dots + \frac{(-1)^p
x^{2p+1}}{(2p+1)!} + o(x^{2p+2}),\\
\frac{1}{1 - x} &= 1 + x + x^2 + \dots + x^n + o(x^n),\\
\ln(1 + x) &= x - \frac{x^2}{2} + \frac{x^3}{3} - \dots +
\frac{(-1)^{n-1} x^n}{n} + o(x^n),\\
(1 + x)^\alpha &= 1 + \alpha x + \frac{\alpha(\alpha-1)}{2!}x^2 +
\dots + \binom{\alpha}{n} x^n + o(x^n),
\end{align*}
$$

where $\binom{\alpha}{n} = \frac{\alpha(\alpha - 1)\cdots(\alpha - n
+ 1)}{n!}$ for real $\alpha$. ($\cosh$ and $\sinh$: same as $\cos$, $\sin$ without the alternating signs.)

**Proof.** Each function is smooth near $0$ with [derivatives](https://one-course.com/books/math/3/en/chapter/14-differentiation#def-b1-derivative-def) easy to evaluate: $(\eu^x)^{(k)} = \eu^x$; the [derivatives](https://one-course.com/books/math/3/en/chapter/14-differentiation#def-b1-derivative-def) of $\sin$ and $\cos$ cycle with period $4$; $\bigl((1+x)^\alpha\bigr)^{(k)} = \alpha(\alpha -
1)\cdots(\alpha - k + 1)(1+x)^{\alpha - k}$; $\bigl(\ln(1+x)\bigr)^{(k)} = \frac{(-1)^{k-1}(k-1)!}{(1+x)^k}$. Apply Taylor–Young at $a = 0$. (The geometric one is exact: $\frac{1}{1-x} - \sum_0^n x^k = \frac{x^{n+1}}{1 - x} = o(x^n)$.) ∎

**Method 16.12 (Computing with expansions).**

1. *Fix the target order* $n$ first, and truncate every intermediate result there — carrying higher terms is wasted work, dropping lower ones is an error.
2. *Sums, products* : expand each factor to order $n$ and multiply, discarding beyond $x^n$ .
3. *Composition* $f(u(x))$ with $u(x) \to 0$ : substitute the expansion of $u$ into that of $f$ , order by order.
4. *Quotients* : write $\frac{1}{1 + v}$ with $v \to 0$ and use the geometric expansion.
5. *Integrate* an expansion term by term (differentiating requires more care — justification: [integral](https://one-course.com/books/math/3/en/chapter/15-integration-on-a-segment#thm-b1-integration-def) of $o(t^n)$ from $0$ to $x$ is $o(x^{n+1})$ , by direct bounding).

**Example 16.13 (Composition, with the bookkeeping shown).**

Expand $\eu^{\sin x}$ to order $3$. Inner expansion: $u = \sin x
= x - \frac{x^3}{6} + o(x^3)$, which indeed tends to $0$. Outer: $\eu^u = 1 + u + \frac{u^2}{2} + \frac{u^3}{6} + o(u^3)$, and $o(u^3) = o(x^3)$ since $u \sim x$. Powers of $u$, truncated at $x^3$:

$$
u^2 = x^2 + o(x^3), \qquad u^3 = x^3 + o(x^3)
$$

(the cross term $2x\cdot(-\frac{x^3}{6})$ is already $x^4$). Assemble:

$$
\eu^{\sin x} = 1 + \Bigl(x - \frac{x^3}{6}\Bigr) + \frac{x^2}{2}
+ \frac{x^3}{6} + o(x^3)
= 1 + x + \frac{x^2}{2} + o(x^3) :
$$

the two $x^3$ contributions cancel exactly. The closing insight: $\eu^{\sin x}$ and $\eu^x$ agree to order $3$ — not because $\sin x \approx x$ crudely, but because the first disagreement of the exponents ($-\frac{x^3}{6}$) enters multiplied by $\eu^0 = 1$ and is then met by the cubic term of the outer exponential; order-by-order bookkeeping detects such conspiracies, eyeballing never does. (The next term is $-\frac{x^4}{8}$: the truce ends at order $4$.)

**Example 16.14.**

Expansion of $\tan$ at order $5$. Write $\tan x =
\sin x \cdot \frac{1}{\cos x}$:

$$
\frac{1}{\cos x} = \frac{1}{1 - \bigl(\frac{x^2}{2} - \frac{x^4}{24}
+ o(x^5)\bigr)}
= 1 + \Bigl(\frac{x^2}{2} - \frac{x^4}{24}\Bigr)
+ \Bigl(\frac{x^2}{2}\Bigr)^{\!2} + o(x^5)
= 1 + \frac{x^2}{2} + \frac{5x^4}{24} + o(x^5),
$$

then

$$
\tan x = \Bigl(x - \frac{x^3}{6} + \frac{x^5}{120}\Bigr)
\Bigl(1 + \frac{x^2}{2} + \frac{5x^4}{24}\Bigr) + o(x^5)
= x + \frac{x^3}{3} + \frac{2x^5}{15} + o(x^5) .
$$

![The sine (dark) against its Taylor polynomials at 0: T_1 = x, T_3 = x - x3/6 (dashed), T_5 = x - x3/6 + x5/120 (dotted). Each new pair of terms hugs the curve over a visibly wider window, but every polynomial eventually tears away: a Taylor expansion is a local contract, sharpened at 0 and silent far away. The Taylor–Lagrange bound |x|n+1(n+1)! quantifies the window; the flat function of shows the contract can even be empty beyond the point itself.](https://one-course.com/images/onecourse/chapters/math-3/b1-taylor/fig-7578dfaf5f91.svg)

*The sine (dark) against its Taylor [polynomials](https://one-course.com/books/math/3/en/chapter/8-polynomials#def-b1-poly-def) at $0$: $T_1 = x$, $T_3 = x - \frac{x^3}{6}$ (dashed), $T_5 = x -
\frac{x^3}{6} + \frac{x^5}{120}$ (dotted). Each new pair of terms hugs the curve over a visibly wider window, but every [polynomial](https://one-course.com/books/math/3/en/chapter/8-polynomials#def-b1-poly-def) eventually tears away: a Taylor expansion is a *local* contract, sharpened at $0$ and silent far away. The Taylor–Lagrange bound $\frac{\abs{x}^{n+1}}{(n+1)!}$ quantifies the window; the flat function of [Problem 16.1](#pb-b1-taylor-1) shows the contract can even be empty beyond the point itself.*

## 16.3 Applications

**Example 16.15 (Limits).**

$$
\lim_{x \to 0} \frac{x - \sin x}{x^3}:
\qquad
x - \sin x = \frac{x^3}{6} + o(x^3) \sim \frac{x^3}{6},
\qquad\text{so the limit is } \frac16
$$

— settling the question raised in [Exercise 4.9](https://one-course.com/books/math/3/en/chapter/4-standard-functions#exo-b1-functions-9). Likewise $\displaystyle\lim_{x\to0}\Bigl(\frac{\sin
x}{x}\Bigr)^{1/x^2}$: the logarithm is

$$
\frac{1}{x^2}\ln\Bigl(1 - \frac{x^2}{6} + o(x^2)\Bigr)
= \frac{1}{x^2}\Bigl(-\frac{x^2}{6} + o(x^2)\Bigr)
\longrightarrow -\frac16,
\qquad\text{limit } \eu^{-1/6}.
$$

**Remark 16.16 (Common pitfalls with expansions).**

(i) *Never add or subtract equivalents*: from $\tan x \sim
x$ and $\sin x \sim x$ one may *not* conclude $\tan x -
\sin x \sim 0$ (meaningless) — the honest route is expansions:

$$
\tan x - \sin x
= \Bigl(x + \frac{x^3}{3}\Bigr) - \Bigl(x - \frac{x^3}{6}\Bigr)
+ o(x^3) = \frac{x^3}{2} + o(x^3) \sim \frac{x^3}{2} .
$$

(ii) *Expand past the massacre*: in the same computation, order $1$ sees only $x - x = 0$; whenever leading terms cancel, raise the order until a nonzero coefficient survives, and only then convert back to an equivalent. (iii) *Equivalents do not pass through exponentials*: $n^2 + n \sim n^2$, yet $\eu^{n^2+n} = \eu^{n}\,\eu^{n^2}$ is *not* equivalent to $\eu^{n^2}$ — exponentiate only *expansions of the exponent* whose error tends to $0$, never equivalents of the exponent. (Logarithms are safer: if $u_n \sim v_n \to \ell \neq
1$, $\ell > 0$, then $\ln u_n \sim \ln v_n$.) (iv) *The $o(\cdot)$ calculus is one-directional*: $o(x^2) + o(x^2) =
o(x^2)$, $5\,o(x^2) = o(x^2)$, $x\cdot o(x^2) = o(x^3)$ — but an $o(x^2)$ is not a specific function, so never cancel two of them against each other: $o(x^2) - o(x^2)$ is $o(x^2)$, not $0$.

**Proposition 16.17 (Local behavior).**

Suppose $f(x) = f(a) + c\,(x - a)^p + o\bigl((x-a)^p\bigr)$ with $c
\neq 0$ (first nonzero term after the constant; $p \geq 2$ at a critical point).

- If $p$ is even: $f$ has a local minimum at $a$ if $c > 0$ , a local maximum if $c < 0$ .
- If $p$ is odd: no extremum ( $f - f(a)$ changes sign); if moreover the expansion starts after a linear term $f'(a)(x  - a)$ , the graph crosses its tangent: an *inflection* .

**Proof.** Near $a$, $f(x) - f(a) = (x-a)^p\bigl(c + o(1)\bigr)$ has the sign of $c\,(x-a)^p$: constant sign for $p$ even, changing for $p$ odd. ∎

**Example 16.18 (Exponents must be expanded to o(1)o(1)o(1)).**

Find an equivalent of $u_n = \bigl(1 + \frac1n\bigr)^{n^2}$. Expand the *exponent* until its error tends to $0$:

$$
n^2 \ln\Bigl(1 + \frac1n\Bigr)
= n^2\Bigl(\frac1n - \frac{1}{2n^2} +
O\Bigl(\frac{1}{n^3}\Bigr)\Bigr)
= n - \frac12 + O\Bigl(\frac1n\Bigr),
$$

so $u_n = \eu^{\,n - 1/2}\,\eu^{O(1/n)}$ with $\eu^{O(1/n)} \to
1$:

$$
u_n \;\sim\; \eu^{\,n - \frac12} .
$$

Note what would have gone wrong with less care: stopping the exponent at $n^2\cdot\frac1n = n + O(1)$ leaves an $\eu^{O(1)}$ factor — bounded but not tending to $1$ — and no equivalent can be asserted. The rule of the pitfalls above, in positive form: an equivalent of $\eu^{a_n}$ requires the expansion of $a_n$ *up to a term tending to zero*, every coefficient before that being kept exactly.

**Example 16.19 (Classifying a flat critical point).**

Study $f(x) = \cos x + \frac{x^2}{2}$ near $0$. Both $f'(0) =
0$ and $f''(0) = -\cos 0 + 1 = 0$: the [second-derivative](https://one-course.com/books/math/3/en/chapter/14-differentiation#def-b1-derivative-def) test is mute. Expand instead:

$$
f(x) = \Bigl(1 - \frac{x^2}{2} + \frac{x^4}{24} +
o(x^4)\Bigr) + \frac{x^2}{2}
= 1 + \frac{x^4}{24} + o(x^4) :
$$

first nonzero term $c\,x^p$ with $p = 4$ even and $c =
\frac{1}{24} > 0$: a local minimum, of unusual flatness (the graph leaves its minimum value like $x^4$, not $x^2$). The closing insight: the expansion sees in one line what iterated differentiation obscures — and [Proposition 16.17](#prop-b1-taylor-local) is the systematic dictionary from “first surviving term” to “local shape”.

**Example 16.20 (Expansions at infinity).**

Two computations where the variable runs to $+\infty$ and the substitution $h = \frac1x \to 0^+$ imports the whole toolbox. First,

$$
x - x^2\ln\Bigl(1 + \frac1x\Bigr)
= x - x^2\Bigl(\frac1x - \frac{1}{2x^2} +
O\Bigl(\frac{1}{x^3}\Bigr)\Bigr)
= \frac12 + O\Bigl(\frac1x\Bigr) \longrightarrow \frac12 .
$$

Second, the [arctangent](https://one-course.com/books/math/3/en/chapter/4-standard-functions#def-b1-functions-arc) at infinity: from $\arctan x +
\arctan\frac1x = \frac\pi2$ for $x > 0$ ([Proposition 4.12](https://one-course.com/books/math/3/en/chapter/4-standard-functions#prop-b1-functions-arcidentities)) and the expansion of $\arctan$ at $0$ ([Exercise 16.3](#exo-b1-taylor-3)),

$$
\arctan x = \frac\pi2 - \arctan\frac1x
= \frac\pi2 - \frac1x + \frac{1}{3x^3} +
o\Bigl(\frac{1}{x^3}\Bigr) :
$$

the graph approaches its asymptote $y = \frac\pi2$ from below at speed $\frac1x$. The closing insight: there is no separate theory of expansions at infinity — one reciprocal substitution reduces them to expansions at $0$, provided every intermediate $O$ and $o$ is carried along honestly.

**Example 16.21 (Asymptote by expansion).**

As $x \to +\infty$,

$$
\sqrt{x^2 + x} = x\sqrt{1 + \tfrac1x}
= x\Bigl(1 + \frac{1}{2x} - \frac{1}{8x^2} + o\bigl(\tfrac{1}{x^2}\bigr)\Bigr)
= x + \frac12 - \frac{1}{8x} + o\bigl(\tfrac 1x\bigr):
$$

the line $y = x + \frac12$ is an asymptote, approached *from below* (the next term $-\frac{1}{8x}$ is negative).

**Remark 16.22 (Where expansions work next).**

Asymptotic expansions are the standing language of the rest of the book: in [Chapter 17](https://one-course.com/books/math/3/en/chapter/17-numerical-series#ch-b1-series) they decide convergence (equivalents feed the comparison tests, and the study of $\sum
\frac{1}{n^\alpha}$ is an expansion in disguise); in the Year 2 volume they become *power series*, where the Taylor [polynomial](https://one-course.com/books/math/3/en/chapter/8-polynomials#def-b1-poly-def) acquires infinitely many terms and a radius of convergence; and every linearization in physics — the pendulum, first-order perturbation — is a Taylor–Young [statement](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-statement) with the $o(\cdot)$ silently dropped. The one warning worth engraving: an expansion describes a function only *near one point* — see the flat function of the weekend problem, whose expansion at $0$ is identically zero without the function being so.

**Remark 16.23 (Perspectives inside this volume).**

Expansions are the working language of the remaining analysis and of the geometry to come. [Chapter 17](https://one-course.com/books/math/3/en/chapter/17-numerical-series#ch-b1-series) converts them into convergence verdicts: an equivalent of the general term is an expansion truncated at its first term, and the finer tests (alternating with error control) consume the second term too. [Chapter 24](https://one-course.com/books/math/3/en/chapter/24-plane-curves#ch-b1-curves) reads local geometry from expansions of the two coordinate functions: whether a parametrized curve crosses, kisses, or cusps at a point is decided by which powers of $t$ survive in $x(t)$ and $y(t)$ — the plane version of [Proposition 16.17](#prop-b1-taylor-local). And [Chapter 25](https://one-course.com/books/math/3/en/chapter/25-functions-of-two-variables#ch-b1-multivar) stops at order one on purpose: the tangent plane is a two-variable Taylor–Young [statement](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-statement), with the full second-order theory (Hessians, saddle points) deferred to the Year 2 volume. The common thread: every “local” question in this book is answered by writing down the first surviving term of an expansion.

## 16.4 Exercises

**Exercise 16.1 ★.**

Give the expansions at $0$: $\eu^{2x}$ to order $3$; $\;\ln(1 - x)$ to order $4$; $\;\sqrt{1 + x}$ to order $3$; $\;\dfrac{1}{1 + x^2}$ to order $6$.

**Solution of Exercise 16.1.**

$$
\eu^{2x} = 1 + 2x + 2x^2 + \frac{4x^3}{3} + o(x^3);
\qquad
\ln(1 - x) = -x - \frac{x^2}{2} - \frac{x^3}{3} - \frac{x^4}{4} +
o(x^4);
$$

$$
\sqrt{1+x} = 1 + \frac x2 - \frac{x^2}{8} + \frac{x^3}{16} + o(x^3);
\qquad
\frac{1}{1 + x^2} = 1 - x^2 + x^4 - x^6 + o(x^6),
$$

the last by substituting $-x^2$ into the geometric expansion.

**Exercise 16.2 ★.**

Compute the limits:

$$
\lim_{x\to 0} \frac{\eu^x - 1 - x}{x^2},
\qquad
\lim_{x\to 0} \frac{\cos x - \sqrt{1 - x^2}}{x^4},
\qquad
\lim_{x\to 0} \frac{\ln(1+x) - \sin x}{x^2}.
$$

**Solution of Exercise 16.2.**

$\eu^x - 1 - x = \frac{x^2}{2} + o(x^2)$: limit $\dfrac12$.

$\cos x = 1 - \frac{x^2}{2} + \frac{x^4}{24} + o(x^4)$ and $\sqrt{1 - x^2} = 1 - \frac{x^2}{2} - \frac{x^4}{8} + o(x^4)$: difference $\frac{x^4}{24} + \frac{x^4}{8} = \frac{x^4}{6} + o(x^4)$: limit $\dfrac16$.

$\ln(1+x) - \sin x = \bigl(x - \frac{x^2}{2}\bigr) - x + o(x^2) =
-\frac{x^2}{2} + o(x^2)$: limit $-\dfrac12$.

**Exercise 16.3 ★.**

Expand $\arctan x$ at $0$ to order $5$ by integrating the expansion of $\frac{1}{1 + x^2}$, and $\arcsin x$ to order $5$ by integrating that of $(1 - x^2)^{-1/2}$.

**Solution of Exercise 16.3.**

$\frac{1}{1+t^2} = 1 - t^2 + t^4 + o(t^5)$; integrating from $0$ to $x$ ([Method 16.12](#met-b1-taylor-compute) (5)):

$$
\arctan x = x - \frac{x^3}{3} + \frac{x^5}{5} + o(x^5)\ \ (\text{even
}o(x^6)\text{, by oddness}).
$$

$(1 - t^2)^{-1/2} = 1 + \frac{t^2}{2} + \frac38 t^4 + o(t^5)$ (binomial expansion with $\alpha = -\frac12$, $x = -t^2$: $\binom{-1/2}{2} = \frac{(-\frac12)(-\frac32)}{2} = \frac38$); integrating:

$$
\arcsin x = x + \frac{x^3}{6} + \frac{3x^5}{40} + o(x^5) .
$$

**Exercise 16.4 ★.**

Using Taylor–Lagrange for $\exp$ on $\intcc{0}{1}$, prove that

$$
\Bigl| \eu - \sum_{k=0}^{n} \frac{1}{k!} \Bigr| \leq
\frac{3}{(n+1)!},
$$

and determine an $n$ guaranteeing $6$ exact decimals of $\eu$.

**Solution of Exercise 16.4.**

Taylor–Lagrange ([Theorem 16.7](#thm-b1-taylor-lagrange)) for $\exp$ at $a =
0$, $x = 1$: the $(n+1)$-st [derivative](https://one-course.com/books/math/3/en/chapter/14-differentiation#def-b1-derivative-def) is $\eu^t \leq \eu < 3$ on $\intcc{0}{1}$, so

$$
\Bigl|\eu - \sum_{k=0}^{n} \frac{1}{k!}\Bigr| \leq
\frac{3}{(n+1)!} .
$$

For $6$ exact decimals, want $\frac{3}{(n+1)!} < 5\times 10^{-7}$, i.e. $(n+1)! > 6\times 10^{6}$: since $10! = 3\,628\,800$ and $11! =
39\,916\,800$, $n + 1 = 11$, i.e. $n = 10$ suffices.

**Exercise 16.5 ★★.**

Expand to order $2$ in $\frac1n$ and deduce the limit and the speed of convergence:

$$
\Bigl(1 + \frac 1n\Bigr)^{\!n}
= \eu\Bigl(1 - \frac{1}{2n} + \frac{11}{24n^2} +
o\Bigl(\frac{1}{n^2}\Bigr)\Bigr).
$$

**Solution of Exercise 16.5.**

$n\ln\bigl(1 + \frac1n\bigr) = n\Bigl(\frac1n - \frac{1}{2n^2} +
\frac{1}{3n^3} + o\bigl(\frac{1}{n^3}\bigr)\Bigr) = 1 - \frac{1}{2n}
+ \frac{1}{3n^2} + o\bigl(\frac{1}{n^2}\bigr)$. Exponentiating, with $u = -\frac{1}{2n} + \frac{1}{3n^2}$ and $\eu^u = 1 + u + \frac{u^2}2
+ o(u^2)$:

$$
\Bigl(1 + \frac1n\Bigr)^n = \eu\cdot \eu^{u}
= \eu\Bigl(1 - \frac{1}{2n} + \frac{1}{3n^2} + \frac{1}{8n^2}
+ o\Bigl(\frac{1}{n^2}\Bigr)\Bigr)
= \eu\Bigl(1 - \frac{1}{2n} + \frac{11}{24n^2} +
o\Bigl(\frac{1}{n^2}\Bigr)\Bigr).
$$

Limit $\eu$; the error is $\sim \dfrac{\eu}{2n}$: slow (one digit per tenfold increase of $n$).

**Exercise 16.6 ★★.**

Study the local behavior at $0$ of $f(x) = x^2 - x^4$ and of $g(x) =
x^3 + x^5$; and find the position of the graph of $h(x) = \eu^x$ relative to its tangent at $a = 1$, locally then globally.

**Solution of Exercise 16.6.**

$f(x) = x^2 - x^4 = x^2(1 + o(1))$: first term $x^2$, $p = 2$ even, coefficient $> 0$: local minimum at $0$ (not global: $f(2) = -12$).

$g(x) = x^3 + x^5$: first term $x^3$, odd $p$: no extremum; $g$ crosses its (horizontal) tangent: inflection at $0$.

$h = \exp$ at $a = 1$: $h(x) = \eu + \eu(x-1) + \frac{\eu}{2}(x-1)^2
+ o((x-1)^2)$; the difference with the tangent is $\frac{\eu}{2}(x-1)^2 + o(\cdot) > 0$ locally. Globally: $\eu^x -
\eu x \geq 0$ for all $x$ by convexity ([Theorem 14.19](https://one-course.com/books/math/3/en/chapter/14-differentiation#thm-b1-derivative-convexchar) (3)): the graph lies above every tangent, with equality only at the contact point.

**Exercise 16.7 ★★.**

Determine the asymptotes at $\pm\infty$ of $f(x) =
\sqrt[3]{x^3 + x^2}$ and the position of the curve relative to them.

**Solution of Exercise 16.7.**

For $x \to +\infty$:

$$
f(x) = x\Bigl(1 + \frac1x\Bigr)^{1/3}
= x\Bigl(1 + \frac{1}{3x} - \frac{1}{9x^2} +
o\Bigl(\frac{1}{x^2}\Bigr)\Bigr)
= x + \frac13 - \frac{1}{9x} + o\Bigl(\frac1x\Bigr):
$$

asymptote $y = x + \frac13$, curve below it near $+\infty$. As $x
\to -\infty$, the same computation is valid (the cube root is defined for all reals, and $\frac1x \to 0$): same asymptote $y = x +
\frac13$, but now $-\frac{1}{9x} > 0$: curve *above* the line.

**Exercise 16.8 ★★.**

Find the equivalent, as $n \to \infty$, of

$$
u_n = \sqrt{n+1} - \sqrt n, \qquad
v_n = \ln(n+1) - \ln n, \qquad
w_n = \sin\frac{1}{n} - \tan\frac{1}{n},
$$

each as a power of $n$ times a constant.

**Solution of Exercise 16.8.**

$u_n = \sqrt n\bigl(\sqrt{1 + \tfrac1n} - 1\bigr) = \sqrt
n\bigl(\frac{1}{2n} + o(\frac1n)\bigr) \sim \dfrac{1}{2\sqrt n}$.

$v_n = \ln\bigl(1 + \frac1n\bigr) \sim \dfrac 1n$.

$w_n$: with $h = \frac1n \to 0$, $\sin h - \tan h = \bigl(h -
\frac{h^3}{6}\bigr) - \bigl(h + \frac{h^3}{3}\bigr) + o(h^3) =
-\frac{h^3}{2} + o(h^3)$, so $w_n \sim -\dfrac{1}{2n^3}$.

**Exercise 16.9 ★★★.**

Let $f$ be $C^2$ on $\R$. Prove that for every $x$ and $h > 0$:

$$
\abs{f'(x)} \leq \frac{\abs{f(x+h) - f(x-h)}}{2h} +
\frac{h}{2}\sup_{\intcc{x-h}{x+h}}\abs{f''} ,
$$

and deduce the Landau–Kolmogorov-type inequality: if $\abs f \leq
M_0$ and $\abs{f''} \leq M_2$ on $\R$, then $\abs{f'} \leq
\sqrt{2 M_0 M_2}$ everywhere. *(Optimize over $h$.)*

**Solution of Exercise 16.9.**

Taylor–Lagrange at order $1$ around $x$, on both sides:

$$
f(x + h) = f(x) + h f'(x) + R_+,\quad
f(x - h) = f(x) - h f'(x) + R_-,
\qquad \abs{R_\pm} \leq \frac{h^2}{2} \sup \abs{f''} .
$$

Subtracting: $f(x+h) - f(x-h) = 2h f'(x) + (R_+ - R_-)$, so

$$
\abs{f'(x)} \leq \frac{\abs{f(x+h) - f(x-h)}}{2h} +
\frac{h}{2}\sup_{\intcc{x-h}{x+h}}\abs{f''} .
$$

With global bounds: $\abs{f'(x)} \leq \frac{M_0}{h} + \frac{M_2
h}{2}$ for every $h > 0$. The right side is minimized at $h =
\sqrt{2M_0/M_2}$ ([derivative](https://one-course.com/books/math/3/en/chapter/14-differentiation#def-b1-derivative-def) zero), with value $\sqrt{2M_0M_2}$ — hence $\abs{f'} \leq \sqrt{2M_0M_2}$. (If $M_2 = 0$, let $h \to
\infty$: $f' = 0$, consistent.)

**Exercise 16.10 ★★★.**

The sequence $u_0 \in \intoo{0}{\pi}$, $u_{n+1} = \sin u_n$ decreases to $0$ (justify briefly). To find its speed, consider $v_n
= \frac{1}{u_n^2}$:

1. using the expansion of $\sin$ , prove $v_{n+1} - v_n \to  \frac13$ ;
2. with Cesàro ( [Exercise 11.10](https://one-course.com/books/math/3/en/chapter/11-sequences#exo-b1-seq-10) ), deduce $\frac{v_n}{n}  \to \frac13$ , then the equivalent $u_n \sim  \sqrt{\dfrac{3}{n}}$ .

**Solution of Exercise 16.10.**

On $\intoo{0}{\pi}$: $0 < \sin u < u$, so $(u_n)$ is strictly decreasing, positive, hence convergent; the limit is a fixed point of $\sin$ in $\intcc{0}{\pi}$, and $\sin \ell = \ell$ forces $\ell =
0$ (since $\sin x < x$ for $x > 0$).

1. Using $\sin u = u - \frac{u^3}{6} + o(u^3)$ as $u \to 0$: $$v_{n+1} - v_n = \frac{1}{\sin^2 u_n} - \frac{1}{u_n^2}  = \frac{1}{u_n^2}\Bigl(\Bigl(1 - \frac{u_n^2}{6} +  o(u_n^2)\Bigr)^{\!-2} - 1\Bigr)  = \frac{1}{u_n^2}\Bigl(\frac{u_n^2}{3} + o(u_n^2)\Bigr)  \longrightarrow \frac13 .$$
2. By [Exercise 11.10](https://one-course.com/books/math/3/en/chapter/11-sequences#exo-b1-seq-10) (3) (Cesàro for differences), $\frac{v_n}{n} \to \frac13$, i.e. $v_n \sim \frac n3$, i.e. $u_n^2 \sim \frac 3n$: since $u_n > 0$, $$u_n \sim \sqrt{\frac{3}{n}} .$$

**Exercise 16.11 ★★.**

(Differences of infinities) Compute

$$
\lim_{x \to 0} \Bigl(\frac{1}{x^2} - \frac{1}{\sin^2 x}\Bigr)
\qquad\text{and}\qquad
\lim_{x \to 0^+} \Bigl(\frac 1x - \frac{1}{\ln(1 + x)}\Bigr)
$$

by reducing to a common denominator and expanding numerator and denominator separately.

**Solution of Exercise 16.11.**

Common denominators. First limit:

$$
\frac{1}{x^2} - \frac{1}{\sin^2 x}
= \frac{\sin^2 x - x^2}{x^2\sin^2 x},
\qquad
\sin^2 x = \Bigl(x - \frac{x^3}{6} + o(x^4)\Bigr)^{\!2}
= x^2 - \frac{x^4}{3} + o(x^5) ,
$$

so the numerator is $-\frac{x^4}{3} + o(x^4)$ while the denominator is $\sim x^4$: the limit is $-\dfrac13$.

Second: $\dfrac1x - \dfrac{1}{\ln(1+x)} = \dfrac{\ln(1+x) -
x}{x\ln(1+x)}$; the numerator is $-\frac{x^2}{2} + o(x^2)$, the denominator $x\bigl(x + o(x)\bigr) \sim x^2$: the limit is $-\dfrac12$.

**Exercise 16.12 ★★★.**

(Asymptotics of implicit roots) Show that for every $k \in
\N^*$ the equation $\tan x = x$ has exactly one solution $x_k$ in $\intoo{k\pi - \frac\pi2}{k\pi + \frac\pi2}$, that $x_k = k\pi
+ \frac\pi2 - \varepsilon_k$ with $\varepsilon_k =
\arctan\frac{1}{x_k}$, and deduce the expansion

$$
x_k = k\pi + \frac\pi2 - \frac{1}{k\pi} + o\Bigl(\frac
1k\Bigr) \qquad (k \to \infty).
$$

**Solution of Exercise 16.12.**

On $I_k = \intoo{k\pi - \frac\pi2}{k\pi + \frac\pi2}$, the function $g(x) = \tan x - x$ has [derivative](https://one-course.com/books/math/3/en/chapter/14-differentiation#def-b1-derivative-def) $\tan^2 x \geq 0$, vanishing only at the single point $k\pi$: $g$ is strictly increasing on $I_k$ ([Corollary 14.12](https://one-course.com/books/math/3/en/chapter/14-differentiation#cor-b1-derivative-monotone) (2)), with limits $-\infty$ and $+\infty$ at the ends: exactly one zero $x_k$. For $k \geq 1$, $g(k\pi) = -k\pi < 0$, so $x_k \in
\intoo{k\pi}{k\pi + \frac\pi2}$: write $x_k = k\pi + \frac\pi2 -
\varepsilon_k$ with $\varepsilon_k \in \intoo{0}{\frac\pi2}$. Then

$$
x_k = \tan x_k = \tan\Bigl(\frac\pi2 - \varepsilon_k\Bigr)
= \frac{1}{\tan\varepsilon_k}
\quad\Longrightarrow\quad
\varepsilon_k = \arctan\frac{1}{x_k} ,
$$

using $\tan\varepsilon_k = \frac{1}{x_k}$ and $\varepsilon_k \in
\intoo{0}{\frac\pi2}$. Since $x_k \geq k\pi \to \infty$: $\varepsilon_k \to 0$, and

$$
\varepsilon_k = \arctan\frac{1}{x_k} = \frac{1}{x_k} +
O\Bigl(\frac{1}{x_k^3}\Bigr)
= \frac{1}{k\pi + O(1)} + O\Bigl(\frac{1}{k^3}\Bigr)
= \frac{1}{k\pi} + O\Bigl(\frac{1}{k^2}\Bigr) ,
$$

whence $x_k = k\pi + \frac\pi2 - \frac{1}{k\pi} +
o\bigl(\frac1k\bigr)$.

## 16.5 Problem: Alternating sums, certified digits, and the irrationality of $\cos 1$

**Problem 16.1.**

Weekend problem — the alternating estimate $\abs{S - S_n} \leq a_{n+1}$: $\ln 2$ and $\pi$ with proven decimals, Machin’s formula, and $\cos 1 \notin \Q$

An alternating sum with decreasing terms is the friendliest object in numerical analysis: its error is bounded by the first omitted term, *with known sign*. This problem proves that principle with the adjacent-sequences theorem, then spends it three ways: certified decimals for $\ln 2$ (three competing routes) and for $\pi$ (Leibniz, then Machin’s 1706 formula, still the idea behind record computations for centuries), the irrationality of $\cos 1$, $\sin 1$ and $\cosh 1$, and, as a counterweight, the Taylor–Lagrange *equality* and the flat function whose Taylor expansion lies. Throughout, “series” language is informal: every sum here is a *sequence of partial sums*, as in [Example 11.12](https://one-course.com/books/math/3/en/chapter/11-sequences#ex-b1-seq-e); the theory proper [opens](https://one-course.com/books/math/3/en/chapter/12-topology-of-the-real-line#def-b1-topology-open) in [Chapter 17](https://one-course.com/books/math/3/en/chapter/17-numerical-series#ch-b1-series).

**Part I — The alternating estimate.** Let $(a_k)_{k \geq 0}$ decrease to $0$ and $S_n = \sum_{k=0}^{n}
(-1)^k a_k$.

1. Show that $(S_{2n+1})$ is nondecreasing, $(S_{2n})$ nonincreasing, and that they are adjacent ([Theorem 11.11](https://one-course.com/books/math/3/en/chapter/11-sequences#thm-b1-seq-adjacent)): both converge to a common $S$ with, for every $n$, $$S_{2n+1} \leq S \leq S_{2n},  \qquad  \abs{S - S_n} \leq a_{n+1} ,$$ the error having the sign of the first omitted term. Show moreover that if the decrease is *strict*, all these inequalities are strict.
2. First dividend: for $x = 1$ in the exponential series, compare with [Theorem 16.7](#thm-b1-taylor-lagrange) at $a = 0$ : show that $T_n = \sum_{k=0}^{n} \frac{(-1)^k}{k!}$ converges to $\eu^{-1}$ with $\abs{\eu^{-1} - T_n} <  \frac{1}{(n+1)!}$ .
3. (Leibniz, 1674) From the *exact* finite identity $$\frac{1}{1 + t^2} = \sum_{k=0}^{n} (-1)^k t^{2k}  + \frac{(-1)^{n+1} t^{2n+2}}{1 + t^2} ,$$ integrated over $\intcc{0}{1}$, prove $$\frac\pi4 = \sum_{k=0}^{n} \frac{(-1)^k}{2k+1} + \rho_n,  \qquad \abs{\rho_n} \leq \frac{1}{2n+3} .$$
4. Slowness: how many terms of Leibniz guarantee six exact decimals of $\pi$ ? (About two million.) Evaluate $4 S_4  = 4\bigl(1 - \frac13 + \frac15 - \frac17 +  \frac19\bigr)$ and its distance to $\pi$ , to feel the pain.

**Part II — $\ln 2$ three ways.**

5. (Way 1: alternating harmonic) From $\frac{1}{1+t} =  \sum_{k=0}^{n-1}(-1)^k t^k + \frac{(-1)^n t^n}{1+t}$ integrated over $\intcc{0}{1}$: $$\ln 2 = \sum_{k=1}^{n} \frac{(-1)^{k-1}}{k} + (-1)^n  R_n, \qquad \frac{1}{2(n+1)} \leq R_n \leq  \frac{1}{n+1} :$$ the error is of exact order $\frac 1n$ — a million terms for six decimals.
6. (Way 2: the fast series) Integrate $\frac{1}{1 - t^2} =  \sum_{k=0}^{n} t^{2k} + \frac{t^{2n+2}}{1-t^2}$ from $0$ to $x \in \intoo{0}{1}$ and evaluate at $x = \frac13$ (note $\frac{1 + 1/3}{1 - 1/3} = 2$): $$\ln 2 = 2\sum_{k=0}^{n} \frac{(1/3)^{2k+1}}{2k+1} +  \tilde\rho_n,  \qquad  0 < \tilde\rho_n \leq \frac{9}{4}\cdot  \frac{(1/3)^{2n+3}}{2n+3} :$$ geometric convergence, roughly one digit per term.
7. (Way 3: [Riemann sums](https://one-course.com/books/math/3/en/chapter/15-integration-on-a-segment#thm-b1-integration-riemann) and a hidden identity) Prove by induction the identity $$\sum_{k=1}^{2n} \frac{(-1)^{k-1}}{k} = H_{2n} - H_n =  \sum_{k=1}^{n} \frac{1}{n + k} ,$$ and recover $\ln 2$ as the Riemann-sum limit of [Example 15.21](https://one-course.com/books/math/3/en/chapter/15-integration-on-a-segment#ex-b1-integration-riemannexample): Ways 1 and 3 are secretly the same number seen twice.
8. Shootout at six terms: compare $\sum_{k=1}^{6}  \frac{(-1)^{k-1}}{k} = 0.6167$ with Way 2 at $n = 5$ , which already gives $\ln 2 = 0.693147$ with error $\leq  1.1\cdot10^{-7}$ . Explain the structural reason (evaluation point deep inside the [interval](https://one-course.com/books/math/3/en/chapter/10-real-numbers#prop-b1-reals-intervals) of convergence versus on its [boundary](https://one-course.com/books/math/3/en/chapter/12-topology-of-the-real-line#def-b1-topology-closure) ).
9. How many terms of Way 2 certify *ten* decimals of $\ln 2$ ? Show $n = 10$ suffices.

**Part III — Machin’s formula.**

10. Compute $(5 + \iu)^4$ and verify the complex identity $$(5 + \iu)^4 = 2\,(1 + \iu)\,(239 + \iu) .$$ Taking arguments ([Chapter 3](https://one-course.com/books/math/3/en/chapter/3-complex-numbers#ch-b1-complex) conventions), deduce Machin’s formula $$\frac\pi4 = 4\arctan\frac15 - \arctan\frac{1}{239} .$$ (Check that no argument leaves $\intoo{-\frac\pi2}{\frac\pi2}$.)
11. As in question 3, establish for $0 < x < 1$: $$\arctan x = \sum_{k=0}^{n} \frac{(-1)^k  x^{2k+1}}{2k+1} + r_n(x), \qquad \abs{r_n(x)} \leq  \frac{x^{2n+3}}{2n+3} .$$
12. Certify $\pi$ to seven decimals with six terms: bound the total error of $$\pi \approx 16\sum_{k=0}^{4}  \frac{(-1)^k (1/5)^{2k+1}}{2k+1}  - 4\Bigl(\frac{1}{239} - \frac{1}{3\cdot239^3}\Bigr)$$ by $16\,\frac{(1/5)^{11}}{11} +  4\,\frac{(1/239)^5}{5} < 5\cdot10^{-8}$, and give the resulting value $3.1415926\dots$
13. Compare the three routes to $\pi$ now available — Leibniz (question 4), the Dalzell [integrals](https://one-course.com/books/math/3/en/chapter/15-integration-on-a-segment#thm-b1-integration-def) of [Problem 15.1](https://one-course.com/books/math/3/en/chapter/15-integration-on-a-segment#pb-b1-integration-1) (error $4^{1-5m}$ ), Machin (error $\approx 16\cdot 5^{-(2n+3)}$ ) — in digits per term, and explain why shrinking the evaluation point beats everything.

**Part IV — The integer trap, alternating edition.**

14. Suppose $\cos 1 = \frac pq$ . Multiply the strict alternating bracketing of $\sum_{k}  \frac{(-1)^k}{(2k)!}$ (question 1) by $(2n)!$ with $2n  \geq q$ , and derive a contradiction: $\cos 1$ is irrational.
15. Adapt to $\sin 1 = \sum_k \frac{(-1)^k}{(2k+1)!}$ (multiply by $(2n+1)!$ ): $\sin 1 \notin \Q$ . Both irrational, yet $\cos^2 1 + \sin^2 1 = 1$ : irrationality is not stable under algebra.
16. The non-alternating cousin: $\cosh 1 = \sum_k  \frac{1}{(2k)!}$ (in the partial-sum sense, with the two-sided tail bound $0 < \cosh 1 - \sum_{k \leq n}  \frac{1}{(2k)!} < \frac{2}{(2n+2)!}$ , to be proved). Conclude $\cosh 1 \notin \Q$ by the same trap.
17. Push to $\cos\frac 1m$ for every integer $m \geq 1$ : multiply by $m^{2n}(2n)!$ and conclude $\cos\frac1m  \notin \Q$ . Where does the same attempt break for $\cos\frac ab$ with $b > 1$ general? (Identify the denominator that no longer clears.)

**Part V — Sharper and darker: the equality form, and a function that fools Taylor.**

18. (Taylor–Lagrange, equality form) Let $f$ be $n + 1$ times [differentiable](https://one-course.com/books/math/3/en/chapter/14-differentiation#def-b1-derivative-def) between $a$ and $x$. Define $g(t) =  f(x) - \sum_{k=0}^{n} \frac{f^{(k)}(t)}{k!}(x - t)^k -  A\,\frac{(x-t)^{n+1}}{(n+1)!}$ with the constant $A$ chosen so that $g(a) = 0$. Compute $g'$ (the sum telescopes), apply Rolle on $\intcc{a}{x}$, and conclude that there exists $c$ strictly between $a$ and $x$ with $$f(x) = \sum_{k=0}^{n} \frac{f^{(k)}(a)}{k!}(x-a)^k  + \frac{f^{(n+1)}(c)}{(n+1)!}\,(x-a)^{n+1} .$$
19. Dividend of the equality: for $x > 0$ show $$\eu^x > 1 + x + \frac{x^2}{2!} + \dots +  \frac{x^n}{n!}$$ (strictly, for every $n$), and locate where the inequality reverses for $x < 0$ according to the parity of $n$.
20. Benchmark the remainders on $\sin(0.5)$ at order $3$ : Young gives only $o(x^3)$ (no number); Lagrange gives $\abs{\sin 0.5 - (0.5 - \frac{0.5^3}{6})} \leq  \frac{0.5^5}{120} = 2.61\cdot10^{-4}$ ; the alternating estimate gives the same bound *plus* the sign information $\sin 0.5 > 0.5 - \frac{0.5^3}{6}$ . Compare with the true error $2.59\cdot10^{-4}$ : the bound is nearly attained. Which tool would you reach for, and when?
21. (The flat function) Let $f(x) = \eu^{-1/x^2}$ for $x  \neq 0$ , $f(0) = 0$ . Show $f$ is [continuous](https://one-course.com/books/math/3/en/chapter/13-limits-and-continuity#def-b1-continuity-continuous) at $0$ , that $f'(0) = 0$ , and more generally — by proving that every [derivative](https://one-course.com/books/math/3/en/chapter/14-differentiation#def-b1-derivative-def) has the form $f^{(k)}(x) =  P_k\bigl(\frac1x\bigr)\eu^{-1/x^2}$ for a [polynomial](https://one-course.com/books/math/3/en/chapter/8-polynomials#def-b1-poly-def) $P_k$ (induction) — that $f^{(k)}(0) = 0$ for all $k$ *([growth comparison](https://one-course.com/books/math/3/en/chapter/4-standard-functions#prop-b1-functions-powerrules) [Proposition 4.6](https://one-course.com/books/math/3/en/chapter/4-standard-functions#prop-b1-functions-powerrules))* . Conclude: all Taylor [polynomials](https://one-course.com/books/math/3/en/chapter/8-polynomials#def-b1-poly-def) of $f$ at $0$ vanish, yet $f(x) > 0$ for $x \neq 0$ : Taylor–Young holds at every order, and says nothing about $f$ away from $0$ . Expansions describe *germs* , not functions.

**Part VI — Synthesis.**

22. Run the trap once more, on $\eu^{-1} = \sum_k  \frac{(-1)^k}{k!}$ : multiply the strict alternating bracketing by $n!$ and conclude $\eu^{-1} \notin \Q$ , hence $\eu \notin \Q$ — the third proof of this fact in the volume. List the three ( [adjacent sequences](https://one-course.com/books/math/3/en/chapter/11-sequences#thm-b1-seq-adjacent) , [Exercise 11.9](https://one-course.com/books/math/3/en/chapter/11-sequences#exo-b1-seq-9) ; [integrals](https://one-course.com/books/math/3/en/chapter/15-integration-on-a-segment#thm-b1-integration-def) , [Problem 15.1](https://one-course.com/books/math/3/en/chapter/15-integration-on-a-segment#pb-b1-integration-1) ; alternating sums, here) and what each needed.
23. The fine print: monotonicity is not decorative. Let $b_k = \frac1k$ for odd $k$ and $b_k = \frac{1}{k^2}$ for even $k$ : the $b_k$ are positive and tend to $0$ , yet the partial sums of $\sum (-1)^k b_k$ diverge to $-\infty$ . Prove it *(split the partial sum into the even part, bounded via [Example 11.22](https://one-course.com/books/math/3/en/chapter/11-sequences#ex-b1-seq-basel), and the odd part, which dominates half the harmonic series, [Exercise 11.5](https://one-course.com/books/math/3/en/chapter/11-sequences#exo-b1-seq-5))* , and say exactly which step of question 1 used monotonicity.
24. Verify Euler’s simpler identity $\arctan\frac12 +  \arctan\frac13 = \frac\pi4$ via $(2 + \iu)(3 + \iu) =  5(1 + \iu)$ , estimate the terms needed for six decimals of $\pi$ by this route ( $n = 10$ suffices), and place it between Leibniz and Machin in the ranking of question 13.
25. Synthesis, one sentence each: (i) state the alternating estimate and its two outputs (bound and sign); (ii) why exact finite identities with explicit remainders beat limit [statements](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-statement) for certified numerics; (iii) inventory of the problem ( $\pi$ to $10^{-7}$ by hand, $\ln 2$ to ten decimals, four irrationality proofs, one equality theorem, one warning example); (iv) which of these threads [Chapter 17](https://one-course.com/books/math/3/en/chapter/17-numerical-series#ch-b1-series) will pick up (the alternating series test, absolute versus conditional convergence, and the rearrangement drama of its weekend problem).

**Solution of Problem 16.1.**

**1.** $S_{2n+1} - S_{2n-1} = a_{2n} - a_{2n+1} \geq 0$ and $S_{2n+2} - S_{2n} = a_{2n+2} - a_{2n+1} \leq 0$, while $S_{2n}
- S_{2n+1} = a_{2n+1} \to 0$: the sequences $(S_{2n+1})$, $(S_{2n})$ are adjacent, converging to a common $S$ ([Theorem 11.11](https://one-course.com/books/math/3/en/chapter/11-sequences#thm-b1-seq-adjacent)) with $S_{2n+1} \leq S \leq S_{2n}$. For even $n$: $S_{n+1} \leq S \leq S_n$ gives $-a_{n+1} \leq S -
S_n \leq 0$; for odd $n$: $0 \leq S - S_n \leq a_{n+1}$. In both cases $\abs{S - S_n} \leq a_{n+1}$ and $S - S_n$ has the sign of $(-1)^{n+1}$, the first omitted term. Strict decrease makes every displayed inequality strict, in particular $0 < \abs{S -
S_n} < a_{n+1}$.

**2.** $a_k = \frac{1}{k!}$ decreases strictly to $0$: question 1 applies. Taylor–Lagrange ([Theorem 16.7](#thm-b1-taylor-lagrange)) for $\exp$ between $-1$ and $0$: $\abs{\eu^{-1} - T_n} \leq \frac{1}{(n+1)!}$ (the [derivative](https://one-course.com/books/math/3/en/chapter/14-differentiation#def-b1-derivative-def) $\eu^t$ is $\leq 1$ there), so $T_n \to \eu^{-1}$, and the limit $S$ of question 1 *is* $\eu^{-1}$, with the strict bounds $0 < \abs{\eu^{-1} - T_n} < \frac{1}{(n+1)!}$.

**3.** Integrating the identity over $\intcc{0}{1}$: the left side is $\arctan 1 = \frac\pi4$ (fundamental theorem), the $k$-th term gives $\frac{(-1)^k}{2k+1}$, and

$$
\rho_n = (-1)^{n+1}\int_0^1 \frac{t^{2n+2}}{1+t^2}\dd t,
\qquad
\abs{\rho_n} \leq \int_0^1 t^{2n+2}\dd t = \frac{1}{2n+3} .
$$

**4.** The error on $\pi$ is $4\abs{\rho_n} \leq
\frac{4}{2n+3}$: below $10^{-6}$ requires $2n + 3 > 4\cdot10^6$, about two million terms. Meanwhile $4S_4 = 4\bigl(1 - \frac13 +
\frac15 - \frac17 + \frac19\bigr) = 4 \times 0.834921 =
3.339683$, almost $0.2$ away from $\pi$: five terms, not even one digit.

**5.** Integrating over $\intcc{0}{1}$: $\ln 2 =
\sum_{k=1}^{n} \frac{(-1)^{k-1}}{k} + (-1)^n R_n$ with $R_n =
\int_0^1 \frac{t^n}{1+t}\dd t$; from $\frac12 \leq \frac{1}{1+t}
\leq 1$: $\frac{1}{2(n+1)} \leq R_n \leq \frac{1}{n+1}$. The error is trapped between two multiples of $\frac1n$: six decimals cost about a million terms.

**6.** Integrating from $0$ to $x$: $\frac12\ln\frac{1 +
x}{1 - x} = \sum_{k=0}^{n} \frac{x^{2k+1}}{2k+1} + \int_0^x
\frac{t^{2n+2}}{1-t^2}\dd t$. At $x = \frac13$: $\frac{1 +
1/3}{1 - 1/3} = 2$, and on $\intcc{0}{\frac13}$, $\frac{1}{1 -
t^2} \leq \frac98$:

$$
\ln 2 = 2\sum_{k=0}^{n} \frac{(1/3)^{2k+1}}{2k+1} +
\tilde\rho_n,
\qquad
0 < \tilde\rho_n \leq \frac{9}{4}\cdot
\frac{(1/3)^{2n+3}}{2n+3} :
$$

each extra term [divides](https://one-course.com/books/math/3/en/chapter/6-integer-arithmetic#def-b1-arith-divides) the error by about $9$.

**7.** Induction: for $n = 1$: $1 - \frac12 = \frac12 = H_2
- H_1$. Step:

$$
H_{2n+2} - H_{n+1} = (H_{2n} - H_n) + \frac{1}{2n+1} +
\frac{1}{2n+2} - \frac{1}{n+1}
= (H_{2n} - H_n) + \frac{1}{2n+1} - \frac{1}{2n+2},
$$

which is exactly the increment of the alternating sum. And $H_{2n} - H_n = \sum_{k=1}^{n}\frac{1}{n+k}$ is the Riemann sum of [Example 15.21](https://one-course.com/books/math/3/en/chapter/15-integration-on-a-segment#ex-b1-integration-riemannexample), converging to $\ln
2$: the even partial sums of Way 1 *are* the [Riemann sums](https://one-course.com/books/math/3/en/chapter/15-integration-on-a-segment#thm-b1-integration-riemann) of Way 3.

**8.** $\sum_{k=1}^{6}\frac{(-1)^{k-1}}{k} = 0.61667$, error $0.0765$; Way 2 at $n = 5$ gives $0.6931471$ with error $\leq \frac94\cdot\frac{(1/3)^{13}}{13} = 1.1\cdot10^{-7}$. The reason: Way 1 evaluates the logarithm series at the [boundary](https://one-course.com/books/math/3/en/chapter/12-topology-of-the-real-line#def-b1-topology-closure) point $x = 1$, where the terms decay like $\frac1k$; Way 2 evaluates at $x = \frac13$, deep inside, where each term carries a fresh factor $\frac19$.

**9.** Ten decimals: want $\tilde\rho_n \leq
5\cdot10^{-11}$. At $n = 10$: $\frac94 \cdot
\frac{(1/3)^{23}}{23} = \frac94\cdot\frac{1.06\cdot10^{-11}}{23}
\approx 1.0\cdot10^{-12} < 5\cdot10^{-11}$: eleven terms suffice.

**10.** $(5+\iu)^2 = 24 + 10\iu$, then $(5+\iu)^4 = (24 +
10\iu)^2 = 476 + 480\iu$; and $2(1+\iu)(239+\iu) = 2(238 +
240\iu) = 476 + 480\iu$: equal. Arguments: $\arg(5 + \iu) =
\arctan\frac15$, so the left side has argument $4\arctan\frac15 \approx 0.79 \in \intoo{0}{\pi}$; the right side has argument $\frac\pi4 + \arctan\frac{1}{239} \in
\intoo{0}{\pi}$. Two equal complex numbers with arguments in the same [interval](https://one-course.com/books/math/3/en/chapter/10-real-numbers#prop-b1-reals-intervals) of length $< 2\pi$:

$$
4\arctan\frac15 = \frac\pi4 + \arctan\frac{1}{239} ,
$$

which is Machin’s formula.

**11.** Integrate $\frac{1}{1+t^2} = \sum_{k=0}^n (-1)^k
t^{2k} + \frac{(-1)^{n+1}t^{2n+2}}{1+t^2}$ from $0$ to $x$:

$$
\arctan x = \sum_{k=0}^{n}\frac{(-1)^k x^{2k+1}}{2k+1} +
r_n(x),
\qquad
\abs{r_n(x)} \leq \int_0^x t^{2n+2}\dd t =
\frac{x^{2n+3}}{2n+3} .
$$

**12.** Errors: $16\,\frac{(1/5)^{11}}{11} = 3.0\cdot
10^{-8}$ and $4\,\frac{(1/239)^5}{5} \approx 1\cdot10^{-12}$: total $< 5\cdot10^{-8}$. The displayed sum evaluates to $3.14159268\dots$, hence $\pi = 3.1415926\dots$ certified to $5\cdot10^{-8}$: seven decimals from six terms (five at $\frac15$, two at $\frac1{239}$ counting generously).

**13.** Leibniz: error $\sim \frac1n$, so each new digit multiplies the workload by ten. Dalzell ([Problem 15.1](https://one-course.com/books/math/3/en/chapter/15-integration-on-a-segment#pb-b1-integration-1), question 22): error $4^{1-5m}$, about three digits per step, each step a heavier [polynomial](https://one-course.com/books/math/3/en/chapter/8-polynomials#def-b1-poly-def). Machin: error ratio $\frac{1}{25}$ per term, about $1.4$ digits per term, each term one division. The moral: the remainder of a geometric-type expansion scales like $x^{2n}$, so making $x$ small buys digits at a *fixed* cost per term — Machin’s complex identity is precisely a machine for shrinking $x$.

**14.** $a_k = \frac{1}{(2k)!}$ decreases strictly to $0$; by question 1 and [Proposition 16.11](#prop-b1-taylor-standard) (Lagrange bound as in question 2), $\sum_{k \leq n}\frac{(-1)^k}{(2k)!}
\to \cos 1$ with the *strict* bracket $0 < \bigl|\cos 1 -
S'_n\bigr| < \frac{1}{(2n+2)!}$. Suppose $\cos 1 = \frac pq$ and take $2n \geq q$: then $(2n)!\,S'_n = \sum_{k\leq n} (-1)^k
\frac{(2n)!}{(2k)!} \in \Z$ and $(2n)!\,\frac pq \in \Z$, while

$$
0 < \Bigl|(2n)!\cos 1 - (2n)!S'_n\Bigr| <
\frac{(2n)!}{(2n+2)!} = \frac{1}{(2n+1)(2n+2)} < 1 :
$$

a nonzero integer of absolute value $< 1$. Contradiction: $\cos
1 \notin \Q$.

**15.** Identically with $a_k = \frac{1}{(2k+1)!}$, multiplying by $(2n+1)!$ with $2n + 1 \geq q$: $\sin 1 \notin
\Q$. Yet $\cos^2 1 + \sin^2 1 = 1 \in \Q$: products and sums of irrationals may be rational — irrationality passes through no algebraic operation for free.

**16.** Tail bound: for $m > n$,

$$
\sum_{k=n+1}^{m} \frac{1}{(2k)!}
\leq \frac{1}{(2n+2)!}\Bigl(1 + \frac12 + \frac14 +
\dots\Bigr) \leq \frac{2}{(2n+2)!} ,
$$

since each successive ratio is $\frac{1}{(2k+1)(2k+2)} \leq
\frac12$; the tail is positive (its first term is). So $0 <
\cosh 1 - \sum_{k\leq n}\frac{1}{(2k)!} < \frac{2}{(2n+2)!}$, and multiplying by $(2n)!$ with $2n \geq q$ traps a nonzero integer in $\intoo{0}{1}$ again: $\cosh 1 \notin \Q$.

**17.** $\cos\frac1m = \sum_k
\frac{(-1)^k}{m^{2k}(2k)!}$: the terms decrease strictly to zero, and $m^{2n}(2n)!\cdot\frac{1}{m^{2k}(2k)!} =
m^{2(n-k)}\frac{(2n)!}{(2k)!} \in \Z$ for $k \leq n$. If $\cos\frac1m = \frac pq$, multiply the strict bracket by $q\,m^{2n}(2n)!$: the error is bounded by $\frac{q}{m^2(2n+1)(2n+2)} < 1$ for large $n$: contradiction. For $\frac ab$ with $a \geq 2$: clearing denominators multiplies the tail by $b^{2n}(2n)!$, but the first omitted term is $\frac{a^{2n+2}}{b^{2n+2}(2n+2)!}$, and the product $\frac{a^{2n+2}}{b^2(2n+1)(2n+2)}$ *explodes*: the numerator $a^{2n+2}$ no longer clears, and the trap jams. (The result is still true — via Niven-style machinery, not this one.)

**18.** At $t = x$ every term of $g$ vanishes except $f(x)
- f(x) = 0$: $g(x) = 0$; $A$ is chosen so $g(a) = 0$. Differentiating, the sum telescopes:

$$
g'(t) = -\frac{f^{(n+1)}(t)}{n!}(x - t)^n +
A\,\frac{(x-t)^n}{n!}
= \frac{(x-t)^n}{n!}\bigl(A - f^{(n+1)}(t)\bigr) .
$$

Rolle on the segment from $a$ to $x$ gives $c$ strictly between with $g'(c) = 0$; since $(x - c)^n \neq 0$: $A =
f^{(n+1)}(c)$. Unfolding $g(a) = 0$ yields the Taylor equality with remainder $\frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}$.

**19.** For $x > 0$, the remainder is $\frac{\eu^{c}}
{(n+1)!}x^{n+1} > 0$: the exponential exceeds each of its Taylor [polynomials](https://one-course.com/books/math/3/en/chapter/8-polynomials#def-b1-poly-def), strictly, at every order. For $x < 0$ the sign of the remainder is that of $x^{n+1}$: $\eu^x$ is *above* the [polynomial](https://one-course.com/books/math/3/en/chapter/8-polynomials#def-b1-poly-def) for $n$ odd, *below* it for $n$ even — alternating sides, as the graphs of $1 + x$ and $1
+ x + \frac{x^2}{2}$ against $\eu^x$ already show.

**20.** True error: $\sin 0.5 - 0.4791667 =
2.59\cdot10^{-4}$, against the bound $\frac{0.5^5}{120} =
2.60\cdot10^{-4}$: nearly attained (the next term dominates the tail). Young: for limits and local analysis, where no constant is needed. Lagrange: for certified decimals. Alternating: when applicable, same bound *plus* the direction of the error — the best of the three, but the rarest.

**21.** [Continuity](https://one-course.com/books/math/3/en/chapter/13-limits-and-continuity#def-b1-continuity-continuous) at $0$: with $u = \frac{1}{x^2} \to
+\infty$, $f(x) = \eu^{-u} \to 0 = f(0)$. [Derivative](https://one-course.com/books/math/3/en/chapter/14-differentiation#def-b1-derivative-def) at $0$: $\bigl|\frac{f(h)}{h}\bigr| = \sqrt u\,\eu^{-u} \to 0$ ([Proposition 4.6](https://one-course.com/books/math/3/en/chapter/4-standard-functions#prop-b1-functions-powerrules)): $f'(0) = 0$. For $x \neq
0$, $f'(x) = \frac{2}{x^3}\eu^{-1/x^2}$: the form $P_1\bigl(\frac1x\bigr)\eu^{-1/x^2}$ with $P_1(X) = 2X^3$; inductively, differentiating $P_k(\frac1x)\eu^{-1/x^2}$ gives $P_{k+1}(X) = 2X^3 P_k(X) - X^2 P_k'(X)$, a [polynomial](https://one-course.com/books/math/3/en/chapter/8-polynomials#def-b1-poly-def). Then

$$
\frac{f^{(k)}(h) - 0}{h} = v\,P_k(v)\,\eu^{-v^2}
\Big|_{v = 1/h} \longrightarrow 0
$$

([polynomial](https://one-course.com/books/math/3/en/chapter/8-polynomials#def-b1-poly-def) against $\eu^{-v^2}$, [growth comparison](https://one-course.com/books/math/3/en/chapter/4-standard-functions#prop-b1-functions-powerrules) at $\pm\infty$): by induction $f^{(k)}(0) = 0$ for all $k$. All Taylor [polynomials](https://one-course.com/books/math/3/en/chapter/8-polynomials#def-b1-poly-def) of $f$ at $0$ vanish, yet $f > 0$ off $0$: Taylor–Young is exact at every order and blind beyond the germ. An expansion is local information only.

**22.** By question 2, $0 < \abs{\eu^{-1} - T_n} <
\frac{1}{(n+1)!}$, strictly. If $\eu^{-1} = \frac pq$, take $n
\geq q$ and multiply by $n!$: $n!\,T_n \in \Z$ and $n!\frac pq
\in \Z$, so a nonzero integer has absolute value $<
\frac{n!}{(n+1)!} = \frac{1}{n+1} < 1$: contradiction. Hence $\eu^{-1} \notin \Q$, and $\eu = \frac{1}{\eu^{-1}}$ is irrational too. The three proofs: [adjacent sequences](https://one-course.com/books/math/3/en/chapter/11-sequences#thm-b1-seq-adjacent) squeezing $q!\,\eu$ ([Exercise 11.9](https://one-course.com/books/math/3/en/chapter/11-sequences#exo-b1-seq-9)); the [integral](https://one-course.com/books/math/3/en/chapter/15-integration-on-a-segment#thm-b1-integration-def) recurrence $A_n =
\eu - nA_{n-1}$ ([Problem 15.1](https://one-course.com/books/math/3/en/chapter/15-integration-on-a-segment#pb-b1-integration-1)); the alternating bracket (here). One trap, three certificates of smallness.

**23.** Group the partial sums in pairs: $\sum_{k=1}^{2n}
(-1)^k b_k = E_n - O_n$ with $E_n = \sum_{j=1}^{n}
\frac{1}{4j^2}$, bounded (by the telescoping bound of [Example 11.22](https://one-course.com/books/math/3/en/chapter/11-sequences#ex-b1-seq-basel), $E_n \leq \frac12$), and $O_n =
\sum_{j=1}^{n}\frac{1}{2j-1} \geq \frac12 H_n \to +\infty$ ([Exercise 11.5](https://one-course.com/books/math/3/en/chapter/11-sequences#exo-b1-seq-5)): the partial sums tend to $-\infty$. Monotonicity was used in question 1 exactly where $S_{2n+1} -
S_{2n-1} = a_{2n} - a_{2n+1}$ needed a sign: without decrease, the even and odd [subsequences](https://one-course.com/books/math/3/en/chapter/11-sequences#def-b1-seq-subsequence) need not be monotone, and adjacency collapses.

**24.** $(2+\iu)(3+\iu) = 5 + 5\iu = 5(1+\iu)$; taking arguments (all in $\intoo{0}{\frac\pi2}$): $\arctan\frac12 +
\arctan\frac13 = \frac\pi4$. Series cost for six decimals: error $\leq 4\bigl(\frac{(1/2)^{2n+3}}{2n+3} +
\frac{(1/3)^{2n+3}}{2n+3}\bigr)$, which at $n = 10$ is $\approx 2\cdot10^{-8} < 5\cdot10^{-7}$: eleven terms. Ranking: better than Leibniz by an exponential margin, behind Machin (whose dominant point $\frac15$ is smaller than $\frac12$): roughly $0.6$ digits per term against Machin’s $1.4$.

**25.** (i) For decreasing $a_k \to 0$, the alternating partial sums converge with $\abs{S - S_n} \leq a_{n+1}$ and the error carries the sign of the first omitted term. (ii) A finite identity with explicit remainder can be *evaluated and bounded at a chosen point*, while a limit [statement](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-statement) only promises eventual closeness — certification needs the former. (iii) Extracted: $\pi$ to $5\cdot10^{-8}$ by Machin, $\ln 2$ to ten decimals by the $\frac13$-series, irrationality of $\cos 1$, $\sin 1$, $\cosh 1$, $\cos\frac1m$ and $\eu^{-1}$, the Taylor–Lagrange equality, and the flat-function warning. (iv) [Chapter 17](https://one-course.com/books/math/3/en/chapter/17-numerical-series#ch-b1-series) upgrades question 1 into the alternating series test, separates absolute from conditional convergence, and its weekend problem stages the rearrangement drama for which the alternating harmonic series of Way 1 is the star witness.