--- title: "Differentiation" book: "University Mathematics — Year 1" subject: math language: en chapter: 14 exercises: 12 source: https://one-course.com/books/math/3/en/chapter/14-differentiation --- # Chapter 14 — Differentiation [Derivatives](#def-b1-derivative-def) were computed throughout the High School volume; what was missing is the chain of theorems that turns computation into information about functions: Rolle’s theorem, the mean value theorem, and their consequences — monotonicity criteria, Lipschitz bounds, convexity. Everything in this chapter concerns functions defined on an [interval](https://one-course.com/books/math/3/en/chapter/10-real-numbers#prop-b1-reals-intervals) $I$. ## 14.1 The derivative **Definition 14.1.** $f \colon I \to \R$ is *differentiable at $x_0 \in I$* when the difference quotient $\frac{f(x) - f(x_0)}{x - x_0}$ has a (finite) limit as $x \to x_0$; the limit is written $f'(x_0)$. Equivalently: $$ f(x_0 + h) = f(x_0) + f'(x_0)\,h + h\,\varepsilon(h), \qquad \varepsilon(h) \xrightarrow[h \to 0]{} 0 , $$ the graph then admitting the tangent line $y = f(x_0) + f'(x_0)(x - x_0)$. Differentiability at $x_0$ implies [continuity](https://one-course.com/books/math/3/en/chapter/13-limits-and-continuity#def-b1-continuity-continuous) at $x_0$ (read the display). $f$ is differentiable on $I$ when it is at every point; $f$ is of class $C^1$ when moreover $f'$ is [continuous](https://one-course.com/books/math/3/en/chapter/13-limits-and-continuity#def-b1-continuity-continuous), and of class $C^k$ when $f$ can be differentiated $k$ times with $f^{(k)}$ [continuous](https://one-course.com/books/math/3/en/chapter/13-limits-and-continuity#def-b1-continuity-continuous). **Example 14.2.** The converse of “[differentiable](#def-b1-derivative-def) $\Rightarrow$ [continuous](https://one-course.com/books/math/3/en/chapter/13-limits-and-continuity#def-b1-continuity-continuous)” fails: $\abs{\,\cdot\,}$ at $0$. More surprisingly, [differentiable](#def-b1-derivative-def) does not imply $C^1$: the function $f(x) = x^2 \sin\frac 1x$ ($f(0) = 0$) is [differentiable](#def-b1-derivative-def) everywhere, with $f'(0) = 0$, but $f'(x) = 2x \sin\frac1x - \cos\frac 1x$ has no limit at $0$ ([Exercise 14.2](#exo-b1-derivative-2)). **Example 14.3 (Differentiable at exactly one point).** Let $f(x) = x^2$ for $x \in \Q$ and $f(x) = 0$ for $x \notin \Q$. At $0$: $\bigl|\frac{f(h) - 0}{h}\bigr| \leq \abs h \to 0$, so $f$ is [differentiable](#def-b1-derivative-def) at $0$ with $f'(0) = 0$. At any $x_0 \neq 0$, $f$ is not even [continuous](https://one-course.com/books/math/3/en/chapter/13-limits-and-continuity#def-b1-continuity-continuous): rational and irrational sequences converging to $x_0$ send $f$ to $x_0^2 \neq 0$ and to $0$ respectively ([density](https://one-course.com/books/math/3/en/chapter/12-topology-of-the-real-line#def-b1-topology-dense), [Theorem 10.14](https://one-course.com/books/math/3/en/chapter/10-real-numbers#thm-b1-reals-density)). So differentiability is a genuinely *pointwise* notion: it can hold at one point of $\R$ and nowhere else. The moral for practice: [statements](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-statement) like the monotonicity criterion or Rolle require the [derivative](#def-b1-derivative-def) *on an [interval](https://one-course.com/books/math/3/en/chapter/10-real-numbers#prop-b1-reals-intervals)* — possessing $f'(x_0)$ at isolated points, however many, supports no global conclusion whatsoever. **Theorem 14.4 (Operations).** If $f, g$ are [differentiable](#def-b1-derivative-def) at $x_0$ (and where the formulas make sense): $$ (f + g)' = f' + g', \qquad (fg)' = f'g + fg', \qquad \Bigl(\frac fg\Bigr)' = \frac{f'g - fg'}{g^2}, $$ and if $g$ is [differentiable](#def-b1-derivative-def) at $f(x_0)$: $\;(g \circ f)'(x_0) = g'\bigl(f(x_0)\bigr)\, f'(x_0)$ (chain rule). **Proof.** Sum: immediate. Product: write $$ f(x)g(x) - f(x_0)g(x_0) = \bigl(f(x) - f(x_0)\bigr) g(x) + f(x_0)\bigl(g(x) - g(x_0)\bigr), $$ divide by $x - x_0$ and let $x \to x_0$ ($g$ is [continuous](https://one-course.com/books/math/3/en/chapter/13-limits-and-continuity#def-b1-continuity-continuous) at $x_0$). Quotient: treat $\frac 1g$ via $\frac{1/g(x) - 1/g(x_0)}{x - x_0} = \frac{-1}{g(x)g(x_0)}\cdot\frac{g(x) - g(x_0)}{x - x_0}$, then apply the product rule. Chain rule: with $y_0 = f(x_0)$, define $\theta(y) = \frac{g(y) - g(y_0)}{y - y_0}$ for $y \neq y_0$ and $\theta(y_0) = g'(y_0)$: $\theta$ is [continuous](https://one-course.com/books/math/3/en/chapter/13-limits-and-continuity#def-b1-continuity-continuous) at $y_0$, and for $x \neq x_0$, $$ \frac{g(f(x)) - g(f(x_0))}{x - x_0} = \theta\bigl(f(x)\bigr)\cdot \frac{f(x) - f(x_0)}{x - x_0} \longrightarrow g'(y_0)\, f'(x_0), $$ the first factor by composition of limits (this device handles the case $f(x) = f(x_0)$ cleanly, where the naive “multiply and divide by $f(x) - f(x_0)$” breaks). ∎ **Theorem 14.5 (Derivative of an inverse function).** Let $f$ be [continuous](https://one-course.com/books/math/3/en/chapter/13-limits-and-continuity#def-b1-continuity-continuous) and strictly monotonic on $I$, [differentiable](#def-b1-derivative-def) at $x_0$ with $f'(x_0) \neq 0$. Then $f^{-1}$ ([Theorem 13.16](https://one-course.com/books/math/3/en/chapter/13-limits-and-continuity#thm-b1-continuity-bijection)) is [differentiable](#def-b1-derivative-def) at $y_0 = f(x_0)$, with $$ (f^{-1})'(y_0) = \frac{1}{f'(x_0)} = \frac{1}{f'\bigl(f^{-1}(y_0)\bigr)} . $$ If $f'(x_0) = 0$, the inverse has a vertical tangent at $y_0$. **Proof.** For $y \to y_0$, set $x = f^{-1}(y)$: [continuity](https://one-course.com/books/math/3/en/chapter/13-limits-and-continuity#def-b1-continuity-continuous) of $f^{-1}$ gives $x \to x_0$, and $$ \frac{f^{-1}(y) - f^{-1}(y_0)}{y - y_0} = \frac{x - x_0}{f(x) - f(x_0)} = \frac{1}{\dfrac{f(x) - f(x_0)}{x - x_0}} \longrightarrow \frac{1}{f'(x_0)} . $$ Vertical-tangent claim: if $f'(x_0) = 0$, the displayed quotient is the reciprocal of a quantity that tends to $0$ while keeping one constant sign (for $f$ strictly increasing, $\frac{f(x) - f(x_0)}{x - x_0} > 0$ for all $x \neq x_0$): the difference quotient of $f^{-1}$ therefore tends to $+\infty$ (to $-\infty$ for $f$ decreasing). The inverse remains [continuous](https://one-course.com/books/math/3/en/chapter/13-limits-and-continuity#def-b1-continuity-continuous) but is not [differentiable](#def-b1-derivative-def) at $y_0$ — its graph, the reflection of $f$’s across the diagonal, stands vertical exactly where $f$’s ran horizontal, as $x^{1/3}$ at $0$ illustrates against $x^3$. ∎ **Example 14.6 (Inverse derivatives, twice).** The theorem recomputes the classical [derivatives](#def-b1-derivative-def) with no limit work. For $\ln = \exp^{-1}$: at $y = \eu^x$, $$ (\ln)'(y) = \frac{1}{\exp'(x)} = \frac{1}{\eu^{x}} = \frac1y , $$ valid for every $y > 0$ since $\exp' = \exp$ never vanishes. For $\arctan = \tan^{-1}$: at $y = \tan x$, $$ (\arctan)'(y) = \frac{1}{1 + \tan^2 x} = \frac{1}{1 + y^2} , $$ using $\tan' = 1 + \tan^2 > 0$. The closing insight: the formula converts knowledge about a function into knowledge about its inverse at the price of one substitution — and the substitution ($x = \ln y$, $x = \arctan y$) is exactly the [statement](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-statement) that the two variables live on opposite sides of the bijection. ## 14.2 Rolle and the mean value theorem **Proposition 14.7 (Interior extremum).** If $f$ is [differentiable](#def-b1-derivative-def) at an *[interior](https://one-course.com/books/math/3/en/chapter/12-topology-of-the-real-line#def-b1-topology-closure)* point $x_0$ of $I$ and has a local extremum there, then $f'(x_0) = 0$. **Proof.** Say a local maximum: there is $r > 0$ with $f(x) \leq f(x_0)$ for $\abs{x - x_0} \leq r$, and interiority guarantees both sides of $x_0$ are available within $I$. For $0 < h \leq r$ the quotient $\frac{f(x_0 + h) - f(x_0)}{h}$ has numerator $\leq 0$ and denominator $> 0$: it is $\leq 0$, and its limit $f'(x_0)$ inherits $\leq 0$ (wide inequalities pass to limits, [Theorem 11.7](https://one-course.com/books/math/3/en/chapter/11-sequences#thm-b1-seq-order)); for $-r \leq h < 0$ the quotient is $\geq 0$, giving $f'(x_0) \geq 0$. Hence $f'(x_0) = 0$. (At an endpoint, only one sign is available: the conclusion fails there — think of $x$ on $\intcc{0}{1}$, maximal at $1$ with [derivative](#def-b1-derivative-def) $1$.) ∎ **Theorem 14.8 (Rolle).** Let $f$ be [continuous](https://one-course.com/books/math/3/en/chapter/13-limits-and-continuity#def-b1-continuity-continuous) on $\intcc{a}{b}$, [differentiable](#def-b1-derivative-def) on $\intoo{a}{b}$, with $f(a) = f(b)$. Then $f'(c) = 0$ for some $c \in \intoo{a}{b}$. **Proof.** By the extreme value theorem ([Theorem 13.13](https://one-course.com/books/math/3/en/chapter/13-limits-and-continuity#thm-b1-continuity-evt)), $f$ attains its maximum and minimum on $\intcc{a}{b}$. If both are attained at endpoints, then (since $f(a) = f(b)$) max $=$ min and $f$ is constant: any [interior](https://one-course.com/books/math/3/en/chapter/12-topology-of-the-real-line#def-b1-topology-closure) $c$ works. Otherwise an extremum is attained at an [interior](https://one-course.com/books/math/3/en/chapter/12-topology-of-the-real-line#def-b1-topology-closure) point $c$, and [Proposition 14.7](#prop-b1-derivative-fermat) gives $f'(c) = 0$. ∎ **Theorem 14.9 (Mean value theorem).** Let $f$ be [continuous](https://one-course.com/books/math/3/en/chapter/13-limits-and-continuity#def-b1-continuity-continuous) on $\intcc{a}{b}$, [differentiable](#def-b1-derivative-def) on $\intoo{a}{b}$. There exists $c \in \intoo{a}{b}$ with $$ f(b) - f(a) = f'(c)\,(b - a) . $$ *Mean value inequality:* if moreover $m \leq f' \leq M$ on $\intoo{a}{b}$, then $m(b-a) \leq f(b) - f(a) \leq M(b-a)$; in particular $\abs{f'} \leq K$ implies that $f$ is $K$-Lipschitz. **Proof.** Apply Rolle to $g(x) = f(x) - \frac{f(b) - f(a)}{b - a}(x - a)$: $g$ is [continuous](https://one-course.com/books/math/3/en/chapter/13-limits-and-continuity#def-b1-continuity-continuous) on $\intcc{a}{b}$, [differentiable](#def-b1-derivative-def) inside, and $g(a) = f(a) = g(b)$. At the point $c$ where $g'(c) = 0$: $f'(c) = \frac{f(b)-f(a)}{b-a}$. The inequality follows by bounding $f'(c)$; the Lipschitz [statement](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-statement) applies it to every pair of points. ∎ ![The mean value theorem: some tangent (dashed) is parallel to the chord (gray). Its abscissa c is where Rolle’s theorem, applied to the function minus its chord, finds a critical point.](https://one-course.com/images/onecourse/chapters/math-3/b1-derivative/fig-a2d4f0542a62.svg) *The mean value theorem: some tangent (dashed) is parallel to the chord (gray). Its abscissa $c$ is where Rolle’s theorem, applied to the function minus its chord, finds a critical point.* **Example 14.10 (Newton’s method is Heron’s).** Newton’s method for solving $f(x) = 0$ replaces the curve by its tangent at the current guess $x_n$ and takes the tangent’s root as the next guess: $$ 0 = f(x_n) + f'(x_n)(x_{n+1} - x_n) \quad\Longrightarrow\quad x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} . $$ Run it on $f(x) = x^2 - 2$: $$ x_{n+1} = x_n - \frac{x_n^2 - 2}{2x_n} = \frac{x_n}{2} + \frac{1}{x_n} = \frac12\Bigl(x_n + \frac{2}{x_n}\Bigr) : $$ exactly Heron’s iteration ([Example 11.24](https://one-course.com/books/math/3/en/chapter/11-sequences#ex-b1-seq-heron)), two millennia early. The quadratic speed observed there is now explained by the tangent picture: near a simple root, curve and tangent differ by a second-order error, so each step roughly squares the error — the general [statement](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-statement) follows from the Taylor bounds of [Chapter 16](https://one-course.com/books/math/3/en/chapter/16-taylor-formulas-and-asymptotic-expansions#ch-b1-taylor). The closing insight: where dichotomy ([Example 13.12](https://one-course.com/books/math/3/en/chapter/13-limits-and-continuity#ex-b1-continuity-dichotomy)) uses only [continuity](https://one-course.com/books/math/3/en/chapter/13-limits-and-continuity#def-b1-continuity-continuous) and gains one bit per step, Newton spends a [derivative](#def-b1-derivative-def) to double the number of correct digits per step. **Example 14.11 (The mean value theorem as an estimator).** How large is $\sqrt{101}$? Apply the theorem to $f(t) = \sqrt t$ on $\intcc{100}{101}$: for some $c \in \intoo{100}{101}$, $$ \sqrt{101} - 10 = \frac{1}{2\sqrt c}, \qquad\text{so}\qquad \frac{1}{2\sqrt{101}} < \sqrt{101} - 10 < \frac{1}{20} = 0.05 , $$ and since $\sqrt{101} < 10.05$, the left bound exceeds $\frac{1}{20.1} > 0.0497$: thus $10.0497 < \sqrt{101} < 10.05$ (true value $10.049875\dots$) — three correct decimals from one [derivative](#def-b1-derivative-def) evaluation. Likewise $\abs{\sin a - \sin b} \leq \abs{a - b}$ (bound $\abs{\cos}\leq 1$): the Lipschitz estimates used since [Chapter 11](https://one-course.com/books/math/3/en/chapter/11-sequences#ch-b1-seq) are all this theorem. The closing insight: the mean value theorem is a zeroth-order Taylor formula — it trades one unknown point $c$ for a hard inequality, and [Chapter 16](https://one-course.com/books/math/3/en/chapter/16-taylor-formulas-and-asymptotic-expansions#ch-b1-taylor) will iterate exactly this trade. **Corollary 14.12 (Monotonicity criterion).** Let $f$ be [continuous](https://one-course.com/books/math/3/en/chapter/13-limits-and-continuity#def-b1-continuity-continuous) on $I$, [differentiable](#def-b1-derivative-def) on the [interior](https://one-course.com/books/math/3/en/chapter/12-topology-of-the-real-line#def-b1-topology-closure). 1. $f' \geq 0$ on the [interior](https://one-course.com/books/math/3/en/chapter/12-topology-of-the-real-line#def-b1-topology-closure) $\iff$ $f$ is increasing; $f' = 0$ $\iff$ $f$ constant. 2. If $f' > 0$ except at finitely many points where it vanishes, $f$ is *strictly* increasing. **Proof.** If $f' \geq 0$: for $x < y$ in $I$, the mean value theorem on $\intcc{x}{y}$ gives $f(y) - f(x) = f'(c)(y - x) \geq 0$. Conversely, difference quotients of an increasing function are $\geq 0$, so their limits are too. The constant case: apply the previous to $f'$ and $-f' \geq 0$. Strict version: $f$ is increasing; equality $f(x) = f(y)$ for $x < y$ would freeze $f$ on $\intcc{x}{y}$, forcing $f' = 0$ there — infinitely many points. ∎ **Example 14.13 (Equal derivatives, unequal functions).** On $\R^* = \intoo{-\infty}{0} \cup \intoo{0}{+\infty}$, both $f(x) = \ln\abs x$ and $g(x) = \ln\abs x + \mathbf{1}_{x>0}$ (add $1$ on the right half-line only) satisfy $f' = g' = \frac1x$. They do not differ by a constant: the criterion “$f' = 0 \implies f$ constant” is an *[interval](https://one-course.com/books/math/3/en/chapter/10-real-numbers#prop-b1-reals-intervals)* [statement](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-statement) — its proof runs the mean value theorem between two points, which requires the whole segment joining them to lie in the domain. On each half-line separately, the primitives of $\frac1x$ are $\ln\abs x + c$ with one constant per half-line, two independent constants in total. [Chapter 15](https://one-course.com/books/math/3/en/chapter/15-integration-on-a-segment#ch-b1-integration) inherits this fine print: “the” primitive of a function is well defined up to a constant *on each [interval](https://one-course.com/books/math/3/en/chapter/10-real-numbers#prop-b1-reals-intervals) of its domain*, and antiderivative tables silently assume connectivity. **Example 14.14 (Strictness for free).** $x \mapsto x^3$ is *strictly* increasing on $\R$ even though its [derivative](#def-b1-derivative-def) vanishes at $0$: the criterion’s clause “$f' > 0$ except at finitely many points” is exactly designed for such flat points. By contrast, $f' \geq 0$ alone only gives increase in the wide sense (a constant function qualifies), and a [derivative](#def-b1-derivative-def) vanishing on a whole subinterval does freeze the function there. The practical rule: to claim strict monotonicity, list the zeros of $f'$; finitely many (or more generally, none on any subinterval) is harmless, an [interval](https://one-course.com/books/math/3/en/chapter/10-real-numbers#prop-b1-reals-intervals) of them is fatal. **Example 14.15 (A complete variation study).** Study $f(x) = x^3 - 3x + 1$ on $\R$. [Derivative](#def-b1-derivative-def): $f'(x) = 3(x^2 - 1)$, positive on $\intoo{-\infty}{-1}$, negative on $\intoo{-1}{1}$, positive on $\intoo{1}{+\infty}$: by the monotonicity criterion, $f$ increases, then decreases, then increases, with a local maximum $f(-1) = 3$ and a local minimum $f(1) = -1$. Limits: $\mp\infty$ at $\mp\infty$. Consequences, read off the variation table with the intermediate value theorem on each monotone branch: $f$ vanishes exactly once in each of $$ \intoo{-\infty}{-1}, \qquad \intoo{-1}{1}, \qquad \intoo{1}{+\infty} $$ (the values at the junctions have opposite signs: $3 > 0 > -1$), so the equation $x^3 - 3x + 1 = 0$ has exactly three real roots; numerically they sit near $-1.88$, $0.35$, $1.53$. The closing insight: a variation table is a *proof device*, not a sketch — monotone branch plus sign change equals exactly one root, and the table enumerates the branches exhaustively. **Theorem 14.16 (Leibniz formula).** If $f, g$ are $n$ times [differentiable](#def-b1-derivative-def), so is $fg$, and $$ (fg)^{(n)} = \sum_{k=0}^{n} \binom nk f^{(k)}\, g^{(n-k)} . $$ **Proof.** Induction on $n$, exactly parallel to the binomial theorem. The case $n = 1$ is the product rule. Assuming the formula at rank $n$, differentiate once more: $$ (fg)^{(n+1)} = \sum_{k=0}^{n} \binom nk \Bigl( f^{(k+1)} g^{(n-k)} + f^{(k)} g^{(n-k+1)} \Bigr), $$ then reindex the first sum with $j = k + 1$ and collect the coefficient of $f^{(j)} g^{(n+1-j)}$: it is $\binom{n}{j-1} + \binom nj = \binom{n+1}{j}$ by Pascal’s rule ([Proposition 2.15](https://one-course.com/books/math/3/en/chapter/2-counting#prop-b1-counting-identities)), the [boundary](https://one-course.com/books/math/3/en/chapter/12-topology-of-the-real-line#def-b1-topology-closure) terms $j = 0$ and $j = n + 1$ carrying $\binom{n+1}{0} = \binom{n+1}{n+1} = 1$ as they should. ∎ **Example 14.17 (Leibniz in action).** Compute $\bigl(x^2 \eu^x\bigr)^{(n)}$ for $n \geq 2$. Take $f = x^2$, whose [derivatives](#def-b1-derivative-def) die quickly ($f' = 2x$, $f'' = 2$, $f^{(k)} = 0$ for $k \geq 3$), and $g = \eu^x$: only three terms of the Leibniz sum survive, $$ \bigl(x^2\eu^x\bigr)^{(n)} = \binom n0 x^2 \eu^x + \binom n1 (2x)\,\eu^x + \binom n2\, 2\,\eu^x = \eu^x\bigl(x^2 + 2nx + n(n-1)\bigr). $$ Sanity check at $n = 1$: $\eu^x(x^2 + 2x)$, which is indeed $(x^2\eu^x)'$. The closing insight: use Leibniz when one factor is a [polynomial](https://one-course.com/books/math/3/en/chapter/8-polynomials#def-b1-poly-def) — the sum then has only $\deg + 1$ terms, and the formula is a closed form, not an abstract identity. (For two infinitely-lively factors like $\eu^x\sin x$, complex exponentials from [Chapter 3](https://one-course.com/books/math/3/en/chapter/3-complex-numbers#ch-b1-complex) are the better tool.) ## 14.3 Convexity **Definition 14.18.** $f \colon I \to \R$ is *convex* when every chord lies above the graph: $$ \forall x, y \in I,\ \forall t \in \intcc{0}{1}, \quad f\bigl(tx + (1-t)y\bigr) \leq t f(x) + (1-t) f(y). $$ ($f$ is *concave* when $-f$ is convex.) **Theorem 14.19 (Differential characterizations).** Let $f$ be [differentiable](#def-b1-derivative-def) on $I$. The following are equivalent: 1. $f$ is [convex](#def-b1-derivative-convex) ; 2. $f'$ is increasing on $I$ ; 3. the graph lies above every tangent: $f(y) \geq f(x) + f'(x)(y - x)$ for all $x, y \in I$ . If $f$ is twice [differentiable](#def-b1-derivative-def): $f$ [convex](#def-b1-derivative-convex) $\iff f'' \geq 0$. **Proof.** (1 $\Rightarrow$ 3) Convexity written as $\frac{f(x + t(y-x)) - f(x)}{t} \leq f(y) - f(x)$ for $t \in \intoc{0}{1}$; let $t \to 0^+$: $f'(x)(y - x) \leq f(y) - f(x)$. (3 $\Rightarrow$ 2) For $x < y$, the two tangent inequalities at $x$ and at $y$ give $f'(x)(y-x) \leq f(y) - f(x) \leq f'(y)(y - x)$, hence $f'(x) \leq f'(y)$. (2 $\Rightarrow$ 1) Fix $x < y$ and $t \in \intoo{0}{1}$, and let $z = tx + (1-t)y \in \intoo{x}{y}$. By the mean value theorem on $\intcc{x}{z}$ and $\intcc{z}{y}$: there are $c_1 < z < c_2$ with $$ \frac{f(z) - f(x)}{z - x} = f'(c_1) \leq f'(c_2) = \frac{f(y) - f(z)}{y - z} , $$ and clearing denominators ($z - x = (1-t)(y-x)$, $y - z = t(y-x)$) rearranges exactly into the convexity inequality. Twice [differentiable](#def-b1-derivative-def) case: $f'' \geq 0 \iff f'$ increasing ([Corollary 14.12](#cor-b1-derivative-monotone)). ∎ ![Convexity, twice: every chord (gray) lies above the graph, and the graph lies above every tangent (dashed).](https://one-course.com/images/onecourse/chapters/math-3/b1-derivative/fig-d730906df39f.svg) *Convexity, twice: every chord (gray) lies above the graph, and the graph lies above every tangent (dashed).* **Example 14.20 (Classical convexity inequalities).** $\exp$ is [convex](#def-b1-derivative-convex) ($\exp'' = \exp > 0$): its tangent at $0$ gives $\eu^x \geq 1 + x$ for all $x$. $\ln$ is concave: its tangent at $1$ gives $\ln x \leq x - 1$; its chords give, for $0 < a \leq b$, the inequality between geometric and arithmetic means: taking $t = \frac12$ in concavity, $$ \ln\frac{a + b}{2} \geq \frac{\ln a + \ln b}{2} = \ln\sqrt{ab}, \qquad\text{so}\qquad \sqrt{ab} \leq \frac{a+b}{2} . $$ The general arithmetic–geometric inequality is [Exercise 14.9](#exo-b1-derivative-9). **Example 14.21 (A convexity inequality from scratch).** The function $f(t) = t\ln t$ is [convex](#def-b1-derivative-convex) on $\intoo{0}{+\infty}$: $f''(t) = \frac1t > 0$. Its midpoint inequality, multiplied by $2$, reads: for all $a, b > 0$, $$ a\ln a + b\ln b \;\geq\; (a + b)\,\ln\frac{a + b}{2} , $$ with equality iff $a = b$ (strict convexity). Test drive: $a = 1$, $b = 3$ gives $3\ln 3 = 3.296$ against $4\ln 2 = 2.773$. This innocuous inequality is the two-point case of the *entropy* comparison that reappears with Jensen’s inequality ([Exercise 14.9](#exo-b1-derivative-9)) and in the information-theoretic asymptotics of the Year 3 volume. The closing insight: to manufacture an inequality, find a function whose second [derivative](#def-b1-derivative-def) has a sign and write down what convexity says — the differential characterization turns one sign check into infinitely many inequalities. **Remark 14.22 (Common pitfalls with derivatives).** (i) *A positive [derivative](#def-b1-derivative-def) at one point does not give monotonicity near it*: $f(x) = \frac x2 + x^2\sin\frac1x$ (with $f(0) = 0$) has $f'(0) = \frac12 > 0$, yet $$ f'(x) = \frac12 + 2x\sin\frac1x - \cos\frac1x $$ equals $-\frac12$ at each $x_n = \frac{1}{2\pi n}$: every [neighborhood](https://one-course.com/books/math/3/en/chapter/12-topology-of-the-real-line#def-b1-topology-open) of $0$ contains descents. Monotonicity needs $f' \geq 0$ *on an [interval](https://one-course.com/books/math/3/en/chapter/10-real-numbers#prop-b1-reals-intervals)* ([Corollary 14.12](#cor-b1-derivative-monotone)); the pointwise sign only controls the crossing of the tangent line. (ii) *Rolle’s three hypotheses are all active*: $\abs x$ on $\intcc{-1}{1}$ (no [interior](https://one-course.com/books/math/3/en/chapter/12-topology-of-the-real-line#def-b1-topology-closure) differentiability), $x$ on $\intcc{0}{1}$ (ends not equal), and $x - \lfloor x\rfloor$ on $\intcc{0}{1}$ ([continuity](https://one-course.com/books/math/3/en/chapter/13-limits-and-continuity#def-b1-continuity-continuous) fails at $1$) each break exactly one hypothesis and the conclusion. (iii) *[Derivatives](#def-b1-derivative-def) may be discontinuous, but not arbitrarily*: $f'$ can oscillate ([Example 14.2](#ex-b1-derivative-notc1)) yet always satisfies the intermediate value property (Darboux, [Exercise 14.10](#exo-b1-derivative-10)): a [derivative](#def-b1-derivative-def) never jumps — if you compute a one-sided “[derivative](#def-b1-derivative-def) limit” with a jump, you have differentiated a [non-differentiable](#def-b1-derivative-def) function. (iv) *The inverse formula needs $f' \neq 0$*: $x \mapsto x^3$ is a smooth strictly increasing bijection whose inverse $x^{1/3}$ has a vertical tangent at $0$ — differentiability of the inverse is lost exactly where $f'$ vanishes ([Theorem 14.5](#thm-b1-derivative-inverse)). **Remark 14.23 (Where the mean value theorem works next).** Nearly every quantitative [statement](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-statement) of the next chapters is this chapter’s mean value theorem in costume: the fundamental theorem of calculus ([Chapter 15](https://one-course.com/books/math/3/en/chapter/15-integration-on-a-segment#ch-b1-integration)) differentiates the area function and concludes with the monotonicity criterion; the Taylor–Lagrange formula ([Chapter 16](https://one-course.com/books/math/3/en/chapter/16-taylor-formulas-and-asymptotic-expansions#ch-b1-taylor)) is the mean value theorem iterated $n$ times; the error analysis of Newton’s method and of fixed-point iterations ([Exercise 14.11](#exo-b1-derivative-11)) is the Lipschitz form; and this chapter’s weekend problem ([Problem 14.1](#pb-b1-derivative-1)) turns the same Lipschitz bound into number theory — a repulsion inequality between [algebraic numbers](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#pb-b1-logic-1) and rationals, yielding the first [transcendental number](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#pb-b1-logic-1) in history. In the Year 2 volume, the mean value *inequality* survives in several variables when the equality does not. **Example 14.24 (Young’s inequality from concavity).** Let $p, q > 1$ with $\frac1p + \frac1q = 1$. For all $a, b > 0$: $$ ab \;\leq\; \frac{a^p}{p} + \frac{b^q}{q} . $$ Proof by one application of the concavity of $\ln$ with weights $\frac1p, \frac1q$ (the two-point Jensen inequality, as in [Exercise 14.9](#exo-b1-derivative-9)): $$ \ln\Bigl(\frac{a^p}{p} + \frac{b^q}{q}\Bigr) \;\geq\; \frac1p \ln(a^p) + \frac1q \ln(b^q) = \ln a + \ln b = \ln(ab), $$ and $\ln$ increasing converts the inequality of logarithms into the claim; equality iff $a^p = b^q$ (strict concavity). The case $p = q = 2$ is the arithmetic-geometric inequality $ab \leq \frac{a^2 + b^2}{2}$ in disguise. The closing insight: Young’s inequality is the algebraic seed of the Hölder and Minkowski inequalities of the Year 2 volume — one concavity [statement](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-statement) about $\ln$, harvested for norms. **Remark 14.25 (Perspectives inside this volume).** The [derivative](#def-b1-derivative-def) acquires three new lives before the volume ends. In [Chapter 16](https://one-course.com/books/math/3/en/chapter/16-taylor-formulas-and-asymptotic-expansions#ch-b1-taylor) it iterates: $n$ [derivatives](#def-b1-derivative-def) at a point compress into one [polynomial](https://one-course.com/books/math/3/en/chapter/8-polynomials#def-b1-poly-def) plus a controlled error, and the mean value theorem becomes the Lagrange remainder. In [Chapter 24](https://one-course.com/books/math/3/en/chapter/24-plane-curves#ch-b1-curves), differentiation turns geometric: for a parametrized curve $t \mapsto (x(t), y(t))$, the pair $(x'(t), y'(t))$ is a velocity *vector*, tangency becomes collinearity, and critical points become cusps to classify. In [Chapter 25](https://one-course.com/books/math/3/en/chapter/25-functions-of-two-variables#ch-b1-multivar), one variable is frozen at a time: partial [derivatives](#def-b1-derivative-def) repeat this chapter twice over, and the tangent line grows into a tangent plane. All three chapters inherit the same grammar — local linear approximation plus an error term — first spoken here. ## 14.4 Exercises **Exercise 14.1 ★.** Differentiate (specifying the domains): $x^x$; $\;\ln\bigl(x + \sqrt{x^2+1}\bigr)$; $\;\arctan\frac{1}{x}$; $\;\sqrt{1 + \eu^{2x}}$. **Solution of Exercise 14.1.** $x^x = \eu^{x\ln x}$ on $\intoo{0}{+\infty}$: [derivative](#def-b1-derivative-def) $(\ln x + 1)\,x^x$. $\ln(x + \sqrt{x^2+1})$ on $\R$ (the argument is always $> 0$): [derivative](#def-b1-derivative-def) $\frac{1}{\sqrt{x^2+1}}$ (computed in [Proposition 4.21](https://one-course.com/books/math/3/en/chapter/4-standard-functions#prop-b1-functions-invhyp) — it is $\operatorname{arsinh}$). $\arctan\frac1x$ on $\R^*$: [derivative](#def-b1-derivative-def) $\frac{-1/x^2}{1 + 1/x^2} = \frac{-1}{1 + x^2}$ (consistent with [Proposition 4.12](https://one-course.com/books/math/3/en/chapter/4-standard-functions#prop-b1-functions-arcidentities) (2): the function is $\pm\frac\pi2 - \arctan x$ on each half-line). $\sqrt{1 + \eu^{2x}}$ on $\R$: [derivative](#def-b1-derivative-def) $\frac{\eu^{2x}}{\sqrt{1 + \eu^{2x}}}$. **Exercise 14.2 ★.** Complete [Example 14.2](#ex-b1-derivative-notc1): prove that $f(x) = x^2 \sin\frac1x$, $f(0) = 0$, is [differentiable](#def-b1-derivative-def) at $0$ with $f'(0) = 0$, and that $f'$ has no limit at $0$. **Solution of Exercise 14.2.** At $0$: $\bigl|\frac{f(h) - 0}{h}\bigr| = \abs{h \sin\frac1h} \leq \abs h \to 0$, so $f'(0) = 0$. For $x \neq 0$, the usual rules give $f'(x) = 2x\sin\frac1x - \cos\frac1x$. Along $x_n = \frac{1}{2\pi n}$: $f'(x_n) = 0 - 1 \to -1$; along $y_n = \frac{1}{(2n+1)\pi}$: $f'(y_n) = 0 + 1 \to 1$. Two sequences tending to $0$ with different limits of $f'$: no limit ([Theorem 13.3](https://one-course.com/books/math/3/en/chapter/13-limits-and-continuity#thm-b1-continuity-seqchar)), so $f'$ is not [continuous](https://one-course.com/books/math/3/en/chapter/13-limits-and-continuity#def-b1-continuity-continuous) at $0$ and $f$ is [differentiable](#def-b1-derivative-def) without being $C^1$. **Exercise 14.3 ★.** Using the mean value theorem or the tangent inequalities, prove that for all $x > 0$: $$ \frac{x}{1 + x} < \ln(1 + x) < x . $$ Deduce $\lim_{n\to\infty} \bigl(1 + \frac xn\bigr)^n = \eu^x$ for every $x > 0$. **Solution of Exercise 14.3.** $\ln(1+x) < x$ for $x > 0$: concavity tangent inequality at $0$ (strict away from the contact point since $\ln$ is strictly concave; or apply the mean value theorem: $\ln(1+x) = \frac{x}{1+c}$ for some $c \in \intoo{0}{x}$, and $\frac{x}{1+c} < x$). The same mean value identity gives the lower bound: $\frac{x}{1+c} > \frac{x}{1+x}$. Consequence: with $x/n$ in place of $x$, $$ \frac{x/n}{1 + x/n} < \ln\Bigl(1 + \frac xn\Bigr) < \frac xn \quad\implies\quad \frac{x}{1 + x/n} < n \ln\Bigl(1 + \frac xn\Bigr) < x . $$ The left member tends to $x$: by the squeeze, $n\ln(1 + \frac xn) \to x$, and by [continuity](https://one-course.com/books/math/3/en/chapter/13-limits-and-continuity#def-b1-continuity-continuous) of $\exp$, $\bigl(1 + \frac xn\bigr)^n = \eu^{n\ln(1 + x/n)} \to \eu^x$. **Exercise 14.4 ★.** Let $P$ be a real [polynomial](https://one-course.com/books/math/3/en/chapter/8-polynomials#def-b1-poly-def) with $k$ distinct real roots. Prove that $P'$ has at least $k - 1$ distinct real roots, interlaced with those of $P$. Deduce that if $P$ has all its roots real, so does $P'$. **Solution of Exercise 14.4.** Let $x_1 < x_2 < \dots < x_k$ be distinct roots of $P$. On each $\intcc{x_i}{x_{i+1}}$, Rolle ([Theorem 14.8](#thm-b1-derivative-rolle)) produces $c_i \in \intoo{x_i}{x_{i+1}}$ with $P'(c_i) = 0$: that is $k - 1$ roots of $P'$, distinct because the [open](https://one-course.com/books/math/3/en/chapter/12-topology-of-the-real-line#def-b1-topology-open) [intervals](https://one-course.com/books/math/3/en/chapter/10-real-numbers#prop-b1-reals-intervals) are disjoint — and interlaced by construction. If $P$ (degree $n$) has all roots real, write them with multiplicities $m_1 + \dots + m_k = n$. Each root of [multiplicity](https://one-course.com/books/math/3/en/chapter/8-polynomials#def-b1-poly-derivative) $m_i \geq 2$ is a root of $P'$ of [multiplicity](https://one-course.com/books/math/3/en/chapter/8-polynomials#def-b1-poly-derivative) $m_i - 1$ ([Proposition 8.11](https://one-course.com/books/math/3/en/chapter/8-polynomials#prop-b1-poly-multiplicity)), contributing $\sum (m_i - 1) = n - k$; Rolle contributes $k - 1$ more, all distinct from these. Total $\geq n - 1 = \deg P'$: all roots of $P'$ are real. **Exercise 14.5 ★★.** Let $f$ be [differentiable](#def-b1-derivative-def) on $\R$ with $f' (x)\to \ell$ as $x \to +\infty$. Prove that $\frac{f(x)}{x} \to \ell$ *(mean value theorem on $\intcc{A}{x}$)*. Does $f(x+1) - f(x) \to \ell$ hold too? **Solution of Exercise 14.5.** Fix $\varepsilon > 0$ and $A$ with $\abs{f'(t) - \ell} \leq \varepsilon$ for $t \geq A$. For $x > A$, the mean value theorem on $\intcc{A}{x}$ gives $c \in \intoo{A}{x}$ with $$ f(x) = f(A) + f'(c)(x - A), \qquad\text{so}\qquad \Bigl|\frac{f(x)}{x} - \ell\Bigr| \leq \frac{\abs{f(A)} + \abs\ell A}{x} + \abs{f'(c) - \ell} \cdot\frac{x - A}{x} \leq \frac{C_A}{x} + \varepsilon . $$ For $x$ large, $\frac{C_A}{x} \leq \varepsilon$: hence $\frac{f(x)}{x} \to \ell$. Yes: $f(x+1) - f(x) = f'(c_x)$ with $c_x \in \intoo{x}{x+1}$ (mean value theorem on $\intcc{x}{x+1}$), and $c_x \to +\infty$, so $f(x+1) - f(x) \to \ell$. **Exercise 14.6 ★★.** (A discrete Rolle) Let $f$ be $n$ times [differentiable](#def-b1-derivative-def) on $I$ and vanish at $n + 1$ distinct points. Prove that $f^{(n)}$ vanishes at least once. Application: a [polynomial](https://one-course.com/books/math/3/en/chapter/8-polynomials#def-b1-poly-def) of degree $\leq n$ vanishing at $n+1$ points is zero (again). **Solution of Exercise 14.6.** Induction on $n$. For $n = 1$: Rolle. If the claim holds for $n - 1$: $f$ vanishes at $n+1$ points, so by Rolle applied on the $n$ gaps, $f'$ vanishes at $n$ distinct points; the induction hypothesis applied to $f'$ ($n-1$ times [differentiable](#def-b1-derivative-def), $n$ zeros) makes $(f')^{(n-1)} = f^{(n)}$ vanish somewhere. Application: if $P$ of degree $\leq n$ vanishes at $n+1$ points, then $P^{(n)}$, a constant equal to $n!$ times the leading coefficient, vanishes: the leading coefficient is $0$, and one concludes by downward induction (or directly: all coefficients vanish). **Exercise 14.7 ★★.** Let $f$ be twice [differentiable](#def-b1-derivative-def) on $\intcc{a}{b}$ with $f(a) = f(b) = 0$ and $f(x_0) > 0$ for some [interior](https://one-course.com/books/math/3/en/chapter/12-topology-of-the-real-line#def-b1-topology-closure) $x_0$. Prove that $f''(c) < 0$ for some $c \in \intoo{a}{b}$. *(Two mean value theorems and a comparison of slopes.)* **Solution of Exercise 14.7.** By the mean value theorem on $\intcc{a}{x_0}$ and on $\intcc{x_0}{b}$: $$ f'(c_1) = \frac{f(x_0) - f(a)}{x_0 - a} = \frac{f(x_0)}{x_0 - a} > 0, \qquad f'(c_2) = \frac{f(b) - f(x_0)}{b - x_0} = \frac{-f(x_0)}{b - x_0} < 0, $$ with $c_1 < x_0 < c_2$. Then the mean value theorem applied to $f'$ on $\intcc{c_1}{c_2}$ gives $c$ with $$ f''(c) = \frac{f'(c_2) - f'(c_1)}{c_2 - c_1} < 0 . \qedhere $$ **Exercise 14.8 ★★.** Study the function $f(x) = \dfrac{\ln x}{x}$ on $\intoo{0}{+\infty}$: variations, limits, maximum. Deduce that $a^b > b^a$ for all reals $\eu \leq a < b$, and settle the famous special case: which of $\eu^\pi$, $\pi^\eu$ is larger? Check against the small integer pairs $(2,3)$ and $(2,4)$: why do they behave differently? **Solution of Exercise 14.8.** $f'(x) = \frac{1 - \ln x}{x^2}$: $f$ increases on $\intoc{0}{\eu}$, decreases on $\intco{\eu}{+\infty}$, with maximum $f(\eu) = \frac1\eu$; limits $-\infty$ at $0^+$ and $0$ at $+\infty$ ([growth comparison](https://one-course.com/books/math/3/en/chapter/4-standard-functions#prop-b1-functions-powerrules)). For $\eu \leq a < b$: $f$ strictly decreasing there gives $\frac{\ln a}{a} > \frac{\ln b}{b}$, i.e. $b \ln a > a \ln b$, i.e. $a^b > b^a$. With $a = \eu < b = \pi$: $\eu^\pi > \pi^\eu$. Small pairs: $(2, 3)$: $2^3 = 8 < 9 = 3^2$ — reversed! The reason: $2 < \eu$, and on $\intoo{0}{\eu}$ the function $f$ is *increasing*, so the comparison flips when both numbers sit below $\eu$, and is unpredictable across $\eu$ ($f(2) = f(4)$ explains the tie $2^4 = 4^2 = 16$). **Exercise 14.9 ★★.** (Arithmetic–geometric inequality) Using the concavity of $\ln$ with general weights (Jensen’s inequality for $n$ points, to be proved by induction on $n$), show that for positive reals $a_1, \dots, a_n$: $$ \sqrt[n]{a_1 a_2 \cdots a_n} \leq \frac{a_1 + \dots + a_n}{n}, $$ with equality iff all $a_i$ are equal. **Solution of Exercise 14.9.** *Jensen for $\ln$, by induction on $n$.* Claim: for positive $x_i$ and weights $\lambda_i > 0$ with $\sum \lambda_i = 1$: $\ln\bigl(\sum \lambda_i x_i\bigr) \geq \sum \lambda_i \ln x_i$. For $n = 2$ this is concavity. Step: with $\Lambda = \lambda_1 + \dots + \lambda_{n-1} = 1 - \lambda_n$ and $y = \sum_{i 0$. By the extreme value theorem, $g$ attains its minimum on $\intcc{a}{b}$ at some $c$. It is not at $a$: since $g'(a) < 0$, points just right of $a$ have $g < g(a)$. It is not at $b$: since $g'(b) > 0$, points just left of $b$ have $g < g(b)$. So $c$ is [interior](https://one-course.com/books/math/3/en/chapter/12-topology-of-the-real-line#def-b1-topology-closure), and [Proposition 14.7](#prop-b1-derivative-fermat) gives $g'(c) = 0$, i.e. $f'(c) = v$. **Exercise 14.11 ★★★.** Let $f \colon \R \to \R$ be [differentiable](#def-b1-derivative-def) with $\abs{f'(x)} \leq k < 1$ for all $x$ (a *contraction*). Prove that $f$ has exactly one fixed point $\ell$, and that every sequence $u_{n+1} = f(u_n)$ converges to $\ell$ with $\abs{u_n - \ell} \leq k^n\abs{u_0 - \ell}$. *(Existence: apply the intermediate value theorem to $f(x) - x$ on a large segment, using the Lipschitz bound; or use completeness with the Cauchy criterion.)* **Solution of Exercise 14.11.** *Uniqueness:* two fixed points $\ell \neq \ell'$ would give $\abs{\ell - \ell'} = \abs{f(\ell) - f(\ell')} \leq k\abs{\ell - \ell'} < \abs{\ell - \ell'}$, absurd. *Existence:* $g(x) = f(x) - x$ satisfies, by the mean value inequality, $f(x) \leq f(0) + k\abs x$; so for $x \geq \frac{\abs{f(0)}}{1 - k}$, $g(x) \leq f(0) + kx - x \leq 0$, and symmetrically $g(-x) \geq 0$ for $x$ large. The intermediate value theorem gives a zero $\ell$ of $g$: a fixed point. *Convergence:* the mean value inequality again: $$ \abs{u_{n+1} - \ell} = \abs{f(u_n) - f(\ell)} \leq k\abs{u_n - \ell}, $$ so by induction $\abs{u_n - \ell} \leq k^n \abs{u_0 - \ell} \to 0$. **Exercise 14.12 ★★★.** (Cauchy’s mean value theorem and l’Hospital’s rule) 1. Let $f, g$ be [continuous](https://one-course.com/books/math/3/en/chapter/13-limits-and-continuity#def-b1-continuity-continuous) on $\intcc{a}{b}$, [differentiable](#def-b1-derivative-def) on $\intoo{a}{b}$, with $g'$ never zero there. Prove that $g(b) \neq g(a)$ and that some $c \in \intoo{a}{b}$ satisfies $$\frac{f(b) - f(a)}{g(b) - g(a)} = \frac{f'(c)}{g'(c)}$$ *(apply Rolle to $h = f - \lambda g$ for the right constant $\lambda$)*. 2. Deduce l’Hospital’s rule in the $\frac00$ form at a point: if $f(a) = g(a) = 0$ and $\frac{f'(x)}{g'(x)} \to \ell$ as $x \to a^+$ , then $\frac{f(x)}{g(x)} \to \ell$ . 3. Show the converse fails: for $f(x) = x^2\sin\frac1x$ ( $f(0) = 0$ ) and $g(x) = x$ , the quotient $\frac{f}{g}$ has a limit at $0$ but $\frac{f'}{g'}$ has none. **Solution of Exercise 14.12.** 1. If $g(b) = g(a)$ , Rolle would give an [interior](https://one-course.com/books/math/3/en/chapter/12-topology-of-the-real-line#def-b1-topology-closure) zero of $g'$ : excluded. Set $\lambda = \frac{f(b) - f(a)}{g(b) - g(a)}$ and $h = f - \lambda g$ : $h$ is [continuous](https://one-course.com/books/math/3/en/chapter/13-limits-and-continuity#def-b1-continuity-continuous) on $\intcc{a}{b}$ , [differentiable](#def-b1-derivative-def) inside, and $h(b) - h(a) = f(b) - f(a) - \lambda(g(b) - g(a)) = 0$ . Rolle provides $c$ with $h'(c) = 0$ , i.e. $f'(c) = \lambda\,g'(c)$ ; divide by $g'(c) \neq 0$ . 2. For $x > a$ close to $a$, part (1) on $\intcc{a}{x}$ (where $g' \neq 0$) gives $g(x) \neq 0$ and $c_x \in \intoo{a}{x}$ with $$\frac{f(x)}{g(x)} = \frac{f(x) - f(a)}{g(x) - g(a)} = \frac{f'(c_x)}{g'(c_x)} .$$ As $x \to a^+$, $c_x \to a^+$ (squeeze), so the right side tends to $\ell$: $\frac{f}{g} \to \ell$. 3. $\frac{f(x)}{g(x)} = x\sin\frac1x \to 0$ , while $\frac{f'(x)}{g'(x)} = 2x\sin\frac1x - \cos\frac1x$ has no limit at $0$ ( [Exercise 14.2](#exo-b1-derivative-2) ): l’Hospital’s rule transfers information only from $\frac{f'}{g'}$ to $\frac fg$ , never back. ## 14.5 Problem: Liouville’s inequality and the first transcendental number **Problem 14.1.** Weekend problem — [algebraic numbers](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#pb-b1-logic-1) repel rationals: $\abs{x - p/q} \geq C/q^d$, and the transcendence of $\sum 10^{-n!}$ A real number is *algebraic* when it is a root of a nonzero [polynomial](https://one-course.com/books/math/3/en/chapter/8-polynomials#def-b1-poly-def) with integer coefficients, and *transcendental* otherwise. In 1844 Liouville produced the first number ever *proved* transcendental, and the engine of his proof is this chapter’s mean value theorem: an [algebraic number](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#pb-b1-logic-1) of degree $d$ cannot be approximated by rationals better than $C/q^d$ — so a number approximable *faster than every power* cannot be algebraic. This problem builds the inequality, constructs Liouville’s number $L = 0.110001000\dots$ (ones at the factorial positions, via the digit machinery of [Problem 10.1](https://one-course.com/books/math/3/en/chapter/10-real-numbers#pb-b1-reals-1)), proves its transcendence, and ends with Cantor’s rival proof and effective bounds for $\sqrt2$ and $2^{1/3}$. **Part I — How well can rationals be approximated?** 1. Show that two distinct rationals $\frac ab \neq \frac pq$ (written with $b, q \geq 1$ ) satisfy $\bigl|\frac ab - \frac pq\bigr| \geq \frac{1}{bq}$ . Deduce: if $x = \frac ab$ and $0 < \bigl|x - \frac pq\bigr| < \frac{1}{bq}$ , no such $\frac pq$ exists — a rational repels all other rationals at scale $\frac 1q$ . 2. Prove that for *every* rational $\frac pq$ ( $q \geq 1$ ): $\bigl|\sqrt2 - \frac pq\bigr| \geq \frac{1}{4q^2}$ *(if the distance exceeds $1$ this is clear; otherwise bound $\abs{\sqrt2 + p/q} < 4$ and use the nonzero integer $\abs{p^2 - 2q^2} \geq 1$)* . 3. In the other direction: check that $(p, q) \mapsto (p + 2q, p + q)$ preserves $\abs{p^2 - 2q^2} = 1$, generate from $(1,1)$ the pairs $(3,2)$, $(7,5)$, $(17,12)$, $(41,29)$, $(99,70)$, and show each satisfies $$\Bigl|\sqrt2 - \frac pq\Bigr| = \frac{1}{q^2\,(\sqrt2 + p/q)} < \frac{1}{2q^2} :$$ infinitely many approximations of order $2$. With question 2: the approximation exponent of $\sqrt 2$ is *exactly* $2$. 4. (Dirichlet) Let $x$ be irrational and $N \in \N^*$ . Consider the $N + 1$ fractional parts of $0, x, 2x, \dots, Nx$ in the $N$ boxes $\intco{\frac kN}{\frac{k + 1}{N}}$ : by the pigeonhole principle ( [Corollary 2.3](https://one-course.com/books/math/3/en/chapter/2-counting#cor-b1-counting-pigeonhole) ), two fall in one box. Deduce $q \leq N$ and $p$ with $\abs{qx - p} < \frac 1N$ , hence infinitely many rationals with $\bigl|x - \frac pq\bigr| < \frac{1}{q^2}$ : *every* irrational is approximable to order $2$ . **Part II — Liouville’s inequality.** Let $x$ be irrational and algebraic. 5. Show that among the nonzero integer [polynomials](https://one-course.com/books/math/3/en/chapter/8-polynomials#def-b1-poly-def) vanishing at $x$ there is one, say $P$ of degree $d$ , with *no rational root* ; and check $d \geq 2$ *(divide out a factor $X - \frac ab$ over $\Q$ and clear denominators; degree $1$ would make $x$ rational)* . 6. Show that for every rational $\frac pq$ ( $q \geq 1$ ): $\bigl|P\bigl(\frac pq\bigr)\bigr| \geq \frac{1}{q^d}$ *($q^d P(p/q)$ is a nonzero integer)* . 7. Let $M = \max_{\intcc{x-1}{x+1}} \abs{P'}$ ([Theorem 13.13](https://one-course.com/books/math/3/en/chapter/13-limits-and-continuity#thm-b1-continuity-evt)). Using the mean value theorem between $x$ and $\frac pq$, prove *Liouville’s inequality*: with $C = \min\bigl(1, \frac 1M\bigr) > 0$, $$\Bigl| x - \frac pq \Bigr| \geq \frac{C}{q^{\,d}} \qquad\text{for every rational } \frac pq,\ q \geq 1 .$$ 8. Call $x$ a *Liouville number* when for every $n \in \N$ there is a rational $\frac pq$ with $q \geq 2$ and $0 < \bigl|x - \frac pq\bigr| < q^{-n}$ . Prove that a Liouville number is irrational *(question 1: choose $n$ with $2^{\,n-1} > b$)* . 9. Prove Liouville’s theorem: *a Liouville number is transcendental* *(combine questions 7 and 8: the inequality $C < q^{\,d-n}$ fails for large $n$)* . **Part III — The number $L$.** 10. Let $L$ be the value (in the sense of [Problem 10.1](https://one-course.com/books/math/3/en/chapter/10-real-numbers#pb-b1-reals-1) ) of the decimal digit string with digit $1$ at the positions $n!$ ( $n = 1, 2, 3, \dots$ ) and $0$ elsewhere, i.e. $L = \sup_k t_k$ with $t_k = \sum_{n=1}^{k} 10^{-n!}$ . Write out the first $25$ digits of $L$ . 11. Prove the tail bracketing, for every $k \geq 1$: $$10^{-(k+1)!} \;\leq\; L - t_k \;\leq\; \frac{10}{9}\,10^{-(k+1)!} \;<\; 2\cdot 10^{-(k+1)!}$$ *(bound every partial sum beyond $t_k$ by a finite geometric sum)*. 12. Write $t_k = \frac{p_k}{q_k}$ with $q_k = 10^{k!}$ . Show $0 < L - \frac{p_k}{q_k} < \frac{2}{q_k^{\,k+1}}$ , and conclude that $L$ is a Liouville number in the sense of question 8. 13. Conclude: $L$ is transcendental — the first explicit example in history (Liouville, 1844). Cross-check its irrationality directly: its digits are not eventually periodic (growing gaps, as in [Problem 10.1](https://one-course.com/books/math/3/en/chapter/10-real-numbers#pb-b1-reals-1) , question 20). 14. Generalize: replace each digit $1$ by an arbitrary nonzero digit $d_n \in \intint{1}{9}$ . Show the value is still a Liouville number, and deduce — by the diagonal argument of [Problem 10.1](https://one-course.com/books/math/3/en/chapter/10-real-numbers#pb-b1-reals-1) (question 22) applied to these digit choices — that there are uncountably many [transcendental numbers](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#pb-b1-logic-1) of this shape. **Part IV — The hierarchy of approximation orders.** Say $x$ is *approximable to order $\mu$* when for some constant $c > 0$ infinitely many rationals satisfy $\bigl|x - \frac pq\bigr| < \frac{c}{q^{\mu}}$. 15. Assemble the hierarchy from Parts I–III: rationals are approximable to order $1$ and no better; $\sqrt 2$ to order $2$ and no better; every irrational to order at least $2$ ; an [algebraic number](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#pb-b1-logic-1) of degree $d$ to no order beyond $d$ ; Liouville numbers to every order. Justify each claim by citing the relevant question. 16. Show that $L + r$ is a Liouville number for every rational $r = \frac ab$ *(translate the approximants: the new denominators are $b\,q_k$)* . Conclude that Liouville — hence transcendental — numbers are [dense](https://one-course.com/books/math/3/en/chapter/12-topology-of-the-real-line#def-b1-topology-dense) in $\R$ . 17. (Cantor, 1874) Prove that the [set](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-sets) of [algebraic numbers](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#pb-b1-logic-1) is countable: there are finitely many integer [polynomials](https://one-course.com/books/math/3/en/chapter/8-polynomials#def-b1-poly-def) with degree plus sum of $\abs{\text{coefficients}}$ bounded by $h$ , each with at most $\deg$ roots; a countable union of [finite sets](https://one-course.com/books/math/3/en/chapter/2-counting#def-b1-counting-card) is countable. Since no sequence exhausts $\R$ ( [Problem 10.1](https://one-course.com/books/math/3/en/chapter/10-real-numbers#pb-b1-reals-1) , question 22), [transcendental numbers](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#pb-b1-logic-1) exist — in fact form an uncountable [set](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-sets) . Compare the two proofs: what does Liouville’s give that Cantor’s cannot? 18. Prove directly from question 2 that $\sqrt 2$ is *not* a Liouville number *(for $n \geq 3$, the inequality $q^{-n} > \frac{1}{4q^2}$ bounds $q$; then only finitely many candidate rationals remain, all at positive distance from $\sqrt2$)* . Generalize: no [algebraic number](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#pb-b1-logic-1) is Liouville. **Part V — Effective constants.** 19. For the Pell pair $(99, 70)$: verify $99^2 - 2\cdot70^2 = 1$ and evaluate the exact error $$\sqrt2 - \frac{99}{70} = \frac{-1}{70^2\,\bigl(\sqrt2 + \frac{99}{70}\bigr)}, \qquad \Bigl|\sqrt 2 - \frac{99}{70}\Bigr| \approx 7.2\cdot 10^{-5} :$$ five correct digits from a three-digit fraction. 20. Run Part II on $x = 2^{1/3}$, $P = X^3 - 2$: check $P$ has no rational root, bound $M = \max_{\intcc{x-1}{x+1}} 3t^2 \leq 3\,(1 + 2^{1/3})^2 < 16$, and conclude the effective inequality $$\Bigl| 2^{1/3} - \frac pq \Bigr| \geq \frac{1}{16\,q^3} \qquad \text{for all } \frac pq .$$ 21. Payoff: show that any rational approximating $2^{1/3}$ within $10^{-6}$ must have denominator $q \geq 40$ . 22. Show that the base $10$ is irrelevant: the binary analogue $\sum_{n\geq1} 2^{-n!}$ (value of the binary string with ones at factorial positions) is also a Liouville number, hence transcendental. **Part VI — Frontiers and synthesis.** 23. Let $x^\dagger$ be the value of the decimal string with ones exactly at the positions $3^k$ ( $k \geq 0$ ). Show $x^\dagger$ is approximable to order $3$ , and deduce from Liouville’s inequality that $x^\dagger$ is neither rational nor a quadratic irrational. Explain why the method stalls there: order $3$ is compatible with algebraicity of degree $\geq 3$ , and closing that gap (any exponent $> 2$ suffices, for every [algebraic number](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#pb-b1-logic-1) ) is Roth’s theorem, far beyond this volume. 24. Quantify Cantor: show that the [algebraic numbers](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#pb-b1-logic-1) of degree $\leq d$ given by [polynomials](https://one-course.com/books/math/3/en/chapter/8-polynomials#def-b1-poly-def) with coefficients in $\intint{-H}{H}$ number at most $d\,(2H + 1)^{d+1}$ . (This finiteness is what made question 17 work.) 25. Synthesis, one sentence each: (i) locate the single analytic ingredient of Liouville’s proof (which theorem of this chapter, used where); (ii) state the tension that powers it (integrality forces $\abs{P(p/q)} \geq q^{-d}$ , smoothness forbids $\abs{P(p/q)} > M\abs{x - p/q}$ ); (iii) contrast Liouville’s and Cantor’s proofs of the existence of [transcendental numbers](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#pb-b1-logic-1) ; (iv) name where this volume meets the theme again — the weekend problem of [Chapter 15](https://one-course.com/books/math/3/en/chapter/15-integration-on-a-segment#ch-b1-integration) proves $\pi$ irrational by the same integrality-versus-smallness squeeze, with integrals in place of [derivatives](#def-b1-derivative-def) . **Solution of Problem 14.1.** **1.** $\bigl|\frac ab - \frac pq\bigr| = \frac{\abs{aq - bp}}{bq}$, and $aq - bp$ is a nonzero integer when the fractions differ: the distance is $\geq \frac{1}{bq}$. So no rational other than $x$ itself enters the punctured [interval](https://one-course.com/books/math/3/en/chapter/10-real-numbers#prop-b1-reals-intervals) of radius $\frac{1}{bq}$ around $x = \frac ab$. **2.** If $\bigl|\sqrt2 - \frac pq\bigr| \geq 1 \geq \frac{1}{4q^2}$, done. Otherwise $\frac pq \in \intoo{\sqrt2 - 1}{\sqrt2 + 1}$, so $0 < \sqrt2 + \frac pq < 2\sqrt2 + 1 < 4$. Since $\sqrt 2 \notin \Q$, $p^2 - 2q^2$ is a nonzero integer, and $$ \Bigl|\sqrt2 - \frac pq\Bigr| = \frac{\abs{2q^2 - p^2}}{q^2\,\bigl(\sqrt2 + \frac pq\bigr)} \geq \frac{1}{4q^2} . $$ **3.** $(p + 2q)^2 - 2(p + q)^2 = -(p^2 - 2q^2)$: the value $\pm1$ propagates. From $(1,1)$: $$ (3,2),\ (7,5),\ (17,12),\ (41,29),\ (99,70), $$ with $p^2 - 2q^2$ alternating $-1, +1, \dots$ For these, $\frac pq \geq 1$, so $\sqrt2 + \frac pq > 2$ and $$ \Bigl|\sqrt2 - \frac pq\Bigr| = \frac{1}{q^2(\sqrt2 + p/q)} < \frac{1}{2q^2} , $$ with $q \to \infty$: infinitely many order-$2$ approximations. With question 2, the exponent $2$ is exact for $\sqrt 2$. **4.** The $N + 1$ numbers $kx - \lfloor kx\rfloor$ ($0 \leq k \leq N$) lie in the $N$ boxes $\intco{\frac jN}{\frac{j+1}{N}}$: two share a box ([Corollary 2.3](https://one-course.com/books/math/3/en/chapter/2-counting#cor-b1-counting-pigeonhole)), say for $i < j$. With $q = j - i \leq N$ and $p = \lfloor jx\rfloor - \lfloor ix\rfloor$: $\abs{qx - p} < \frac1N$, so $\bigl|x - \frac pq\bigr| < \frac{1}{Nq} \leq \frac{1}{q^2}$. Letting $N \to \infty$: since $x$ is irrational, each fixed fraction has positive distance to $x$, while $\frac{1}{Nq} \leq \frac 1N \to 0$ forces new fractions to appear: infinitely many distinct $\frac pq$ with $\bigl|x - \frac pq\bigr| < \frac{1}{q^2}$. **5.** Start from any nonzero integer $P_0$ with $P_0(x) = 0$. If $P_0$ has a rational root $\frac ab$, the factor theorem ([Theorem 8.7](https://one-course.com/books/math/3/en/chapter/8-polynomials#thm-b1-poly-factor)) writes $P_0 = \bigl(X - \frac ab\bigr)Q$ with $Q \in \Q[X]$; since $x \neq \frac ab$ ($x$ irrational), $Q(x) = 0$, and clearing denominators gives a nonzero *integer* [polynomial](https://one-course.com/books/math/3/en/chapter/8-polynomials#def-b1-poly-def) of smaller degree vanishing at $x$. The degree drops at each step, so the process stops: we reach $P \in \Z[X]$, $P(x) = 0$, with no rational root, of some degree $d$. If $d \leq 1$, $P = uX + v$ would make $x = -\frac vu$ rational: so $d \geq 2$. **6.** $q^d\,P\bigl(\frac pq\bigr) = a_d p^d + a_{d-1} p^{d-1} q + \dots + a_0 q^d$ is an integer, and it is nonzero because $P$ has no rational root: $\bigl|P\bigl(\frac pq\bigr)\bigr| \geq q^{-d}$. **7.** Note $M > 0$: $P'$ is a nonzero [polynomial](https://one-course.com/books/math/3/en/chapter/8-polynomials#def-b1-poly-def) ($d \geq 2$), so it cannot vanish identically on $\intcc{x-1}{x+1}$. If $\bigl|x - \frac pq\bigr| > 1$, then it exceeds $\frac{C}{q^d}$ trivially. Otherwise $\frac pq \in \intcc{x-1}{x+1}$ and the mean value theorem ([Theorem 14.9](#thm-b1-derivative-mvt)) gives $c$ between $x$ and $\frac pq$ with $$ \Bigl|P\Bigl(\frac pq\Bigr)\Bigr| = \Bigl|P\Bigl(\frac pq\Bigr) - P(x)\Bigr| = \abs{P'(c)}\,\Bigl|x - \frac pq\Bigr| \leq M\,\Bigl|x - \frac pq\Bigr| , $$ so with question 6: $\bigl|x - \frac pq\bigr| \geq \frac{1}{Mq^d} \geq \frac{C}{q^d}$. **8.** Suppose $x = \frac ab$ is Liouville. Pick $n$ with $2^{n-1} > b$ and the corresponding $\frac pq$, $q \geq 2$: $$ 0 < \Bigl|x - \frac pq\Bigr| < \frac{1}{q^n} = \frac{1}{q^{n-1}\,q} \leq \frac{1}{2^{n-1} q} < \frac{1}{bq} , $$ contradicting question 1. So Liouville numbers are irrational. **9.** If a Liouville $x$ were algebraic: it is irrational (question 8), so questions 5–7 provide $d \geq 2$ and $C > 0$ with $\bigl|x - \frac pq\bigr| \geq \frac{C}{q^d}$ always. For each $n$, the Liouville approximant gives $\frac{C}{q^d} < q^{-n}$, i.e. $C < q^{d-n} \leq 2^{d-n}$ (as $q \geq 2$). For $n$ large, $2^{d-n} < C$: contradiction. Liouville numbers are transcendental. **10.** Ones at positions $1, 2, 6, 24$; all other digits among the first $25$ vanish: $$ L = 0.1100010000\,0000000000\,00010\dots $$ **11.** For $m > k$, the positions $n!$ with $n > k$ are distinct integers $\geq (k+1)!$, so the finite geometric sum gives $$ t_m - t_k = \sum_{n=k+1}^{m} 10^{-n!} \leq \sum_{j = (k+1)!}^{m!} 10^{-j} < 10^{-(k+1)!}\,\frac{1}{1 - \frac1{10}} = \frac{10}{9}\,10^{-(k+1)!} ; $$ taking the [supremum](https://one-course.com/books/math/3/en/chapter/10-real-numbers#def-b1-reals-bounds) over $m$: $L - t_k \leq \frac{10}{9}10^{-(k+1)!} < 2\cdot10^{-(k+1)!}$. Lower bound: $L \geq t_{k+1} = t_k + 10^{-(k+1)!}$. **12.** $p_k = 10^{k!}\,t_k \in \N$, $q_k = 10^{k!}$, and $(k+1)! = (k+1)\,k!$ gives $10^{-(k+1)!} = q_k^{-(k+1)}$: question 11 reads $$ 0 < L - \frac{p_k}{q_k} < \frac{2}{q_k^{\,k+1}} . $$ Given $n$: for $k \geq n$, $2\,q_k^{-(k+1)} \leq q_k^{-n}$ (indeed $q_k^{\,k+1-n} \geq q_k \geq 10 > 2$), and $q_k \geq 2$: the definition of question 8 is met. $L$ is a Liouville number. **13.** By question 9, $L$ is transcendental — the first number in history proved transcendental (Liouville, 1844). Digit cross-check: the string has infinitely many ones with consecutive gaps $(k+1)! - k! = k\cdot k! \to \infty$, so it is not eventually periodic, and $L \notin \Q$ by the periodicity criterion of [Problem 10.1](https://one-course.com/books/math/3/en/chapter/10-real-numbers#pb-b1-reals-1) (question 18) — consistent. **14.** With digits $d_n \in \intint{1}{9}$ at the factorial positions: the tail bound of question 11 scales by at most $9$: $0 < L' - t'_k \leq 9\cdot\frac{10}{9}\,10^{-(k+1)!} = 10\,q_k^{-(k+1)}$ (positivity because the digit at position $(k+1)!$ is nonzero). For $k \geq n$: $10\,q_k^{-(k+1)} \leq q_k^{-n}$ since $q_k^{\,k+1-n} \geq 10$: again a Liouville number, hence transcendental. These values are pairwise distinct for distinct digit choices (the strings are proper — zeros abound — and proper strings determine their value, [Problem 10.1](https://one-course.com/books/math/3/en/chapter/10-real-numbers#pb-b1-reals-1), question 10). Given any list $k \mapsto x_k$ of them, choose the $k$-th factorial digit in $\intint{1}{9}$ different from that of $x_k$: a number of the same shape missing from the list. Uncountably many explicit transcendentals. **15.** First a lemma: *if $\bigl|x - \frac pq\bigr| \geq \frac{C}{q^s}$ for all $\frac pq \neq x$, then $x$ is not approximable to any order $\mu > s$.* Indeed infinitely many $\frac pq \neq x$ with $\bigl|x - \frac pq\bigr| < \frac{c}{q^\mu}$ would force $\frac{C}{q^s} < \frac{c}{q^\mu}$, i.e. $q^{\mu - s} < \frac cC$: the $q$ are bounded, and boundedly many fractions lie within distance $1$ of $x$ — finitely many candidates, not infinitely many. Now the hierarchy: rationals are approximable to order $1$ ($\frac pq$ with $p = \lfloor qx\rfloor + 1$ gives error $\leq \frac1q < \frac2q$) and to no order $\mu > 1$ (question 1 gives the hypothesis of the lemma with $s = 1$, $C = \frac1b$); $\sqrt2$: order $2$ (question 3) and no more (question 2 and the lemma); every irrational: at least $2$ (question 4); algebraic of degree $d$: at most $d$ (question 7 and the lemma); Liouville numbers: every order (question 12’s display, with $c = 2$). **16.** With $r = \frac ab$: $\frac{p_k}{q_k} + \frac ab = \frac{b p_k + a q_k}{b q_k} =: \frac{P_k}{Q_k}$, $Q_k = b q_k \geq 2$, and $$ \Bigl|(L + r) - \frac{P_k}{Q_k}\Bigr| = L - \frac{p_k}{q_k} < 2\,q_k^{-(k+1)} = 2\,b^{\,k+1} Q_k^{-(k+1)} . $$ Given $n$: for large $k$, $Q_k^{\,k+1-n} \geq Q_k = b\,10^{k!} \geq 2\,b^{\,k+1}$ (the factorial crushes the power), so the error is $< Q_k^{-n}$: $L + r$ is Liouville. Since $\Q$ is [dense](https://one-course.com/books/math/3/en/chapter/12-topology-of-the-real-line#def-b1-topology-dense) and each $L + r$ is transcendental, [transcendental numbers](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#pb-b1-logic-1) are [dense](https://one-course.com/books/math/3/en/chapter/12-topology-of-the-real-line#def-b1-topology-dense) in $\R$. **17.** For $h \geq 1$ there are finitely many $P \in \Z[X]$ with $\deg P + \sum_i \abs{a_i} \leq h$ (degree $\leq h$ and each coefficient in $\intint{-h}{h}$: at most $(2h+1)^{h+1}$). Every nonzero integer [polynomial](https://one-course.com/books/math/3/en/chapter/8-polynomials#def-b1-poly-def) has such a height, and has at most $\deg P$ real roots: the [algebraic numbers](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#pb-b1-logic-1) form a countable union (over $h$) of [finite sets](https://one-course.com/books/math/3/en/chapter/2-counting#def-b1-counting-card), hence can be listed as a single sequence. If the transcendentals could also be listed, interleaving the two lists would list $\R$, contradicting [Problem 10.1](https://one-course.com/books/math/3/en/chapter/10-real-numbers#pb-b1-reals-1) (question 22). So [transcendental numbers](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#pb-b1-logic-1) form an uncountable [set](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-sets). Comparison: Cantor proves *most* reals are transcendental yet exhibits none; Liouville exhibits one, with effective constants (Part V) — existence by abundance versus existence by construction. **18.** From question 2, the lemma hypothesis holds with $s = 2$, $C = \frac14$. If $\sqrt2$ were Liouville, then for $n = 3$: $\frac{1}{4q^2} < q^{-3}$ forces $q < 4$, so $q \in \{2, 3\}$; only finitely many $\frac pq$ with these $q$ lie within $1$ of $\sqrt2$, each at some positive distance $\geq \varepsilon_0$ ($\sqrt2$ irrational); choosing $n$ with $2^{-n} < \varepsilon_0$ leaves no admissible $\frac pq$ at all: contradiction. The same argument with $\frac{C}{q^d}$ shows no [algebraic number](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#pb-b1-logic-1) is Liouville — question 9 in effective clothing. **19.** $99^2 - 2\cdot70^2 = 9801 - 9800 = 1$. Hence $$ \sqrt2 - \frac{99}{70} = \frac{2 - (99/70)^2}{\sqrt2 + 99/70} = \frac{-1}{4900\,\bigl(\sqrt2 + \tfrac{99}{70}\bigr)} , \qquad \Bigl|\sqrt2 - \frac{99}{70}\Bigr| = \frac{1}{4900 \times 2.8284\dots} \approx 7.2\cdot10^{-5} : $$ $\frac{99}{70} = 1.414285\dots$ against $\sqrt2 = 1.414213\dots$ — five correct digits. **20.** Rational-root test for $P = X^3 - 2$: candidates $\pm1, \pm2, \pm\frac12$, none a root. So $d = 3$ and Part II applies to $x = 2^{1/3} = 1.2599\dots$ On $\intcc{x - 1}{x + 1} \subseteq \intcc{0.25}{2.26}$: $\abs{P'(t)} = 3t^2 \leq 3\,(1 + 2^{1/3})^2 < 3\times(2.26)^2 = 15.32 < 16$, so $M < 16$ and $C \geq \frac{1}{16}$: $$ \Bigl|2^{1/3} - \frac pq\Bigr| \geq \frac{1}{16\,q^3} \qquad\text{for all rationals.} $$ **21.** If $\bigl|2^{1/3} - \frac pq\bigr| < 10^{-6}$, then $\frac{1}{16 q^3} < 10^{-6}$, i.e. $q^3 > \frac{10^6}{16} = 62\,500$; since $39^3 = 59\,319 < 62\,500 \leq 64\,000 = 40^3$: $q \geq 40$. **22.** Run Part III in base $2$: $B = \sup_k \sum_{n\leq k} 2^{-n!}$, $q_k = 2^{k!}$, and the geometric tail (ratio $\frac12$) gives $0 < B - \frac{p_k}{q_k} \leq 2\cdot2^{-(k+1)!} = 2\,q_k^{-(k+1)} \leq q_k^{-n}$ for $k \geq n$. So $B$ is Liouville, hence transcendental: nothing in the argument is decimal. **23.** With ones at positions $3^k$: $q_k = 10^{3^k}$ and the tail bound gives $0 < x^\dagger - \frac{p_k}{q_k} < 2\cdot10^{-3^{k+1}} = 2\,q_k^{-3}$ (as $3^{k+1} = 3\cdot3^k$): infinitely many approximations of order $3$. By the lemma of question 15: order $3 > 1$ rules out rationality, and order $3 > 2$ rules out being a quadratic irrational (whose Liouville inequality has $s = d = 2$). But an [algebraic number](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#pb-b1-logic-1) of degree $\geq 3$ is only repelled at order $d \geq 3$: Liouville’s method cannot separate $x^\dagger$ from the cubics. The gap is closed by Roth’s theorem — every algebraic irrational has approximation order exactly $2$ — a twentieth-century result far beyond this volume; granted it, $x^\dagger$ too is transcendental. **24.** There are at most $(2H+1)^{d+1}$ tuples $(a_0, \dots, a_d)$ with entries in $\intint{-H}{H}$, and each nonzero [polynomial](https://one-course.com/books/math/3/en/chapter/8-polynomials#def-b1-poly-def) among them has at most $d$ real roots: at most $d\,(2H+1)^{d+1}$ [algebraic numbers](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#pb-b1-logic-1) arise — the finiteness that let question 17 enumerate them all. **25.** (i) The single analytic ingredient is the mean value theorem, in question 7, converting the vanishing $P(x) = 0$ into the Lipschitz repulsion $\abs{P(p/q)} \leq M\abs{x - p/q}$. (ii) The tension: integrality pushes $\abs{P(p/q)}$ up to $q^{-d}$, smoothness pulls it down to $M\abs{x - p/q}$ — a rational too close to $x$ would be crushed between the two. (iii) Liouville constructs one transcendental with effective constants; Cantor shows almost all reals are transcendental without naming one: construction versus [cardinality](https://one-course.com/books/math/3/en/chapter/2-counting#def-b1-counting-card). (iv) The weekend problem of [Chapter 15](https://one-course.com/books/math/3/en/chapter/15-integration-on-a-segment#ch-b1-integration) proves the irrationality of $\pi$ by the same squeeze — an integral that would be a positive integer yet is trapped in $\intoo{0}{1}$ — with integration replacing differentiation as the analytic half.