---
title: "Standard Functions"
book: "University Mathematics — Year 1"
subject: math
language: en
chapter: 4
exercises: 12
source: https://one-course.com/books/math/3/en/chapter/4-standard-functions
---

# Chapter 4 — Standard Functions

Analysis is only as useful as the stock of functions one masters. To the collection inherited from secondary school — powers, exponential, logarithm, trigonometric functions — this chapter adds their inverse functions ($\arcsin$, $\arccos$, $\arctan$) and the hyperbolic family. Derivatives are used freely at High School level; the theory behind inverse functions is completed in Chapters [13](https://one-course.com/books/math/3/en/chapter/13-limits-and-continuity#ch-b1-continuity) and [14](https://one-course.com/books/math/3/en/chapter/14-differentiation#ch-b1-derivative).

## 4.1 Exponential, logarithm, powers

**Proposition 4.1 (Recap and characterization).**

$\exp \colon \R \to \intoo{0}{+\infty}$ and $\ln \colon
\intoo{0}{+\infty} \to \R$ are reciprocal bijections, strictly increasing, with

$$
\exp(x + y) = \exp x \exp y, \qquad \ln(xy) = \ln x + \ln y,
\qquad \exp' = \exp, \qquad \ln'(x) = \frac 1x .
$$

Moreover $\exp$ is the *only* differentiable function $f \colon
\R \to \R$ with $f' = f$ and $f(0) = 1$.

**Proof.** The recap is High School material. For uniqueness, let $f' = f$, $f(0) = 1$, and set $g(x) = f(x)\,\eu^{-x}$. Then $g' = f'\eu^{-x} -
f\eu^{-x} = 0$, so $g$ is constant equal to $g(0) = 1$: $f = \exp$. ∎

**Definition 4.2 (General powers).**

For $x > 0$ and $\alpha \in \R$: $\;x^\alpha = \eu^{\alpha \ln x}$. For $a > 0$, $a \neq 1$, the *logarithm to base $a$* is $\log_a x =
\frac{\ln x}{\ln a}$, the inverse of $x \mapsto a^x$.

**Example 4.3 (Solving exponential equations).**

Solve $2^x = 5^{\,x-1}$ in $\R$. Both sides are positive, so take logarithms — a reversible step:

$$
x\ln2 = (x - 1)\ln5
\iff x(\ln2 - \ln5) = -\ln5
\iff x = \frac{\ln5}{\ln5 - \ln2} = \frac{\ln 5}{\ln\frac52}
\approx 1.756 .
$$

Next, solve $x^{\sqrt2} = 3$ for $x > 0$: raise to the power $\frac1{\sqrt2}$ (i.e. apply the reciprocal bijection): $x = 3^{1/\sqrt2} = \eu^{(\ln 3)/\sqrt2} \approx 2.175$. The insight: every equation mixing powers unwinds through $\ln$ and $\exp$, because the definition $x^\alpha = \eu^{\alpha\ln x}$ reduces all power manipulations to arithmetic of exponents — but only on the domain $x > 0$ where that definition lives.

**Example 4.4 (Doubling times).**

A quantity grows by $3\%$ per step: after $n$ steps it is multiplied by $(1.03)^n$. When does it double? Solve $(1.03)^n
\geq 2$:

$$
n\ln(1.03) \geq \ln2
\iff n \geq \frac{\ln 2}{\ln 1.03}
= \frac{0.6931}{0.02956} \approx 23.4 ,
$$

so the first doubling occurs at step $24$. (The financiers’ “rule of $72$”, which estimates the doubling time as $72$ divided by the rate in percent, is this computation with the approximation $\ln(1 + x) \approx x$, quantified in [Chapter 16](https://one-course.com/books/math/3/en/chapter/16-taylor-formulas-and-asymptotic-expansions#ch-b1-taylor).) Exponential processes are best reasoned about through their logarithms: on that scale, growth is linear and questions become divisions.

**Example 4.5 (How long is 220262^{2026}22026?).**

The number of decimal digits of an integer $N \geq 1$ is $\floor{\log_{10} N} + 1$ (indeed $N$ has $d$ digits exactly when $10^{d-1} \leq N < 10^d$, i.e. $d - 1 \leq \log_{10}N < d$). For $N = 2^{2026}$:

$$
\log_{10} 2^{2026} = 2026\,\log_{10}2
= 2026 \times 0.301030 = 609.887 ,
$$

so $2^{2026}$ has $610$ digits. The fractional part carries a bonus: $10^{0.887} \approx 7.7$, so the number *begins* with $7$. One multiplication answered a question about a number no one will ever write out — logarithms compress multiplicative size into additive size, which is their whole historical point ([Example 7.12](https://one-course.com/books/math/3/en/chapter/7-algebraic-structures#ex-b1-structures-expmorphism) makes that sentence precise).

**Proposition 4.6 (Power rules and growth comparison).**

For $x, y > 0$ and $\alpha, \beta \in \R$:

$$
x^{\alpha+\beta} = x^\alpha x^\beta, \quad
(x^\alpha)^\beta = x^{\alpha\beta}, \quad
(xy)^\alpha = x^\alpha y^\alpha, \quad
(x^\alpha)' = \alpha\, x^{\alpha - 1}.
$$

Growth scale as $x \to +\infty$, for every $\alpha > 0$ and $\beta > 0$:

$$
\frac{(\ln x)^\beta}{x^\alpha} \longrightarrow 0,
\qquad
\frac{x^\alpha}{\eu^{\beta x}} \longrightarrow 0 .
$$

**Proof.** The identities transcribe those of $\exp$ and $\ln$ through the definition. In detail for the second (the least obvious): $x^\alpha
> 0$ and $\ln(x^\alpha) = \alpha\ln x$ (apply $\ln$ to the definition), so

$$
(x^\alpha)^\beta = \eu^{\beta\ln(x^\alpha)}
= \eu^{\beta\alpha\ln x} = x^{\alpha\beta} ;
$$

the first and third are the same two-line unwindings via $\exp(u + v) = \exp u\exp v$ and $\ln(xy) = \ln x + \ln y$. The derivative is the chain rule: $(\eu^{\alpha\ln x})' =
\frac{\alpha}{x} \eu^{\alpha\ln x}$.

Comparisons: from $\frac{\ln t}{t} \to 0$ (proved in the High School volume), substitute $t = x^{\alpha/\beta}$: $\frac{\ln x}{x^{\alpha/\beta}} \to 0$, then raise to the power $\beta$. For the second, substitute $x = \ln u$ in the first. ∎

**Example 4.7 (Two limits every reader should own).**

What are $\lim_{x \to 0^+} x^x$ and $\lim_{x \to +\infty}
x^{1/x}$? Both powers are defined through the exponential, so settle the exponent first:

$$
x^x = \eu^{x\ln x}, \qquad
x\ln x = -\frac{\ln(1/x)}{1/x} \longrightarrow 0
\quad (x \to 0^+),
\qquad\text{so } x^x \longrightarrow \eu^0 = 1 ;
$$

and $x^{1/x} = \eu^{(\ln x)/x} \to \eu^0 = 1$ as $x \to +\infty$, directly by the [growth comparison](#prop-b1-functions-powerrules). The insight: an indeterminate power (of shape $0^0$ or $\infty^0$) is *always* handled by rewriting $u^v = \eu^{v\ln u}$ and analyzing the product $v \ln u$ — never by guessing from base and exponent separately. The function $x \mapsto \frac{\ln x}{x}$ that decided both limits is studied exhaustively in the weekend problem of this chapter.

![The growth scale of : near the right edge, x has barely passed 1.6 while x has left the frame. Every ratio (log over power, power over exponential) tends to 0 — the picture only hints at what the substitutions in the proof make exact.](https://one-course.com/images/onecourse/chapters/math-3/b1-functions/fig-e212b0476b06.svg)

*The growth scale of [Proposition 4.6](#prop-b1-functions-powerrules): near the right edge, $\ln x$ has barely passed $1.6$ while $\eu^x$ has left the frame. Every ratio (log over power, power over exponential) tends to $0$ — the picture only hints at what the substitutions in the proof make exact.*

**Remark 4.8 (Common pitfalls with powers and logarithms).**

1. *Domains.* $x^\alpha$ for irrational $\alpha$ needs $x > 0$ ; $(-8)^{1/3}$ is best avoided in favor of “the real cube root of $-8$ ”, because the rule $(x^\alpha)^\beta = x^{\alpha\beta}$ silently fails on negatives: $\bigl((-8)^2\bigr)^{1/6} = 2 \neq -2$ .
2. *$\sqrt{x^2} = \abs x$* , not $x$ : forgetting the absolute value is the classic source of lost negative solutions.
3. *$\ln(xy) = \ln x + \ln y$ requires $x, y > 0$.* For negative arguments, $\ln(xy)$ can be defined while the right-hand side is not.
4. *Which function grows?* In $x^\alpha$ the *base* varies; in $a^x$ the *exponent* does. Their growths are wildly different ( [Proposition 4.6](#prop-b1-functions-powerrules) ), and hybrid expressions like $x^{\ln x}$ or $x^{1/x}$ must be rewritten as $\eu^{(\cdot)}$ before any reasoning — as in [Example 4.7](#ex-b1-functions-xx) .

## 4.2 Inverse trigonometric functions

**Definition 4.9 (arcsin⁡\arcsinarcsin, arccos⁡\arccosarccos, arctan⁡\arctanarctan).**

The restrictions

$$
\sin \colon \intcc{-\tfrac\pi2}{\tfrac\pi2} \to \intcc{-1}{1},
\qquad
\cos \colon \intcc{0}{\pi} \to \intcc{-1}{1},
\qquad
\tan \colon \intoo{-\tfrac\pi2}{\tfrac\pi2} \to \R
$$

are strictly monotonic bijections. Their inverse [maps](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-map) are written $\arcsin$, $\arccos$ and $\arctan$. Thus, for instance, $y = \arcsin x$ ($x \in \intcc{-1}{1}$) is *the* angle in $\intcc{-\frac\pi2}{\frac\pi2}$ whose sine is $x$.

![From left to right: , , . Each inherits its graph from the restricted direct function by reflection in the line y = x.](https://one-course.com/images/onecourse/chapters/math-3/b1-functions/fig-2fc622239438.svg)

*From left to right: $\arcsin$, $\arccos$, $\arctan$. Each inherits its graph from the restricted direct function by reflection in the line $y = x$.*

**Proposition 4.10 (Derivatives).**

On the interior of their domains:

$$
\arcsin' x = \frac{1}{\sqrt{1 - x^2}},
\qquad
\arccos' x = \frac{-1}{\sqrt{1 - x^2}},
\qquad
\arctan' x = \frac{1}{1 + x^2}.
$$

**Proof.** Anticipating the inverse-function rule proved in [Chapter 14](https://one-course.com/books/math/3/en/chapter/14-differentiation#ch-b1-derivative): if $f$ is a differentiable bijection and $f' \neq 0$, then $(f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))}$. For $\arcsin$: with $y = \arcsin x$,

$$
\arcsin' x = \frac{1}{\cos y}
= \frac{1}{\sqrt{1 - \sin^2 y}} = \frac{1}{\sqrt{1 - x^2}},
$$

where $\cos y = +\sqrt{1 - \sin^2 y}$ because $y \in
\intoo{-\frac\pi2}{\frac\pi2}$ forces $\cos y > 0$. Similarly $\arccos' x = \frac{1}{-\sin y}$ with $\sin y > 0$ on $\intoo{0}{\pi}$, and $\arctan' x = \frac{1}{1 + \tan^2 y} =
\frac{1}{1 + x^2}$ using $\tan' = 1 + \tan^2$. ∎

**Example 4.11 (Recognizing a hidden constant).**

Study $g(x) = \arctan\dfrac{1 - x}{1 + x}$ on $\intoo{-1}
{+\infty}$. Chain rule and a short computation:

$$
g'(x) = \frac{1}{1 + \bigl(\frac{1-x}{1+x}\bigr)^2}\cdot
\frac{-(1+x) - (1-x)}{(1+x)^2}
= \frac{-2}{(1+x)^2 + (1-x)^2}
= \frac{-1}{1 + x^2} ,
$$

since $(1+x)^2 + (1-x)^2 = 2 + 2x^2$. So $g' = -\arctan'$: the function $g + \arctan$ has zero derivative on the *interval* $\intoo{-1}{+\infty}$, hence is constant there; its value at $x =
0$ is $\arctan 1 + \arctan 0 = \frac\pi4$. Conclusion:

$$
\arctan\frac{1 - x}{1 + x} = \frac\pi4 - \arctan x
\qquad (x > -1) .
$$

On $\intoo{-\infty}{-1}$ the same derivative computation holds but the constant is different ($-\frac{3\pi}4$: evaluate the limit as $x \to -\infty$). The insight: “zero derivative implies constant” is an interval-by-interval [statement](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-statement) — exactly the subtlety exploited in [Exercise 4.6](#exo-b1-functions-6) and [Exercise 4.10](#exo-b1-functions-10).

**Proposition 4.12 (Standard identities).**

1. For $x \in \intcc{-1}{1}$ : $\arcsin x + \arccos x =  \dfrac{\pi}{2}$ .
2. For $x > 0$ : $\arctan x + \arctan\dfrac 1x = \dfrac{\pi}{2}$ (and $-\frac\pi2$ for $x < 0$ ).
3. $\sin(\arccos x) = \cos(\arcsin x) = \sqrt{1 - x^2}$ ; $\;\tan(\arcsin x) = \dfrac{x}{\sqrt{1 - x^2}}$ for $\abs x < 1$ .

**Proof.** (1) The derivative of $x \mapsto \arcsin x + \arccos x$ is zero on $\intoo{-1}{1}$ ([Proposition 4.10](#prop-b1-functions-arcderiv)), so the function is constant there, equal to its value $\frac\pi2$ at $0$; the endpoint values $\pm 1$ are checked directly ($\frac\pi2 + 0$ and $-\frac\pi2 +
\pi$).

(2) Same method on $\intoo{0}{+\infty}$: the derivative is $\frac{1}{1+x^2} + \frac{-1/x^2}{1 + 1/x^2} = \frac{1}{1+x^2} -
\frac{1}{x^2 + 1} = 0$, and at $x = 1$ the sum is $2\arctan 1 =
\frac\pi2$. For $x < 0$, use oddness of $\arctan$.

(3) With $y = \arccos x \in \intcc{0}{\pi}$: $\sin y \geq 0$, so $\sin y = \sqrt{1 - \cos^2 y} = \sqrt{1 - x^2}$; likewise for $\cos(\arcsin x)$, and the tangent formula is the quotient. ∎

**Remark 4.13.**

$\arcsin(\sin\theta) = \theta$ holds *only* for $\theta \in
\intcc{-\frac\pi2}{\frac\pi2}$: for instance $\arcsin(\sin\pi) = 0 \neq \pi$. The composition in the other direction, $\sin(\arcsin x) = x$, is valid on all of $\intcc{-1}{1}$.

**Example 4.14 (Wrapping back to the principal interval).**

Compute

$$
\arctan\Bigl(\tan\frac{3\pi}4\Bigr)
\qquad\text{and}\qquad
\arctan\Bigl(\tan\frac{17\pi}5\Bigr).
$$

The recipe: replace the angle by the unique angle of $\intoo{-\frac\pi2}{\frac\pi2}$ with the same tangent, i.e. subtract the right multiple of $\pi$ (the period of $\tan$). First: $\frac{3\pi}4 - \pi =
-\frac\pi4$, so the answer is $-\frac\pi4$. Second: $\frac{17\pi}
5 - 3\pi = \frac{2\pi}5 \in \intoo{-\frac\pi2}{\frac\pi2}$, so the answer is $\frac{2\pi}5$. The computation is a Euclidean division of the angle by $\pi$ in disguise — and the analogous recipes for $\arcsin$ (reflect into $\intcc{-\frac\pi2}{\frac\pi2}$, period $2\pi$) and $\arccos$ (reflect into $\intcc0\pi$) drive [Exercise 4.1](#exo-b1-functions-1) and the piecewise answer of [Exercise 4.10](#exo-b1-functions-10).

**Example 4.15 (Adding arctangents safely).**

Let us prove

$$
\arctan\frac12 + \arctan\frac15 + \arctan\frac18 = \frac\pi4 .
$$

Two ingredients: the tangent addition formula, and — the step beginners forget — a *localization* of the sum. First, $\tan(u + v) = \frac{\tan u + \tan v}{1 - \tan u\tan v}$ with $u = \arctan\frac12$, $v = \arctan\frac15$ gives

$$
\tan(u + v) = \frac{\frac12 + \frac15}{1 - \frac1{10}}
= \frac{7/10}{9/10} = \frac79,
\qquad\text{then}\qquad
\tan\Bigl(u + v + \arctan\frac18\Bigr)
= \frac{\frac79 + \frac18}{1 - \frac7{72}}
= \frac{65/72}{65/72} = 1 .
$$

Second, each of the three angles lies in $\intoo0{\frac\pi4}$ (their arguments are less than $1$), so the sum lies in $\intoo0{\frac{3\pi}4}$; the only angle there with tangent $1$ is $\frac\pi4$. Without the localization the conclusion would be “$\frac\pi4$ up to a multiple of $\pi$” — half a proof. The same two-step discipline runs [Exercise 4.8](#exo-b1-functions-8) and [Exercise 4.12](#exo-b1-functions-12).

**Example 4.16 (An arctangent equation).**

Solve $\arctan x + \arctan 2x = \dfrac\pi4$. First localize: the left side has the sign of $x$ (both terms do), so any solution has $x > 0$, and then the sum lies in $\intoo0\pi$. Take tangents ([injective](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-inj) on no interval of length $\pi$, but combined with the localization this will suffice): the addition formula gives

$$
\tan\bigl(\arctan x + \arctan 2x\bigr)
= \frac{3x}{1 - 2x^2} = 1
\iff 2x^2 + 3x - 1 = 0
\iff x = \frac{-3 \pm \sqrt{17}}4 .
$$

The negative root is discarded by the localization; for the positive candidate $x_0 = \frac{\sqrt{17} - 3}4 \approx 0.28$, the sum lies in $\intoo0\pi$ with tangent $1$, and the only such angle is $\frac\pi4$ (in $\intoo{\frac\pi2}\pi$ the tangent is negative): $x_0$ is the unique solution. Note the shape of the argument: taking tangents may create solutions, never lose them, so one solves the polynomial equation *then* filters by localization — the same forward-check discipline as squaring an equation.

## 4.3 Hyperbolic functions

**Definition 4.17 (cosh⁡\coshcosh, sinh⁡\sinhsinh, tanh⁡\tanhtanh).**

For $x \in \R$:

$$
\cosh x = \frac{\eu^x + \eu^{-x}}{2},
\qquad
\sinh x = \frac{\eu^x - \eu^{-x}}{2},
\qquad
\tanh x = \frac{\sinh x}{\cosh x}
$$

(*hyperbolic cosine, sine, tangent*). $\cosh$ is even, $\sinh$ and $\tanh$ are odd.

**Proposition 4.18 (Basic properties).**

For all $x, y \in \R$:

1. $\cosh^2 x - \sinh^2 x = 1$ ;
2. $\cosh' = \sinh$ , $\;\sinh' = \cosh$ , $\;\tanh' = 1 - \tanh^2  = \dfrac{1}{\cosh^2}$ ;
3. $\cosh(x+y) = \cosh x\cosh y + \sinh x \sinh y$ and $\sinh(x+y) = \sinh x \cosh y + \cosh x \sinh y$ ;
4. $\sinh$ is a strictly increasing bijection of $\R$ onto $\R$ ; $\cosh$ restricted to $\R_+$ is a strictly increasing bijection onto $\intco{1}{+\infty}$ ; $\tanh$ is a strictly increasing bijection of $\R$ onto $\intoo{-1}{1}$ .

**Proof.** (1) $\bigl(\frac{\eu^x + \eu^{-x}}{2}\bigr)^2 - \bigl(\frac{\eu^x -
\eu^{-x}}{2}\bigr)^2 = \frac{(\eu^{2x} + 2 + \eu^{-2x}) - (\eu^{2x} -
2 + \eu^{-2x})}{4} = 1$.

(2) Differentiate the defining formulas; for $\tanh$, the quotient rule gives $\frac{\cosh^2 - \sinh^2}{\cosh^2}$, which is both $1 - \tanh^2$ and $\frac{1}{\cosh^2}$ by (1).

(3) Expand the right-hand sides using the definitions. In detail for the first formula:

$$
\cosh x\cosh y + \sinh x\sinh y
= \frac{(\eu^x + \eu^{-x})(\eu^y + \eu^{-y})
+ (\eu^x - \eu^{-x})(\eu^y - \eu^{-y})}{4} ;
$$

of the eight products, the four “mixed” ones ($\eu^{x-y}$, $\eu^{y-x}$) cancel in pairs, while $\eu^{x+y}$ and $\eu^{-(x+y)}$ each appear twice: the total is $\frac{2\eu^{x+y} + 2\eu^{-(x+y)}}{4} = \cosh(x+y)$. The sine formula is identical with the mixed terms surviving instead.

(4) $\sinh' = \cosh \geq 1 > 0$, so $\sinh$ is strictly increasing, with limits $\pm\infty$ (dominant term $\pm\frac{\eu^{\abs
x}}{2}$); the bijection [statement](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-statement) then follows from the intermediate value theorem (used at High School level; systematic treatment in [Chapter 13](https://one-course.com/books/math/3/en/chapter/13-limits-and-continuity#ch-b1-continuity)). On $\R_+$, $\cosh' = \sinh \geq 0$ vanishing only at $0$: strictly increasing from $\cosh 0 = 1$ to $+\infty$. And $\tanh' > 0$ with limits $\pm 1$ at $\pm\infty$: in detail,

$$
\tanh x = \frac{\eu^x - \eu^{-x}}{\eu^x + \eu^{-x}}
= \frac{1 - \eu^{-2x}}{1 + \eu^{-2x}}
\xrightarrow[x \to +\infty]{} 1 ,
$$

after dividing numerator and denominator by $\eu^x$, and oddness gives the limit $-1$ at $-\infty$; the strictly increasing function $\tanh$ therefore [maps](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-map) $\R$ onto $\intoo{-1}1$. ∎

![Left: and , asymptotically glued to x/2 (dashed). Right: , increasing from -1 to 1.](https://one-course.com/images/onecourse/chapters/math-3/b1-functions/fig-714927f508af.svg)

*Left: $\cosh$ and $\sinh$, asymptotically glued to $\frac{\eu^x}{2}$ (dashed). Right: $\tanh$, increasing from $-1$ to $1$.*

**Remark 4.19 (Why “hyperbolic”?).**

The point $(\cosh t, \sinh t)$ runs along the branch $x > 0$ of the hyperbola $x^2 - y^2 = 1$ (by [Proposition 4.18](#prop-b1-functions-hyprules) (1)), exactly as $(\cos t, \sin
t)$ runs along the circle $x^2 + y^2 = 1$. Every circular identity has a hyperbolic sibling, with sign changes governed by $\sinh^2 \leftrightarrow -\sin^2$.

**Example 4.20 (The addition formula for tanh⁡\tanhtanh).**

Dividing the two addition formulas of [Proposition 4.18](#prop-b1-functions-hyprules) (3) by $\cosh x\cosh y$:

$$
\tanh(x + y)
= \frac{\sinh x\cosh y + \cosh x\sinh y}
{\cosh x\cosh y + \sinh x\sinh y}
= \frac{\tanh x + \tanh y}{1 + \tanh x\,\tanh y} ,
$$

the hyperbolic sibling of the tangent addition formula — with a $+$ where trigonometry has a $-$. A dividend: since $\abs{\tanh}
< 1$, the right-hand side is a “velocity addition” rule that never leaves $\intoo{-1}1$: if $u, v \in \intoo{-1}1$ then $\frac{u + v}{1 + uv} \in \intoo{-1}1$ as well (write $u = \tanh
a$, $v = \tanh b$, possible by bijectivity, and read the formula backwards). Checking that algebraically, without [hyperbolic functions](#def-b1-functions-hyperbolic), is a slightly painful exercise; parametrizing by $\tanh$ makes it one line — the same strategy that circular functions provide for the unit circle.

**Proposition 4.21 (Inverse hyperbolic functions).**

The inverses granted by [Proposition 4.18](#prop-b1-functions-hyprules) (4) have closed forms:

$$
\operatorname{arsinh} x = \ln\bigl(x + \sqrt{x^2 + 1}\bigr)
\ (x \in \R),
\qquad
\operatorname{arcosh} x = \ln\bigl(x + \sqrt{x^2 - 1}\bigr)
\ (x \geq 1),
$$

$$
\operatorname{artanh} x = \frac 12 \ln\frac{1 + x}{1 - x}
\ (\abs x < 1),
$$

with derivatives $\frac{1}{\sqrt{x^2+1}}$, $\frac{1}{\sqrt{x^2-1}}$ ($x > 1$) and $\frac{1}{1 - x^2}$ respectively.

**Proof.** For $\operatorname{arsinh}$: solve $x = \sinh y = \frac{\eu^y -
\eu^{-y}}{2}$. Setting $u = \eu^y > 0$: $u^2 - 2xu - 1 = 0$, so $u = x
+ \sqrt{x^2 + 1}$ (the root $x - \sqrt{x^2+1}$ is negative), and $y =
\ln(x + \sqrt{x^2+1})$. The other two are identical computations ($u^2 - 2xu + 1 = 0$ for $\operatorname{arcosh}$, keeping the root $\geq 1$; a two-line rearrangement for $\operatorname{artanh}$). Derivatives: differentiate the logarithmic expressions, e.g.

$$
\bigl(\ln(x + \sqrt{x^2+1})\bigr)'
= \frac{1 + \frac{x}{\sqrt{x^2+1}}}{x + \sqrt{x^2+1}}
= \frac{1}{\sqrt{x^2 + 1}} . \qedhere
$$

∎

**Example 4.22 (Closed forms at work).**

The solution of $\cosh t = 2$ with $t \geq 0$ is, by the closed form, $t = \operatorname{arcosh} 2 = \ln(2 + \sqrt3) \approx
1.317$; the other solution is $-t$, by evenness — and indeed $\ln(2 - \sqrt3) = \ln\frac1{2 + \sqrt3} = -\ln(2 + \sqrt3)$: the two roots $u = \eu^t$ of the quadratic $u^2 - 4u + 1 = 0$ are reciprocals, as their product $1$ (Vieta) demands. This tiny computation displays the general pattern: hyperbolic equations convert to quadratics in $\eu^t$, and the symmetry $t \mapsto -t$ appears as the symmetry $u \mapsto 1/u$ of the quadratic — worth remembering when solving [Exercise 4.7](#exo-b1-functions-7).

**Method 4.23 (Choosing the right primitive form).**

The three derivative patterns to memorize for integration ([Chapter 15](https://one-course.com/books/math/3/en/chapter/15-integration-on-a-segment#ch-b1-integration)):

$$
\int \frac{\dd x}{1 + x^2} = \arctan x + C,
\qquad
\int \frac{\dd x}{\sqrt{1 - x^2}} = \arcsin x + C,
\qquad
\int \frac{\dd x}{\sqrt{x^2 + 1}} = \operatorname{arsinh} x + C,
$$

and for a general quadratic, reduce to these by completing the square and rescaling.

**Remark 4.24 (Where this chapter is used).**

This chapter is the working vocabulary of all the analysis to come. The growth scale of [Proposition 4.6](#prop-b1-functions-powerrules) decides convergence questions throughout Chapters [11](https://one-course.com/books/math/3/en/chapter/11-sequences#ch-b1-seq) and [17](https://one-course.com/books/math/3/en/chapter/17-numerical-series#ch-b1-series); the derivative formulas of [Proposition 4.10](#prop-b1-functions-arcderiv) and [Proposition 4.21](#prop-b1-functions-invhyp) are the primitives most often needed in [Chapter 15](https://one-course.com/books/math/3/en/chapter/15-integration-on-a-segment#ch-b1-integration), via [Method 4.23](#met-b1-functions-primitive); [hyperbolic functions](#def-b1-functions-hyperbolic) parametrize the solutions of the equation $y'' = y$ in [Chapter 5](https://one-course.com/books/math/3/en/chapter/5-linear-differential-equations#ch-b1-diffeq) exactly as circular functions parametrize those of $y'' = -y$. The characterization of $\exp$ by $f' = f$, $f(0) = 1$ ([Proposition 4.1](#prop-b1-functions-expln)) is the one-dimensional seed of the theory of linear differential equations, and the catenary curve $y = \cosh x$ returns among the plane curves of [Chapter 24](https://one-course.com/books/math/3/en/chapter/24-plane-curves#ch-b1-curves).

**Remark 4.25 (Interlude: the inverse-function program).**

This chapter ran one program four times: restrict a function until it becomes a strictly monotonic bijection, name the inverse, transport every formula through it. What was *used* at each step — that a continuous strictly monotonic function on an interval is a bijection onto an interval, and that its inverse is continuous, then differentiable away from critical points — was borrowed on High School credit. The debt is repaid inside this volume: [Chapter 13](https://one-course.com/books/math/3/en/chapter/13-limits-and-continuity#ch-b1-continuity) proves the bijection [statement](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-statement) (the monotone inverse theorem, resting on the intermediate value theorem), and [Chapter 14](https://one-course.com/books/math/3/en/chapter/14-differentiation#ch-b1-derivative) proves the derivative rule $(f^{-1})' = 1/(f' \circ f^{-1})$ that silently produced every formula of [Proposition 4.10](#prop-b1-functions-arcderiv) and [Proposition 4.21](#prop-b1-functions-invhyp). Reading those chapters with $\arcsin$ and $\operatorname{arsinh}$ in mind — as the worked examples the theory was built to justify — is the intended way around the apparent circularity.

## 4.4 Exercises

**Exercise 4.1 ★.**

Compute without calculator: $\arcsin\frac{\sqrt 3}{2}$; $\;\arccos\bigl(-\frac 12\bigr)$; $\;\arctan(-1)$; $\;\arcsin\bigl(\sin\frac{5\pi}{6}\bigr)$; $\;\arccos\bigl(\cos\frac{7\pi}{4}\bigr)$.

**Solution of Exercise 4.1.**

$\arcsin\frac{\sqrt 3}{2} = \frac\pi3$; $\;\arccos\bigl(-\frac12\bigr) = \frac{2\pi}{3}$; $\;\arctan(-1) = -\frac\pi4$.

$\arcsin\bigl(\sin\frac{5\pi}{6}\bigr)$: $\sin\frac{5\pi}{6} =
\frac12$, and the angle of $\intcc{-\frac\pi2}{\frac\pi2}$ with sine $\frac12$ is $\frac\pi6$ (not $\frac{5\pi}{6}$).

$\arccos\bigl(\cos\frac{7\pi}{4}\bigr)$: $\cos\frac{7\pi}{4} =
\frac{\sqrt 2}{2}$, and the angle of $\intcc{0}{\pi}$ with that cosine is $\frac\pi4$.

**Exercise 4.2 ★.**

Give the domain of definition of $f(x) = \arcsin(2x - 1)$ and compute $f'$ where defined. Same questions for $g(x) = \arctan\sqrt{x}$.

**Solution of Exercise 4.2.**

$f(x) = \arcsin(2x - 1)$ requires $-1 \leq 2x - 1 \leq 1$, i.e. $x
\in \intcc{0}{1}$. On $\intoo{0}{1}$, the chain rule and [Proposition 4.10](#prop-b1-functions-arcderiv) give

$$
f'(x) = \frac{2}{\sqrt{1 - (2x-1)^2}} = \frac{2}{\sqrt{4x - 4x^2}}
= \frac{1}{\sqrt{x(1 - x)}} .
$$

$g(x) = \arctan\sqrt x$ is defined for $x \geq 0$, and for $x > 0$:

$$
g'(x) = \frac{1}{1 + x} \cdot \frac{1}{2\sqrt x}
= \frac{1}{2\sqrt x\,(1 + x)} .
$$

**Exercise 4.3 ★.**

Prove that for all $x \in \R$: $\cosh x + \sinh x = \eu^x$ and $(\cosh x + \sinh x)^n = \cosh nx + \sinh nx$ for every $n \in \Z$ (a hyperbolic de Moivre formula).

**Solution of Exercise 4.3.**

$\cosh x + \sinh x = \frac{\eu^x + \eu^{-x}}{2} + \frac{\eu^x -
\eu^{-x}}{2} = \eu^x$. Hence for $n \in \Z$:

$$
(\cosh x + \sinh x)^n = (\eu^x)^n = \eu^{nx}
= \cosh nx + \sinh nx .
$$

**Exercise 4.4 ★.**

Simplify $\cos(2\arcsin x)$ and $\sin(2\arctan x)$ into algebraic expressions of $x$.

**Solution of Exercise 4.4.**

With $y = \arcsin x$: $\cos 2y = 1 - 2\sin^2 y = 1 - 2x^2$, so $\cos(2\arcsin x) = 1 - 2x^2$.

With $y = \arctan x$: $\sin 2y = 2 \sin y \cos y = 2 \tan y \cos^2 y =
\frac{2\tan y}{1 + \tan^2 y}$, so $\sin(2\arctan x) =
\dfrac{2x}{1 + x^2}$.

**Exercise 4.5 ★.**

Order, for large $x$, the functions $x^{100}$, $\;\eu^{\sqrt x}$, $\;(\ln x)^{1000}$, $\;\eu^{x/100}$, $\;x^{\ln x}$, from slowest to fastest growth, with justifications based on [Proposition 4.6](#prop-b1-functions-powerrules).

**Solution of Exercise 4.5.**

From slowest to fastest:

$$
(\ln x)^{1000} \;\ll\; x^{100} \;\ll\; x^{\ln x} \;\ll\;
\eu^{\sqrt x} \;\ll\; \eu^{x/100} .
$$

Justifications: $(\ln x)^{1000}/x^{100} \to 0$ by [Proposition 4.6](#prop-b1-functions-powerrules) (logs lose to powers). $x^{100} = \eu^{100\ln x}$ and $x^{\ln x} = \eu^{(\ln x)^2}$: since $(\ln x)^2 - 100\ln x \to +\infty$, the second wins. $x^{\ln x} =
\eu^{(\ln x)^2} \ll \eu^{\sqrt x}$ because $(\ln x)^2/\sqrt x \to 0$ (logs lose to the power $x^{1/4}$, squared). Finally $\sqrt x - x/100
\to -\infty$, so $\eu^{\sqrt x} \ll \eu^{x/100}$.

**Exercise 4.6 ★★.**

Study the function $f(x) = \arctan\dfrac{2x}{1 - x^2}$ on its domain: compute $f'$, compare with $(2\arctan x)'$, and express $f(x)$ in terms of $\arctan x$ on each of the three intervals of the domain.

**Solution of Exercise 4.6.**

Domain: $x \neq \pm 1$, three intervals. On each,

$$
f'(x) = \frac{\bigl(\frac{2x}{1-x^2}\bigr)'}{1 +
\frac{4x^2}{(1-x^2)^2}}
= \frac{\frac{2(1-x^2) + 4x^2}{(1-x^2)^2}}
{\frac{(1-x^2)^2 + 4x^2}{(1-x^2)^2}}
= \frac{2(1 + x^2)}{(1 + x^2)^2} = \frac{2}{1 + x^2}
= (2\arctan x)' .
$$

So $f(x) - 2\arctan x$ is constant on each interval. Values: at $x =
0$, $f(0) = 0$: the constant is $0$ on $\intoo{-1}{1}$. As $x \to
+\infty$, $\frac{2x}{1-x^2} \to 0^-$ so $f \to 0$, while $2\arctan x
\to \pi$: the constant is $-\pi$ on $\intoo{1}{+\infty}$. By oddness, it is $+\pi$ on $\intoo{-\infty}{-1}$. Summary: $f = 2\arctan x$ on $\intoo{-1}{1}$, $f = 2\arctan x - \pi$ for $x > 1$, $f = 2\arctan x +
\pi$ for $x < -1$. (This is the double-angle formula for the tangent, read through $\arctan$.)

**Exercise 4.7 ★★.**

Solve in $\R$: $\;\cosh x = 2$; then $5\cosh x - 4 \sinh x = 3$ (express the solutions with logarithms). *Hint for the second: write everything with $u = \eu^x$.*

**Solution of Exercise 4.7.**

$\cosh x = 2$: with $u = \eu^x > 0$, $u + u^{-1} = 4$, so $u^2 - 4u +
1 = 0$, $u = 2 \pm \sqrt 3$: $x = \ln(2 + \sqrt 3)$ or $x = \ln(2 -
\sqrt 3) = -\ln(2 + \sqrt 3)$ (the two solutions are opposite, as $\cosh$ is even; both are valid). Equivalently $x = \pm
\operatorname{arcosh} 2$.

$5\cosh x - 4\sinh x = 3$: substituting the exponential definitions, $\frac{5(u + u^{-1}) - 4(u - u^{-1})}{2} = 3$, i.e. $u + 9u^{-1} =
6$, i.e. $u^2 - 6u + 9 = (u - 3)^2 = 0$: $u = 3$, one solution $x =
\ln 3$.

**Exercise 4.8 ★★.**

Prove the identity $\arctan\frac{1}{2} + \arctan\frac{1}{3} =
\frac{\pi}{4}$, then Machin’s formula

$$
4\arctan\frac 15 - \arctan\frac{1}{239} = \frac{\pi}{4}.
$$

*Hint: compute the tangent of both sides using the addition formula, and control in which interval each side lies.*

**Solution of Exercise 4.8.**

Let $\alpha = \arctan\frac12 + \arctan\frac13$. Addition formula:

$$
\tan\alpha = \frac{\frac12 + \frac13}{1 - \frac12\cdot\frac13}
= \frac{5/6}{5/6} = 1 .
$$

Both [arctangents](#def-b1-functions-arc) lie in $\intoo{0}{\frac\pi4}$ (their arguments are in $\intoo{0}{1}$), so $\alpha \in \intoo{0}{\frac\pi2}$; the only angle there with tangent $1$ is $\frac\pi4$.

Machin: let $\beta = \arctan\frac15$. Double angle twice:

$$
\tan 2\beta = \frac{2/5}{1 - 1/25} = \frac{5}{12},
\qquad
\tan 4\beta = \frac{2 \cdot 5/12}{1 - 25/144} = \frac{120}{119}.
$$

Then

$$
\tan\Bigl(4\beta - \frac\pi4\Bigr)
= \frac{\frac{120}{119} - 1}{1 + \frac{120}{119}}
= \frac{1/119}{239/119} = \frac{1}{239}.
$$

Location: $\beta < \arctan 1 = \frac\pi4$ and in fact $\tan 4\beta =
\frac{120}{119}$ close to $1$ with $4\beta \in \intoo{0}{\frac\pi2}$ (since $\beta < \frac\pi8$, as $\tan\frac\pi8 = \sqrt 2 - 1 > \frac15$); so $4\beta - \frac\pi4 \in \intoo{-\frac\pi4}{\frac\pi4}$, where $\arctan$ inverts $\tan$: $4\beta - \frac\pi4 =
\arctan\frac{1}{239}$, which is Machin’s formula.

**Exercise 4.9 ★★.**

Prove that for all $x \geq 0$: $\;x - \dfrac{x^3}{6} \leq \sin x \leq
x$ *(study the successive derivatives of the differences)*, and deduce $\lim_{x \to 0^+} \frac{x - \sin x}{x^3} = \frac 16$ is plausible — the limit itself is established in [Chapter 16](https://one-course.com/books/math/3/en/chapter/16-taylor-formulas-and-asymptotic-expansions#ch-b1-taylor).

**Solution of Exercise 4.9.**

Let $f(x) = x - \sin x$: $f(0) = 0$ and $f'(x) = 1 - \cos x \geq 0$, so $f \geq 0$ on $\R_+$: $\sin x \leq x$.

Let $g(x) = \sin x - x + \frac{x^3}{6}$: $g(0) = 0$, $g'(x) = \cos x -
1 + \frac{x^2}{2}$, $g'(0) = 0$, $g''(x) = -\sin x + x = f(x) \geq 0$ on $\R_+$. So $g'$ is increasing with $g'(0) = 0$, hence $g' \geq 0$, hence $g$ is increasing with $g(0) = 0$: $g \geq 0$, i.e. $x -
\frac{x^3}{6} \leq \sin x$ on $\R_+$.

Consequently $0 \leq \frac{x - \sin x}{x^3} \leq \frac 16$ for $x >
0$: the ratio is trapped near $\frac16$’s scale, and [Chapter 16](https://one-course.com/books/math/3/en/chapter/16-taylor-formulas-and-asymptotic-expansions#ch-b1-taylor) shows its limit is exactly $\frac 16$.

**Exercise 4.10 ★★★.**

For $x \in \intcc{-1}{1}$, set $h(x) = \arcsin\bigl(2x\sqrt{1 - x^2}\,\bigr) - 2\arcsin x$. One might expect $h = 0$ from the identity $\sin 2y = 2\sin y\cos y$ — but $h$ is *not* identically zero. Compute $h'$ on the open intervals where it exists, evaluate $h$ at well-chosen points, and give the full piecewise-constant description of $h$ on $\intcc{-1}{1}$.

**Solution of Exercise 4.10.**

Write $y = \arcsin x$, so $2x\sqrt{1 - x^2} = 2\sin y\cos y = \sin
2y$ and $h(x) = \arcsin(\sin 2y) - 2y$.

For $\abs x \leq \frac{\sqrt 2}{2}$: $2y \in
\intcc{-\frac\pi2}{\frac\pi2}$, so $\arcsin(\sin 2y) = 2y$ and $h =
0$.

For $x \in \intoc{\frac{\sqrt 2}{2}}{1}$: $2y \in
\intoc{\frac\pi2}{\pi}$, and the angle of $\intcc{-\frac\pi2}{\frac\pi2}$ with sine $\sin 2y$ is $\pi - 2y$: so $h(x) = \pi - 4\arcsin x$. (Check by derivative: there $h'(x) = \frac{2(1 - 2x^2)}{\abs{1 - 2x^2}\sqrt{1 - x^2}} -
\frac{2}{\sqrt{1-x^2}} = \frac{-4}{\sqrt{1 - x^2}}$, the derivative of $-4\arcsin x$; and at $x = 1$, $h(1) = \arcsin 0 - 2\cdot\frac\pi2
= -\pi = \pi - 4\cdot\frac\pi2$.)

For $x \in \intco{-1}{-\frac{\sqrt 2}{2}}$, by oddness of $h$: $h(x) = -\pi - 4\arcsin x$.

**Exercise 4.11 ★★★.**

(Gudermannian) Let $g(x) = \arctan(\sinh x)$ for $x \in \R$. Prove that $g$ is an odd, strictly increasing bijection from $\R$ onto $\intoo{-\frac\pi2}{\frac\pi2}$, that $g'(x) = \frac{1}{\cosh x}$, and that $\tan g(x) = \sinh x$, $\;\sin g(x) = \tanh x$, $\;\cos g(x) = \frac{1}{\cosh x}$: the function $g$ links circular and hyperbolic trigonometry without complex numbers.

**Solution of Exercise 4.11.**

$g = \arctan \circ \sinh$ is a composition of odd, strictly increasing functions, so it is odd and strictly increasing; as $x \to
+\infty$, $\sinh x \to +\infty$ so $g(x) \to \frac\pi2$, and $g$ is a bijection onto $\intoo{-\frac\pi2}{\frac\pi2}$ (continuity plus the intermediate value theorem). Chain rule with [Proposition 4.18](#prop-b1-functions-hyprules) (1):

$$
g'(x) = \frac{\cosh x}{1 + \sinh^2 x} = \frac{\cosh x}{\cosh^2 x}
= \frac{1}{\cosh x} .
$$

By construction $\tan g(x) = \sinh x$. Then, since $g(x) \in
\intoo{-\frac\pi2}{\frac\pi2}$ has positive cosine,

$$
\begin{align*}
\cos g(x) &= \frac{1}{\sqrt{1 + \tan^2 g(x)}}
= \frac{1}{\sqrt{1 + \sinh^2 x}} = \frac{1}{\cosh x},\\
\sin g(x) &= \tan g(x) \cos g(x) = \frac{\sinh x}{\cosh x} = \tanh x .
\end{align*}
$$

**Exercise 4.12 ★★.**

Prove that

$$
\arctan 1 + \arctan 2 + \arctan 3 = \pi .
$$

*Hint: compute $\arctan 2 + \arctan 3$ first, localizing the sum as in [Example 4.15](#ex-b1-functions-arctansum); beware that the addition formula’s denominator is negative here.*

**Solution of Exercise 4.12.**

Set $u = \arctan 2$ and $v = \arctan 3$; both lie in $\intoo{\frac\pi4}{\frac\pi2}$ (their arguments exceed $1$), so $u
+ v \in \intoo{\frac\pi2}{\pi}$. The addition formula gives

$$
\tan(u + v) = \frac{2 + 3}{1 - 6} = -1 ,
$$

and the unique angle of $\intoo{\frac\pi2}{\pi}$ with tangent $-1$ is $\frac{3\pi}4$: so $\arctan 2 + \arctan 3 = \frac{3\pi}4$. (Blindly applying $\arctan$ to the tangent would give $-\frac\pi4$, off by $\pi$ — the localization is what saves the computation, and the negative denominator $1 - 6$ is precisely the signal that the sum has left $\intoo{-\frac\pi2}{\frac\pi2}$.) Adding $\arctan 1 = \frac\pi4$:

$$
\arctan 1 + \arctan 2 + \arctan 3 = \frac\pi4 + \frac{3\pi}4 = \pi .
$$

## 4.5 Problem: The equation $x^y = y^x$

**Problem 4.1.**

Which pairs of positive numbers satisfy $x^y = y^x$? Everyone knows one accidental-looking example, $2^4 = 4^2 = 16$; this problem shows that nothing about it is accidental. The whole equation is governed by the variations of the single function

$$
f(t) = \frac{\ln t}{t} \qquad (t > 0),
$$

whose study yields: the complete solution [set](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-sets) (a diagonal plus one curved branch through $(\eu, \eu)$), a rational parametrization of the branch, the fact that $(2, 4)$ is its *only* integer point and the classification of all its rational points — plus, as dividends, the comparison of $\eu^\pi$ with $\pi^\eu$ and the monotone convergence of $\bigl(1 + \frac1n\bigr)^n$ to $\eu$.

**Part I — The function $f(t) = \ln t / t$.**

1. Justify that $f$ is differentiable on $\intoo0{+\infty}$ , compute $f'$ , and draw up the variation table: $f$ increases strictly on $\intoc0\eu$ , decreases strictly on $\intco\eu{+\infty}$ , with maximum $f(\eu) = \frac1\eu$ .
2. Determine the limits of $f$ at $0^+$ and $+\infty$ ( [Proposition 4.6](#prop-b1-functions-powerrules) ), the sign of $f$ (negative on $\intoo01$ , zero at $1$ , positive beyond), and sketch the graph.
3. Check by direct computation that $f(2) = f(4)$ . (Keep this equality in mind: the entire problem grows out of it.)
4. Let $c \in \R$ . Discuss, according to the value of $c$ , the number of solutions of $f(t) = c$ : exactly one for $c \leq  0$ ; exactly two (one in $\intoo1\eu$ , one in $\intoo\eu{+\infty}$ ) for $0 < c < \frac1\eu$ ; exactly one for $c = \frac1\eu$ ; none for $c > \frac1\eu$ .
5. Deduce that $\eu^x > x^\eu$ for every $x > 0$ with $x \neq  \eu$ , and in particular settle which of $\eu^\pi$ and $\pi^\eu$ is larger.

**Part II — The equation and its curve.** In this part $x, y > 0$.

6. Show that $x^y = y^x \iff f(x) = f(y)$ .
7. Deduce the structure of the solution [set](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-sets) : all diagonal pairs $(x, x)$ ; and the *nontrivial* pairs ( $x \neq  y$ ), which satisfy: both coordinates are $> 1$ , and one lies in $\intoo1\eu$ while the other lies in $\intoo\eu{+\infty}$ .
8. Parametrize the nontrivial pairs: writing $y = tx$ with $t  > 0$, $t \neq 1$, show that $x^y = y^x$ forces $$x(t) = t^{\frac1{t-1}},  \qquad  y(t) = t\,x(t) = t^{\frac{t}{t-1}},$$ and that conversely every such pair is a solution.
9. Verify that $t = 2$ gives $(2, 4)$ , and prove the symmetry $x(1/t) = y(t)$ , $y(1/t) = x(t)$ : inverting the parameter swaps the two coordinates.
10. Determine the limits of $x(t)$ and $y(t)$ as $t \to 1$ (both tend to $\eu$ ), as $t \to +\infty$ ( $x \to 1$ , $y  \to +\infty$ ) and as $t \to 0^+$ ( $x \to +\infty$ , $y \to  1$ ). Describe the resulting branch: a curve asymptotic to the lines $x = 1$ and $y = 1$ , crossing the diagonal at $(\eu, \eu)$ .
11. Show that $t \mapsto x(t)$ is strictly decreasing on $\intoo1{+\infty}$ . *(Study $h(t) = 1 - \frac1t - \ln  t$, whose sign controls the derivative of $\ln x(t) =  \frac{\ln t}{t - 1}$.)*
12. Conclude that the nontrivial solutions define a strictly decreasing bijection $\varphi \colon \intoo1\eu \to  \intoo\eu{+\infty}$ with $x^{\varphi(x)} = \varphi(x)^x$ , and that $\varphi$ is an involution of the branch: $\varphi(\varphi(x)) = x$ wherever both sides are defined.

**Part III — Integer and rational points.**

13. Prove that $(2, 4)$ and $(4, 2)$ are the only nontrivial *integer* solutions of $x^y = y^x$ .
14. For $n \in \N^*$, apply the parametrization with $t = 1 +  \frac1n$ and show that $$x_n = \Bigl(1 + \frac1n\Bigr)^{n},  \qquad  y_n = \Bigl(1 + \frac1n\Bigr)^{n+1}$$ is a nontrivial *rational* solution for every $n$, with $(x_1, y_1) = (2, 4)$.
15. Conversely, let $(x, y)$ be a nontrivial rational solution with $y > x$ , and write $t = y/x = 1 + \frac rs$ in lowest terms ( $r, s \in \N^*$ ). Using $x = t^{s/r}$ and admitting the uniqueness of prime factorization (familiar from school; proved in [Chapter 6](https://one-course.com/books/math/3/en/chapter/6-integer-arithmetic#ch-b1-arith) ), show that $x$ rational forces both $s$ and $s + r$ to be $r$ -th powers of integers.
16. Show with the binomial theorem that $a^r - b^r = r$ has no integer solutions $a > b \geq 1$ when $r \geq 2$ , and conclude: the rational solutions of $x^y = y^x$ are exactly the pairs $(x_n, y_n)$ of question 14 (and their swaps).
17. Verify the case $n = 2$ numerically: compute $f(9/4)$ and $f(27/8)$ to four decimal places and check they agree.

**Part IV — Dividends.**

18. Use questions 7 and 11 to prove, with no further computation, that the sequence $x_n = \bigl(1 +  \frac1n\bigr)^n$ is strictly increasing with $x_n < \eu$ , that $y_n = \bigl(1 + \frac1n\bigr)^{n+1}$ is strictly decreasing with $y_n > \eu$ , and that both converge to $\eu$ . (The parameter $t_n = 1 + \frac1n$ decreases to $1$ .)
19. Establish the general comparison rule for $1 < a < b$ : if $\eu \leq a$ then $a^b > b^a$ ; if $b \leq \eu$ then $a^b < b^a$ ; and show by the two examples $(2, 3)$ and $(2, 5)$ that in the mixed case $a < \eu < b$ both outcomes really occur.
20. Suppose the involution $\varphi$ of question 12 is differentiable at $\eu$ (it is). Differentiate the identity $\varphi(\varphi(x)) = x$ at $x = \eu$ and deduce $\varphi'(\eu) = -1$ : the branch crosses the diagonal at right angles to it.
21. Find the unique solution pair with $y = 3x$ , in closed form, and check it numerically to four decimals via $f$ .
22. Among the numbers $\sqrt2$ , $\sqrt[3]3$ , $\sqrt[4]4$ , $\sqrt[5]5$ , determine the largest and identify the two that are equal. *(Compare $n^{1/n} =  \eu^{f(n)}$.)*

**Part V — Synthesis.**

23. Describe the complete solution [set](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-sets) of $x^y = y^x$ in the quarter-plane $x, y > 0$ — diagonal plus branch, their intersection $(\eu, \eu)$ , the asymptotes, the integer point $(2, 4)$ , the rational points accumulating at $(\eu, \eu)$ — in a form you could sketch from memory.
24. Where exactly did the problem use: (i) the [growth comparisons](#prop-b1-functions-powerrules) of [Proposition 4.6](#prop-b1-functions-powerrules) ; (ii) the intermediate value theorem (through the bijection [statements](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-statement) ); (iii) the admitted uniqueness of prime factorization? One sentence each.
25. Moral, in a short paragraph: a single variation table solved an equation in two unknowns, classified its rational points, and proved the monotone convergence of $\bigl(1 + \frac1n\bigr)^n$ — comment on this economy, and name where each thread is industrialized later in the volume ( [Chapter 11](https://one-course.com/books/math/3/en/chapter/11-sequences#ch-b1-seq) for the sequence, [Chapter 14](https://one-course.com/books/math/3/en/chapter/14-differentiation#ch-b1-derivative) for variation tables, [Chapter 16](https://one-course.com/books/math/3/en/chapter/16-taylor-formulas-and-asymptotic-expansions#ch-b1-taylor) for the precision the table lacks).

**Solution of Problem 4.1.**

**1.** $f$ is a quotient of differentiable functions with nonvanishing denominator on $\intoo0{+\infty}$, and

$$
f'(t) = \frac{\frac1t \cdot t - \ln t}{t^2} = \frac{1 - \ln
t}{t^2} ,
$$

positive for $t < \eu$, zero at $t = \eu$, negative for $t > \eu$: $f$ increases strictly on $\intoc0\eu$, decreases strictly on $\intco\eu{+\infty}$, with maximum $f(\eu) = \frac1\eu$.

**2.** As $t \to 0^+$: $\ln t \to -\infty$ and $\frac1t \to
+\infty$, so $f(t) \to -\infty$. As $t \to +\infty$: $f(t) \to 0$ by the [growth comparison](#prop-b1-functions-powerrules) of [Proposition 4.6](#prop-b1-functions-powerrules) ($\beta = \alpha = 1$). Sign: that of $\ln t$, so $f < 0$ on $\intoo01$, $f(1) = 0$, $f > 0$ on $\intoo1{+\infty}$. The graph climbs from $-\infty$, crosses zero at $1$, peaks at $(\eu,
\frac1\eu)$, then decays to $0^+$.

**3.** $f(4) = \frac{\ln 4}4 = \frac{2\ln 2}4 = \frac{\ln 2}2
= f(2)$.

**4.** By the variation table and the intermediate value theorem (used at High School level; formalized in [Chapter 13](https://one-course.com/books/math/3/en/chapter/13-limits-and-continuity#ch-b1-continuity)). For $c < 0$: solutions exist only where $f < 0$, i.e. in $\intoo01$ where $f$ is a strictly increasing bijection onto $\intoo{-\infty}0$: exactly one. For $c = 0$: only $t = 1$. For $0 < c < \frac1\eu$: on $\intoo1\eu$, $f$ increases from $0$ to $\frac1\eu$: one solution; on $\intoo\eu{+\infty}$, $f$ decreases from $\frac1\eu$ to $0$: one more; total two. For $c = \frac1\eu$: only the maximum point $t = \eu$. For $c >
\frac1\eu$: none.

**5.** For $x \neq \eu$, strict maximality gives $f(x) <
f(\eu)$, i.e. $\frac{\ln x}x < \frac1\eu$, i.e. $\eu\ln x < x$, i.e. $\ln(x^\eu) < \ln(\eu^x)$: $x^\eu < \eu^x$. With $x = \pi$: $\eu^\pi > \pi^\eu$ (numerically $23.14 > 22.46$).

**6.** For $x, y > 0$, both sides are positive, so

$$
x^y = y^x \iff y\ln x = x\ln y \iff \frac{\ln x}x = \frac{\ln y}y
\iff f(x) = f(y) ,
$$

dividing by $xy > 0$.

**7.** Let $f(x) = f(y) = c$ with $x \neq y$. If $c \leq 0$ or $c = \frac1\eu$, question 4 says the equation $f = c$ has a single solution: impossible. So $0 < c < \frac1\eu$, and again by question 4 the two solutions are one point of $\intoo1\eu$ and one of $\intoo\eu{+\infty}$: both coordinates exceed $1$ and they straddle $\eu$.

**8.** Substituting $y = tx$ in $y\ln x = x\ln y$:

$$
tx\ln x = x(\ln t + \ln x)
\iff (t - 1)\ln x = \ln t
\iff \ln x = \frac{\ln t}{t - 1} ,
$$

so $x = t^{1/(t-1)}$ and $y = tx = t^{1 + \frac1{t-1}} =
t^{t/(t-1)}$. Conversely, for these values, $\ln y = \frac{t\ln
t}{t-1} = t\ln x$ and $y = tx$, so $\frac{\ln y}y = \frac{t\ln
x}{tx} = \frac{\ln x}x$: a solution. Every nontrivial solution has some ratio $t = y/x \in \intoo0{+\infty} \setminus \{1\}$, so the parametrization is complete.

**9.** $t = 2$: $x = 2^{1/1} = 2$, $y = 2^{2/1} = 4$. And

$$
x(1/t) = (1/t)^{\frac1{\frac1t - 1}}
= (t^{-1})^{\frac{t}{1 - t}}
= t^{\frac{t}{t - 1}} = y(t) ,
$$

then $y(1/t) = \frac1t\,x(1/t) = \frac{y(t)}t = x(t)$: the parameter change $t \mapsto 1/t$ swaps the coordinates, as the symmetry of the equation demands.

**10.** As $t \to 1$: $\frac{\ln t}{t-1} \to 1$ (it is the difference quotient of $\ln$ at $1$), so $x \to \eu^1 = \eu$ and $y = tx \to \eu$. As $t \to +\infty$: $\ln x = \frac{\ln t}{t-1}
\to 0$ so $x \to 1$, while $\ln y = \frac{t\ln t}{t-1} \to
+\infty$ so $y \to +\infty$. As $t \to 0^+$: $\ln x = \frac{\ln
t}{t-1} \to \frac{-\infty}{-1} = +\infty$ so $x \to +\infty$, while $\ln y = \frac{t\ln t}{t-1} \to \frac{0}{-1} = 0$ so $y \to
1$ (using $t\ln t \to 0$, [Example 4.7](#ex-b1-functions-xx)). The branch therefore runs from the asymptote $y = 1$ (far right), up through $(\eu, \eu)$ on the diagonal, and off along the asymptote $x = 1$ (far top) — symmetric about the diagonal by question 9.

**11.** $\ln x(t) = g(t) = \frac{\ln t}{t-1}$, and

$$
g'(t) = \frac{\frac{t-1}t - \ln t}{(t-1)^2}
= \frac{h(t)}{(t-1)^2},
\qquad h(t) = 1 - \frac1t - \ln t .
$$

$h(1) = 0$ and $h'(t) = \frac1{t^2} - \frac1t = \frac{1 -
t}{t^2} < 0$ for $t > 1$: $h < 0$ on $\intoo1{+\infty}$, so $g' <
0$ and $x = \eu^g$ is strictly decreasing there (from $\eu$ to $1$, by question 10).

**12.** By question 11, $t \mapsto x(t)$ is a strictly decreasing bijection from $\intoo1{+\infty}$ onto $\intoo1\eu$; write $t(x)$ for its inverse (also strictly decreasing) and set $\varphi(x) = y(t(x))$. Note $h < 0$ also on $\intoo01$ (there $h' > 0$ and $h(1) = 0$), so $x(\cdot)$ is decreasing on $\intoo01$ as well; hence $y(t) = x(1/t)$ (question 9) is *increasing* in $t$ on $\intoo1{+\infty}$, from $\eu$ to $+\infty$. Composing: $\varphi = y \circ t(\cdot)$ is strictly decreasing from $\intoo1\eu$ onto $\intoo\eu{+\infty}$, and $x^{\varphi(x)} = \varphi(x)^x$ by question 8. Finally, for a common value $c = f(x) \in \intoo0{\frac1\eu}$, the pair $\{x,
\varphi(x)\}$ is *the* two-element solution [set](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-sets) of $f = c$ (question 4); extending $\varphi$ to $\intoo\eu{+\infty}$ as the inverse [map](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-map), $\varphi(\varphi(x))$ returns to the other (i.e. original) element: $\varphi \circ \varphi = \mathrm{id}$.

**13.** If $(x, y)$ is a nontrivial integer solution with $x
< y$, question 7 puts $x \in \intoo1\eu$: the only integer there is $x = 2$. Then $f(y) = f(2) = \frac{\ln2}2$ with $y > \eu$; by question 4 the equation $f = \frac{\ln2}2$ has exactly one solution beyond $\eu$, and question 3 exhibits it: $y = 4$. Hence $(2, 4)$, and its swap, are the only ones.

**14.** $t = 1 + \frac1n = \frac{n+1}n$ gives $\frac1{t - 1}
= n$ and $\frac t{t-1} = n + 1$, so

$$
x_n = \Bigl(\frac{n+1}n\Bigr)^{n},
\qquad
y_n = \Bigl(\frac{n+1}n\Bigr)^{n+1},
$$

manifestly rational, distinct ($y_n = t\,x_n \neq x_n$), and $n =
1$ gives $x_1 = 2$, $y_1 = 4$.

**15.** $t = y/x$ is rational and $> 1$; write $t - 1 =
\frac rs$ in lowest terms, so $t = \frac{s + r}s$ with $\gcd(s +
r, s) = \gcd(r, s) = 1$. Question 8 gives $x = t^{s/r}$, hence

$$
x^r = t^s = \frac{(s+r)^s}{s^s} ,
$$

a fraction in lowest terms (no prime divides both $s + r$ and $s$). Writing $x = \frac pq$ in lowest terms, $x^r = \frac{p^r}
{q^r}$ is also in lowest terms, and by uniqueness of the reduced representation: $p^r = (s+r)^s$ and $q^r = s^s$. Compare the exponent of any prime $\ell$ in $q^r = s^s$: $r \cdot
v_\ell(q) = s \cdot v_\ell(s)$, so $r$ divides $s\,v_\ell(s)$; since $\gcd(r, s) = 1$, $r$ divides $v_\ell(s)$ for every prime $\ell$ (unique factorization, admitted; proved in [Chapter 6](https://one-course.com/books/math/3/en/chapter/6-integer-arithmetic#ch-b1-arith)), so $s = b^r$ for an integer $b$. The same argument on $p^r = (s+r)^s$ gives $s + r = a^r$.

**16.** Suppose $a^r - b^r = r$ with integers $a > b \geq 1$ and $r \geq 2$. Then $a \geq b + 1$, so by the binomial theorem

$$
a^r - b^r \geq (b+1)^r - b^r
= \sum_{k=0}^{r-1}\binom rk b^k
\geq \binom r0 + \binom r{r-1} b^{r-1}
\geq 1 + r > r ,
$$

a contradiction. From question 15, $a^r - b^r = (s + r) - s = r$ forces $r = 1$: $t = 1 + \frac1s$, and the solution is $(x_s,
y_s)$ of question 14. Together with the swapped pairs, the classification is complete.

**17.** $f\bigl(\tfrac94\bigr) = \frac{\ln 2.25}{2.25} =
\frac{0.81093}{2.25} = 0.3604$ and $f\bigl(\tfrac{27}8\bigr) =
\frac{\ln 3.375}{3.375} = \frac{1.21640}{3.375} = 0.3604$ (four decimals): equal, as the construction promises — so $\bigl(\tfrac94\bigr)^{27/8} = \bigl(\tfrac{27}8\bigr)^{9/4}$.

**18.** The parameters $t_n = 1 + \frac1n$ decrease strictly to $1$. Since $x(\cdot)$ is strictly decreasing on $\intoo1{+\infty}$ (question 11), $x_n = x(t_n)$ is strictly *increasing*; since $y(\cdot)$ is strictly increasing there (question 12), $y_n = y(t_n)$ is strictly *decreasing*. By question 7, $x_n \in \intoo1\eu$ and $y_n \in
\intoo\eu{+\infty}$: so $x_n < \eu < y_n$ for every $n$. Finally $t_n \to 1$ and question 10 give $x_n \to \eu$ and $y_n \to \eu$. The classical monotone convergence of $\bigl(1 +
\frac1n\bigr)^n$ falls out of the geometry of the branch, with no new inequality.

**19.** If $\eu \leq a < b$: $f$ strictly decreasing on $\intco\eu{+\infty}$ gives $f(a) > f(b)$, i.e. $b\ln a > a\ln
b$, i.e. $a^b > b^a$. If $1 < a < b \leq \eu$: $f$ strictly increasing gives $f(a) < f(b)$, hence $a^b < b^a$. Mixed case: $2 < \eu < 3$ and $2^3 = 8 < 9 = 3^2$; but $2 < \eu < 5$ and $2^5 = 32 > 25 = 5^2$: both outcomes occur, decided by which side of the branch the point $(a, b)$ falls.

**20.** The branch meets the diagonal at $\eu$ and $\varphi$ extends continuously there with $\varphi(\eu) = \eu$. Differentiating $\varphi(\varphi(x)) = x$ at $x = \eu$ by the chain rule: $\varphi'(\varphi(\eu))\,\varphi'(\eu) =
\varphi'(\eu)^2 = 1$, so $\varphi'(\eu) = \pm1$; $\varphi$ is decreasing, so $\varphi'(\eu) = -1$. The branch crosses the diagonal with slope $-1$: perpendicularly.

**21.** $y = 3x$ is the case $t = 3$: $x = 3^{1/2} = \sqrt3$ and $y = 3^{3/2} = 3\sqrt3$. Check: $f(\sqrt3) =
\frac{0.54931}{1.73205} = 0.3171$ and $f(3\sqrt3) =
\frac{\ln 5.19615}{5.19615} = \frac{1.64792}{5.19615} = 0.3171$: equal to four decimals, so $(\sqrt3)^{3\sqrt3} =
(3\sqrt3)^{\sqrt3}$.

**22.** $n^{1/n} = \eu^{f(n)}$, and $\exp$ is increasing, so the ordering is that of $f(n)$: $f(3) = \frac{\ln3}3 \approx
0.3662$ beats $f(2) = f(4) = \frac{\ln2}2 \approx 0.3466$ (question 3) and $f(5) \approx 0.3219$. The largest is $\sqrt[3]3$, and the equal pair is $\sqrt2 = \sqrt[4]4$ — the integer pair $(2, 4)$ in yet another disguise.

**23.** The solution [set](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-sets) is the union of the diagonal $\{(x,
x) : x > 0\}$ and a single branch, symmetric about the diagonal, strictly decreasing from the asymptote $x = 1$ (as $y \to
+\infty$) to the asymptote $y = 1$ (as $x \to +\infty$), crossing the diagonal exactly once, at $(\eu, \eu)$, with slope $-1$ there. On the branch sit the integer points $(2, 4)$ and $(4, 2)$ — the only ones — and the rational points $(x_n, y_n)$, which march monotonically along the branch toward $(\eu, \eu)$ without ever reaching it ($\eu$ is irrational; the branch’s rational points accumulate at an irrational corner).

**24.** (i) [Growth comparisons](#prop-b1-functions-powerrules) gave the limits of $f$ at $+\infty$ (question 2) and $t\ln t \to 0$ (question 10), shaping both the variation table and the asymptotes. (ii) The intermediate value theorem, through the bijection [statements](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-statement), converted the variation table into exact solution counts (question 4) and into the existence of the inverse [map](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-map) $t(x)$ (question 12). (iii) Unique factorization powered the two lowest-terms identifications of question 15, the arithmetic heart of the rational-point classification.

**25.** One derivative computation — the sign of $1 - \ln
t$ — generated everything: the solution count for each level $c$, the shape and asymptotes of the branch, the extremal inequality $\eu^x \geq x^\eu$, the classification of integer and rational solutions, and the monotone convergence of $\bigl(1 +
\frac1n\bigr)^n$. This is the economy of thinking with variation tables: a one-dimensional study resolves a two-variable equation because the equation factors through a single function. [Chapter 11](https://one-course.com/books/math/3/en/chapter/11-sequences#ch-b1-seq) will redo the convergence of $(x_n)$ with the general theory of monotone sequences; [Chapter 14](https://one-course.com/books/math/3/en/chapter/14-differentiation#ch-b1-derivative) founds variation tables rigorously on the mean value theorem; and [Chapter 16](https://one-course.com/books/math/3/en/chapter/16-taylor-formulas-and-asymptotic-expansions#ch-b1-taylor) supplies what the table cannot — the *speed* of the convergence $x_n \to \eu$ (it is of order $1/n$).