Chapter 2Continuity of Functions, the Mean Value Theorem, and Hyperbolic Functions

Proof: $\displaystyle\lim_{x\to0}\frac{\sin x}{x}=1$ (the case $0 \lt x \lt \frac\pi2$)

Computing the areas of triangle $OBP$, sector $OBP$, and triangle $OBT$ by the formulas “$\frac12\times$ base $\times$ height” and “$\frac12\times$ radius$^2\times$ central angle,” respectively, we get

$$ \underbrace{\frac12\cdot1\cdot1\cdot\sin x}_{\text{triangle }OBP} \ \le\ \underbrace{\frac12\cdot1\cdot1\cdot x}_{\text{sector }OBP} \ \le\ \underbrace{\frac12\cdot1\cdot1\cdot\tan x}_{\text{triangle }OBT} $$

(The area of triangle $OBP$ is $\frac12\cdot1\cdot\sin x$, because its base is $OB=1$ and its height is the $y$-coordinate $\sin x$ of $P$. The area of the sector is $\frac12 x$, obtained by substituting $r=1,\theta=x$ into the area formula $\frac12 r^2\theta$ for a sector with central angle $x$ and radius $1$. The area of triangle $OBT$ is $\frac12\tan x$, because its base is $OB=1$ and its height is $BT=\tan x$.) Multiplying both sides by $2$, we obtain

$$ \sin x \le x \le \tan x $$

We split this into the two inequalities

$$ \begin{cases}\sin x\le x&\cdots① \\ x\le\tan x&\cdots②\end{cases} $$

and treat them separately. Dividing both sides of ① by $x$ (since $x \gt 0$ for $0 \lt x \lt \frac\pi2$), we get

$$ \frac{\sin x}{x}\le1 $$

For ②, substituting $\tan x=\dfrac{\sin x}{\cos x}$ gives $x\le\dfrac{\sin x}{\cos x}$, so multiplying both sides by $\cos x$ (since $\cos x \gt 0$ for $0 \lt x \lt \frac\pi2$), we get

$$ x\cos x\le\sin x\quad\text{that is,}\quad\cos x\le\frac{\sin x}{x} $$

Combining the two inequalities $\dfrac{\sin x}{x}\le1$ and $\cos x\le\dfrac{\sin x}{x}$ obtained from ① and ②, we get

$$ \cos x \le \frac{\sin x}{x} \le 1 \qquad\cdots③ $$

This is an inequality that sandwiches $\dfrac{\sin x}{x}$ between two quantities. Taking $\displaystyle\lim_{x\to0}$ of each side of ③, we obtain

$$ \lim_{x\to0}\cos x \ \le\ \lim_{x\to0}\frac{\sin x}{x} \ \le\ \lim_{x\to0}1 $$

Since $\cos x$ is a function continuous at $x=0$, we have $\displaystyle\lim_{x\to0}\cos x=\cos0=1$ (the definition of continuity is stated precisely in Section 2.2; here we use it as a fact already learned in high school). Thus the leftmost and rightmost sides are both $1$, so by the squeeze theorem proved in Chapter 1, Section 1.3 (the theorem stating that a quantity squeezed from both sides by quantities that approach the same value converges to that value), we obtain

$$ \therefore\ \lim_{x\to0}\frac{\sin x}{x} = 1 $$

Thus the stated limit holds. (For the case $0 \gt x \gt -\frac\pi2$, the same result follows from $\sin(-x)/(-x)=\sin x/x$, so the limit holds as a two-sided limit.)

Proof: $\displaystyle\lim_{x\to0}\frac{\tan x}{x}=1$

Since $\tan x=\dfrac{\sin x}{\cos x}$,

$$ \lim_{x\to0}\frac{\tan x}{x} = \lim_{x\to0}\frac{\dfrac{\sin x}{\cos x}}{x} = \lim_{x\to0}\underbrace{\frac{\sin x}{x}}_{\to\,1}\cdot\frac{1}{\cos x} $$

(We separated $\dfrac{\sin x}{\cos x\cdot x}=\dfrac{\sin x}{x}\cdot\dfrac1{\cos x}$ into a product.) Since $\cos x$ is continuous at $x=0$ and $\cos0=1\neq0$, we have $\displaystyle\lim_{x\to0}\frac{1}{\cos x}=\frac{1}{\lim_{x\to0}\cos x}=\frac11=1$. Therefore, by the product rule for limits (if each limit exists, the limit of a product equals the product of the limits),

$$ \lim_{x\to0}\frac{\tan x}{x} = 1\cdot1 = 1 \qquad\therefore\ \lim_{x\to0}\frac{\tan x}{x}=1 $$

Proof: $\displaystyle\lim_{x\to0}\frac{1-\cos x}{x^2}=\frac12$

We multiply the numerator and denominator by $1+\cos x$ (this is an equivalent transformation, because $1+\cos x\neq0$ is guaranteed near $x=0$).

$$ \lim_{x\to0}\frac{1-\cos x}{x^2} = \lim_{x\to0}\frac{(1-\cos x)(1+\cos x)}{x^2(1+\cos x)} = \lim_{x\to0}\frac{1-\cos^2x}{x^2(1+\cos x)} $$

By the basic trigonometric identity $\sin^2x+\cos^2x=1$ (already learned in high school), $1-\cos^2x=\sin^2x$, so

$$ = \lim_{x\to0}\frac{\sin^2x}{x^2(1+\cos x)} = \lim_{x\to0}\left(\frac{\sin x}{x}\right)^2\cdot\frac{1}{1+\cos x} $$

(We separated $\dfrac{\sin^2x}{x^2}=\left(\dfrac{\sin x}{x}\right)^2$.) Using the result $\lim_{x\to0}\frac{\sin x}{x}=1$ of Section 2.1.1 and the continuity of $\cos x$, $\lim_{x\to0}\cos x=1$, we get

$$ = 1^2\cdot\frac{1}{1+1} = 1\cdot\frac12 = \frac12 \qquad\therefore\ \lim_{x\to0}\frac{1-\cos x}{x^2}=\frac12 $$

Derivation: $\displaystyle\sin\theta=\frac{e^{i\theta}-e^{-i\theta}}{2i}$

By Euler's formula, $e^{i\theta}=\cos\theta+i\sin\theta\ \cdots①$. Also, by de Moivre's theorem, $e^{-i\theta}=\cos\theta-i\sin\theta\ \cdots②$. Computing ①$-$②, the $\cos\theta$ terms cancel each other, and we obtain

$$ e^{i\theta}-e^{-i\theta} = 2i\sin\theta $$

Dividing both sides by $2i$ (this is allowed because $i\neq0$), we get

$$ \frac{e^{i\theta}-e^{-i\theta}}{2i} = \sin\theta \qquad\therefore\ \sin\theta = \frac{e^{i\theta}-e^{-i\theta}}{2i} $$

Derivation: $\displaystyle\cos\theta=\frac{e^{i\theta}+e^{-i\theta}}{2}$

We use ① and ② again. This time, computing ①$+$②, the $i\sin\theta$ terms cancel each other, and we obtain

$$ e^{i\theta}+e^{-i\theta} = 2\cos\theta $$

Dividing both sides by $2$, we get

$$ \frac{e^{i\theta}+e^{-i\theta}}{2} = \cos\theta \qquad\therefore\ \cos\theta = \frac{e^{i\theta}+e^{-i\theta}}{2} $$

Derivation: $\displaystyle\tan\theta=\frac{e^{i\theta}-e^{-i\theta}}{i(e^{i\theta}+e^{-i\theta})}$

Substituting the two preceding results into $\tan\theta=\dfrac{\sin\theta}{\cos\theta}$, we get

$$ \tan\theta = \frac{\dfrac{e^{i\theta}-e^{-i\theta}}{2i}}{\dfrac{e^{i\theta}+e^{-i\theta}}{2}} $$

This is a complex fraction (a fraction whose numerator and denominator are themselves fractions). Multiplying the numerator and the denominator by $2i$ and simplifying (the numerator becomes $\frac{e^{i\theta}-e^{-i\theta}}{2i}\times2i=e^{i\theta}-e^{-i\theta}$, and the denominator becomes $\frac{e^{i\theta}+e^{-i\theta}}{2}\times2i=i(e^{i\theta}+e^{-i\theta})$), we get

$$ = \frac{e^{i\theta}-e^{-i\theta}}{i(e^{i\theta}+e^{-i\theta})} \qquad\therefore\ \tan\theta = \frac{e^{i\theta}-e^{-i\theta}}{i(e^{i\theta}+e^{-i\theta})} $$

Proof

Since $f(x)$ is continuous on $[a,b]$ and differentiable on $(a,b)$, its graph is connected smoothly (with no breaks or sharp corners). We divide the proof into three cases according to how $f(x)$ compares with $f(a)$.

[1] The case where $f(x) \gt f(a)$ for some $x\in(a,b)$.

Since $f(x)$ is continuous on $[a,b]$, the extreme value theorem (the theorem that a continuous function on a bounded closed interval always has a maximum and a minimum), which we accept as a prerequisite, shows that $f(x)$ has a maximum on $[a,b]$. By the hypothesis $f(x) \gt f(a)$ ($=f(b)$) of Case [1], the point where this maximum is attained is not an endpoint $a,b$ but some point $c$ ($a \lt c \lt b$) in the interior of the open interval. That is, $f(x)$ has a maximum value $f(c)$ ($a \lt c \lt b$). Even if we shift from the point $c$ where the maximum is attained by any amount $h$ ($h\neq0$) in either direction, the value must be at most the maximum, so the following inequality holds:

$$ f(c) \geqq f(c+h) \qquad\cdots① $$

We now restate this inequality in terms of the derivative at a point, splitting into cases according to the sign of $h$.

[i] The case $h \gt 0$. By ①, $f(c+h)-f(c)\leqq0$. Dividing this by $h$ (a positive number) does not change the direction of the inequality, so

$$ \frac{f(c+h)-f(c)}{h} \leqq 0 $$

Taking the limit $h\to0$ (since $f$ is differentiable at $c$, the limit $f'(c)$ exists, and taking the limit preserves the direction of $\leqq$), we get

$$ \lim_{h\to0^+}\frac{f(c+h)-f(c)}{h} = f'(c) \leqq 0 $$

[ii] The case $h \lt 0$. By ① we likewise have $f(c+h)-f(c)\leqq0$. This time we divide by $h$ (a negative number), so the direction of the inequality is reversed, giving

$$ \frac{f(c+h)-f(c)}{h} \geqq 0 $$

Taking the limit $h\to0$,

$$ \lim_{h\to0^-}\frac{f(c+h)-f(c)}{h} = f'(c) \geqq 0 $$

Combining [i] and [ii], we obtain $0\leqq f'(c)\leqq0$, and hence $f'(c)=0$.

[2] The case where $f(x) \lt f(a)$ for some $x\in(a,b)$.

We argue symmetrically to [1]. By the extreme value theorem, $f(x)$ has a minimum on $[a,b]$. By the hypothesis of [2], the point where this minimum is attained is also not an endpoint but an interior point $c$ ($a \lt c \lt b$). A point shifted from $c$ by $h\ (h\neq0)$ always satisfies

$$ f(c) \leqq f(c+h) \qquad\cdots② $$

[i] When $h \gt 0$, ② gives $f(c+h)-f(c)\geqq0$, so dividing by the positive number $h$ we get $\dfrac{f(c+h)-f(c)}{h}\geqq0$, and taking the limit gives $f'(c)\geqq0$. [ii] When $h \lt 0$, ② again gives $f(c+h)-f(c)\geqq0$, but dividing by the negative number $h$ reverses the inequality, so $\dfrac{f(c+h)-f(c)}{h}\leqq0$, and taking the limit gives $f'(c)\leqq0$. Combining [i] and [ii], we obtain $0\leqq f'(c)\leqq0$, so once again $f'(c)=0$ holds.

[3] The case where $f(x)=f(a)$ for all $x\in(a,b)$.

In this case $f(x)$ is a constant function that takes the constant value $f(a)$ throughout $(a,b)$. The derivative of a constant function is identically $0$, so $f'(x)=0$ throughout the whole range $a \lt x \lt b$. Therefore, no matter which point of the interval we choose as $c$, $f'(c)=0$ holds.

The cases [1], [2], and [3] are mutually exclusive, and every $x\in(a,b)$ necessarily belongs to one of the cases (combining the negation of [1] and the negation of [2] gives [3]). Thus, in each case we have shown that there exists a $c$ satisfying $a \lt c \lt b$ for which $f'(c)=0$.

∎

Proof

The equation of the line through the point $A(a,f(a))$ and the point $B(b,f(b))$, as a line of slope $\dfrac{f(b)-f(a)}{b-a}$, can be written as

$$ y = \frac{f(b)-f(a)}{b-a}(x-a) + f(a) $$

Let $F(x)$ denote the difference function between the curve $y=f(x)$ and this line $y$.

$$ F(x) = f(x) - y = f(x) - \frac{f(b)-f(a)}{b-a}(x-a) - f(a) \qquad\cdots① $$

Substituting $x=a,b$ into ①, we check the values at the two endpoints.

[1] When $x=a$,

$$ F(a) = f(a) - \frac{f(b)-f(a)}{b-a}(a-a) - f(a) = f(a) - 0 - f(a) = 0 $$

(Since $(a-a)=0$, the middle term vanishes entirely.)

[2] When $x=b$,

$$ F(b) = f(b) - \frac{f(b)-f(a)}{b-a}(b-a) - f(a) = f(b) - \bigl(f(b)-f(a)\bigr) - f(a) = f(b)-f(b)+f(a)-f(a) = 0 $$

(We used the fact that $\dfrac{f(b)-f(a)}{b-a}\times(b-a)=f(b)-f(a)$ by cancellation.)

$F(x)$ is the difference between the continuous function $f(x)$ and a linear function, so it is continuous on $[a,b]$; and it is the difference between the differentiable function $f(x)$ and a linear function, so it is differentiable on $(a,b)$; thus its graph is a smooth curve. Moreover, by [1] and [2] we have $F(a)=F(b)(=0)$, so Rolle's theorem (Theorem 2.1) applies directly, and there necessarily exists a $c$ satisfying

$$ F'(c) = 0 \qquad(a \lt c \lt b) $$

Now we differentiate ① with respect to $x$. The factor $\dfrac{f(b)-f(a)}{b-a}$ in the second term on the right-hand side is a constant that does not depend on $x$ (since we are regarding $a,b$ as fixed), and noting that differentiating $(x-a)$ with respect to $x$ gives $1$, we get

$$ F'(x) = f'(x) - \frac{f(b)-f(a)}{b-a} $$

Substituting $x=c$ into this and using $F'(c)=0$, we obtain

$$ F'(c) = f'(c) - \frac{f(b)-f(a)}{b-a} = 0 \qquad(a \lt c \lt b) $$

Rearranging this (we simply move $-\dfrac{f(b)-f(a)}{b-a}$ to the other side), we get

$$ \frac{f(b)-f(a)}{b-a} = f'(c) \qquad(a \lt c \lt b) $$

which is the desired conclusion of the mean value theorem.

∎

Proof

Let $X=g(x)$, $Y=f(x)$ ($a\le x\le b$), and regard this as a curve in the $(X,Y)$ plane. The equation $Y$ of the line through the point $A(g(a),f(a))$ and the point $B(g(b),f(b))$ can be written, as the line with slope $\dfrac{f(b)-f(a)}{g(b)-g(a)}$ (since $g(a)\neq g(b)$, the slope is defined), as

$$ Y = \frac{f(b)-f(a)}{g(b)-g(a)}\bigl\{X-g(a)\bigr\} + f(a) $$

Let $F(x)$ be the difference function between $f(x)$ and this line $Y$ (with $X=g(x)$ substituted). Then

$$ F(x) = f(x) - Y = f(x) - \frac{f(b)-f(a)}{g(b)-g(a)}\bigl\{g(x)-g(a)\bigr\} - f(a) \qquad(a\le x\le b) $$

(We substituted $X=g(x)$.) Substitute $x=a,b$ into $F(x)$.

[1] When $x=a$,

$$ F(a) = f(a) - \frac{f(b)-f(a)}{g(b)-g(a)}\bigl\{g(a)-g(a)\bigr\} - f(a) = f(a) - 0 - f(a) = 0 $$

[2] When $x=b$,

$$ F(b) = f(b) - \frac{f(b)-f(a)}{g(b)-g(a)}\bigl\{g(b)-g(a)\bigr\} - f(a) = f(b) - \bigl(f(b)-f(a)\bigr) - f(a) = 0 $$

From [1] and [2], $F(a)=F(b)(=0)$. Also, because $f(x),g(x)$ are continuous and differentiable, $F(x)$ is continuous on $[a,b]$ and differentiable on $(a,b)$, so by Rolle's theorem there exists $c$ satisfying

$$ F'(c) = 0 \qquad(a \lt c \lt b) $$

Differentiate $F(x)$ with respect to $x$. (Note that $\dfrac{f(b)-f(a)}{g(b)-g(a)}$ in the second term on the right-hand side is a constant that does not depend on $x$, because we regard $a,b$ as fixed, and that when $\{g(x)-g(a)\}$ is differentiated with respect to $x$, the derivative of the constant $g(a)$ is $0$, so that $g'(x)$ remains.) We get

$$ F'(x) = f'(x) - \frac{f(b)-f(a)}{g(b)-g(a)}\cdot g'(x) $$

Substituting $x=c$, we obtain from $F'(c)=0$

$$ F'(c) = f'(c) - \frac{f(b)-f(a)}{g(b)-g(a)}\cdot g'(c) = 0 $$

Rearranging this (since $g'(c)\neq0$, we may divide both sides by $g'(c)$), we get

$$ f'(c) = \frac{f(b)-f(a)}{g(b)-g(a)}\cdot g'(c) \quad\Longrightarrow\quad \frac{f'(c)}{g'(c)} = \frac{f(b)-f(a)}{g(b)-g(a)} $$ $$ \therefore\ \frac{f(b)-f(a)}{g(b)-g(a)} = \frac{f'(c)}{g'(c)} $$

∎

Proof (I)

In the statement $\dfrac{f(b)-f(a)}{g(b)-g(a)}=\dfrac{f'(c)}{g'(c)}$ of Cauchy's mean value theorem (Theorem 2.3), replace one of the endpoints, $b$, by the variable $x$ (Cauchy's mean value theorem applies without change to the interval $[a,x]$ as well).

$$ \frac{f(x)-f(a)}{g(x)-g(a)} = \frac{f'(c)}{g'(c)} \qquad(a \lt c \lt x\ \text{or}\ x \lt c \lt a) $$

(If $x$ lies to the right of $a$, then $a \lt c \lt x$; if it lies to the left, then $x \lt c \lt a$.) Now, by hypothesis $f(a)=0,g(a)=0$, so the terms $-f(a),-g(a)$ on the left-hand side vanish, and we obtain the simple form

$$ \frac{f(x)}{g(x)} = \frac{f'(c)}{g'(c)} $$

Taking $\displaystyle\lim_{x\to a}$ of both sides of this equation, we get

$$ \lim_{x\to a}\frac{f(x)}{g(x)} = \lim_{x\to a}\frac{f'(c)}{g'(c)} $$

Here, since $c$ is a number lying between $a$ and $x$, as $x\to a$ the number $c$ also approaches $a$ ($c\to a$). Therefore the limit on the right-hand side is the limit of $\dfrac{f'(c)}{g'(c)}$ in the situation where $c\to a$ accompanies $x\to a$, and this has the same value as $\displaystyle\lim_{c\to a}\frac{f'(c)}{g'(c)}$, in which $c$ is read as the variable. If we rename the letter $c$ to $x$,

$$ \lim_{x\to a}\frac{f(x)}{g(x)} = \lim_{c\to a}\frac{f'(c)}{g'(c)} = \lim_{\substack{x\to a\\(c\to a)}}\frac{f'(c)}{g'(c)} $$ $$ \therefore\ \lim_{x\to a}\frac{f(x)}{g(x)} = \lim_{x\to a}\frac{f'(x)}{g'(x)} $$

∎

Proof (II)

Suppose $\displaystyle\lim_{x\to a}\frac{f'(x)}{g'(x)}=\alpha$. Written with the $\varepsilon$-$\delta$ argument, this reads

$$ \forall\varepsilon \gt 0,\ \exists\delta \gt 0,\ \text{s.t.}\quad 0 \lt |x-a| \lt \delta \ \Rightarrow\ \left|\frac{f'(x)}{g'(x)}-\alpha\right| \lt \varepsilon \qquad\cdots① $$

Fix a single such $\delta$. Now take $x$ between $a$ and $b$ (the point $b$ is fixed as a point in the $\delta$-neighborhood, on the side farther from $a$ than $x$, and $x$ moves on the side closer to $a$ than that $b$). Applying Cauchy's mean value theorem to the interval $[b,x]$ (or $[x,b]$), we obtain

$$ \frac{f(x)-f(b)}{g(x)-g(b)} = \frac{f'(c)}{g'(c)} \qquad(b \lt c \lt x \lt a\ \text{or}\ a \lt x \lt c \lt b) $$

for some $c$ lying between $b$ and $x$. Forcibly transforming the left-hand side into a form in which $f(x),g(x)$ are factored out (dividing the numerator and denominator by $f(x),g(x)$, respectively), we get

$$ \frac{f(x)-f(b)}{g(x)-g(b)} = \frac{f(x)}{g(x)}\cdot\frac{1-\dfrac{f(b)}{f(x)}}{1-\dfrac{g(b)}{g(x)}} = \frac{f'(c)}{g'(c)} $$

Now, $c$ lies between $b$ and $x$, and $b,x$ both lie in the $\delta$-neighborhood of $a$, so $c$ also lies in the $\delta$-neighborhood of $a$: $0 \lt |c-a| \lt \delta$. Therefore ① applies directly to $c$, and we have

$$ \left|\frac{f'(c)}{g'(c)}-\alpha\right| \lt \varepsilon $$

Combining this with the equality above, we obtain

$$ \left|\frac{f(x)}{g(x)}\cdot\frac{1-\dfrac{f(b)}{f(x)}}{1-\dfrac{g(b)}{g(x)}}-\alpha\right| \lt \varepsilon \qquad\cdots(\star) $$

Now recall that $b$ was fixed as a single constant (inside the $\delta$-neighborhood) from the very beginning. With that in mind, consider the limit $x\to a$, in which $x$ (as a variable moving on the side closer to $a$ than $b$) approaches $a$. Since $b$ is fixed, $f(b),g(b)$ do not move; they are just constants. On the other hand, by hypothesis $\displaystyle\lim_{x\to a}f(x)=\pm\infty$ and $\displaystyle\lim_{x\to a}g(x)=\pm\infty$, so as $x\to a$ the denominators become arbitrarily large, and

$$ \frac{f(b)}{f(x)} \longrightarrow 0, \qquad \frac{g(b)}{g(x)} \longrightarrow 0 $$

Therefore the correction factor $\dfrac{1-f(b)/f(x)}{1-g(b)/g(x)}$ in $(\star)$ approaches $1$ as $x\to a$. Carrying out the limiting operation anew within the range $0 \lt |x-a| \lt \delta$, we then obtain

$$ \left|\frac{f(x)}{g(x)}\cdot\frac{1-\dfrac{f(b)}{f(x)}}{1-\dfrac{g(b)}{g(x)}}-\alpha\right| \longrightarrow \left|\frac{f(x)}{g(x)}-\alpha\right| \lt \varepsilon \qquad\cdots② $$

(Because the correction factor converges to $1$, the inequality in $(\star)$ passes directly, in the limit where $x$ is brought sufficiently close to $a$, to $\left|\frac{f(x)}{g(x)}-\alpha\right| \lt \varepsilon$.) Combining ① and ②, we have shown that for any $\varepsilon \gt 0$ there exists a $\delta$ satisfying $\left|\dfrac{f(x)}{g(x)}-\alpha\right| \lt \varepsilon$—and this is precisely the definition of a limit by the $\varepsilon$-$\delta$ argument, so

$$ \lim_{x\to a}\frac{f(x)}{g(x)} = \alpha = \lim_{x\to a}\frac{f'(x)}{g'(x)} $$ $$ \therefore\ \lim_{x\to a}\frac{f(x)}{g(x)} = \lim_{x\to a}\frac{f'(x)}{g'(x)} $$

∎

Proof: $\sinh(\alpha+\beta)=\sinh\alpha\cosh\beta+\cosh\alpha\sinh\beta$

Write the right-hand side in terms of exponential functions according to Definition 2.2.

$$ \sinh\alpha\cosh\beta+\cosh\alpha\sinh\beta = \frac14(e^\alpha-e^{-\alpha})(e^\beta+e^{-\beta}) + \frac14(e^\alpha+e^{-\alpha})(e^\beta-e^{-\beta}) $$

(We substituted $\sinh\alpha=\frac12(e^\alpha-e^{-\alpha})$, $\cosh\beta=\frac12(e^\beta+e^{-\beta})$, and so on directly, and used $\frac12\times\frac12=\frac14$.) Expanding each product, we get

$$ = \frac14\left(e^{\alpha+\beta}+e^{\alpha-\beta}-e^{-\alpha+\beta}-e^{-(\alpha+\beta)}\right) + \frac14\left(e^{\alpha+\beta}-e^{\alpha-\beta}+e^{-\alpha+\beta}-e^{-(\alpha+\beta)}\right) $$

When the two parenthesized expressions are added, $e^{\alpha-\beta}$ and $-e^{\alpha-\beta}$ cancel, as do $-e^{-\alpha+\beta}$ and $e^{-\alpha+\beta}$, and only the terms $e^{\alpha+\beta}$ and $-e^{-(\alpha+\beta)}$ remain, doubled.

$$ = \frac12\left(e^{\alpha+\beta}-e^{-(\alpha+\beta)}\right) $$

This is exactly $\sinh x=\frac12(e^x-e^{-x})$ of Definition 2.2 with $x=\alpha+\beta$ substituted, so

$$ = \sinh(\alpha+\beta) $$

Hence $\sinh(\alpha+\beta)=\sinh\alpha\cosh\beta+\cosh\alpha\sinh\beta$ is proved. (The case $\alpha-\beta$ gives the same conclusion as well: replace $\beta$ by $-\beta$ and use $\sinh(-\beta)=-\sinh\beta$ and $\cosh(-\beta)=\cosh\beta$, which are immediate from the definition, since that replacement merely interchanges $e^{-\beta}$ and $e^{\beta}$.)

Proof: $\cosh(\alpha+\beta)=\cosh\alpha\cosh\beta+\sinh\alpha\sinh\beta$

Likewise, we expand the right-hand side.

$$ \cosh\alpha\cosh\beta+\sinh\alpha\sinh\beta = \frac14(e^\alpha+e^{-\alpha})(e^\beta+e^{-\beta}) + \frac14(e^\alpha-e^{-\alpha})(e^\beta-e^{-\beta}) $$ $$ = \frac14\left(e^{\alpha+\beta}+e^{\alpha-\beta}+e^{-\alpha+\beta}+e^{-(\alpha+\beta)}\right) + \frac14\left(e^{\alpha+\beta}-e^{\alpha-\beta}-e^{-\alpha+\beta}+e^{-(\alpha+\beta)}\right) $$

This time $e^{\alpha-\beta}$ and $-e^{\alpha-\beta}$ cancel, as do $e^{-\alpha+\beta}$ and $-e^{-\alpha+\beta}$, and the terms $e^{\alpha+\beta}$ and $e^{-(\alpha+\beta)}$ remain, doubled.

$$ = \frac12\left(e^{\alpha+\beta}+e^{-(\alpha+\beta)}\right) = \cosh(\alpha+\beta) $$

Hence $\cosh(\alpha+\beta)=\cosh\alpha\cosh\beta+\sinh\alpha\sinh\beta$ is proved.

Proof: $\tanh(\alpha+\beta)=\dfrac{\tanh\alpha+\tanh\beta}{1+\tanh\alpha\tanh\beta}$

We use the definition $\tanh x=\dfrac{\sinh x}{\cosh x}$ and the two addition formulas just proved.

$$ \tanh(\alpha+\beta) = \frac{\sinh(\alpha+\beta)}{\cosh(\alpha+\beta)} = \frac{\sinh\alpha\cosh\beta+\cosh\alpha\sinh\beta}{\cosh\alpha\cosh\beta+\sinh\alpha\sinh\beta} $$

Dividing both the numerator and the denominator by $\cosh\alpha\cosh\beta$ (this is allowed because $\cosh$ is always positive), we get

$$ = \frac{\dfrac{\sinh\alpha\cosh\beta}{\cosh\alpha\cosh\beta}+\dfrac{\cosh\alpha\sinh\beta}{\cosh\alpha\cosh\beta}}{\dfrac{\cosh\alpha\cosh\beta}{\cosh\alpha\cosh\beta}+\dfrac{\sinh\alpha\sinh\beta}{\cosh\alpha\cosh\beta}} = \frac{\dfrac{\sinh\alpha}{\cosh\alpha}+\dfrac{\sinh\beta}{\cosh\beta}}{1+\dfrac{\sinh\alpha}{\cosh\alpha}\cdot\dfrac{\sinh\beta}{\cosh\beta}} = \frac{\tanh\alpha+\tanh\beta}{1+\tanh\alpha\tanh\beta} $$

∎

Derivation: Maclaurin expansions of $\sinh x,\cosh x$ and comparison with the trigonometric functions

The Maclaurin expansions of $\sin x,\cos x$ (a result beyond the scope of high school mathematics, but formally proved in Chapter 4, Section 4.2.2; here we borrow the result) are

$$ \sin x = \frac{x}{1!}-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\frac{x^9}{9!}-\cdots \qquad\cdots① $$ $$ \cos x = 1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\frac{x^8}{8!}-\cdots \qquad\cdots② $$

Also, the Maclaurin expansions of $e^x,e^{-x}$ are

$$ e^x = 1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\frac{x^5}{5!}+\frac{x^6}{6!}+\cdots \qquad\cdots③ $$ $$ e^{-x} = 1-\frac{x}{1!}+\frac{x^2}{2!}-\frac{x^3}{3!}+\frac{x^4}{4!}-\frac{x^5}{5!}+\frac{x^6}{6!}-\cdots \qquad\cdots④ $$

(④ is obtained from ③ by replacing $x$ with $-x$ and tidying up using $(-x)^n=(-1)^nx^n$.) Substituting ③ and ④ into $\sinh x=\frac12(e^x-e^{-x})$ of Definition 2.2, we obtain

$$ \sinh x = \frac12\left\{\left(1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\cdots\right)-\left(1-\frac{x}{1!}+\frac{x^2}{2!}-\frac{x^3}{3!}+\frac{x^4}{4!}-\cdots\right)\right\} $$

The even-degree terms ($1,\frac{x^2}{2!},\frac{x^4}{4!},\ldots$) have the same sign, so they cancel in the subtraction, whereas the odd-degree terms ($\frac{x}{1!},\frac{x^3}{3!},\ldots$) have opposite signs, so they survive the subtraction, doubled.

$$ = \frac12\left\{2\left(\frac{x}{1!}+\frac{x^3}{3!}+\frac{x^5}{5!}+\frac{x^7}{7!}+\cdots\right)\right\} = \frac{x}{1!}+\frac{x^3}{3!}+\frac{x^5}{5!}+\frac{x^7}{7!}+\frac{x^9}{9!}+\cdots \qquad\cdots⑤ $$

Similarly, when we substitute ③ and ④ into $\cosh x=\frac12(e^x+e^{-x})$ of Definition 2.2, this time the odd-degree terms cancel and the even-degree terms survive, doubled.

$$ \cosh x = \frac12\left\{\left(1+\frac{x}{1!}+\frac{x^2}{2!}+\cdots\right)+\left(1-\frac{x}{1!}+\frac{x^2}{2!}-\cdots\right)\right\} = 1+\frac{x^2}{2!}+\frac{x^4}{4!}+\frac{x^6}{6!}+\frac{x^8}{8!}+\cdots \qquad\cdots⑥ $$

Now we line up the four expansions ①, ②, ⑤, ⑥ side by side and compare them.

$$ \sin x = x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\frac{x^9}{9!}-\cdots,\qquad \sinh x = x+\frac{x^3}{3!}+\frac{x^5}{5!}+\frac{x^7}{7!}+\frac{x^9}{9!}+\cdots $$ $$ \cos x = 1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\frac{x^8}{8!}-\cdots,\qquad \cosh x = 1+\frac{x^2}{2!}+\frac{x^4}{4!}+\frac{x^6}{6!}+\frac{x^8}{8!}+\cdots $$

The only difference from the expansions of $\sin x,\cos x$ is that all the negative signs have been changed to positive signs; the absolute values of the terms (the coefficient and the degree of each term) are exactly the same.

Proof: $\cosh^2x-\sinh^2x=1$ (direct computation from the definition)

Based on Definition 2.2, we compute $\cosh^2x,\sinh^2x$ using the expansion formula for squares, $(A\pm B)^2=A^2\pm2AB+B^2$.

$$ \cosh^2x = \left\{\frac12(e^x+e^{-x})\right\}^2 = \frac14\left(e^{2x}+2\cdot e^x\cdot e^{-x}+e^{-2x}\right) = \frac14(e^{2x}+e^{-2x}+2) $$

(We used $e^x\cdot e^{-x}=e^{x-x}=e^0=1$.) Similarly,

$$ \sinh^2x = \left\{\frac12(e^x-e^{-x})\right\}^2 = \frac14\left(e^{2x}-2+e^{-2x}\right) = \frac14(e^{2x}+e^{-2x}-2) $$

Taking the difference, the parts $e^{2x}+e^{-2x}$ cancel, and

$$ \cosh^2x-\sinh^2x = \frac14\left\{(e^{2x}+e^{-2x}+2)-(e^{2x}+e^{-2x}-2)\right\} = \frac14\times4 = 1 $$ $$ \therefore\ \cosh^2x-\sinh^2x=1 $$

Alternative solution: Proof using the addition formulas

Substituting $\alpha=\beta=x$ into $\cosh(\alpha-\beta)=\cosh\alpha\cosh\beta-\sinh\alpha\sinh\beta$ of Theorem 2.5, we obtain

$$ \cosh(x-x) = \cosh x\cosh x - \sinh x\sinh x = \cosh^2x-\sinh^2x $$

On the other hand, the left-hand side is $\cosh(x-x)=\cosh0=\dfrac{e^0+e^0}{2}=\dfrac{1+1}{2}=1$. Therefore,

$$ \cosh^2x-\sinh^2x = \cosh0 = 1 \qquad\therefore\ \cosh^2x-\sinh^2x=1 $$

(If the addition formulas have already been proved, this is faster than going back to the definition and recomputing—it is exactly the same idea as deriving $\cos^2x+\sin^2x=1$ from $\cos(\alpha-\alpha)=1$ for the trigonometric functions.)

Proof: $\sinh2x=2\sinh x\cosh x$

Substitute $\alpha=\beta=x$ into the addition formula (Theorem 2.5).

$$ \sinh(x+x) = \sinh x\cosh x+\cosh x\sinh x = 2\sinh x\cosh x $$

Since the left-hand side is $\sinh(x+x)=\sinh2x$,

$$ \therefore\ \sinh2x = 2\sinh x\cosh x $$

(This is the double-angle formula for the hyperbolic functions—the counterpart of $\sin2x=2\sin x\cos x$ for the trigonometric functions.)

Proof: $\cosh2x=\cosh^2x+\sinh^2x$

Likewise, substitute $\alpha=\beta=x$ into the addition formula.

$$ \cosh(x+x) = \cosh x\cosh x+\sinh x\sinh x = \cosh^2x+\sinh^2x $$

Since the left-hand side is $\cosh(x+x)=\cosh2x$,

$$ \therefore\ \cosh2x = \cosh^2x+\sinh^2x $$

(This is also a double-angle formula for the hyperbolic functions; note that, compared with $\cos2x=\cos^2x-\sin^2x$ for the trigonometric functions, the sign has changed from $-$ to $+$. The phenomenon we saw in Section 2.5.3, “all the signs become $+$,” appears here too.)

Proof

$y=\sinh^{-1}x$ means the expression obtained by solving $x=\sinh y$ for $y$. So we first write $x$ as a function of $y$, and then, conversely, take the approach of expressing $y$ in terms of $x$.

$$ y=\sinh^{-1}x \ \Longleftrightarrow\ x=\sinh y $$

Here, note that, by the definitions $\cosh y=\frac12(e^y+e^{-y})$ and $\sinh y=\frac12(e^y-e^{-y})$, adding the two leaves only $e^y$ standing alone.

$$ \cosh y + \sinh y = \frac12(e^y+e^{-y}) + \frac12(e^y-e^{-y}) = e^y $$

That is, $e^y=\cosh y+\sinh y$. Now we rewrite the identity $\cosh^2y-\sinh^2y=1$, proved in Section 2.5.4, as $\cosh y=\sqrt{1+\sinh^2y}$ (since $\cosh y$ is always positive, the sign $+$ suffices for the square root) and substitute it, obtaining

$$ e^y = \sqrt{1+\sinh^2y}+\sinh y $$

Since we had set $x=\sinh y$, this can be rewritten directly as

$$ e^y = \sqrt{1+x^2}+x $$

Taking the natural logarithm of both sides (the logarithm can be taken because $\sqrt{1+x^2}+x$ is always positive: from $\sqrt{1+x^2} \gt |x|\ge-x$ we get $\sqrt{1+x^2}+x \gt 0$), we obtain

\begin{equation} y = \ln\!\left(\sqrt{x^2+1}+x\right) \label{eq:2-asinh} \end{equation}

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Proof

$y=\cosh^{-1}x$ is the expression obtained by solving $x=\cosh y$ for $y$. As in Section 2.6.2, we use the relation $e^y=\cosh y+\sinh y$. This time we rewrite $\cosh^2y-\sinh^2y=1$ in the form $\sinh y=\pm\sqrt{\cosh^2y-1}$ and substitute it.

$$ e^y = \cosh y+\sinh y = \cosh y\ \pm\sqrt{\cosh^2y-1} $$

Since we had set $x=\cosh y$,

$$ e^y = x\pm\sqrt{x^2-1} $$

Taking the natural logarithm of both sides, we obtain

\begin{equation} y = \ln\!\left(x\pm\sqrt{x^2-1}\right) \label{eq:2-acosh} \end{equation}

Proof

$y=\tanh^{-1}x$ is the expression obtained by solving $x=\tanh y$ for $y$. Using $\tanh y=\dfrac{e^y-e^{-y}}{e^y+e^{-y}}$ from Definition 2.2, we have

$$ x = \frac{e^y-e^{-y}}{e^y+e^{-y}} $$

Multiplying both sides by $(e^y+e^{-y})$ to clear the denominator, we obtain

$$ x(e^y+e^{-y}) = e^y-e^{-y} $$

Rearranging so as to collect the $e^y$ terms together and the $e^{-y}$ terms together on the left-hand and right-hand sides, respectively (move $xe^y-e^y$ to the left-hand side and $-e^{-y}-xe^{-y}$ to the right-hand side), we obtain

$$ xe^{-y}+e^{-y} = e^y-xe^y \quad\Longrightarrow\quad (x+1)e^{-y} = (1-x)e^{y} $$

Dividing both sides by $(1-x)e^{-y}$ (we may divide because $1-x \gt 0$ from $-1 \lt x \lt 1$, and $e^{-y} \gt 0$), we obtain

$$ \frac{x+1}{1-x} = \frac{e^y}{e^{-y}} = e^{2y} $$

That is, $e^{2y}=\dfrac{1+x}{1-x}$. Taking the square root of both sides (the right-hand side is positive because $-1 \lt x \lt 1$, so the square root can be taken), we obtain

$$ e^y = \sqrt{\frac{1+x}{1-x}} $$

Taking the natural logarithm of both sides and using $\ln\sqrt{A}=\frac12\ln A$ (a property of logarithms, already learned in high school), we obtain

\begin{equation} y = \ln\sqrt{\frac{1+x}{1-x}} = \frac12\ln\!\left(\frac{1+x}{1-x}\right) \label{eq:2-atanh} \end{equation}

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