IX The Logarithmic and Exponential Functions of a Real Variable

199. The manner in which

log x

tends to inﬁnity with

x

. It will be remembered that in Ex. XXXVI. 6 we deﬁned certain diﬀerent ways in which a function of

x

may tend to inﬁnity with

x

, distinguishing between functions which, when

x

is large, are of the ﬁrst, second, third, … orders of greatness. A function

f (x)

was said to be of the

k

th order of greatness when

f (x) / x^{k}

tends to a limit diﬀerent from zero as

x

tends to inﬁnity.

It is easy to deﬁne a whole series of functions which tend to inﬁnity with

x

, but whose order of greatness is smaller than the ﬁrst. Thus

\sqrt{x}

\sqrt[3]{x}

\sqrt[4]{x}

, … are such functions. We may say generally that

x^{α}

, where

α

is any positive rational number, is of the

α

th order of greatness when

x

is large. We may suppose

α

as small as we please, e.g. less than

. 000 000 1

. And it might be thought that by giving

α

all possible values we should exhaust the possible ‘orders of inﬁnity’ of

f (x)

. At any rate it might be supposed that if

f (x)

tends to inﬁnity with

x

, however slowly, we could always ﬁnd a value of

α

so small that

x^{α}

would tend to inﬁnity more slowly still; and, conversely, that if

f (x)

tends to inﬁnity with

x

, however rapidly, we could always ﬁnd a value of

α

so great that

x^{α}

would tend to inﬁnity more rapidly still.

Perhaps the most interesting feature of the function

log x

is its behaviour as

x

tends to inﬁnity. It shows that the presupposition stated above, which seems so natural, is unfounded. The logarithm of $x$ tends to inﬁnity with $x$ , but more slowly than any positive power of $x$ , integral or fractional. In other words

log x \to \infty

but

\frac{log x}{x^{α}} \to 0

200. Proof that

(log x) / x^{α} \to 0

x \to \infty

. Let

β

be any positive number. Then

1 / t < 1 / t^{1 - β}

when

t > 1

, and so

log x = \int_{1}^{x} \frac{d t}{t} < \int_{1}^{x} \frac{d t}{t^{1 - β}},

log x < (x^{β} - 1) / β < x^{β} / β,

0 < (log x) / x^{α} < x^{β - α} / β (x > 1) .

(log x) / x^{α} \to 0 .

(log x) / x^{α} = - y^{α} log y

lim_{y \to + 0} y^{α} log y = - lim_{x \to + \infty} (log x) / x^{α} = 0 .

202. Scales of inﬁnity. The logarithmic scale. Let us consider once more the series of functions

x, \sqrt{x}, \sqrt[3]{x}, \dots, \sqrt[n]{x}, \dots,

x, \sqrt{x}, \sqrt[3]{x}, \dots, \sqrt[n]{x}, \dots log x, \sqrt[]{log x}, \sqrt[3]{log x}, \dots \sqrt[n]{log x}, \dots

(log log y) / {(log y)}^{α} = (log x) / x^{α} \to 0 .

and it will by now be obvious that by introducing the functions

log log log x

log log log log x

, … we can prolong the series to any extent we like. By putting

x = 1 / y

we obtain a similar scale of inﬁnity for functions of

y

which tend to

\infty

y

tends to

0

by positive values.⁹⁸

Examples LXXXIV. 1. Between any two terms

f (x)

F (x)

of the series we can insert a new term

φ (x)

such that

φ (x)

tends to

\infty

more slowly than

f (x)

and more rapidly than

F (x)

. [Thus between

\sqrt{x}

and

\sqrt[3]{x}

we could insert

x^{5 / 12}

: between

\sqrt[]{log x}

and

\sqrt[3]{log x}

we could insert

{(log x)}^{5 / 12}

. And, generally,

φ (x) = \sqrt[]{f (x) F (x)}

satisﬁes the conditions stated.]

2. Find a function which tends to

\infty

more slowly than

\sqrt{x}

, but more rapidly than

x^{α}

, where

α

is any rational number less than

1 / 2

. [

\sqrt{x} / (log x)

is such a function; or

\sqrt{x} / {(log x)}^{β}

, where

β

is any positive rational number.]

3. Find a function which tends to

\infty

more slowly than

\sqrt{x}

, but more rapidly than

\sqrt{x} / {(log x)}^{α}

, where

α

is any rational number. [The function

\sqrt{x} / (log log x)

is such a function. It will be gathered from these examples that incompleteness is an inherent characteristic of the logarithmic scale of inﬁnity.]

f (x) = {x^{α} {(log x)}^{α'} {(log log x)}^{α''}} / {x^{β} {(log x)}^{β'} {(log log x)}^{β''}}

f (x) = x^{α - β} {(log x)}^{α' - β'} {(log log x)}^{α'' - β''}

5. Arrange the functions

x / \sqrt[]{log x}

x \sqrt[]{log x} / log log x

x log log x / \sqrt[]{log x}

(x log log log x) / \sqrt[]{log log x}

according to the rapidity with which they tend to inﬁnity as

x \to \infty

log log x / (x log x), (log x) / x, x log log x / \sqrt[]{x^{2} + 1}, {\sqrt[]{x + 1}} / x {(log x)}^{2}

x log log (1 / x), \sqrt{x} / {log (1 / x)}, \sqrt[]{x sin x log (1 / x)}, (1 - cos x) log (1 / x)

D_{x} log log x = 1 / (x log x), D_{x} log log log x = 1 / (x log x log log x),

D_{x} {(log x)}^{α} = α / {x {(log x)}^{1 - α}}, D_{x} {(log log x)}^{α} = α / {x log x {(log log x)}^{1 - α}},

203. The number

e

. We shall now introduce a number, usually denoted by

e

, which is of immense importance in higher mathematics. It is, like

π

, one of the fundamental constants of analysis.

We deﬁne

e

as the number whose logarithm is $1$ . In other words

e

is deﬁned by the equation

1 = \int_{1}^{e} \frac{d t}{t} .

log x^{2} = 2 log x, log x^{3} = 3 log x, \dots, log x^{n} = n log x,

log e^{n} = n log e = n .

p = log e^{p} = log {(e^{p / q})}^{q} = q log e^{p / q},

and

log e^{- y} = - log e^{y} = - y

. Hence the equation (1) is true for all rational values of

y

, positive or negative. In other words the equations

are consequences of one another so long as

y

is rational and

e^{y}

has its positive value. At present we have not given any deﬁnition of a power such as

e^{y}

in which the index is irrational, and the function

e^{y}

is deﬁned for rational values of

y

only.

\int_{1}^{2} \frac{d t}{t} < 1,

\int_{1}^{3} \frac{d t}{t} = \int_{1}^{2} \frac{d t}{t} + \int_{2}^{3} \frac{d t}{t} = \int_{0}^{1} \frac{d u}{2 - u} + \int_{0}^{1} \frac{d u}{2 + u} = 4 \int_{0}^{1} \frac{d u}{4 - u^{2}} > 1,

204. The exponential function. We now deﬁne the exponential function

e^{y}

for all real values of

y

as the inverse of the logarithmic function. In other words we write

We saw that, as

x

varies from

0

towards

\infty

y

increases steadily, in the stricter sense, from

- \infty

towards

\infty

. Thus to one value of

x

corresponds one value of

y

, and conversely. Also

y

is a continuous function of

x

, and it follows from § 109 that

x

is likewise a continuous function of

y

It is easy to give a direct proof of the continuity of the exponential function. For if

x = e^{y}

and

x + ξ = e^{y + η}

then

η = \int_{x}^{x + ξ} \frac{d t}{t} .

Thus

e^{y}

is a positive and continuous function of

y

which increases steadily from

0

towards

\infty

y

increases from

- \infty

towards

\infty

. Moreover

e^{y}

is the positive

y

th power of the number

e

, in accordance with the elementary deﬁnitions, whenever

y

is a rational number. In particular

e^{y} = 1

when

y = 0

. The general form of the graph of

e^{y}

is as shown in Fig. 53.

205. The principal properties of the exponential function. (1) If

x = e^{y}

, so that

y = log x

, then

d y / d x = 1 / x

and

\frac{d x}{d y} = x = e^{y} .

f (y + z) = f (y) f (z) .

This follows, when

y

and

z

are rational, from the ordinary rules of indices. If

y

z

, or both, are irrational then we can choose two sequences

y_{1}

y_{2}

, …,

y_{n}

, … and

z_{1}

z_{2}

, …,

z_{n}

, … of rational numbers such that

lim y_{n} = y

lim z_{n} = z

. Then, since the exponential function is continuous, we have

e^{y} \times e^{z} = lim e^{y_{n}} \times lim e^{z_{n}} = lim e^{y_{n} + z_{n}} = e^{y + z} .

We may also deduce the functional equation satisﬁed by

e^{y}

from that satisﬁed by

log x

. For if

y_{1} = log x_{1}

y_{2} = log x_{2}

, so that

x_{1} = e^{y_{1}}

x_{2} = e^{y_{2}}

, then

y_{1} + y_{2} = log x_{1} + log x_{2} = log x_{1} x_{2}

and

e^{y_{1} + y_{2}} = e^{log x_{1} x_{2}} = x_{1} x_{2} = e^{y_{1}} \times e^{y_{2}} .

Examples LXXXV. 1. If

d x / d y = a x

then

x = K e^{a y}

, where

K

is a constant.

2. There is no solution of the equation

f (y + z) = f (y) f (z)

fundamentally distinct from the exponential function. [We assume that

f (y)

has a diﬀerential coeﬃcient. Diﬀerentiating the equation with respect to

y

and

z

in turn, we obtain

f' (y + z) = f' (y) f (z), f' (y + z) = f (y) f' (z)

3. Prove that

(e^{a y} - 1) / y \to a

y \to 0

. [Applying the Mean Value Theorem, we obtain

e^{a y} - 1 = a y e^{a η}

, where

0 < ∣ η ∣ < ∣ y ∣

206. (3) The function $e^{y}$ tends to inﬁnity with $y$ more rapidly than any power of $y$ , or

lim y^{α} / e^{y} = lim e^{- y} y^{α} = 0 as y \to \infty, for all
values of α however great.We saw that (log x) / x^{β} \to 0 as x \to \infty, for any positive
value of β however
small. Writing α for 1 / β, we see that {(log x)}^{α} / x \to 0 for any value
                                                                                                              

                                                                                                              
of α however large. The
result follows on putting x = e^{y} .
It is clear also that e^{γ y} tends
to \infty if γ > 0, and
to 0 if γ < 0,
and in each case more rapidly than any power
of y .From this result it follows that we can construct a ‘scale of inﬁnity’ similar to that
constructed in §202, but extending in the opposite direction; i.e. a scale of functions which tend
to \infty more and more
rapidly as x \to \infty . 99 The
scale isx, x^{2}, x^{3}, \dots e^{x}, e^{2 x}, \dots e^{x^{2}}, \dots, e^{x^{3}}, \dots, e^{e^{x}}, \dots,where
of coursee^{x^{2}}, \dots,e^{e^{x}}, \dots denotee^{(x^{2})}, \dots,e^{(e^{x})}, \dots.The reader should try to apply the remarks about the logarithmic scale,
made in §202 and Exs. LXXXIV, to this ‘exponential scale’ also. The two scales
may of course (if the order of one is reversed) be combined into one scale\dots log log x, \dots log x, \dots x, \dots e^{x}, \dots e^{e^{x}}, \dots .207. The general power a^{x} . The function a^{x} has been deﬁned only for rational values
of x, except in the
particular case when a = e .
                                                                                                              

                                                                                                              
We shall now consider the case in which a is any positive number.
Suppose that x is a positive
rational number p / q . Then
the positive value y of
the power a^{p / q} is given
by y^{q} = a^{p}; from which it
follows thatq log y = p log a, log y = (p / q) log a = x log a,and
soy = e^{x log a} .We take this as
ourdeﬁnitionofa^{x}whenxis irrational.
Thus1 0^{\sqrt{2}} = e^{\sqrt{2} log 10}. It is to be
observed thata^{x},
whenxis
irrational,   is   deﬁned   only   for   positive   values
ofa,
and   is   itself   essentially   positive;   and   thatlog a^{x} = x log a. The most important
properties of the functiona^{x}are as follows.(1) Whatever value a may have, a^{x} \times a^{y} = a^{x + y} and {(a^{x})}^{y} = a^{x y} . In other
words the laws of indices hold for irrational no less than for rational indices. For, in the ﬁrst
place,a^{x} \times a^{y} = e^{x log a} \times e^{y log a} = e^{(x + y) log a} = a^{x + y};and in
the second{(a^{x})}^{y} = e^{y log a^{x}} = e^{x y log a} = a^{x y} .(2) If a > 1 then a^{x} = e^{x log a} = e^{α x}, where α is positive. The graph
of a^{x} is in this case
similar to that of e^{x},
and a^{x} \to \infty as x \to \infty, more rapidly than
                                                                                                              

                                                                                                              
any power of x .If a < 1 then a^{x} = e^{x log a} = e^{- β x}, where β is positive. The graph
of a^{x} is then similar in shape
to that of e^{x}, but reversed as
regards right and left, and a^{x} \to 0 as x \to \infty, more rapidly
than any power of 1 / x .(3) a^{x} is a continuous
function of x,
andD_{x} a^{x} = D_{x} e^{x log a} = e^{x log a} log a = a^{x} log a .(4) a^{x} is also a
continuous function of a,
andD_{a} a^{x} = D_{a} e^{x log a} = e^{x log a} (x / a) = x a^{x - 1} .(5) (a^{x} - 1) / x \to log a as x \to 0 .
This of course is a mere corollary from the fact that D_{x} a^{x} = a^{x} log a, but the particular
form of the result is often useful; it is of course equivalent to the result (Ex. LXXXV . 3)
that (e^{α x} - 1) / x \to α as x \to 0 .In the course of the preceding chapters a great many results involving the
function a^{x} have been stated
with the limitation that x is
rational. The deﬁnition and theorems given in this section enable us to remove this
restriction.208. The representation of e^{x} as a limit. In Ch. IV, § 73, we proved that {1 + (1 / n)}^{n} tends, as n \to \infty,
to a limit which we denoted provisionally
by e . We shall now identify
this limit with the number e of the preceding sections. We can however establish a more general result,
viz. that expressed by the equationslim_{n \to \infty} {(1 + \frac{x}{n})}^{n} = lim_{n \to \infty} {(1 - \frac{x}{n})}^{- n} = e^{x} . (1)As the result is of very great importance, we shall indicate alternative lines of
proof.(1) Since\frac{d}{d t} log (1 + x t) = \frac{x}{1 + x t},it follows thatlim_{h \to 0} \frac{log (1 + x h)}{h} = x .If we puth = 1 / ξ,
we see thatlim ξ log (1 + \frac{x}{ξ}) = xasξ \to \inftyorξ \to - \infty.
Since the exponential function is continuous it follows that{(1 + \frac{x}{ξ})}^{ξ} = e^{ξ log {1 + (x / ξ)}} \to e^{x}asξ \to \inftyorξ \to - \infty:i.e.thatlim_{ξ \to \infty} {(1 + \frac{x}{ξ})}^{ξ} = lim_{ξ \to - \infty} {(1 + \frac{x}{ξ})}^{ξ} = e^{x} . (2)If we suppose that ξ \to \infty or ξ \to - \infty through integral values only, we obtain the result expressed by the equations (1).(2) If n is any positive
integer, however large, and x > 1,
we have\int_{1}^{x} \frac{d t}{t^{1 + (1 / n)}} < \int_{1}^{x} \frac{d t}{t} < \int_{1}^{x} \frac{d t}{t^{1 - (1 / n)}},orn (1 - x^{- 1 / n}) < log x < n (x^{1 / n} - 1) . (3)Writing y for log x, so
that y is
positive and x = e^{y},
we obtain, after some simple transformations,{(1 + \frac{y}{n})}^{n} < x < {(1 - \frac{y}{n})}^{- n} . (4)Now let1 + \frac{y}{n} = η_{1}, 1 - \frac{y}{n} = \frac{1}{η_{2}} .Then0 < η_{1} < η_{2}, at any rate for suﬃciently
large values ofn; and,
                                                                                                              

                                                                                                              
by (9) of§74,η_{2}^{n} - η_{1}^{n} < n η_{2}^{n - 1} (η_{2} - η_{1}) = y^{2} η_{2}^{n} / n,which
evidently tends to0asn \to \infty. The result
now follows from the inequalities (4). The more general result (2) may be proved in the same way, if
we replace1 / nby a
continuous variableh.209. The representation of log x as a limit. We can also prove (cf. §75) thatlim n (1 - x^{- 1 / n}) = lim n (x^{1 / n} - 1) = log x .Forn (x^{1 / n} - 1) - n (1 - x^{- 1 / n}) = n (x^{1 / n} - 1) (1 - x^{- 1 / n}),which
tends to zero asn \to \infty,
sincen (x^{1 / n} - 1)tends to a
limit (§75) andx^{- 1 / n}to1(Ex. XXVII. 10). The result now follows from the inequalities (3) of§208.Examples LXXXVI. 1.  Prove, by taking y = 1 and n = 6 in the inequalities (4)
of §208, that 2.5 < e < 2.9 .2.  Prove that if t > 1 then (t^{1 / n} - t^{- 1 / n}) / (t - t^{- 1}) < 1 / n, and
so that if x > 1 then\int_{1}^{x} \frac{d t}{t^{1 - (1 / n)}} - \int_{1}^{x} \frac{d t}{t^{1 + (1 / n)}} < \frac{1}{n} \int_{1}^{x} (t - \frac{1}{t}) \frac{d t}{t} = \frac{1}{n} (x + \frac{1}{x} - 2) .Hence deduce the results of§209.3.  If ξ_{n} is
a function of n such that n ξ_{n} \to l as n \to \infty, then {(1 + ξ_{n})}^{n} \to e^{l} . [Writing n log (1 + ξ_{n}) in the forml (\frac{n ξ_{n}}{l}) \frac{log (1 + ξ_{n})}{ξ_{n}},and usingEx. LXXXII. 4,
we see thatn log (1 + ξ_{n}) \to l.]4.  If n ξ_{n} \to \infty,
then {(1 + ξ_{n})}^{n} \to \infty;
and if 1 + ξ_{n} > 0 and n ξ_{n} \to - \infty,
                                                                                                              

                                                                                                              
then{(1 + ξ_{n})}^{n} \to 0 .5.  Deduce from (1) of §208 the theorem that e^{y} tends to inﬁnity more
rapidly than any power of y .210. Common logarithms. The reader is probably familiar with the idea of a logarithm and its use
in numerical calculation. He will remember that in elementary algebra {log}_{a} x, the logarithm
of x to the
base a, is deﬁned
by the equationsx = a^{y}, y = {log}_{a} x .This deﬁnition is of course applicable only whenyis
rational, though this point is often passed over in silence.Our logarithms are therefore logarithms to the
base e . For numerical work
logarithms to the base 10 are used. Ify = log x = {log}_{e} x, z = {log}_{10} x,thenx = e^{y}and
alsox = 1 0^{z} = e^{z log 10}, so
that{log}_{10} x = ({log}_{e} x) / ({log}_{e} 10) .Thus it is easy to pass from one system to the other when once{log}_{e} 10has
been calculated.It is no part of our purpose in this book to go into details concerning the
practical uses of logarithms. If the reader is not familiar with them he should
consult some text-book on Elementary Algebra or Trigonometry. 100Examples LXXXVII. 1.  Show thatD_{x} e^{a x} cos b x = r e^{a x} cos (b x + θ), D_{x} e^{a x} sin b x = r e^{a x} sin (b x + θ)wherer = \sqrt[]{a^{2} + b^{2}},cos θ = a / r,sin θ = b / r. Hence determine
thenth derivatives
of the functionse^{a x} cos b x,e^{a x} sin b x, and show in
particular thatD_{x}^{n} e^{a x} = a^{n} e^{a x}.2.  Trace the curve y = e^{- a x} sin b x, where a and b are positive. Show that y has
an inﬁnity of maxima whose values form a geometrical progression and which lie on the
curvey = \frac{b}{\sqrt[]{a^{2} + b^{2}}} e^{- a x} .(Math. Trip. 1912.)3. Integrals containing the exponential function. Prove that\begin{array}{l} \int e^{a x} cos b x d x & = \frac{a cos b x + b sin b x}{a^{2} + b^{2}} e^{a x}, \\ \int e^{a x} sin b x d x & = \frac{a sin b x - b cos b x}{a^{2} + b^{2}} e^{a x} . \end{array}[Denoting  the  two  integrals  by I, J,
and  integrating  by  parts,  we  obtaina I = e^{a x} cos b x + b J, a J = e^{a x} sin b x - b I .Solve these equations forIandJ.]4.  Prove that the successive areas bounded by the curve of Ex. 2 and the positive half of
the axis of x form a geometrical progression, and that their sum is\frac{b}{a^{2} + b^{2}} \frac{1 + e^{- a π / b}}{1 - e^{- a π / b}} .5.  Prove that if a > 0 then\int_{0}^{\infty} e^{- a x} cos b x d x = \frac{a}{a^{2} + b^{2}}, \int_{0}^{\infty} e^{- a x} sin b x d x = \frac{b}{a^{2} + b^{2}} .6.  If I_{n} = \int e^{a x} x^{n} d x then a I_{n} = e^{a x} x^{n} - n I_{n - 1} . [Integrate by parts. It follows
that I_{n} can be calculated for all
positive integral values of n .]7.  Prove that, if n is
                                                                                                              

                                                                                                              
a positive integer, then\int_{0}^{ξ} e^{- x} x^{n} d x = n! e^{- ξ} (e^{ξ} - 1 - ξ - \frac{ξ^{2}}{2!} - \dots - \frac{ξ^{n}}{n!})and\int_{0}^{\infty} e^{- x} x^{n} d x = n! .8.  Show  how  to  ﬁnd  the  integral  of  any  rational  function
of e^{x} .
[Put x = log u,
when e^{x} = u, d x / d u = 1 / u,
and  the  integral  is  transformed  into  that  of  a  rational  function
of u .]9.  Integrate\frac{e^{2 x}}{(c^{2} e^{x} + a^{2} e^{- x}) (c^{2} e^{x} + b^{2} e^{- x})},distinguishing
the cases in whichais
and is not equal tob.10. Prove that we can integrate any function of the form P (x, e^{a x}, e^{b x}, \dots), where P denotes a polynomial. [This
follows from the fact that P can
be expressed as the sum of a number of terms of the type A x^{m} e^{k x}, where m is a
positive integer.]11. Show how to integrate any function of the formP (x, e^{a x}, e^{b x}, \dots, cos l x, cos m x, \dots, sin l x, sin m x, \dots) .12. Prove that \int_{a}^{\infty} e^{- λ x} R (x) d x,
where λ > 0 and a is
greater than the greatest root of the denominator
                                                                                                              

                                                                                                              
of R (x),
is convergent. [This follows from the fact that e^{λ x} tends to inﬁnity more
rapidly than any power of x .]13. Prove that \int_{- \infty}^{\infty} e^{- λ x^{2} + μ x} d x,
where λ > 0, is convergent
for all values of μ, and
that the same is true of \int_{- \infty}^{\infty} e^{- λ x^{2} + μ x} x^{n} d x,
where n is
any positive integer.14. Draw the graphs of e^{x^{2}}, e^{- x^{2}}, x e^{x}, x e^{- x}, x e^{x^{2}}, x e^{- x^{2}}, and x log x,
determining any maxima and minima of the functions and any points of inﬂexion on
their graphs.15. Show that the equation e^{a x} = b x, where a and b are positive, has two real roots, one, or none, according as b > a e, b = a e, or b < a e . [The tangent
to the curve y = e^{a x} at
the point (ξ, e^{a ξ}) isy - e^{a ξ} = a e^{a ξ} (x - ξ),which passes through
the origin ifa ξ = 1, so that
the liney = a e xtouches the
curve at the point(1 / a, e).
The result now becomes obvious when we draw the liney = b x.
The reader should discuss the cases in whichaorbor both are negative.]16. Show that the equation e^{x} = 1 + x has no real root except x = 0,
and that e^{x} = 1 + x + \frac{1}{2} x^{2} has three real roots.17. Draw the graphs of the functions\begin{matrix} log (x + \sqrt[]{x^{2} + 1}), log (\frac{1 + x}{1 - x}), e^{- a x} {cos}^{2} b x, \\ e^{- {(1 / x)}^{2}}, e^{- {(1 / x)}^{2}} \sqrt[]{1 / x}, e^{- cot x}, e^{- {cot}^{2} x} . \end{matrix}18. Determine roughly the positions of the real roots of the equationslog (x + \sqrt[]{x^{2} + 1}) = \frac{x}{100}, e^{x} - \frac{2 + x}{2 - x} = \frac{1}{10, 000}, e^{x} sin x = 7, e^{x^{2}} sin x = 10, 000 .19. The hyperbolic functions. The hyperbolic functions cosh x, 101 sinh x, \dots are
deﬁned by the equations\begin{matrix} cosh x = \frac{1}{2} (e^{x} + e^{- x}), sinh x = \frac{1}{2} (e^{x} - e^{- x}), \\ tanh x = (sinh x) / (cosh x), coth x = (cosh x) / (sinh x), \\ sech x = 1 / (cosh x), cosech x = 1 / (sinh x) . \end{matrix}Draw the graphs of these functions.20. Establish the formulae\begin{matrix} cosh (- x) = cosh x, sinh (- x) = - sinh x, tanh (- x) = - tanh x, \\ {cosh}^{2} x - {sinh}^{2} x = 1, {sech}^{2} x + {tanh}^{2} x = 1, {coth}^{2} x - {cosech}^{2} x = 1, \\ cosh 2 x = {cosh}^{2} x + {sinh}^{2} x, sinh 2 x = 2 sinh x cosh x, \\ \begin{matrix} 2 cosh (x + y) & = cosh x cosh y & + sinh x sinh y, \\ sinh (x + y) & = sinh x cosh y & + cosh x sinh y . \end{matrix} \end{matrix}21. Verify that these formulae may be deduced from the corresponding formulae in cos x and sin x, by
writing cosh x for cos x and i sinh x for sin x .[It follows that the same is true of all the formulae involving cos n x and sin n x which are deduced from the corresponding elementary properties of cos x and sin x .
The reason of this analogy will appear in Ch. X .]22. Express cosh x and sinh x in terms
(a) of cosh 2 x (b) of sinh 2 x .
Discuss any ambiguities of sign that may occur.(Math. Trip. 1908.)23. Prove that\begin{matrix} D_{x} cosh x = sinh x, D_{x} sinh x = cosh x, \\ D_{x} tanh x = {sech}^{2} x, D_{x} coth x = - {cosech}^{2} x, \\ D_{x} sech x = - sech x tanh x, D_{x} cosech x = - cosech x coth x, \\ D_{x} log cosh x = tanh x, D_{x} log ∣ sinh x ∣ = coth x, \\ D_{x} arc \end{matrix}

Chapter IX
The Logarithmic and Exponential Functions of a Real Variable

Miscellaneous Examples on Chapter IX¹⁰⁷

Chapter IXThe Logarithmic and Exponential Functions of a Real Variable

Miscellaneous Examples on Chapter IX107

Chapter IX
The Logarithmic and Exponential Functions of a Real Variable

Miscellaneous Examples on Chapter IX¹⁰⁷