Chapter IXThe Logarithmic and Exponential Functions of a Real Variable

196

The number of essentially different types of functions with which we have been concerned in the foregoing chapters is not very large. Among those which have occurred the most important for ordinary purposes are polynomials, rational functions, algebraical functions, explicit or implicit, and trigonometrical functions, direct or inverse.

We are however far from having exhausted the list of functions which are important in mathematics. The gradual expansion of the range of mathematical knowledge has been accompanied by the introduction into analysis of one new class of function after another. These new functions have generally been introduced because it appeared that some problem which was occupying the attention of mathematicians was incapable of solution by means of the functions already known. The process may fairly be compared with that by which the irrational and complex numbers were first introduced, when it was found that certain algebraical equations could not be solved by means of the numbers already recognised. One of the most fruitful sources of new functions has been the problem of integration. Attempts have been made to integrate some function $f (x)$ in terms of functions already known. These attempts have failed; and after a certain number of failures it has begun to appear probable that the problem is insoluble. Sometimes it has been proved that this is so; but as a rule such a strict proof has not been forthcoming until later on. Generally it has happened that mathematicians have taken the impossibility for granted as soon as they have become reasonably convinced of it, and have introduced a new function $F (x)$ defined by its possessing the required property, viz. that $F^{'} (x) = f (x)$ . Starting from this definition, they have investigated the properties of $F (x)$ ; and it has then appeared that $F (x)$ has properties which no finite combination of the functions previously known could possibly have; and thus the correctness of the assumption that the original problem could not possibly be solved has been established. One such case occurred in the preceding pages, when in Ch. VI we defined the function $lo g x$ by means of the equation

lo g x = \int \frac{d x}{x} .

Let us consider what grounds we have for supposing $lo g x$ to be a really new function. We have seen already (Ex. xlii. 4) that it cannot be a rational function, since the derivative of a rational function is a rational function whose denominator contains only repeated factors. The question whether it can be an algebraical or trigonometrical function is more difficult. But it is very easy to become convinced by a few experiments that differentiation will never get rid of algebraical irrationalities. For example, the result of differentiating $1 + x$ any number of times is always the product of $1 + x$ by a rational function, and so generally. The reader should test the correctness of the statement by experimenting with a number of examples. Similarly, if we differentiate a function which involves $sin x$ or $cos x$ , one or other of these functions persists in the result.

We have, therefore, not indeed a strict proof that $lo g x$ is a new function—that we do not profess to give¹—but a reasonable presumption that it is. We shall therefore treat it as such, and we shall find on examination that its properties are quite unlike those of any function which we have as yet encountered.

Footnotes

For such a proof see the author’s tract quoted on p. 236. ↩

197Definition of $\log x$

We define $lo g x$ , the logarithm of $x$ , by the equation

lo g x = \int_{1}^{x} \frac{d t}{t} .

We must suppose that $x$ is positive, since (Ex. lxxvi. 2) the integral has no meaning if the range of integration includes the point $x = 0$ . We might have chosen a lower limit other than $1$ ; but $1$ proves to be the most convenient. With this definition $lo g 1 = 0$ .

We shall now consider how $lo g x$ behaves as $x$ varies from $0$ towards $\infty$ . It follows at once from the definition that $lo g x$ is a continuous function of $x$ which increases steadily with $x$ and has a derivative

D_{x} lo g x = 1/ x;

and it follows from § 175 that $lo g x$ tends to $\infty$ as $x \to \infty$ .

If $x$ is positive but less than $1$ , then $lo g x$ is negative. For

lo g x = \int_{1}^{x} \frac{d t}{t} = - \int_{x}^{1} \frac{d t}{t} < 0.

Moreover, if we make the substitution $t = 1/ u$ in the integral, we obtain

lo g x = \int_{1}^{x} \frac{d t}{t} = - \int_{1}^{1/ x} \frac{d u}{u} = - lo g (1/ x) .

Thus $lo g x$ tends steadily to $- \infty$ as $x$ decreases from $1$ to $0$ .

The general form of the graph of the logarithmic function is shown in Fig. 52. Since the derivative of $lo g x$ is $1/ x$ , the slope of

the curve is very gentle when $x$ is very large, and very steep when $x$ is very small.

Examples LXXXII.

1. Prove from the definition that if $u > 0$ then

u / (1 + u) < lo g (1 + u) < u .

[For $lo g (1 + u) = \int_{0}^{u} \frac{d t}{1 + t}$ , and the subject of integration lies between $1$ and $1/ (1 + u)$ .]

2. Prove that $lo g (1 + u)$ lies between $u - \frac{u ^{2}}{2}$ and $u - \frac{u ^{2}}{2 ( 1 + u )}$ when $u$ is positive. [Use the fact that $lo g (1 + u) = u - \int_{0}^{u} \frac{t d t}{1 + t}$ .]

3. If $0 < u < 1$ then $u < - lo g (1 - u) < u / (1 - u)$ .

4. Prove that

x \to 1 lim \frac{lo g x}{x - 1} = t \to 0 lim \frac{lo g ( 1 + t )}{t} = 1.

[Use Ex. 1.]

198The functional equation satisfied by $\log x$

The function $lo g x$ satisfies the functional equation

f (x y) = f (x) + f (y) . (1)

For, making the substitution $t = y u$ , we see that

lo g x y = \int_{1}^{x y} \frac{d t}{t} = \int_{1/ y}^{x} \frac{d u}{u} = \int_{1}^{x} \frac{d u}{u} - \int_{1}^{1/ y} \frac{d u}{u} = lo g x - lo g (1/ y) = lo g x + lo g y,

which proves the theorem.

Examples LXXXIII.

1. It can be shown that there is no solution of the equation (1) which possesses a differential coefficient and is fundamentally distinct from $lo g x$ . For when we differentiate the functional equation, first with respect to $x$ and then with respect to $y$ , we obtain the two equations

y f^{'} (x y) = f^{'} (x), x f^{'} (x y) = f^{'} (y);

and so, eliminating $f^{'} (x y)$ , $x f^{'} (x) = y f^{'} (y)$ . But if this is true for every pair of values of $x$ and $y$ , then we must have $x f^{'} (x) = C$ , or $f^{'} (x) = C / x$ , where $C$ is a constant. Hence

f (x) = \int \frac{C}{x} d x + C^{'} = C lo g x + C^{'},

and it is easy to see that $C^{'} = 0$ . Thus there is no solution fundamentally distinct from $lo g x$ , except the trivial solution $f (x) = 0$ , obtained by taking $C = 0$ .

2. Show in the same way that there is no solution of the equation

f (x) + f (y) = f (\frac{x + y}{1 - x y})

which possesses a differential coefficient and is fundamentally distinct from $arc tan x$ .

199The 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 $x$ , $3 x$ , $4 x$ , … 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 $.0000001$ . 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 $lo g x$ 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 $lo g x \to \infty$ but

\frac{lo g x}{x ^{α}} \to 0

for all positive values of $α$ . This fact is sometimes expressed loosely by saying that the ‘order of infinity of $lo g x$ is infinitely small’; but the reader will hardly require at this stage to be warned against such modes of expression.

200Proof that $(\log x)/x^{\alpha} \to 0$ as $x \to \infty$

Let $β$ be any positive number. Then $1/ t < 1/ t^{1 - β}$ when $t > 1$ , and so

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

lo g x < (x^{β} - 1) / β < x^{β} / β,

when $x > 1$ . Now if $α$ is any positive number we can choose a smaller positive value of $β$ . And then

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

But, since $α > β$ , $x^{β - α} / β \to 0$ as $x \to \infty$ , and therefore

(lo g x) / x^{α} \to 0.

201The behaviour of $\log x$ as $x \to +0$

Since

(lo g x) / x^{α} = - y^{α} lo g y

if $x = 1/ y$ , it follows from the theorem proved above that

y \to + 0 lim y^{α} lo g y = - x \to + \infty lim (lo g x) / x^{α} = 0.

Thus $lo g x$ tends to $- \infty$ and $lo g (1/ x) = - lo g x$ to $\infty$ as $x$ tends to zero by positive values, but $lo g (1/ x)$ tends to $\infty$ more slowly than any positive power of $1/ x$ , integral or fractional.

202Scales of infinity. The logarithmic scale

Let us consider once more the series of functions

x, x, 3 x, \dots, n x, \dots,

which possesses the property that, if $f (x)$ and $ϕ (x)$ are any two of the functions contained in it, then $f (x)$ and $ϕ (x)$ both tend to $\infty$ as $x \to \infty$ , while $f (x) / ϕ (x)$ tends to $0$ or to $\infty$ according as $f (x)$ occurs to the right or the left of $ϕ (x)$ in the series. We can now continue this series by the insertion of new terms to the right of all those already written down. We can begin with $lo g x$ , which tends to infinity more slowly than any of the old terms. Then $lo g x$ tends to $\infty$ more slowly than $lo g x$ , $3 lo g x$ than $lo g x$ , and so on. Thus we obtain a series

x, x, 3 x, \dots, n x, \dots lo g x, lo g x, 3 lo g x, \dots n lo g x, \dots

formed of two simply infinite series arranged one after the other. But this is not all. Consider the function $lo g lo g x$ , the logarithm of $lo g x$ . Since $(lo g x) / x^{α} \to 0$ , for all positive values of $α$ , it follows on putting $x = lo g y$ that

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

Thus $lo g lo g y$ tends to $\infty$ with $y$ , but more slowly than any power of $lo g y$ . Hence we may continue our series in the form

x, x, 3 x, \dots lo g x, lo g x, 3 lo g x, \dots lo g lo g x, lo g lo g x, \dots n lo g lo g x, \dots;

and it will by now be obvious that by introducing the functions $lo g lo g lo g x$ , $lo g lo g lo g lo g x$ , … 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 $\infty$ as $y$ tends to $0$ by positive values.¹

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 $x$ and $3 x$ we could insert $x^{5/12}$ : between $lo g x$ and $3 lo g x$ we could insert $(lo g x)^{5/12}$ . And, generally, $ϕ (x) = f (x) F (x)$ satisfies the conditions stated.]

2. Find a function which tends to $\infty$ more slowly than $x$ , but more rapidly than $x^{α}$ , where $α$ is any rational number less than $1/2$ . [ $x / (lo g x)$ is such a function; or $x / (lo g x)^{β}$ , where $β$ is any positive rational number.]

3. Find a function which tends to $\infty$ more slowly than $x$ , but more rapidly than $x / (lo g x)^{α}$ , where $α$ is any rational number. [The function $x / (lo g lo g x)$ is such a function. It will be gathered from these examples that incompleteness is an inherent characteristic of the logarithmic scale of infinity.]

4. How does the function

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

behave as $x$ tends to $\infty$ ? [If $α \neq = β$ then the behaviour of

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

is dominated by that of $x^{α - β}$ . If $α = β$ then the power of $x$ disappears and the behaviour of $f (x)$ is dominated by that of $(lo g x)^{α^{'} - β^{'}}$ , unless $α^{'} = β^{'}$ , when it is dominated by that of $(lo g lo g x)^{α^{''} - β^{''}}$ . Thus $f (x) \to \infty$ if $α > β$ , or $α = β$ , $α^{'} > β^{'}$ , or $α = β$ , $α^{'} = β^{'}$ , $α^{''} > β^{''}$ , and $f (x) \to 0$ if $α < β$ , or $α = β$ , $α^{'} < β^{'}$ , or $α = β$ , $α^{'} = β^{'}$ , $α^{''} < β^{''}$ .]

5. Arrange the functions $x / lo g x$ , $x lo g x / lo g lo g x$ , $x lo g lo g x / lo g x$ , $(x lo g lo g lo g x) / lo g lo g x$ according to the rapidity with which they tend to infinity as $x \to \infty$ .

6. Arrange

lo g lo g x / (x lo g x), (lo g x) / x, x lo g lo g x / x^{2} + 1, {x + 1} / x (lo g x)^{2}

according to the rapidity with which they tend to zero as $x \to \infty$ .

7. Arrange

x lo g lo g (1/ x), x / {lo g (1/ x)}, x sin x lo g (1/ x), (1 - cos x) lo g (1/ x)

according to the rapidity with which they tend to zero as $x \to + 0$ .

8. Show that

D_{x} lo g lo g x = 1/ (x lo g x), D_{x} lo g lo g lo g x = 1/ (x lo g x lo g lo g x),

and so on.

9. Show that

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

and so on.

Footnotes

For fuller information as to ‘scales of infinity’ see the author’s tract ‘Orders of Infinity’, Camb. Math. Tracts, No. 12. ↩

203The 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} .

Since $lo g x$ is an increasing function of $x$ , in the stricter sense of § 95, it can only pass once through the value $1$ . Hence our definition does in fact define one definite number.

Now $lo g x y = lo g x + lo g y$ and so

lo g x^{2} = 2 lo g x, lo g x^{3} = 3 lo g x, \dots, lo g x^{n} = n lo g x,

where $n$ is any positive integer. Hence

lo g e^{n} = n lo g e = n .

Again, if $p$ and $q$ are any positive integers, and $e^{p / q}$ denotes the positive $q$ th root of $e^{p}$ , we have

p = lo g e^{p} = lo g (e^{p / q})^{q} = q lo g e^{p / q},

so that $lo g e^{p / q} = p / q$ . Thus, if $y$ has any positive rational value, and $e^{y}$ denotes the positive $y$ th power of $e$ , we have

lo g e^{y} = y, (1)

and $lo g e^{- y} = - lo g e^{y} = - y$ . Hence the equation (1) is true for all rational values of $y$ , positive or negative. In other words the equations

y = lo g x, x = e^{y} (2)

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.

Example. Prove that $2 < e < 3$ . [In the first place it is evident that

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

and so $2 < e$ . Also

\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,

so that $e < 3$ .]

204The 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

x = e^{y}

if $y = lo g x$ .

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 $∣ η ∣$ is greater than $ξ / (x + ξ)$ if $ξ > 0$ , and than $∣ ξ ∣/ x$ if $ξ < 0$ ; and if $η$ is very small $ξ$ must also be very small.

Thus $e^{y}$ is a positive and continuous function of $y$ which increases steadily from $0$ towards $\infty$ as $y$ 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.

205The principal properties of the exponential function

(1) If $x = e^{y}$ , so that $y = lo g x$ , then $d y / d x = 1/ x$ and

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

Thus the derivative of the exponential function is equal to the function itself. More generally, if $x = e^{a y}$ then $d x / d y = a e^{a y}$ .

(2) The exponential function satisfies the functional equation

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

This follows, when $y$ and $z$ are rational, from the ordinary rules of indices. If $y$ or $z$ , 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} .

In particular $e^{y} \times e^{- y} = e^{0} = 1$ , or $e^{- y} = 1/ e^{y}$ .

We may also deduce the functional equation satisfied by $e^{y}$ from that satisfied by $lo g x$ . For if $y_{1} = lo g x_{1}$ , $y_{2} = lo g x_{2}$ , so that $x_{1} = e^{y_{1}}$ , $x_{2} = e^{y_{2}}$ , then $y_{1} + y_{2} = lo g x_{1} + lo g x_{2} = lo g x_{1} x_{2}$ and

e^{y_{1} + y_{2}} = e^{l o g 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)

and so $f^{'} (y) / f (y) = f^{'} (z) / f (z)$ , and therefore each is constant. Thus if $x = f (y)$ then $d x / d y = a x$ , where $a$ is a constant, so that $x = K e^{a y}$ (Ex. 1).]

3. Prove that $(e^{a y} - 1) / y \to a$ as $y \to 0$ . [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 $(lo g x) / x^{β} \to 0$ as $x \to \infty$ , for any positive value of $β$ however small. Writing $α$ for $1/ β$ , we see that $(lo g x)^{α} / x \to 0$ 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 $x \to \infty$ .¹ The scale is

x, 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 course $e^{x^{2}}$ , …, $e^{e^{x}}$ , … denote $e^{(x^{2})}$ , …, $e^{(e^{x})}$ , ….

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 lo g lo g x, \dots lo g x, \dots x, \dots e^{x}, \dots e^{e^{x}}, \dots .

Footnotes

The exponential function was introduced by inverting the equation $y = lo g x$ into $x = e^{y}$ ; and we have accordingly, up to the present, used $y$ as the independent and $x$ as the dependent variable in discussing its properties. We shall now revert to the more natural plan of taking $x$ as the independent variable, except when it is necessary to consider a pair of equations of the type $y = lo g x$ , $x = e^{y}$ simultaneously, or when there is some other special reason to the contrary. ↩

207The 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 that

q lo g y = p lo g a, lo g y = (p / q) lo g a = x lo g a,

and so

y = e^{x l o g a} .

We take this as our definition of $a^{x}$ when $x$ is irrational. Thus $1 0^{2} = e^{2 l o g 10}$ . It is to be observed that $a^{x}$ , when $x$ is irrational, is defined only for positive values of $a$ , and is itself essentially positive; and that $lo g a^{x} = x lo g a$ . The most important properties of the function $a^{x}$ 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 l o g a} \times e^{y l o g a} = e^{(x + y) l o g a} = a^{x + y};

and in the second

(a^{x})^{y} = e^{y l o g a^{x}} = e^{x y l o g a} = a^{x y} .

(2) If $a > 1$ then $a^{x} = e^{x l o g a} = e^{α x}$ , 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 $a^{x} = e^{x l o g a} = e^{- β x}$ , 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 $x$ , and

D_{x} a^{x} = D_{x} e^{x l o g a} = e^{x l o g a} lo g a = a^{x} lo g a .

(4) $a^{x}$ is also a continuous function of $a$ , and

D_{a} a^{x} = D_{a} e^{x l o g a} = e^{x l o g a} (x / a) = x a^{x - 1} .

(5) $(a^{x} - 1) / x \to lo g a$ as $x \to 0$ . This of course is a mere corollary from the fact that $D_{x} a^{x} = a^{x} lo g a$ , 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.

208The 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 equations

n \to \infty lim (1 + \frac{x}{n})^{n} = n \to \infty lim (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} lo g (1 + x t) = \frac{x}{1 + x t},

it follows that

h \to 0 lim \frac{lo g ( 1 + x h )}{h} = x .

If we put $h = 1/ ξ$ , we see that

lim ξ lo g (1 + \frac{x}{ξ}) = x

as $ξ \to \infty$ or $ξ \to - \infty$ . Since the exponential function is continuous it follows that

(1 + \frac{x}{ξ})^{ξ} = e^{ξ l o g {1 + (x / ξ)}} \to e^{x}

as $ξ \to \infty$ or $ξ \to - \infty$ : i.e. that

ξ \to \infty lim (1 + \frac{x}{ξ})^{ξ} = ξ \to - \infty lim (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)}},

n (1 - x^{- 1/ n}) < lo g x < n (x^{1/ n} - 1) . (3)

Writing $y$ for $lo g x$ , 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 let

1 + \frac{y}{n} = η_{1}, 1 - \frac{y}{n} = \frac{1}{η _{2}} .

Then $0 < η_{1} < η_{2}$ , at any rate for sufficiently large values of $n$ ; 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 to $0$ as $n \to \infty$ . The result now follows from the inequalities (4). The more general result (2) may be proved in the same way, if we replace $1/ n$ by a continuous variable $h$ .

209The representation of $\log x$ as a limit

We can also prove (cf. § 75) that

lim n (1 - x^{- 1/ n}) = lim n (x^{1/ n} - 1) = lo g x .

For

n (x^{1/ n} - 1) - n (1 - x^{- 1/ n}) = n (x^{1/ n} - 1) (1 - x^{- 1/ n}),

which tends to zero as $n \to \infty$ , since $n (x^{1/ n} - 1)$ tends to a limit (§ 75) and $x^{- 1/ n}$ to $1$ (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 $n lo g (1 + ξ_{n})$ in the form

l (\frac{n ξ _{n}}{l}) \frac{lo g ( 1 + ξ _{n} )}{ξ _{n}},

and using Ex. lxxxii. 4, we see that $n lo g (1 + ξ_{n}) \to l$ .]

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 $y$ .

210Common 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 $lo g_{a} x$ , the logarithm of $x$ to the base $a$ , is defined by the equations

x = a^{y}, y = lo g_{a} x .

This definition is of course applicable only when $y$ is 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. If

y = lo g x = lo g_{e} x, z = lo g_{10} x,

then $x = e^{y}$ and also $x = 1 0^{z} = e^{z l o g 10}$ , so that

lo g_{10} x = (lo g_{e} x) / (lo g_{e} 10) .

Thus it is easy to pass from one system to the other when once $lo g_{e} 10$ has 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.¹

Examples LXXXVII.

1. Show that

D_{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)

where $r = a^{2} + b^{2}$ , $cos θ = a / r$ , $sin θ = b / r$ . Hence determine the $n$ th derivatives of the functions $e^{a x} cos b x$ , $e^{a x} sin b x$ , and show in particular that $D_{x}^{n} e^{a x} = a^{n} e^{a x}$ .

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 curve

y = \frac{b}{a ^{2} + b ^{2}} e^{- a x} .

(Math. Trip. 1912.)

3. Integrals containing the exponential function. Prove that

\int e^{a x} cos b x d x \int e^{a x} sin b x d x = \frac{a cos b x + b sin b x}{a ^{2} + b ^{2}} e^{a x}, = \frac{a sin b x - b cos b x}{a ^{2} + b ^{2}} e^{a x} .

[Denoting the two integrals by $I$ , $J$ , and integrating by parts, we obtain

a I = e^{a x} cos b x + b J, a J = e^{a x} sin b x - b I .

Solve these equations for $I$ and $J$ .]

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 ^{- aπ / b}}{1 - e ^{- aπ / 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 $x = lo g u$ , 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 which $a$ is and is not equal to $b$ .

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 form

P (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 $\int_{- \infty}^{\infty} e^{- λ x^{2} + μx} d x$ , where $λ > 0$ , is convergent for all values of $μ$ , and that the same is true of $\int_{- \infty}^{\infty} e^{- λ x^{2} + μx} x^{n} d x$ , 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 $x lo g x$ , 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 $(ξ, e^{a ξ})$ is

y - e^{a ξ} = a e^{a ξ} (x - ξ),

which passes through the origin if $a ξ = 1$ , so that the line $y = a e x$ touches the curve at the point $(1/ a, e)$ . The result now becomes obvious when we draw the line $y = b x$ . The reader should discuss the cases in which $a$ or $b$ or 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

lo g (x + x^{2} + 1), lo g (\frac{1 + x}{1 - x}), e^{- a x} cos^{2} b x, e^{- (1/ x)^{2}}, e^{- (1/ x)^{2}} 1/ x, e^{- c o t x}, e^{- c o t^{2} x} .

18. Determine roughly the positions of the real roots of the equations

lo g (x + x^{2} + 1) = \frac{x}{100}, e^{x} - \frac{2 + x}{2 - x} = \frac{1}{10000}, e^{x} sin x = 7, e^{x^{2}} sin x = 10000.

19. The hyperbolic functions. The hyperbolic functions $cosh x$ ,² $sinh x$ , … are defined by the equations

cosh x = \frac{1}{2} (e^{x} + e^{- x}), sinh x = \frac{1}{2} (e^{x} - e^{- x}), [1] tanh x = (sinh x) / (cosh x), coth x = (cosh x) / (sinh x), [1] sech x = 1/ (cosh x), cosech x = 1/ (sinh x) .

Draw the graphs of these functions.

20. Establish the formulae

cosh (- x) = cosh x, sinh (- x) = - sinh x, tanh (- x) = - tanh x, [1] cosh^{2} x - sinh^{2} x = 1, sech^{2} x + tanh^{2} x = 1, coth^{2} x - cosech^{2} x = 1, [1] cosh 2 x = cosh^{2} x + sinh^{2} x, sinh 2 x = 2 sinh x cosh x, [1] cosh (x + y) sinh (x + y) = cosh x cosh y = sinh x cosh y + sinh x sinh y, + cosh x sinh y .

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

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} lo g cosh x = tanh x, D_{x} lo g ∣ sinh x ∣ = coth x, D_{x} arc tan e^{x} = \frac{1}{2} sech x, D_{x} lo g ∣ tanh \frac{1}{2} x ∣ = cosech x .

[All these formulae may of course be transformed into formulae in integration.]

24. Prove that $cosh x > 1$ and $- 1 < tanh x < 1$ .

25. Prove that if $y = cosh x$ then $x = lo g {y \pm y^{2} - 1}$ , if $y = sinh x$ then $x = lo g {y + y^{2} + 1}$ , and if $y = tanh x$ then $x = \frac{1}{2} lo g {(1 + y) / (1 - y)}$ . Account for the ambiguity of sign in the first case.

26. We shall denote the functions inverse to $cosh x$ , $sinh x$ , $tanh x$ by $arg cosh x$ , $arg sinh x$ , $arg tanh x$ . Show that $arg cosh x$ is defined only when $x ≧ 1$ , and is in general two-valued, while $arg sinh x$ is defined for all real values of $x$ , and $arg tanh x$ when $- 1 < x < 1$ , and both of the two latter functions are one-valued. Sketch the graphs of the functions.

27. Show that if $- \frac{1}{2} π < x < \frac{1}{2} π$ and $y$ is positive, and $cos x cosh y = 1$ , then

y = lo g (sec x + tan x), D_{x} y = sec x, D_{y} x = sech y .

28. Prove that if $a > 0$ then $\int \frac{d x}{x ^{2} + a ^{2}} = arg sinh (x / a)$ , and $\int \frac{d x}{x ^{2} - a ^{2}}$ is equal to $arg cosh (x / a)$ or to $- arg cosh (- x / a)$ , according as $x > 0$ or $x < 0$ .

29. Prove that if $a > 0$ then $\int \frac{d x}{x ^{2} - a ^{2}}$ is equal to $- (1/ a) arg tanh (x / a)$ or to $- (1/ a) arg coth (x / a)$ , according as $∣ x ∣$ is less than or greater than $a$ . [The results of Exs. 28 and 29 furnish us with an alternative method of writing a good many of the formulae of Ch. VI.]

30. Prove that

\int \frac{d x}{( x - a ) ( x - b )} \int \frac{d x}{( a - x ) ( b - x )} \int \frac{d x}{( x - a ) ( b - x )} = = - = 2 lo g {x - a + x - b} 2 lo g {a - x + b - x} 2 arc tan \frac{x - a}{b - x} (a < b < x), (x < a < b), (a < x < b) .

31. Prove that

\int_{0}^{1} x lo g (1 + \frac{1}{2} x) d x = \frac{3}{4} - \frac{3}{2} lo g \frac{3}{2} < \frac{1}{2} \int_{0}^{1} x^{2} d x = \frac{1}{6} .

(Math. Trip. 1913.)

32. Solve the equation $a cosh x + b sinh x = c$ , where $c > 0$ , showing that it has no real roots if $b^{2} + c^{2} - a^{2} < 0$ , while if $b^{2} + c^{2} - a^{2} > 0$ it has two, one, or no real roots according as $a + b$ and $a - b$ are both positive, of opposite signs, or both negative. Discuss the case in which $b^{2} + c^{2} - a^{2} = 0$ .

33. Solve the simultaneous equations $cosh x cosh y = a$ , $sinh x sinh y = b$ .

34. $x^{1/ x} \to 1$ as $x \to \infty$ . [For $x^{1/ x} = e^{(l o g x) / x}$ , and $(lo g x) / x \to 0$ . Cf. Ex. xxvii. 11.] Show also that the function $x^{1/ x}$ has a maximum when $x = e$ , and draw the graph of the function for positive values of $x$ .

35. $x^{x} \to 1$ as $x \to + 0$ .

36. If ${f (n + 1)} / {f (n)} \to l$ , where $l > 0$ , as $n \to \infty$ , then $n f (n) \to l$ . [For $lo g f (n + 1) - lo g f (n) \to lo g l$ , and so $(1/ n) lo g f (n) \to lo g l$ (Ch. IV, Misc. Ex. 27).]

37. $n n! / n \to 1/ e$ as $n \to \infty$ .

[If $f (n) = n^{- n} n!$ then ${f (n + 1)} / {f (n)} = {1 + (1/ n)}^{- n} \to 1/ e$ . Now use Ex. 36.]

38. $n (2 n)! / (n!)^{2} \to 4$ as $n \to \infty$ .

39. Discuss the approximate solution of the equation $e^{x} = x^{1000000}$ .

[It is easy to see by general graphical considerations that the equation has two positive roots, one a little greater than $1$ and one very large,³ and one negative root a little greater than $- 1$ . To determine roughly the size of the large positive root we may proceed as follows. If $e^{x} = x^{1000000}$ then

x = 1 0^{6} lo g x, lo g x = 13.82 + lo g lo g x, lo g lo g x = 2.63 + lo g (1 + \frac{lo g lo g x}{13.82}),

roughly, since $13.82$ and $2.63$ are approximate values of $lo g 1 0^{6}$ and $lo g lo g 1 0^{6}$ respectively. It is easy to see from these equations that the ratios $lo g x : 13.82$ and $lo g lo g x : 2.63$ do not differ greatly from unity, and that

x = 1 0^{6} (13.82 + lo g lo g x) = 1 0^{6} (13.82 + 2.63) = 16450000

gives a tolerable approximation to the root, the error involved being roughly measured by $1 0^{6} (lo g lo g x - 2.63)$ or $(1 0^{6} lo g lo g x) /13.82$ or $(1 0^{6} \times 2.63) /13.82$ , which is less than $200, 000$ . The approximations are of course very rough, but suffice to give us a good idea of the scale of magnitude of the root.]

40. Discuss similarly the equations

e^{x} = 1000000 x^{1000000}, e^{x^{2}} = x^{1000000000} .

Footnotes

See for example Chrystal’s Algebra, vol. i, ch. xxi. The value of $lo g_{e} 10$ is $2.302 \dots$ and that of its reciprocal $.434 \dots$ . ↩
‘Hyperbolic cosine’: for an explanation of this phrase see Hobson’s Trigonometry, ch. xvi. ↩
The phrase ‘very large’ is of course not used here in the technical sense explained in Ch. IV. It means ‘a good deal larger than the roots of such equations as usually occur in elementary mathematics’. The phrase ‘a little greater than’ must be interpreted similarly. ↩

211Logarithmic tests of convergence for series and integrals

We showed in Ch. VIII (§§ 175 et seq.) that

1 \sum \infty \frac{1}{n ^{s}}, \int_{a}^{\infty} \frac{d x}{x ^{s}} (a > 0)

are convergent if $s > 1$ and divergent if $s ≦ 1$ . Thus $\sum (1/ n)$ is divergent, but $\sum n^{- 1 - α}$ is convergent for all positive values of $α$ .

We saw however in § 200 that with the aid of logarithms we can construct functions which tend to zero, as $n \to \infty$ , more rapidly than $1/ n$ , yet less rapidly than $n^{- 1 - α}$ , however small $α$ may be, provided of course that it is positive. For example $1/ (n lo g n)$ is such a function, and the question as to whether the series

\sum \frac{1}{n lo g n}

is convergent or divergent cannot be settled by comparison with any series of the type $\sum n^{- s}$ .

The same is true of such series as

\sum \frac{1}{n ( lo g n ) ^{2}}, \sum \frac{lo g lo g n}{n lo g n} .

It is a question of some interest to find tests which shall enable us to decide whether series such as these are convergent or divergent; and such tests are easily deduced from the Integral Test of § 174.

For since

D_{x} (lo g x)^{1 - s} = \frac{1 - s}{x ( lo g x ) ^{s}}, D_{x} lo g lo g x = \frac{1}{x lo g x},

we have

\int_{a}^{ξ} \frac{d x}{x ( lo g x ) ^{s}} = \frac{( lo g ξ ) ^{1 - s} - ( lo g a ) ^{1 - s}}{1 - s}, \int_{a}^{ξ} \frac{d x}{x lo g x} = lo g lo g ξ - lo g lo g a,

if $a > 1$ . The first integral tends to the limit $- (lo g a)^{1 - s} / (1 - s)$ as $ξ \to \infty$ , if $s > 1$ , and to $\infty$ if $s < 1$ . The second integral tends to $\infty$ . Hence

the series and integral

n_{0} \sum \infty \frac{1}{n ( lo g n ) ^{s}}, \int_{a}^{\infty} \frac{d x}{x ( lo g x ) ^{s}},

where $n_{0}$ and $a$ are greater than unity, are convergent if $s > 1$ , divergent if $s ≦ 1$ .

It follows, of course, that $\sum ϕ (n)$ is convergent if $ϕ (n)$ is positive and less than $K / {n (lo g n)^{s}}$ , where $s > 1$ , for all values of $n$ greater than some definite value, and divergent if $ϕ (n)$ is positive and greater than $K / (n lo g n)$ for all values of $n$ greater than some definite value. And there is a corresponding theorem for integrals which we may leave to the reader.

Examples LXXXVIII.

1. The series

\sum \frac{1}{n ( lo g n ) ^{2}}, \sum \frac{( lo g n ) ^{100}}{n ^{101/100}}, \sum \frac{n ^{2} - 1}{n ^{2} + 1} \frac{1}{n ( lo g n ) ^{7/6}}

are convergent. [The convergence of the first series is a direct consequence of the theorem of the preceding section. That of the second follows from the fact that $(lo g n)^{100}$ is less than $n^{β}$ for sufficiently large values of $n$ , however small $β$ may be, provided that it is positive. And so, taking $β = 1/200$ , $(lo g n)^{100} n^{- 101/100}$ is less than $n^{- 201/200}$ for sufficiently large values of $n$ . The convergence of the third series follows from the comparison test at the end of the last section.]

2. The series

\sum \frac{1}{n ( lo g n ) ^{6/7}}, \sum \frac{1}{n ^{100/101} ( lo g n ) ^{100}}, \sum \frac{n lo g n}{( n lo g n ) ^{2} + 1}

are divergent.

3. The series

\sum \frac{( lo g n ) ^{p}}{n ^{1 + s}}, \sum \frac{( lo g n ) ^{p} ( lo g lo g n ) ^{q}}{n ^{1 + s}}, \sum \frac{( lo g lo g n ) ^{p}}{n ( lo g n ) ^{1 + s}},

where $s > 0$ , are convergent for all values of $p$ and $q$ ; similarly the series

\sum \frac{1}{n ^{1 - s} ( lo g n ) ^{p}}, \sum \frac{1}{n ^{1 - s} ( lo g n ) ^{p} ( lo g lo g n ) ^{q}}, \sum \frac{1}{n ( lo g n ) ^{1 - s} ( lo g lo g n ) ^{p}}

are divergent.

4. The question of the convergence or divergence of such series as

\sum \frac{1}{n lo g n lo g lo g n}, \sum \frac{lo g lo g lo g n}{n lo g n lo g lo g n}

cannot be settled by the theorem of p. 375, since in each case the function under the sign of summation tends to zero more rapidly than $1/ (n lo g n)$ yet less rapidly than $n^{- 1} (lo g n)^{- 1 - α}$ , where $α$ is any positive number however small. For such series we need a still more delicate test. The reader should be able, starting from the equations

D_{x} (lo g_{k} x)^{1 - s} D_{x} lo g_{k + 1} x = \frac{1 - s}{x lo g x lo g _{2} x \dots lo g _{k - 1} x ( lo g _{k} x ) ^{s}}, = \frac{1}{x lo g x lo g _{2} x \dots lo g _{k - 1} x lo g _{k} x},

where $lo g_{2} x = lo g lo g x$ , $lo g_{3} x = lo g lo g lo g x$ , …, to prove the following theorem: the series and integral

n_{0} \sum \infty \frac{1}{n lo g n lo g _{2} n \dots lo g _{k - 1} n ( lo g _{k} n ) ^{s}}, \int_{a}^{\infty} \frac{d x}{x lo g x lo g _{2} x \dots lo g _{k - 1} x ( lo g _{k} x ) ^{s}}

are convergent if $s > 1$ and divergent if $s ≦ 1$ , $n_{0}$ and $a$ being any numbers sufficiently great to ensure that $lo g_{k} n$ and $lo g_{k} x$ are positive when $n ≧ n_{0}$ or $x ≧ a$ . These values of $n_{0}$ and $a$ increase very rapidly as $k$ increases: thus $lo g x > 0$ requires $x > 1$ , $lo g_{2} x > 0$ requires $x > e$ , $lo g_{3} x > 0$ requires $x > e^{e}$ , and so on; and it is easy to see that $e^{e} > 10$ , $e^{e^{e}} > e^{10} > 20, 000$ , $e^{e^{e^{e}}} > e^{20, 000} > 1 0^{8000}$ .

The reader should observe the extreme rapidity with which the higher exponential functions, such as $e^{e^{x}}$ and $e^{e^{e^{x}}}$ , increase with $x$ . The same remark of course applies to such functions as $a^{a^{x}}$ and $a^{a^{a^{x}}}$ , where $a$ has any value greater than unity It has been computed that $9^{9^{9}}$ has $369, 693, 100$ figures, while $1 0^{1 0^{10}}$ has of course $10, 000, 000, 000$ . Conversely, the rate of increase of the higher logarithmic functions is extremely slow. Thus to make $lo g lo g lo g lo g x > 1$ we have to suppose $x$ a number with over $8000$ figures.¹

5. Prove that the integral $\int_{0}^{a} \frac{1}{x} {lo g (\frac{1}{x})}^{s} d x$ , where $0 < a < 1$ , is convergent if $s < - 1$ , divergent if $s ≧ - 1$ . [Consider the behaviour of

\int_{ϵ}^{a} \frac{1}{x} {lo g (\frac{1}{x})}^{s} d x

as $ϵ \to + 0$ . This result also may be refined upon by the introduction of higher logarithmic factors.]

6. Prove that $\int_{0}^{1} \frac{1}{x} {lo g (\frac{1}{x})}^{s} d x$ has no meaning for any value of $s$ . [The last example shows that $s < - 1$ is a necessary condition for convergence at the lower limit: but ${lo g (1/ x)}^{s}$ tends to $\infty$ like $(1 - x)^{s}$ , as $x \to 1 - 0$ , if $s$ is negative, and so the integral diverges at the upper limit when $s < - 1$ .]

7. The necessary and sufficient conditions for the convergence of $\int_{0}^{1} x^{a - 1} {lo g (\frac{1}{x})}^{s} d x$ are $a > 0$ , $s > - 1$ .

Examples LXXXIX.

1. Euler’s limit. Show that

ϕ (n) = 1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n - 1} - lo g n

tends to a limit $γ$ as $n \to \infty$ , and that $0 < γ ≦ 1$ . [This follows at once from § 174. The value of $γ$ is in fact $.577 \dots$ , and $γ$ is usually called Euler’s constant.]

2. If $a$ and $b$ are positive then

\frac{1}{a} + \frac{1}{a + b} + \frac{1}{a + 2 b} + \dots + \frac{1}{a + ( n - 1 ) b} - \frac{1}{b} lo g (a + nb)

tends to a limit as $n \to \infty$ .

3. If $0 < s < 1$ then

ϕ (n) = 1 + 2^{- s} + 3^{- s} + \dots + (n - 1)^{- s} - \frac{n ^{1 - s}}{1 - s}

tends to a limit as $n \to \infty$ .

4. Show that the series

\frac{1}{1} + \frac{1}{2 ( 1 + \frac{1}{2} )} + \frac{1}{3 ( 1 + \frac{1}{2} + \frac{1}{3} )} + \dots

is divergent. [Compare the general term of the series with $1/ (n lo g n)$ .] Show also that the series derived from $\sum n^{- s}$ , in the same way that the above series is derived from $\sum (1/ n)$ , is convergent if $s > 1$ and otherwise divergent.

5. Prove generally that if $\sum u_{n}$ is a series of positive terms, and

s_{n} = u_{1} + u_{2} + \dots + u_{n},

then $\sum (u_{n} / s_{n - 1})$ is convergent or divergent according as $\sum u_{n}$ is convergent or divergent. [If $\sum u_{n}$ is convergent then $s_{n - 1}$ tends to a positive limit $l$ , and so $\sum (u_{n} / s_{n - 1})$ is convergent. If $\sum u_{n}$ is divergent then $s_{n - 1} \to \infty$ , and

u_{n} / s_{n - 1} > lo g {1 + (u_{n} / s_{n - 1})} = lo g (s_{n} / s_{n - 1})

(Ex. lxxxii. 1); and it is evident that

lo g (s_{2} / s_{1}) + lo g (s_{3} / s_{2}) + \dots + lo g (s_{n} / s_{n - 1}) = lo g (s_{n} / s_{1})

tends to $\infty$ as $n \to \infty$ .]

6. Prove that the same result holds for the series $\sum (u_{n} / s_{n})$ . [The proof is the same in the case of convergence. If $\sum u_{n}$ is divergent, and $u_{n} < s_{n - 1}$ from a certain value of $n$ onwards, then $s_{n} < 2 s_{n - 1}$ , and the divergence of $\sum (u_{n} / s_{n})$ follows from that of $\sum (u_{n} / s_{n - 1})$ . If on the other hand $u_{n} ≧ s_{n - 1}$ for an infinity of values of $n$ , as might happen with a rapidly divergent series, then $u_{n} / s_{n} ≧ \frac{1}{2}$ for all these values of $n$ .]

7. Sum the series $1 - \frac{1}{2} + \frac{1}{3} - \dots$ . [We have

1 + \frac{1}{2} + \dots + \frac{1}{2 n} = lo g (2 n + 1) + γ + ϵ_{n}, 2 (\frac{1}{2} + \frac{1}{4} + \dots + \frac{1}{2 n}) = lo g (n + 1) + γ + ϵ_{n}^{'},

by Ex. 1, $γ$ denoting Euler’s constant, and $ϵ_{n}$ , $ϵ_{n}^{'}$ being numbers which tend to zero as $n \to \infty$ . Subtracting and making $n \to \infty$ we see that the sum of the given series is $lo g 2$ . See also § 213.]

8. Prove that the series

0 \sum \infty (- 1)^{n} (1 + \frac{1}{2} + \dots + \frac{1}{n + 1} - lo g n - C)

oscillates finitely except when $C = γ$ , when it converges.

Footnotes

See the footnote to p. 362. ↩

212Series connected with the exponential and logarithmic functions. Expansion of $e^{x}$ by Taylor’s Theorem

Since all the derivatives of the exponential function are equal to the function itself, we have

e^{x} = 1 + x + \frac{x ^{2}}{2 !} + \dots + \frac{x ^{n - 1}}{( n - 1 )!} + \frac{x ^{n}}{n !} e^{θ x}

where $0 < θ < 1$ . But $x^{n} / n! \to 0$ as $n \to \infty$ , whatever be the value of $x$ (Ex. xxvii. 12); and $e^{θ x} < e^{x}$ . Hence, making $n$ tend to $\infty$ , we have

e^{x} = 1 + x + \frac{x ^{2}}{2 !} + \dots + \frac{x ^{n}}{n !} + \dots . (1)

The series on the right-hand side of this equation is known as the exponential series. In particular we have

e = 1 + 1 + \frac{1}{2 !} + \dots + \frac{1}{n !} + \dots; (2)

and so

(1 + 1 + \frac{1}{2 !} + \dots + \frac{1}{n !} + \dots)^{x} = 1 + x + \frac{x ^{2}}{2 !} + \dots + \frac{x ^{n}}{n !} + \dots, (3)

a result known as the exponential theorem. Also

a^{x} = e^{x l o g a} = 1 + (x lo g a) + \frac{( x lo g a ) ^{2}}{2 !} + \dots (4)

for all positive values of $a$ .

The reader will observe that the exponential series has the property of reproducing itself when every term is differentiated, and that no other series of powers of $x$ would possess this property: for some further remarks in this connection see Appendix II.

The power series for $e^{x}$ is so important that it is worth while to investigate it by an alternative method which does not depend upon Taylor’s Theorem. Let

E_{n} (x) = 1 + x + \frac{x ^{2}}{2 !} + \dots + \frac{x ^{n}}{n !},

and suppose that $x > 0$ . Then

(1 + \frac{x}{n})^{n} = 1 + n (\frac{x}{n}) + \frac{n ( n - 1 )}{1 \cdot 2} (\frac{x}{n})^{2} + \dots + \frac{n ( n - 1 ) \dots 1}{1 \cdot 2 \dots n} (\frac{x}{n})^{n}

which is less than $E_{n} (x)$ . And, provided $n > x$ , we have also, by the binomial theorem for a negative integral exponent,

(1 - \frac{x}{n})^{- n} = 1 + n (\frac{x}{n}) + \frac{n ( n + 1 )}{1 \cdot 2} (\frac{x}{n})^{2} + \dots > E_{n} (x) .

Thus

(1 + \frac{x}{n})^{n} < E_{n} (x) < (1 - \frac{x}{n})^{- n} .

But (§ 208) the first and last functions tend to the limit $e^{x}$ as $n \to \infty$ , and therefore $E_{n} (x)$ must do the same. From this the equation (1) follows when $x$ is positive; its truth when $x$ is negative follows from the fact that the exponential series, as was shown in Ex. lxxxi. 7, satisfies the functional equation $f (x) f (y) = f (x + y)$ , so that $f (x) f (- x) = f (0) = 1$ .

Examples XC.

1. Show that

cosh x = 1 + \frac{x ^{2}}{2 !} + \frac{x ^{4}}{4 !} + \dots, sinh x = x + \frac{x ^{3}}{3 !} + \frac{x ^{5}}{5 !} + \dots .

2. If $x$ is positive then the greatest term in the exponential series is the $([x] + 1)$ -th, unless $x$ is an integer, when the preceding term is equal to it.

3. Show that $n! > (n / e)^{n}$ . [For $n^{n} / n!$ is one term in the series for $e^{n}$ .]

4. Prove that $e^{n} = (n^{n} / n!) (2 + S_{1} + S_{2})$ , where

S_{1} = \frac{1}{1 + ν} + \frac{1}{( 1 + ν ) ( 1 + 2 ν )} + \dots, S_{2} = (1 - ν) + (1 - ν) (1 - 2 ν) + \dots,

and $ν = 1/ n$ ; and deduce that $n!$ lies between $2 (n / e)^{n}$ and $2 (n + 1) (n / e)^{n}$ .

5. Employ the exponential series to prove that $e^{x}$ tends to infinity more rapidly than any power of $x$ . [Use the inequality $e^{x} > x^{n} / n!$ .]

6. Show that $e$ is not a rational number. [If $e = p / q$ , where $p$ and $q$ are integers, we must have

\frac{p}{q} = 1 + 1 + \frac{1}{2 !} + \frac{1}{3 !} + \dots + \frac{1}{q !} + \dots

or, multiplying up by $q!$ ,

q! (\frac{p}{q} - 1 - 1 - \frac{1}{2 !} - \dots - \frac{1}{q !}) = \frac{1}{q + 1} + \frac{1}{( q + 1 ) ( q + 2 )} + \dots

and this is absurd, since the left-hand side is integral, and the right-hand side less than ${1/ (q + 1)} + {1/ (q + 1)}^{2} + \dots = 1/ q$ .]

7. Sum the series $0 \sum \infty P_{r} (n) \frac{x ^{n}}{n !}$ , where $P_{r} (n)$ is a polynomial of degree $r$ in $n$ . [We can express $P_{r} (n)$ in the form

A_{0} + A_{1} n + A_{2} n (n - 1) + \dots + A_{r} n (n - 1) \dots (n - r + 1),

and

0 \sum \infty P_{r} (n) \frac{x ^{n}}{n !} = A_{0} 0 \sum \infty \frac{x ^{n}}{n !} + A_{1} 1 \sum \infty \frac{x ^{n}}{( n - 1 )!} + \dots + A_{r} r \sum \infty \frac{x ^{n}}{( n - r )!} = (A_{0} + A_{1} x + A_{2} x^{2} + \dots + A_{r} x^{r}) e^{x} .]

8. Show that

1 \sum \infty \frac{n ^{3}}{n !} x^{n} = (x + 3 x^{2} + x^{3}) e^{x}, 1 \sum \infty \frac{n ^{4}}{n !} x^{n} = (x + 7 x^{2} + 6 x^{3} + x^{4}) e^{x};

and that if $S_{n} = 1^{3} + 2^{3} + \dots + n^{3}$ then

1 \sum \infty S_{n} \frac{x ^{n}}{n !} = \frac{1}{4} (4 x + 14 x^{2} + 8 x^{3} + x^{4}) e^{x} .

In particular the last series is equal to zero when $x = - 2$ . (Math. Trip. 1904.)

9. Prove that $\sum (n / n!) = e$ , $\sum (n^{2} / n!) = 2 e$ , $\sum (n^{3} / n!) = 5 e$ , and that $\sum (n^{k} / n!)$ , where $k$ is any positive integer, is a positive integral multiple of $e$ .

10. Prove that $1 \sum \infty \frac{( n - 1 ) x ^{n}}{( n + 2 ) n !} = {(x^{2} - 3 x + 3) e^{x} + \frac{1}{2} x^{2} - 3} / x^{2}$ .

[Multiply numerator and denominator by $n + 1$ , and proceed as in Ex. 7.]

11. Determine $a$ , $b$ , $c$ so that ${(x + a) e^{x} + (b x + c)} / x^{3}$ tends to a limit as $x \to 0$ , evaluate the limit, and draw the graph of the function $e^{x} + \frac{b x + c}{x + a}$ .

12. Draw the graphs of $1 + x$ , $1 + x + \frac{1}{2} x^{2}$ , $1 + x + \frac{1}{2} x^{2} + \frac{1}{6} x^{3}$ , and compare them with that of $e^{x}$ .

13. Prove that $e^{- x} - 1 + x - \frac{x ^{n}}{2 !} + \dots - (- 1)^{n} \frac{x ^{n}}{n !}$ is positive or negative according as $n$ is odd or even. Deduce the exponential theorem.

14. If

X_{0} = e^{x}, X_{1} = e^{x} - 1, X_{2} = e^{x} - 1 - x, X_{3} = e^{x} - 1 - x - (x^{2} /2!), \dots,

then $d X_{ν} / d x = X_{ν - 1}$ . Hence prove that if $t > 0$ then

X_{1} (t) = \int_{0}^{t} X_{0} d x < t e^{t}, X_{2} (t) = \int_{0}^{t} X_{1} d x < \int_{0}^{t} x e^{x} d x < e^{t} \int_{0}^{t} x d x = \frac{t ^{2}}{2 !} e^{t},

and generally $X_{ν} (t) < \frac{t ^{ν}}{ν !} e^{t}$ . Deduce the exponential theorem.

15. Show that the expansion in powers of $p$ of the positive root of $x^{2 + p} = a^{2}$ begins with the terms

a {1 - \frac{1}{2} p lo g a + \frac{1}{8} p^{2} lo g a (2 + lo g a)} .

(Math. Trip. 1909.)

213The logarithmic series

Another very important expansion in powers of $x$ is that for $lo g (1 + x)$ . Since

lo g (1 + x) = \int_{0}^{x} \frac{d t}{1 + t},

and $1/ (1 + t) = 1 - t + t^{2} - \dots$ if $t$ is numerically less than unity, it is natural to expect¹ that $lo g (1 + x)$ will be equal, when $- 1 < x < 1$ , to the series obtained by integrating each term of the series $1 - t + t^{2} - \dots$ from $t = 0$ to $t = x$ , i.e. to the series $x - \frac{1}{2} x^{2} + \frac{1}{3} x^{3} - \dots$ . And this is in fact the case. For

1/ (1 + t) = 1 - t + t^{2} - \dots + (- 1)^{m - 1} t^{m - 1} + \frac{( - 1 ) ^{m} t ^{m}}{1 + t},

and so, if $x > - 1$ ,

lo g (1 + x) = \int_{0}^{x} \frac{d t}{1 + t} = x - \frac{x ^{2}}{2} + \dots + (- 1)^{m - 1} \frac{x ^{m}}{m} + (- 1)^{m} R_{m},

where

R_{m} = \int_{0}^{x} \frac{t ^{m} d t}{1 + t} .

We require to show that the limit of $R_{m}$ , when $m$ tends to $\infty$ , is zero. This is almost obvious when $0 < x ≦ 1$ ; for then $R_{m}$ is positive and less than

\int_{0}^{x} t^{m} d t = \frac{x ^{m + 1}}{m + 1},

and therefore less than $1/ (m + 1)$ . If on the other hand $- 1 < x < 0$ , we put $t = - u$ and $x = - ξ$ , so that

R_{m} = (- 1)^{m} \int_{0}^{ξ} \frac{u ^{m} d u}{1 - u},

which shows that $R_{m}$ has the sign of $(- 1)^{m}$ . Also, since the greatest value of $1/ (1 - u)$ in the range of integration is $1/ (1 - ξ)$ , we have

0 < ∣ R_{m} ∣ < \frac{1}{1 - ξ} \int_{0}^{ξ} u^{m} d u = \frac{ξ ^{m}}{( m + 1 ) ( 1 - ξ )} < \frac{1}{( m + 1 ) ( 1 - ξ )} :

and so $R_{m} \to 0$ .

Hence

lo g (1 + x) = x - \frac{1}{2} x^{2} + \frac{1}{3} x^{3} - \dots,

provided that $- 1 < x ≦ 1$ . If $x$ lies outside these limits the series is not convergent. If $x = 1$ we obtain

lo g 2 = 1 - \frac{1}{2} + \frac{1}{3} - \dots,

a result already proved otherwise (Ex. lxxxix. 7).

Footnotes

See Appendix II for some further remarks on this subject. ↩

214The series for the inverse tangent

It is easy to prove in a similar manner that

arc tan x = \int_{0}^{x} \frac{d t}{1 + t ^{2}} = \int_{0}^{x} (1 - t^{2} + t^{4} - \dots) d t = x - \frac{1}{3} x^{3} + \frac{1}{5} x^{5} - \dots,

provided that $- 1 ≦ x ≦ 1$ . The only difference is that the proof is a little simpler; for, since $arc tan x$ is an odd function of $x$ , we need only consider positive values of $x$ . And the series is convergent when $x = - 1$ as well as when $x = 1$ . We leave the discussion to the reader. The value of $arc tan x$ which is represented by the series is of course that which lies between $- \frac{1}{4} π$ and $\frac{1}{4} π$ when $- 1 ≦ x ≦ 1$ , and which we saw in Ch. VII (Ex. lxiii. 3) to be the value represented by the integral. If $x = 1$ , we obtain the formula

\frac{1}{4} π = 1 - \frac{1}{3} + \frac{1}{5} - \dots .

Examples XCI.

1. $lo g (\frac{1}{1 - x}) = x + \frac{1}{2} x^{2} + \frac{1}{3} x^{3} + \dots$ if $- 1 ≦ x < 1$ .

2. $arg tanh x = \frac{1}{2} lo g (\frac{1 + x}{1 - x}) = x + \frac{1}{3} x^{3} + \frac{1}{5} x^{5} + \dots$ if $- 1 < x < 1$ .

3. Prove that if $x$ is positive then

lo g (1 + x) = \frac{x}{1 + x} + \frac{1}{2} (\frac{x}{1 + x})^{2} + \frac{1}{3} (\frac{x}{1 + x})^{3} + \dots .

(Math. Trip. 1911.)

4. Obtain the series for $lo g (1 + x)$ and $arc tan x$ by means of Taylor’s theorem.

[A difficulty presents itself in the discussion of the remainder in the first series when $x$ is negative, if Lagrange’s form $R_{n} = (- 1)^{n - 1} x^{n} / {n (1 + θ x)^{n}}$ is used; Cauchy’s form, viz.

R_{n} = (- 1)^{n - 1} (1 - θ)^{n - 1} x^{n} / (1 + θ x)^{n},

should be used (cf. the corresponding discussion for the Binomial Series, Ex. lvi. 2 and § 163).

In the case of the second series we have

D_{x}^{n} arc tan x = D_{x}^{n - 1} {1/ (1 + x^{2})} = (- 1)^{n - 1} (n - 1)! (x^{2} + 1)^{- n /2} sin {n arc tan (1/ x)}

(Ex. xlv. 11), and there is no difficulty about the remainder, which is obviously not greater in absolute value than $1/ n$ .¹]

5. If $y > 0$ then

lo g y = 2 {\frac{y - 1}{y + 1} + \frac{1}{3} (\frac{y - 1}{y + 1})^{3} + \frac{1}{5} (\frac{y - 1}{y + 1})^{5} + \dots} .

[Use the identity $y = (1 + \frac{y - 1}{y + 1}) / (1 - \frac{y - 1}{y + 1})$ . This series may be used to calculate $lo g 2$ , a purpose for which the series $1 - \frac{1}{2} + \frac{1}{3} - \dots$ , owing to the slowness of its convergence, is practically useless. Put $y = 2$ and find $lo g 2$ to $3$ places of decimals.]

6. Find $lo g 10$ to $3$ places of decimals from the formula

lo g 10 = 3 lo g 2 + lo g (1 + \frac{1}{4}) .

7. Prove that

lo g (\frac{x + 1}{x}) = 2 {\frac{1}{2 x + 1} + \frac{1}{3 ( 2 x + 1 ) ^{3}} + \frac{1}{5 ( 2 x + 1 ) ^{5}} + \dots}

if $x > 0$ , and that

lo g \frac{( x - 1 ) ^{2} ( x + 2 )}{( x + 1 ) ^{2} ( x - 2 )} = 2 {\frac{2}{x ^{3} - 3 x} + \frac{1}{3} (\frac{2}{x ^{3} - 3 x})^{3} + \frac{1}{5} (\frac{2}{x ^{3} - 3 x})^{5} + \dots}

if $x > 2$ . Given that $lo g 2 = .6931471 \dots$ and $lo g 3 = 1.0986123 \dots$ , show, by putting $x = 10$ in the second formula, that $lo g 11 = 2.397895 \dots$ . (Math. Trip. 1912.)

8. Show that if $lo g 2$ , $lo g 5$ , and $lo g 11$ are known, then the formula

lo g 13 = 3 lo g 11 + lo g 5 - 9 lo g 2

gives $lo g 13$ with an error practically equal to $.00015$ . (Math. Trip. 1910.)

9. Show that

\frac{1}{2} lo g 2 = 7 a + 5 b + 3 c, \frac{1}{2} lo g 3 = 11 a + 8 b + 5 c, \frac{1}{2} lo g 5 = 16 a + 12 b + 7 c,

where $a = arg tanh (1/31)$ , $b = arg tanh (1/49)$ , $c = arg tanh (1/161)$ .

[These formulae enable us to find $lo g 2$ , $lo g 3$ , and $lo g 5$ rapidly and with any degree of accuracy.]

10. Show that

\frac{1}{4} π = arc tan (1/2) + arc tan (1/3) = 4 arc tan (1/5) - arc tan (1/239),

and calculate $π$ to $6$ places of decimals.

11. Show that the expansion of $(1 + x)^{1 + x}$ in powers of $x$ begins with the terms $1 + x + x^{2} + 1/2 x^{3}$ . (Math. Trip. 1910.)

12. Show that

lo g_{10} e - x (x + 1) lo g_{10} (\frac{1 + x}{x}) = \frac{lo g _{10} e}{24 x ^{2}},

approximately, for large values of $x$ . Apply the formula, when $x = 10$ , to obtain an approximate value of $lo g_{10} e$ , and estimate the accuracy of the result. (Math. Trip. 1910.)

13. Show that

\frac{1}{1 - x} lo g (\frac{1}{1 - x}) = x + (1 + \frac{1}{2}) x^{2} + (1 + \frac{1}{2} + \frac{1}{3}) x^{3} + \dots,

if $- 1 < x < 1$ . [Use Ex. lxxxi. 2.]

14. Using the logarithmic series and the facts that $lo g_{10} 2.3758 = .3758099 \dots$ and $lo g_{10} e = .4343 \dots$ , show that an approximate solution of the equation $x = 100 lo g_{10} x$ is $237.58121$ . (Math. Trip. 1910.)

15. Expand $lo g cos x$ and $lo g (sin x / x)$ in powers of $x$ as far as $x^{4}$ , and verify that, to this order,

lo g sin x = lo g x - \frac{1}{45} lo g cos x + \frac{64}{45} lo g cos \frac{1}{2} x .

(Math. Trip. 1908.)

16. Show that

\int_{0}^{x} \frac{d t}{1 + t ^{4}} = x - \frac{1}{5} x^{5} + \frac{1}{9} x^{9} - \dots

if $- 1 ≦ x ≦ 1$ . Deduce that

1 - \frac{1}{5} + \frac{1}{9} - \dots = {π + 2 lo g (2 + 1)} /4 2 .

(Math. Trip. 1896.)

[Proceed as in § 214 and use the result of Ex. xlviii. 7.]

17. Prove similarly that

\frac{1}{3} - \frac{1}{7} + \frac{1}{11} - \dots = \int_{0}^{1} \frac{t ^{2} d t}{1 + t ^{4}} = {π - 2 lo g (2 + 1)} /4 2 .

18. Prove generally that if $a$ and $b$ are positive integers then

\frac{1}{a} - \frac{1}{a + b} + \frac{1}{a + 2 b} - \dots = \int_{0}^{1} \frac{t ^{a - 1} d t}{1 + t ^{b}},

and so that the sum of the series can be found. Calculate in this way the sums of $1 - \frac{1}{4} + \frac{1}{7} - \dots$ and $\frac{1}{2} - \frac{1}{5} + \frac{1}{8} - \dots$ .

Footnotes

The formula for $D_{x}^{n} arc tan x$ fails when $x = 0$ , as $arc tan (1/ x)$ is then undefined. It is easy to see (cf. Ex. xlv. 11) that $arc tan (1/ x)$ must then be interpreted as meaning $\frac{1}{2} π$ . ↩

215The Binomial Series

We have already (§ 163) investigated the Binomial Theorem

(1 + x)^{m} = 1 + (1 m) x + (2 m) x^{2} + \dots,

assuming that $- 1 < x < 1$ and that $m$ is rational. When $m$ is irrational we have

(1 + x)^{m} = e^{m l o g (1 + x)}, D_{x} (1 + x)^{m} = {m / (1 + x)} e^{m l o g (1 + x)} = m (1 + x)^{m - 1},

so that the rule for the differentiation of $(1 + x)^{m}$ remains the same, and the proof of the theorem given in § 163 retains its validity. We shall not discuss the question of the convergence of the series when $x = 1$ or $x = - 1$ .¹

Examples XCII.

1. Prove that if $- 1 < x < 1$ then

\frac{1}{1 + x ^{2}} = 1 - \frac{1}{2} x^{2} + \frac{1 \cdot 3}{2 \cdot 4} x^{4} - \dots, \frac{1}{1 - x ^{2}} = 1 + \frac{1}{2} x^{2} + \frac{1 \cdot 3}{2 \cdot 4} x^{4} + \dots .

2. Approximation to quadratic and other surds. Let $M$ be a quadratic surd whose numerical value is required. Let $N^{2}$ be the square nearest to $M$ ; and let $M = N^{2} + x$ or $M = N^{2} - x$ , $x$ being positive. Since $x$ cannot be greater than $N$ , $x / N^{2}$ is comparatively small and the surd $M = N 1 \pm (x / N^{2})$ can be expressed in a series

= N {1 \pm \frac{1}{2} (\frac{x}{N ^{2}}) - \frac{1 \cdot 1}{2 \cdot 4} (\frac{x}{N ^{2}})^{2} \pm \dots},

which is at any rate fairly rapidly convergent, and may be very rapidly so. Thus

67 = 64 + 3 = 8 {1 + \frac{1}{2} (\frac{3}{64}) - \frac{1 \cdot 1}{2 \cdot 4} (\frac{3}{64})^{2} + \dots} .

Let us consider the error committed in taking $8 \frac{3}{16}$ (the value given by the first two terms) as an approximate value. After the second term the terms alternate in sign and decrease. Hence the error is one of excess, and is less than $3^{2} /6 4^{2}$ , which is less than $.003$ .

3. If $x$ is small compared with $N^{2}$ then

N^{2} + x = N + \frac{x}{4 N} + \frac{N x}{2 ( 2 N ^{2} + x )},

the error being of the order $x^{4} / N^{7}$ . Apply the process to $907$ .

[Expanding by the binomial theorem, we have

N^{2} + x = N + \frac{x}{2 N} - \frac{x ^{2}}{8 N ^{3}} + \frac{x ^{3}}{16 N ^{5}},

the error being less than the numerical value of the next term, viz. $5 x^{4} /128 N^{7}$ . Also

\frac{N x}{2 ( 2 N ^{2} + x )} = \frac{x}{4 N} (1 + \frac{x}{2 N ^{2}})^{- 1} = \frac{x}{4 N} - \frac{x ^{2}}{8 N ^{3}} + \frac{x ^{3}}{16 N ^{5}},

the error being less than $x^{4} /32 N^{7}$ . The result follows. The same method may be applied to surds other than quadratic surds, e.g. to $31031$ .]

4. If $M$ differs from $N^{3}$ by less than $1$ per cent. of either then $3 M$ differs from $\frac{2}{3} N + \frac{1}{3} (M / N^{2})$ by less than $N /90000$ . (Math. Trip. 1882.)

5. If $M = N^{4} + x$ , and $x$ is small compared with $N$ , then a good approximation for $4 M$ is

\frac{51}{56} N + \frac{5}{56} \frac{M}{N ^{3}} + \frac{27 N x}{14 ( 7 M + 5 N ^{4} )} .

Show that when $N = 10$ , $x = 1$ , this approximation is accurate to $16$ places of decimals. (Math. Trip. 1886.)

6. Show how to sum the series

0 \sum \infty P_{r} (n) (n m) x^{n},

where $P_{r} (n)$ is a polynomial of degree $r$ in $n$ .

[Express $P_{r} (n)$ in the form $A_{0} + A_{1} n + A_{2} n (n - 1) + \dots$ as in Ex. xc. 7.]

7. Sum the series $0 \sum \infty n (n m) x^{n}$ , $0 \sum \infty n^{2} (n m) x^{n}$ and prove that

0 \sum \infty n^{3} (n m) x^{n} = {m^{3} x^{3} + m (3 m - 1) x^{2} + m x} (1 + x)^{m - 3} .

Footnotes

See Bromwich, Infinite Series, pp. 150 et seq.; Hobson, Plane Trigonometry (3rd edition), p. 271. ↩

216An alternative method of development of the theory of the exponential and logarithmic functions

We shall now give an outline of a method of investigation of the properties of $e^{x}$ and $lo g x$ entirely different in logical order from that followed in the preceding pages. This method starts from the exponential series $1 + x + \frac{x ^{2}}{2 !} + \dots$ . We know that this series is convergent for all values of $x$ , and we may therefore define the function $exp x$ by the equation

exp x = 1 + x + \frac{x ^{2}}{2 !} + \dots . (1)

We then prove, as in Ex. lxxxi. 7, that

exp x \times exp y = exp (x + y) . (2)

Again

\frac{exp h - 1}{h} = 1 + \frac{h}{2 !} + \frac{h ^{2}}{3 !} + \dots = 1 + ρ (h),

where $ρ (h)$ is numerically less than

∣ \frac{1}{2} h ∣ + ∣ \frac{1}{2} h ∣^{2} + ∣ \frac{1}{2} h ∣^{3} + \dots = ∣ \frac{1}{2} h ∣/ (1 - ∣ \frac{1}{2} h ∣),

so that $ρ (h) \to 0$ as $h \to 0$ . And so

\frac{exp ( x + h ) - exp x}{h} = exp x (\frac{exp h - 1}{h}) \to exp x

as $h \to 0$ , or

D_{x} exp x = exp x . (3)

Incidentally we have proved that $exp x$ is a continuous function.

We have now a choice of procedure. Writing $y = exp x$ and observing that $exp 0 = 1$ , we have

\frac{d y}{d x} = y, x = \int_{1}^{y} \frac{d t}{t},

and, if we define the logarithmic function as the function inverse to the exponential function, we are brought back to the point of view adopted earlier in this chapter.

But we may proceed differently. From (2) it follows that if $n$ is a positive integer then

(exp x)^{n} = exp n x, (exp 1)^{n} = exp n .

If $x$ is a positive rational fraction $m / n$ , then

{exp (m / n)}^{n} = exp m = (exp 1)^{m},

and so $exp (m / n)$ is equal to the positive value of $(exp 1)^{m / n}$ . This result may be extended to negative rational values of $x$ by means of the equation

exp x exp (- x) = 1;

and so we have

exp x = (exp 1)^{x} = e^{x},

say, where

e = exp 1 = 1 + 1 + \frac{1}{2 !} + \frac{1}{3 !} + \dots,

for all rational values of $x$ . Finally we define $e^{x}$ , when $x$ is irrational, as being equal to $exp x$ . The logarithm is then defined as the function inverse to $exp x$ or $e^{x}$ .

Example. Develop the theory of the binomial series

1 + (1 m) x + (2 m) x^{2} + \dots = f (m, x),

where $- 1 < x < 1$ , in a similar manner, starting from the equation

f (m, x) f (m^{'}, x) = f (m + m^{'} x)

(Ex. lxxxi. 6).

Misc.Miscellaneous Examples on Chapter IX

1. ¹ Given that $lo g_{10} e = .4343$ and that $2^{10}$ and $3^{21}$ are nearly equal to powers of $10$ , calculate $lo g_{10} 2$ and $lo g_{10} 3$ to four places of decimals. (Math. Trip. 1905.)

2. Determine which of $(\frac{1}{2} e)^{3}$ and $(2)^{\frac{1}{2} π}$ is the greater. [Take logarithms and observe that $3 / (3 + \frac{1}{4} π) < \frac{2}{5} 3 < .6929 < lo g 2$ .]

3. Show that $lo g_{10} n$ cannot be a rational number if $n$ is any positive integer not a power of $10$ . [If $n$ is not divisible by $10$ , and $lo g_{10} n = p / q$ , we have $1 0^{p} = n^{q}$ , which is impossible, since $1 0^{p}$ ends with $0$ and $n^{q}$ does not. If $n = 1 0^{a} N$ , where $N$ is not divisible by $10$ , then $lo g_{10} N$ and therefore

lo g_{10} n = a + lo g_{10} N

cannot be rational.]

4. For what values of $x$ are the functions $lo g x$ , $lo g lo g x$ , $lo g lo g lo g x$ , … (a) equal to $0$ (b) equal to $1$ (c) not defined? Consider also the same question for the functions $l x$ , $l l x$ , $l l l x$ , …, where $l x = lo g ∣ x ∣$ .

5. Show that

lo g x - (1 n) lo g (x + 1) + (2 n) lo g (x + 2) - \dots + (- 1)^{n} lo g (x + n)

is negative and increases steadily towards $0$ as $x$ increases from $0$ towards $\infty$ .

[The derivative of the function is

0 \sum n (- 1)^{r} (r n) \frac{1}{x + r} = \frac{n !}{x ( x + 1 ) \dots ( x + n )},

as is easily seen by splitting up the right-hand side into partial fractions. This expression is positive, and the function itself tends to zero as $x \to \infty$ , since

lo g (x + r) = lo g x + ϵ_{x},

where $ϵ_{x} \to 0$ , and $1 - (1 n) + (2 n) - \dots = 0$ .]

6. Prove that

(\frac{d}{d x})^{n} \frac{lo g x}{x} = \frac{( - 1 ) ^{n} n !}{x ^{n + 1}} (lo g x - 1 - \frac{1}{2} - \dots - \frac{1}{n}) .

(Math. Trip. 1909.)

7. If $x > - 1$ then $x^{2} > (1 + x) {lo g (1 + x)}^{2}$ . (Math. Trip. 1906.)

[Put $1 + x = e^{ξ}$ , and use the fact that $sinh ξ > ξ$ when $ξ > 0$ .]

8. Show that ${lo g (1 + x)} / x$ and $x / {(1 + x) lo g (1 + x)}$ both decrease steadily as $x$ increases from $0$ towards $\infty$ .

9. Show that, as $x$ increases from $- 1$ towards $\infty$ , the function $(1 + x)^{- 1/ x}$ assumes once and only once every value between $0$ and $1$ . (Math. Trip. 1910.)

10. Show that $\frac{1}{lo g ( 1 + x )} - \frac{1}{x} \to \frac{1}{2}$ as $x \to 0$ .

11. Show that $\frac{1}{lo g ( 1 + x )} - \frac{1}{x}$ decreases steadily from $1$ to $0$ as $x$ increases from $- 1$ towards $\infty$ . [The function is undefined when $x = 0$ , but if we attribute to it the value $\frac{1}{2}$ when $x = 0$ it becomes continuous for $x = 0$ . Use Ex. 7 to show that the derivative is negative.]

12. Show that the function $(lo g ξ - lo g x) / (ξ - x)$ , where $ξ$ is positive, decreases steadily as $x$ increases from $0$ to $ξ$ , and find its limit as $x \to ξ$ .

13. Show that $e^{x} > M x^{N}$ , where $M$ and $N$ are large positive numbers, if $x$ is greater than the greater of $2 lo g M$ and $16 N^{2}$ .

[It is easy to prove that $lo g x < 2 x$ ; and so the inequality given is certainly satisfied if

x > lo g M + 2 N x,

and therefore certainly satisfied if $\frac{1}{2} x > lo g M$ , $\frac{1}{2} x > 2 N x$ .]

14. If $f (x)$ and $ϕ (x)$ tend to infinity as $x \to \infty$ , and $f^{'} (x) / ϕ^{'} (x) \to \infty$ , then $f (x) / ϕ (x) \to \infty$ . [Use the result of Ch. VI, Misc. Ex. 33.] By taking $f (x) = x^{α}$ , $ϕ (x) = lo g x$ , prove that $(lo g x) / x^{α} \to 0$ for all positive values of $α$ .

15. If $p$ and $q$ are positive integers then

\frac{1}{p n + 1} + \frac{1}{p n + 2} + \dots + \frac{1}{q n} \to lo g (\frac{q}{p})

as $n \to \infty$ . [Cf. Ex. lxxviii. 6.]

16. Prove that if $x$ is positive then $n lo g {\frac{1}{2} (1 + x^{1/ n})} \to - \frac{1}{2} lo g x$ as $n \to \infty$ . [We have

n lo g {\frac{1}{2} (1 + x^{1/ n})} = n lo g {1 - \frac{1}{2} (1 - x^{1/ n})} = \frac{1}{2} n (1 - x^{1/ n}) \frac{lo g ( 1 - u )}{u}

where $u = \frac{1}{2} (1 - x^{1/ n})$ . Now use § 209 and Ex. lxxxii. 4.]

17. Prove that if $a$ and $b$ are positive then

{\frac{1}{2} (a^{1/ n} + b^{1/ n})}^{n} \to ab .

[Take logarithms and use Ex. 16.]

18. Show that

1 + \frac{1}{3} + \frac{1}{5} + \dots + \frac{1}{2 n - 1} = \frac{1}{2} lo g n + lo g 2 + \frac{1}{2} γ + ϵ_{n},

where $γ$ is Euler’s constant (Ex. lxxxix. 1) and $ϵ_{n} \to 0$ as $n \to \infty$ .

19. Show that

1 + \frac{1}{3} - \frac{1}{2} + \frac{1}{5} + \frac{1}{7} - \frac{1}{4} + \frac{1}{9} + \dots = \frac{3}{2} lo g 2,

the series being formed from the series $1 - \frac{1}{2} + \frac{1}{3} - \dots$ by taking alternately two positive terms and then one negative. [The sum of the first $3 n$ terms is

1 + \frac{1}{3} + \frac{1}{5} + \dots + \frac{1}{4 n - 1} - \frac{1}{2} (1 + \frac{1}{2} + \dots + \frac{1}{n}) = \frac{1}{2} lo g 2 n + lo g 2 + \frac{1}{2} γ + ϵ_{n} - \frac{1}{2} (lo g n + γ + ϵ_{n}^{'}),

where $ϵ_{n}$ and $ϵ_{n}^{'}$ tend to $0$ as $n \to \infty$ . (Cf. Ex. lxxviii. 6).]

20. Show that $1 - \frac{1}{2} - \frac{1}{4} + \frac{1}{3} - \frac{1}{6} - \frac{1}{8} + \frac{1}{5} - \frac{1}{10} - \dots = \frac{1}{2} lo g 2$ .

21. Prove that

1 \sum n \frac{1}{ν ( 36 ν ^{2} - 1 )} = - 3 + 3 Σ_{3 n + 1} - Σ_{n} - S_{n}

where $S_{n} = 1 + \frac{1}{2} + \dots + \frac{1}{n}$ , $Σ_{n} = 1 + \frac{1}{3} + \dots + \frac{1}{2 n - 1}$ . Hence prove that the sum of the series when continued to infinity is

- 3 + \frac{3}{2} lo g 3 + 2 lo g 2.

(Math. Trip. 1905.)

22. Show that

1 \sum \infty \frac{1}{n ( 4 n ^{2} - 1 )} = 2 lo g 2 - 1, 1 \sum \infty \frac{1}{n ( 9 n ^{2} - 1 )} = \frac{3}{2} (lo g 3 - 1) .

23. Prove that the sums of the four series

1 \sum \infty \frac{1}{4 n ^{2} - 1}, 1 \sum \infty \frac{( - 1 ) ^{n - 1}}{4 n ^{2} - 1}, 1 \sum \infty \frac{1}{( 2 n + 1 ) ^{2} - 1}, 1 \sum \infty \frac{( - 1 ) ^{n - 1}}{( 2 n + 1 ) ^{2} - 1}

are $\frac{1}{2}$ , $\frac{1}{4} π - \frac{1}{2}$ , $\frac{1}{4}$ , $\frac{1}{2} lo g 2 - \frac{1}{4}$ respectively.

24. Prove that $n! (a / n)^{n}$ tends to $0$ or to $\infty$ according as $a < e$ or $a > e$ .

[If $u_{n} = n! (a / n)^{n}$ then $u_{n + 1} / u_{n} = a {1 + (1/ n)}^{- n} \to a / e$ . It can be shown that the function tends to $\infty$ when $a = e$ : for a proof, which is rather beyond the scope of the theorems of this chapter, see Bromwich’s Infinite Series, pp. 461 et seq.]

25. Find the limit as $x \to \infty$ of

(\frac{a _{0} + a _{1} x + \dots + a _{r} x ^{r}}{b _{0} + b _{1} x + \dots + b _{r} x ^{r}})^{λ_{0} + λ_{1} x},

distinguishing the different cases which may arise. (Math. Trip. 1886.)

26. Prove that

\sum lo g (1 + \frac{x}{n}) (x > 0)

diverges to $\infty$ . [Compare with $\sum (x / n)$ .] Deduce that if $x$ is positive then

(1 + x) (2 + x) \dots (n + x) / n! \to \infty

as $n \to \infty$ . [The logarithm of the function is $1 \sum n lo g (1 + \frac{x}{ν})$ .]

27. Prove that if $x > - 1$ then

\frac{1}{( x + 1 ) ^{2}} = \frac{1}{( x + 1 ) ( x + 2 )} + \frac{1 !}{( x + 1 ) ( x + 2 ) ( x + 3 )} + \frac{2 !}{( x + 1 ) ( x + 2 ) ( x + 3 ) ( x + 4 )} + \dots .

(Math. Trip. 1908.)

[The difference between $1/ (x + 1)^{2}$ and the sum of the first $n$ terms of the series is

\frac{1}{( x + 1 ) ^{2}} \frac{n !}{( x + 2 ) ( x + 3 ) \dots ( x + n + 1 )} .]

28. No equation of the type

A e^{α x} + B e^{β x} + \dots = 0,

where $A$ , $B$ , … are polynomials and $α$ , $β$ , … different real numbers, can hold for all values of $x$ . [If $α$ is the algebraically greatest of $α$ , $β$ , …, then the term $A e^{α x}$ outweighs all the rest as $x \to \infty$ .]

29. Show that the sequence

a_{1} = e, a_{2} = e^{e^{2}}, a_{3} = e^{e^{e^{3}}}, \dots

tends to infinity more rapidly than any member of the exponential scale.

[Let $e_{1} (x) = e^{x}$ , $e_{2} (x) = e^{e_{1} (x)}$ , and so on. Then, if $e_{k} (x)$ is any member of the exponential scale, $a_{n} > e_{k} (n)$ when $n > k$ .]

30. Prove that

\frac{d}{d x} {ϕ (x)}^{ψ (x)} = \frac{d}{d x} {ϕ (x)}^{α} + \frac{d}{d x} {β^{ψ (x)}}

where $α$ is to be put equal to $ψ (x)$ and $β$ to $ϕ (x)$ after differentiation. Establish a similar rule for the differentiation of $ϕ (x)^{[{ψ (x)}^{χ (x)}]}$ .

31. Prove that if $D_{x}^{n} e^{- x^{2}} = e^{- x^{2}} ϕ_{n} (x)$ then (i) $ϕ_{n} (x)$ is a polynomial of degree $n$ , (ii) $ϕ_{n + 1} = - 2 x ϕ_{n} + ϕ_{n}^{'}$ , and (iii) all the roots of $ϕ_{n} = 0$ are real and distinct, and separated by those of $ϕ_{n - 1} = 0$ . [To prove (iii) assume the truth of the result for $κ = 1$ , $2$ , … $n$ , and consider the signs of $ϕ_{n + 1}$ for the $n$ values of $x$ for which $ϕ_{n} = 0$ and for large (positive or negative) values of $x$ .]

32. The general solution of $f (x y) = f (x) f (y)$ , where $f$ is a differentiable function, is $x^{a}$ , where $a$ is a constant: and that of

f (x + y) + f (x - y) = 2 f (x) f (y)

is $cosh a x$ or $cos a x$ , according as $f^{''} (0)$ is positive or negative. [In proving the second result assume that $f$ has derivatives of the first three orders. Then

2 f (x) + y^{2} {f^{''} (x) + ϵ_{y}} = 2 f (x) [f (0) + y f^{'} (0) + \frac{1}{2} y^{2} {f^{''} (0) + ϵ_{y}^{'}}],

where $ϵ_{y}$ and $ϵ_{y}^{'}$ tend to zero with $y$ . It follows that $f (0) = 1$ , $f^{'} (0) = 0$ , $f^{''} (x) = f^{''} (0) f (x)$ , so that $a = f^{''} (0)$ or $a = - f^{''} (0)$ .]

33. How do the functions $x^{s i n (1/ x)}$ , $x^{s i n^{2} (1/ x)}$ , $x^{cosec (1/ x)}$ behave as $x \to + 0$ ?

34. Trace the curves $y = tan x e^{t a n x}$ , $y = sin x lo g tan \frac{1}{2} x$ .

35. The equation $e^{x} = a x + b$ has one real root if $a < 0$ or $a = 0$ , $b > 0$ . If $a > 0$ then it has two real roots or none, according as $a lo g a > b - a$ or $a lo g a < b - a$ .

36. Show by graphical considerations that the equation $e^{x} = a x^{2} + 2 b x + c$ has one, two, or three real roots if $a > 0$ , none, one, or two if $a < 0$ ; and show how to distinguish between the different cases.

37. Trace the curve $y = \frac{1}{x} lo g (\frac{e ^{x} - 1}{x})$ , showing that the point $(0, \frac{1}{2})$ is a centre of symmetry, and that as $x$ increases through all real values, $y$ steadily increases from $0$ to $1$ . Deduce that the equation

\frac{1}{x} lo g (\frac{e ^{x} - 1}{x}) = α

has no real root unless $0 < α < 1$ , and then one, whose sign is the same as that of $α - \frac{1}{2}$ . [In the first place

y - \frac{1}{2} = \frac{1}{x} {lo g (\frac{e ^{x} - 1}{x}) - lo g e^{\frac{1}{2} x}} = \frac{1}{x} lo g (\frac{sinh \frac{1}{2} x}{\frac{1}{2} x})

is clearly an odd function of $x$ . Also

\frac{d y}{d x} = \frac{1}{x ^{2}} {\frac{1}{2} x coth \frac{1}{2} x - 1 - lo g (\frac{sinh \frac{1}{2} x}{\frac{1}{2} x})} .

The function inside the large bracket tends to zero as $x \to 0$ ; and its derivative is

\frac{1}{x} {1 - (\frac{\frac{1}{2} x}{sinh \frac{1}{2} x})^{2}},

which has the sign of $x$ . Hence $d y / d x > 0$ for all values of $x$ .]

38. Trace the curve $y = e^{1/ x} x^{2} + 2 x$ , and show that the equation

e^{1/ x} x^{2} + 2 x = α

has no real roots if $α$ is negative, one negative root if

0 < α < a = e^{1/ 2} 2 + 22,

and two positive roots and one negative if $α > a$ .

39. Show that the equation $f_{n} (x) = 1 + x + \frac{x ^{2}}{2 !} + \dots + \frac{x ^{n}}{n !} = 0$ has one real root if $n$ is odd and none if $n$ is even.

[Assume this proved for $n = 1$ , $2$ , … $2 k$ . Then $f_{2 k + 1} (x) = 0$ has at least one real root, since its degree is odd, and it cannot have more since, if it had, $f_{2 k + 1}^{'} (x)$ or $f_{2 k} (x)$ would have to vanish once at least. Hence $f_{2 k + 1} (x) = 0$ has just one root, and so $f_{2 k + 2} (x) = 0$ cannot have more than two. If it has two, say $α$ and $β$ , then $f_{2 k + 2}^{'} (x)$ or $f_{2 k + 1} (x)$ must vanish once at least between $α$ and $β$ , say at $γ$ . And

f_{2 k + 2} (γ) = f_{2 k + 1} (γ) + \frac{γ ^{2 k + 2}}{( 2 k + 2 )!} > 0.

But $f_{2 k + 2} (x)$ is also positive when $x$ is large (positively or negatively), and a glance at a figure will show that these results are contradictory. Hence $f_{2 k + 2} (x) = 0$ has no real roots.]

40. Prove that if $a$ and $b$ are positive and nearly equal then

lo g \frac{a}{b} = \frac{1}{2} (a - b) (\frac{1}{a} + \frac{1}{b}),

approximately, the error being about $\frac{1}{6} {(a - b) / a}^{3}$ . [Use the logarithmic series. This formula is interesting historically as having been employed by Napier for the numerical calculation of logarithms.]

41. Prove by multiplication of series that if $- 1 < x < 1$ then

\frac{1}{2} {lo g (1 + x)}^{2} \frac{1}{2} (arc tan x)^{2} = \frac{1}{2} x^{2} - \frac{1}{3} (1 + \frac{1}{2}) x^{3} + \frac{1}{4} (1 + \frac{1}{2} + \frac{1}{3}) x^{4} - \dots, = \frac{1}{2} x^{2} - \frac{1}{4} (1 + \frac{1}{3}) x^{4} + \frac{1}{6} (1 + \frac{1}{3} + \frac{1}{5}) x^{6} - \dots .

42. Prove that

(1 + α x)^{1/ x} = e^{α} {1 - \frac{1}{2} a^{2} x + \frac{1}{24} (8 + 3 a) a^{3} x^{2} (1 + ϵ_{x})},

where $ϵ_{x} \to 0$ with $x$ .

43. The first $n + 2$ terms in the expansion of $lo g (1 + x + \frac{x ^{2}}{2 !} + \dots + \frac{x ^{n}}{n !})$ in powers of $x$ are

x - \frac{x ^{n + 1}}{n !} {\frac{1}{n + 1} - \frac{x}{1 ! ( n + 2 )} + \frac{x ^{2}}{2 ! ( n + 3 )} - \dots + (- 1)^{n} \frac{x ^{n}}{n ! ( 2 n + 1 )}} .

(Math. Trip. 1899.)

44. Show that the expansion of

exp (- x - \frac{x ^{2}}{2} - \dots - \frac{x ^{n}}{n})

in powers of $x$ begins with the terms

1 - x + \frac{x ^{n + 1}}{n + 1} - s = 1 \sum n \frac{x ^{n + s + 1}}{( n + s ) ( n + s + 1 )} .

(Math. Trip. 1909.)

45. Show that if $- 1 < x < 1$ then

\frac{1}{3} x + \frac{1 \cdot 4}{3 \cdot 6} 2^{2} x^{2} + \frac{1 \cdot 4 \cdot 7}{3 \cdot 6 \cdot 9} 3^{2} x^{3} + \dots \frac{1}{3} x + \frac{1 \cdot 4}{3 \cdot 6} 2^{3} x^{2} + \frac{1 \cdot 4 \cdot 7}{3 \cdot 6 \cdot 9} 3^{3} x^{3} + \dots = \frac{x ( x + 3 )}{9 ( 1 - x ) ^{7/3}}, = \frac{x ( x ^{2} + 18 x + 9 )}{27 ( 1 - x ) ^{10/3}} .

[Use the method of Ex. xcii. 6. The results are more easily obtained by differentiation; but the problem of the differentiation of an infinite series is beyond our range.]

46. Prove that

\int_{0}^{\infty} \frac{d x}{( x + a ) ( x + b )} \int_{0}^{\infty} \frac{d x}{( x + a ) ( x + b ) ^{2}} \int_{0}^{\infty} \frac{x d x}{( x + a ) ( x + b ) ^{2}} \int_{0}^{\infty} \frac{d x}{( x + a ) ( x ^{2} + b ^{2} )} \int_{0}^{\infty} \frac{x d x}{( x + a ) ( x ^{2} + b ^{2} )} = \frac{1}{a - b} lo g (\frac{a}{b}), = \frac{1}{( a - b ) ^{2} b} {a - b - b lo g (\frac{a}{b})}, = \frac{1}{( a - b ) ^{2}} {a lo g (\frac{a}{b}) - a + b}, = \frac{1}{( a ^{2} + b ^{2} ) b} {\frac{1}{2} π a - b lo g (\frac{a}{b})}, = \frac{1}{a ^{2} + b ^{2}} {\frac{1}{2} π b + a lo g (\frac{a}{b})},

provided that $a$ and $b$ are positive. Deduce, and verify independently, that each of the functions

a - 1 - lo g a, a lo g a - a + 1, \frac{1}{2} π a - lo g a, \frac{1}{2} π + a lo g a

is positive for all positive values of $a$ .

47. Prove that if $α$ , $β$ , $γ$ are all positive, and $β^{2} > α γ$ , then

\int_{0}^{\infty} \frac{d x}{α x ^{2} + 2 β x + γ} = \frac{1}{β ^{2} - α γ} lo g {\frac{β + β ^{2} - α γ}{α γ}};

while if $α$ is positive and $α γ > β^{2}$ the value of the integral is

\frac{1}{α γ - β ^{2}} arc tan {\frac{α γ - β ^{2}}{β}},

that value of the inverse tangent being chosen which lies between $0$ and $π$ . Are there any other really different cases in which the integral is convergent?

48. Prove that if $a > - 1$ then

\int_{1}^{\infty} \frac{d x}{( x + a ) x ^{2} - 1} = \int_{0}^{\infty} \frac{d t}{cosh t + a} = 2 \int_{1}^{\infty} \frac{d u}{u ^{2} + 2 a u + 1};

and deduce that the value of the integral is

\frac{2}{1 - a ^{2}} arc tan \frac{1 - a}{1 + a}

if $- 1 < a < 1$ , and

\frac{1}{a ^{2} - 1} lo g \frac{a + 1 + a - 1}{a + 1 - a - 1} = \frac{2}{a ^{2} - 1} arg tanh \frac{a - 1}{a + 1}

if $a > 1$ . Discuss the case in which $a = 1$ .

49. Transform the integral $\int_{0}^{\infty} \frac{d x}{( x + a ) x ^{2} + 1}$ , where $a > 0$ , in the same ways, showing that its value is

\frac{1}{a ^{2} + 1} lo g \frac{a + 1 + a ^{2} + 1}{a + 1 - a ^{2} + 1} = \frac{2}{a ^{2} + 1} arg tanh \frac{a ^{2} + 1}{a + 1}

50. Prove that

\int_{0}^{1} arc tan x d x = \frac{1}{4} π - \frac{1}{2} lo g 2.

51. If $0 < α < 1$ , $0 < β < 1$ , then

\int_{- 1}^{1} \frac{d x}{( 1 - 2 α x + α ^{2} ) ( 1 - 2 β x + β ^{2} )} = \frac{1}{α β} lo g \frac{1 + α β}{1 - α β} .

52. Prove that if $a > b > 0$ then

\int_{- \infty}^{\infty} \frac{d θ}{a cosh θ + b sinh θ} = \frac{π}{a ^{2} - b ^{2}}

53. Prove that

\int_{0}^{1} \frac{lo g x}{1 + x ^{2}} d x = - \int_{1}^{\infty} \frac{lo g x}{1 + x ^{2}} d x, \int_{0}^{\infty} \frac{lo g x}{1 + x ^{2}} d x = 0

and deduce that if $a > 0$ then

\int_{0}^{\infty} \frac{lo g x}{a ^{2} + x ^{2}} d x = \frac{π}{2 a} lo g a .

[Use the substitutions $x = 1/ t$ and $x = a u$ .]

54. Prove that

\int_{0}^{\infty} lo g (1 + \frac{a ^{2}}{x ^{2}}) d x = π a

if $a > 0$ . [Integrate by parts.]

Footnotes

A considerable number of these examples are taken from Bromwich’s Infinite Series. ↩

Chapter IXThe Logarithmic and Exponential Functions of a Real Variable

197Definition of log⁡x\log xlogx

198The functional equation satisfied by log⁡x\log xlogx

199The manner in which log⁡x\log xlogx tends to infinity with xxx

200Proof that (log⁡x)/xα→0(\log x)/x^{\alpha} \to 0(logx)/xα→0 as x→∞x \to \inftyx→∞

201The behaviour of log⁡x\log xlogx as x→+0x \to +0x→+0

202Scales of infinity. The logarithmic scale

203The number eee

204The exponential function

205The principal properties of the exponential function

207The general power axa^{x}ax

208The representation of exe^{x}ex as a limit

209The representation of log⁡x\log xlogx as a limit

210Common logarithms

211Logarithmic tests of convergence for series and integrals

212Series connected with the exponential and logarithmic functions. Expansion of exe^{x}ex by Taylor’s Theorem

213The logarithmic series

214The series for the inverse tangent

215The Binomial Series

216An alternative method of development of the theory of the exponential and logarithmic functions

Misc.Miscellaneous Examples on Chapter IX

197Definition of $\log x$

198The functional equation satisfied by $\log x$

199The manner in which $\log x$ tends to infinity with $x$

200Proof that $(\log x)/x^{\alpha} \to 0$ as $x \to \infty$

201The behaviour of $\log x$ as $x \to +0$

203The number $e$

207The general power $a^{x}$

208The representation of $e^{x}$ as a limit

209The representation of $\log x$ as a limit

212Series connected with the exponential and logarithmic functions. Expansion of $e^{x}$ by Taylor’s Theorem