Chapter VIIAdditional Theorems in the Differential and Integral Calculus

147Higher Mean Value Theorems

In the preceding chapter (§ 125) we proved that if $f (x)$ has a derivative $f^{'} (x)$ throughout the interval $(a, b)$ then

f (b) - f (a) = (b - a) f^{'} (ξ),

where $a < ξ < b$ ; or that, if $f (x)$ has a derivative throughout $(a, a + h)$ , then

f (a + h) - f (a) = h f^{'} (a + θ_{1} h), (1)

where $0 < θ_{1} < 1$ . This we proved by considering the function

f (b) - f (x) - \frac{b - x}{b - a} {f (b) - f (a)}

which vanishes when $x = a$ and when $x = b$ .

Let us now suppose that $f (x)$ has also a second derivative $f^{''} (x)$ throughout $(a, b)$ , an assumption which of course involves the continuity of the first derivative $f^{'} (x)$ , and consider the function

f (b) - f (x) - (b - x) f^{'} (x) - (\frac{b - x}{b - a})^{2} {f (b) - f (a) - (b - a) f^{'} (a)} .

This function also vanishes when $x = a$ and when $x = b$ ; and its derivative is

\frac{2 ( b - x )}{( b - a ) ^{2}} {f (b) - f (a) - (b - a) f^{'} (a) - \frac{1}{2} (b - a)^{2} f^{''} (x)},

and this must vanish (§ 121) for some value of $x$ between $a$ and $b$ (exclusive of $a$ and $b$ ). Hence there is a value $ξ$ of $x$ , between $a$ and $b$ , and therefore capable of representation in the form $a + θ_{2} (b - a)$ , where $0 < θ_{2} < 1$ , for which

f (b) = f (a) + (b - a) f^{'} (a) + \frac{1}{2} (b - a)^{2} f^{''} (ξ) .

If we put $b = a + h$ we obtain the equation

f (a + h) = f (a) + h f^{'} (a) + \frac{1}{2} h^{2} f^{''} (a + θ_{2} h), (2)

which is the standard form of what may be called the Mean Value Theorem of the second order.

The analogy suggested by (1) and (2) at once leads us to formulate the following theorem:

Taylor’s or the General Mean Value Theorem. If $f (x)$ is a function of $x$ which has derivatives of the first $n$ orders throughout the interval $(a, b)$ , then

f (b) = f (a) + (b - a) f^{'} (a) + \frac{( b - a ) ^{2}}{2 !} f^{''} (a) + \dots + \frac{( b - a ) ^{n - 1}}{( n - 1 )!} f^{(n - 1)} (a) + \frac{( b - a ) ^{n}}{n !} f^{(n)} (ξ),

where $a < ξ < b$ ; and if $b = a + h$ then

f (a + h) = f (a) + h f^{'} (a) + \frac{1}{2} h^{2} f^{''} (a) + \dots + \frac{h ^{n - 1}}{( n - 1 )!} f^{(n - 1)} (a) + \frac{h ^{n}}{n !} f^{(n)} (a + θ_{n} h),

where $0 < θ_{n} < 1$ .

The proof proceeds on precisely the same lines as were adopted before in the special cases in which $n = 1$ and $n = 2$ . We consider the function

F_{n} (x) - (\frac{b - x}{b - a})^{n} F_{n} (a),

where

F_{n} (x) = f (b) - f (x) - (b - x) f^{'} (x) - \frac{( b - x ) ^{2}}{2 !} f^{''} (x) - \dots - \frac{( b - x ) ^{n - 1}}{( n - 1 )!} f^{(n - 1)} (x) .

This function vanishes for $x = a$ and $x = b$ ; its derivative is

\frac{n ( b - x ) ^{n - 1}}{( b - a ) ^{n}} {F_{n} (a) - \frac{( b - a ) ^{n}}{n !} f^{(n)} (x)};

and there must be some value of $x$ between $a$ and $b$ for which the derivative vanishes. This leads at once to the desired result.

In view of the great importance of this theorem we shall give at the end of this chapter another proof, not essentially distinct from that given above, but different in form and depending on the method of integration by parts.

Examples LV.

1. Suppose that $f (x)$ is a polynomial of degree $r$ . Then $f^{(n)} (x)$ is identically zero when $n > r$ , and the theorem leads to the algebraical identity

f (a + h) = f (a) + h f^{'} (a) + \frac{h ^{2}}{2 !} f^{''} (a) + \dots + \frac{h ^{r}}{r !} f^{(r)} (a) .

2. By applying the theorem to $f (x) = 1/ x$ , and supposing $x$ and $x + h$ positive, obtain the result

\frac{1}{x + h} = \frac{1}{x} - \frac{h}{x ^{2}} + \frac{h ^{2}}{x ^{3}} - \dots + \frac{( - 1 ) ^{n - 1} h ^{n - 1}}{x ^{n}} + \frac{( - 1 ) ^{n} h ^{n}}{( x + θ _{n} h ) ^{n + 1}} .

[Since

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

we can verify the result by showing that $x^{n} (x + h)$ can be put in the form $(x + θ_{n} h)^{n + 1}$ , or that $x^{n + 1} < x^{n} (x + h) < (x + h)^{n + 1}$ , as is evidently the case.]

3. Obtain the formula

sin (x + h) = sin x + h cos x - \frac{h ^{2}}{2 !} sin x - h^{3} \frac{3 !}{cos} x + \dots + (- 1)^{n - 1} \frac{h ^{2 n - 1}}{( 2 n - 1 )!} cos x + (- 1)^{n} h^{2 n} \frac{2 n !}{sin} (x + θ_{2 n} h),

the corresponding formula for $cos (x + h)$ , and similar formulae involving powers of $h$ extending up to $h^{2 n + 1}$ .

4. Show that if $m$ is a positive integer, and $n$ a positive integer not greater than $m$ , then

(x + h)^{m} = x^{m} + (1 m) x^{m - 1} h + \dots + (n - 1 m) x^{m - n + 1} h^{n - 1} + (n m) (x + θ_{n} h)^{m - n} h^{n} .

Show also that, if the interval $(x, x + h)$ does not include $x = 0$ , the formula holds for all real values of $m$ and all positive integral values of $n$ ; and that, even if $x < 0 < x + h$ or $x + h < 0 < x$ , the formula still holds if $m - n$ is positive.

5. The formula $f (x + h) = f (x) + h f^{'} (x + θ_{1} h)$ is not true if $f (x) = 1/ x$ and $x < 0 < x + h$ . [For $f (x + h) - f (x) > 0$ and $h f^{'} (x + θ_{1} h) = - h / (x + θ_{1} h)^{2} < 0$ ; it is evident that the conditions for the truth of the Mean Value Theorem are not satisfied.]

6. If $x = - a$ , $h = 2 a$ , $f (x) = x^{1/3}$ , then the equation

f (x + h) = f (x) + h f^{'} (x + θ_{1} h)

is satisfied by $θ_{1} = \frac{1}{2} \pm \frac{1}{18} 3$ . [This example shows that the result of the theorem may hold even if the conditions under which it was proved are not satisfied.]

7. Newton’s method of approximation to the roots of equations. Let $ξ$ be an approximation to a root of an algebraical equation $f (x) = 0$ , the actual root being $ξ + h$ . Then

0 = f (ξ + h) = f (ξ) + h f^{'} (ξ) + \frac{1}{2} h^{2} f^{''} (ξ + θ_{2} h),

so that

h = - \frac{f ( ξ )}{f ^{'} ( ξ )} - \frac{1}{2} h^{2} \frac{f ^{''} ( ξ + θ _{2} h )}{f ^{'} ( ξ )} .

It follows that in general a better approximation than $x = ξ$ is

x = ξ - \frac{f ( ξ )}{f ^{'} ( ξ )} .

If the root is a simple root, so that $f^{'} (ξ + h) \neq = 0$ , we can, when $h$ is small enough, find a positive constant $K$ such that $∣ f^{'} (x) ∣ > K$ for all the values of $x$ which we are considering, and then, if $h$ is regarded as of the first order of smallness, $f (ξ)$ is of the first order of smallness, and the error in taking $ξ - {f (ξ) / f^{'} (ξ)}$ as the root is of the second order.

8. Apply this process to the equation $x^{2} = 2$ , taking $ξ = 3/2$ as the first approximation. [We find $h = - 1/12$ , $ξ + h = 17/12 = 1.417 \dots$ , which is quite a good approximation, in spite of the roughness of the first. If now we repeat the process, taking $ξ = 17/12$ , we obtain $ξ + h = 577/408 = 1.414215 \dots$ , which is correct to $5$ places of decimals.

9. By considering in this way the equation $x^{2} - 1 - y = 0$ , where $y$ is small, show that $1 + y = 1 + \frac{1}{2} y - {\frac{1}{4} y^{2} / (2 + y)}$ approximately, the error being of the fourth order.

10. Show that the error in taking the root to be $ξ - (f / f^{'}) - \frac{1}{2} (f^{2} f^{''} / f^{' 3})$ , where $ξ$ is the argument of every function, is in general of the third order.

11. The equation $sin x = α x$ , where $α$ is small, has a root nearly equal to $π$ . Show that $(1 - α) π$ is a better approximation, and $(1 - α + α^{2}) π$ a better still. [The method of Exs. 7–10 does not depend on $f (x) = 0$ being an algebraical equation, so long as $f^{'}$ and $f^{''}$ are continuous.]

12. Show that the limit when $h \to 0$ of the number $θ_{n}$ which occurs in the general Mean Value Theorem is $1/ (n + 1)$ , provided that $f^{(n + 1)} (x)$ is continuous.

[For $f (x + h)$ is equal to each of

f (x) + \dots + \frac{h ^{n}}{n !} f^{(n)} (x + θ_{n} h), f (x) + \dots + \frac{h ^{n}}{n !} f^{(n)} (x) + \frac{h ^{n + 1}}{( n + 1 )!} f^{(n + 1)} (x + θ_{n + 1} h),

where $θ_{n + 1}$ as well as $θ_{n}$ lies between $0$ and $1$ . Hence

f^{(n)} (x + θ_{n} h) = f^{(n)} (x) + \frac{h f ^{(n + 1)} ( x + θ _{n + 1} h )}{n + 1}

But if we apply the original Mean Value Theorem to the function $f^{(n)} (x)$ , taking $θ_{n} h$ in place of $h$ , we find

f^{(n)} (x + θ_{n} h) = f^{(n)} (x) + θ_{n} h f^{(n + 1)} (x + θ θ_{n} h),

where $θ$ also lies between $0$ and $1$ . Hence

θ_{n} f^{(n + 1)} (x + θ θ_{n} h) = \frac{f ^{(n + 1)} ( x + θ _{n + 1} h )}{n + 1},

from which the result follows, since $f^{(n + 1)} (x + θ θ_{n} h)$ and $f^{(n + 1)} (x + θ_{n + 1} h)$ tend to the same limit $f^{(n + 1)} (x)$ as $h \to 0$ .]

13. Prove that ${f (x + 2 h) - 2 f (x + h) + f (x)} / h^{2} \to f^{''} (x)$ as $h \to 0$ , provided that $f^{''} (x)$ is continuous. [Use equation (2) of § 147.]

14. Show that, if the $f^{(n)} (x)$ is continuous for $x = 0$ , then

f (x) = a_{0} + a_{1} x + a_{2} x^{2} + \dots + (a_{n} + ϵ_{x}) x^{n},

where $a_{r} = f^{(r)} (0) / r!$ and $ϵ_{x} \to 0$ as $x \to 0$ .¹

15. Show that if

a_{0} + a_{1} x + a_{2} x^{2} + \dots + (a_{n} + ϵ_{x}) x^{n} = b_{0} + b_{1} x + b_{2} x^{2} + \dots + (b_{n} + η_{x}) x^{n},

where $ϵ_{x}$ and $η_{x}$ tend to zero as $x \to 0$ , then $a_{0} = b_{0}$ , $a_{1} = b_{1}$ , …, $a_{n} = b_{n}$ . [Making $x \to 0$ we see that $a_{0} = b_{0}$ . Now divide by $x$ and afterwards make $x \to 0$ . We thus obtain $a_{1} = b_{1}$ ; and this process may be repeated as often as is necessary. It follows that if $f (x) = a_{0} + a_{1} x + a_{2} x^{2} + \dots + (a_{n} + ϵ_{x}) x^{n}$ , and the first $n$ derivatives of $f (x)$ are continuous, then $a_{r} = f^{(r)} (0) / r!$ .]

Footnotes

It is in fact sufficient to suppose that $f^{(n)} (0)$ exists. See R. H. Fowler, “The elementary differential geometry of plane curves” (Cambridge Tracts in Mathematics, No. 20, p. 104). ↩

148Taylor’s Series

Suppose that $f (x)$ is a function all of whose differential coefficients are continuous in an interval $(a - η, a + η)$ surrounding the point $x = a$ . Then, if $h$ is numerically less than $η$ , we have

f (a + h) = f (a) + h f^{'} (a) + \dots + \frac{h ^{n - 1}}{( n - 1 )!} f^{(n - 1)} (a) + \frac{h ^{n}}{n !} f^{(n)} (a + θ_{n} h),

where $0 < θ_{n} < 1$ , for all values of $n$ . Or, if

S_{n} = 0 \sum n - 1 \frac{h ^{ν}}{ν !} f^{(ν)} (a), R_{n} = \frac{h ^{n}}{n !} f^{(n)} (a + θ_{n} h),

we have

f (a + h) - S_{n} = R_{n} .

Now let us suppose, in addition, that we can prove that $R_{n} \to 0$ as $n \to \infty$ . Then

f (a + h) = n \to \infty lim S_{n} = f (a) + h f^{'} (a) + \frac{h ^{2}}{2 !} f^{''} (a) + \dots .

This expansion of $f (a + h)$ is known as Taylor’s Series. When $a = 0$ the formula reduces to

f (h) = f (0) + h f^{'} (0) + \frac{h ^{2}}{2 !} f^{''} (0) + \dots,

which is known as Maclaurin’s Series. The function $R_{n}$ is known as Lagrange’s form of the remainder.

The reader should be careful to guard himself against supposing that the continuity of all the derivatives of $f (x)$ is a sufficient condition for the validity of Taylor’s series. A direct discussion of the behaviour of $R_{n}$ is always essential.

Examples LVI.

1. Let $f (x) = sin x$ . Then all the derivatives of $f (x)$ are continuous for all values of $x$ . Also $∣ f^{n} (x) ∣ ≦ 1$ for all values of $x$ and $n$ . Hence in this case $∣ R_{n} ∣ ≦ h^{n} / n!$ , which tends to zero as $n \to \infty$ (Ex. xxvii. 12) whatever value $h$ may have. It follows that

sin (x + h) = sin x + h cos x - \frac{h ^{2}}{2 !} sin x - \frac{h ^{3}}{3 !} cos x + \frac{h ^{4}}{4 !} sin x + \dots,

for all values of $x$ and $h$ . In particular

sin h = h - \frac{h ^{3}}{3 !} + \frac{h ^{5}}{5 !} - \dots,

for all values of $h$ . Similarly we can prove that

cos (x + h) = cos x - h sin x - \frac{h ^{2}}{2 !} cos x + \frac{h ^{3}}{3 !} sin x + \dots, cos h = 1 - \frac{h ^{2}}{2 !} + \frac{h ^{4}}{4 !} - \dots .

2. The Binomial Series. Let $f (x) = (1 + x)^{m}$ , where $m$ is any rational number, positive or negative. Then $f^{(n)} (x) = m (m - 1) \dots (m - n + 1) (1 + x)^{m - n}$ and Maclaurin’s Series takes the form

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

When $m$ is a positive integer the series terminates, and we obtain the ordinary formula for the Binomial Theorem with a positive integral exponent. In the general case

R_{n} = \frac{x ^{n}}{n !} f^{(n)} (θ_{n} x) = (n m) x^{n} (1 + θ_{n} x)^{m - n},

and in order to show that Maclaurin’s Series really represents $(1 + x)^{m}$ for any range of values of $x$ when $m$ is not a positive integer, we must show that $R_{n} \to 0$ for every value of $x$ in that range. This is so in fact if $- 1 < x < 1$ , and may be proved, when $0 ≦ x < 1$ , by means of the expression given above for $R_{n}$ , since $(1 + θ_{n} x)^{m - n} < 1$ if $n > m$ , and $(n m) x^{n} \to 0$ as $n \to \infty$ (Ex. xxvii. 13). But a difficulty arises if $- 1 < x < 0$ , since $1 + θ_{n} x < 1$ and $(1 + θ_{n} x)^{m - n} > 1$ if $n > m$ ; knowing only that $0 < θ_{n} < 1$ , we cannot be assured that $1 + θ_{n} x$ is not quite small and $(1 + θ_{n} x)^{m - n}$ quite large.

In fact, in order to prove the Binomial Theorem by means of Taylor’s Theorem, we need some different form for $R_{n}$ , such as will be given later (§ 162).

149Applications of Taylor’s Theorem

A. Maxima and minima. Taylor’s Theorem may be applied to give greater theoretical completeness to the tests of Ch. VI, §§ 122–123, though the results are not of much practical importance. It will be remembered that, assuming that $ϕ (x)$ has derivatives of the first two orders, we stated the following as being sufficient conditions for a maximum or minimum of $ϕ (x)$ at $x = ξ$ : for a maximum, $ϕ^{'} (ξ) = 0$ , $ϕ^{''} (ξ) < 0$ ; for a minimum, $ϕ^{'} (ξ) = 0$ , $ϕ^{''} (ξ) > 0$ . It is evident that these tests fail if $ϕ^{''} (ξ)$ as well as $ϕ^{'} (ξ)$ is zero.

Let us suppose that the first $n$ derivatives

ϕ^{'} (x), ϕ^{''} (x), \dots, ϕ^{(n)} (x)

are continuous, and that all save the last vanish when $x = ξ$ . Then, for sufficiently small values of $h$ ,

ϕ (ξ + h) - ϕ (ξ) = \frac{h ^{n}}{n !} ϕ^{(n)} (ξ + θ_{n} h) .

In order that there should be a maximum or a minimum this expression must be of constant sign for all sufficiently small values of $h$ , positive or negative. This evidently requires that $n$ should be even. And if $n$ is even there will be a maximum or a minimum according as $ϕ^{(n)} (ξ)$ is negative or positive.

Thus we obtain the test:

if there is to be a maximum or minimum the first derivative which does not vanish must be an even derivative, and there will be a maximum if it is negative, a minimum if it is positive.

Examples LVII.

1. Verify the result when $ϕ (x) = (x - a)^{m}$ , $m$ being a positive integer, and $ξ = a$ .

2. Test the function $(x - a)^{m} (x - b)^{n}$ , where $m$ and $n$ are positive integers, for maxima and minima at the points $x = a$ , $x = b$ . Draw graphs of the different possible forms of the curve $y = (x - a)^{m} (x - b)^{n}$ .

3. Test the functions $sin x - x$ , $sin x - x + \frac{x ^{3}}{6}$ , $sin x - x + \frac{x ^{3}}{6} - \frac{x ^{5}}{120}$ , …, $cos x - 1$ , $cos x - 1 + \frac{x ^{2}}{2}$ , $cos x - 1 + \frac{x ^{2}}{2} - \frac{x ^{4}}{24}$ , … for maxima or minima at $x = 0$ .

150B. The calculation of certain limits

Suppose that $f (x)$ and $ϕ (x)$ are two functions of $x$ whose derivatives $f^{'} (x)$ and $ϕ^{'} (x)$ are continuous for $x = ξ$ and that $f (ξ)$ and $ϕ (ξ)$ are both equal to zero. Then the function

ψ (x) = f (x) / ϕ (x)

is not defined when $x = ξ$ . But of course it may well tend to a limit as $x \to ξ$ .

Now

f (x) = f (x) - f (ξ) = (x - ξ) f^{'} (x_{1}),

where $x_{1}$ lies between $ξ$ and $x$ ; and similarly $ϕ (x) = (x - ξ) ϕ^{'} (x_{2})$ , where $x_{2}$ also lies between $ξ$ and $x$ . Thus

ψ (x) = f^{'} (x_{1}) / ϕ^{'} (x_{2}) .

We must now distinguish four cases.

(1) If neither $f^{'} (ξ)$ nor $ϕ^{'} (ξ)$ is zero, then

f (x) / ϕ (x) \to f^{'} (ξ) / ϕ^{'} (ξ) .

(2) If $f^{'} (ξ) = 0$ , $ϕ^{'} (ξ) \neq = 0$ , then

f (x) / ϕ (x) \to 0.

(3) If $f^{'} (ξ) \neq = 0$ , $ϕ^{'} (ξ) = 0$ , then $f (x) / ϕ (x)$ becomes numerically very large as $x \to ξ$ : but whether $f (x) / ϕ (x)$ tends to $\infty$ or $- \infty$ , or is sometimes large and positive and sometimes large and negative, we cannot say, without further information as to the way in which $ϕ^{'} (x) \to 0$ as $x \to ξ$ .

(4) If $f^{'} (ξ) = 0$ , $ϕ^{'} (ξ) = 0$ , then we can as yet say nothing about the behaviour of $f (x) / ϕ (x)$ as $x \to 0$ .

But in either of the last two cases it may happen that $f (x)$ and $ϕ (x)$ have continuous second derivatives. And then

f (x) ϕ (x) = f (x) - f (ξ) - (x - ξ) f^{'} (ξ) = \frac{1}{2} (x - ξ)^{2} f^{''} (x_{1}), = ϕ (x) - ϕ (ξ) - (x - ξ) ϕ^{'} (ξ) = \frac{1}{2} (x - ξ)^{2} ϕ^{''} (x_{2}),

where again $x_{1}$ and $x_{2}$ lie between $ξ$ and $x$ ; so that

ψ (x) = f^{''} (x_{1}) / ϕ^{''} (x_{2}) .

We can now distinguish a variety of cases similar to those considered above. In particular, if neither second derivative vanishes for $x = ξ$ , we have

f (x) / ϕ (x) \to f^{''} (ξ) / ϕ^{''} (ξ) .

It is obvious that this argument can be repeated indefinitely, and we obtain the following theorem:

suppose that $f (x)$ and $ϕ (x)$ and their derivatives, so far as may be wanted, are continuous for $x = ξ$ . Suppose further that $f^{(p)} (x)$ and $ϕ^{(q)} (x)$ are the first derivatives of $f (x)$ and $ϕ (x)$ which do not vanish when $x = ξ$ . Then

(1) if $p = q$ , $f (x) / ϕ (x) \to f^{(p)} (ξ) / ϕ^{(p)} (ξ)$ ;

(2) if $p > q$ , $f (x) / ϕ (x) \to 0$ ;

(3) if $p < q$ , and $q - p$ is even, either $f (x) / ϕ (x) \to + \infty$ or $f (x) / ϕ (x) \to - \infty$ , the sign being the same as that of $f^{(p)} (ξ) / ϕ^{(q)} (ξ)$ ;

(4) if $p < q$ and $q - p$ is odd, either $f (x) / ϕ (x) \to + \infty$ or $f (x) / ϕ (x) \to - \infty$ , as $x \to ξ + 0$ , the sign being the same as that of $f^{(p)} (ξ) / ϕ^{(q)} (ξ)$ , while if $x \to ξ - 0$ the sign must be reversed.

This theorem is in fact an immediate corollary from the equations

f (x) = \frac{( x - ξ ) ^{p}}{p !} f^{(p)} (x_{1}), ϕ (x) = \frac{( x - ξ ) ^{q}}{q !} ϕ^{(q)} (x_{2}) .

Examples LVIII.

1. Find the limit of

{x - (n + 1) x^{n + 1} + n x^{n + 2}} / (1 - x)^{2},

as $x \to 1$ . [Here the functions and their first derivatives vanish for $x = 1$ , and $f^{''} (1) = n (n + 1)$ , $ϕ^{''} (1) = 2$ .]

2. Find the limits as $x \to 0$ of

(tan x - x) / (x - sin x), (tan n x - n tan x) / (n sin x - sin n x) .

3. Find the limit of $x {x^{2} + a^{2} - x}$ as $x \to \infty$ . [Put $x = 1/ y$ .]

4. Prove that

x \to n lim (x - n) cosec x π = \frac{( - 1 ) ^{n}}{π}, x \to n lim \frac{1}{x - n} {cosec x π - \frac{( - 1 ) ^{n}}{( x - n ) π}} = \frac{( - 1 ) ^{n} π}{6},

$n$ being any integer; and evaluate the corresponding limits involving $cot x π$ .

5. Find the limits as $x \to 0$ of

\frac{1}{x ^{3}} (cosec x - \frac{1}{x} - \frac{x}{6}), \frac{1}{x ^{3}} (cot x - \frac{1}{x} + \frac{x}{3}) .

6. $(sin x arc sin x - x^{2}) / x^{6} \to \frac{1}{18}$ , $(tan x arc tan x - x^{2}) / x^{6} \to \frac{2}{9}$ , as $x \to 0$ .

151C. The contact of plane curves

Two curves are said to intersect (or cut) at a point if the point lies on each of them. They are said to touch at the point if they have the same tangent at the point.

Let us suppose now that $f (x)$ , $ϕ (x)$ are two functions which possess derivatives of all orders continuous for $x = ξ$ , and let us consider the curves $y = f (x)$ , $y = ϕ (x)$ . In general $f (ξ)$ and $ϕ (ξ)$ will not be equal. In this case the abscissa $x = ξ$ does not correspond to a point of intersection of the curves. If however $f (ξ) = ϕ (ξ)$ , the curves intersect in the point $x = ξ$ , $y = f (ξ) = ϕ (ξ)$ . Let us suppose this to be the case. Then in order that the curves should not only cut but touch at this point it is obviously necessary and sufficient that the first derivatives $f^{'} (x)$ , $ϕ^{'} (x)$ should also have the same value when $x = ξ$ .

The contact of the curves in this case may be regarded from a different point of view. In the figure the two

curves are drawn touching at $P$ , and $QR$ is equal to $ϕ (ξ + h) - f (ξ + h)$ , or, since $ϕ (ξ) = f (ξ)$ , $ϕ^{'} (ξ) = f^{'} (ξ)$ , to

\frac{1}{2} h^{2} {ϕ^{''} (ξ + θ h) - f^{''} (ξ + θ h)},

where $θ$ lies between $0$ and $1$ . Hence

lim \frac{QR}{h ^{2}} = \frac{1}{2} {ϕ^{''} (ξ) - f^{''} (ξ)},

when $h \to 0$ . In other words, when the curves touch at the point whose abscissa is $ξ$ , the difference of their ordinates at the point whose abscissa is $ξ + h$ is at least of the second order of smallness when $h$ is small.

The reader will easily verify that $lim (QR / h) = ϕ^{'} (ξ) - f^{'} (ξ)$ when the curves cut and do not touch, so that $QR$ is then of the first order of smallness only.

It is evident that the degree of smallness of $QR$ may be taken as a kind of measure of the closeness of the contact of the curves. It is at once suggested that if the first $n - 1$ derivatives of $f$ and $ϕ$ have equal values when $x = ξ$ , then $QR$ will be of $n$ th order of smallness; and the reader will have no difficulty in proving that this is so and that

lim \frac{QR}{h ^{n}} = \frac{1}{n !} {ϕ^{(n)} (ξ) - f^{(n)} (ξ)} .

We are therefore led to frame the following definition:

Contact of the $n$ th order. If $f (ξ) = ϕ (ξ)$ , $f^{'} (ξ) = ϕ^{'} (ξ)$ , …, $f^{(n)} (ξ) = ϕ^{(n)} (ξ)$ , but $f^{(n + 1)} (ξ) \neq = ϕ^{(n + 1)} (ξ)$ , then the curves $y = f (x)$ , $y = ϕ (x)$ will be said to have contact of the $n$ th order at the point whose abscissa is $ξ$ .

The preceding discussion makes the notion of contact of the $n$ th order dependent on the choice of axes, and fails entirely when the tangent to the curves is parallel to the axis of $y$ . We can deal with this case by taking $y$ as the independent and $x$ as the dependent variable. It is better, however, to consider $x$ and $y$ as functions of a parameter $t$ . An excellent account of the theory will be found in Mr Fowler’s tract referred to on p. 266, or in de la Vallée Poussin’s Cours d’Analyse, vol. ii, pp. 396 et seq.

Examples LIX.

1. Let $ϕ (x) = a x + b$ , so that $y = ϕ (x)$ is a straight line. The conditions for contact at the point for which $x = ξ$ are $f (ξ) = a ξ + b$ , $f^{'} (ξ) = a$ . If we determine $a$ and $b$ so as to satisfy these equations we find $a = f^{'} (ξ)$ , $b = f (ξ) - ξ f^{'} (ξ)$ , and the equation of the tangent to $y = f (x)$ at the point $x = ξ$ is

y = x f^{'} (ξ) + {f (ξ) - ξ f^{'} (ξ)},

or $y - f (ξ) = (x - ξ) f^{'} (ξ)$ . Cf. Ex. xxxix. 5.

2. The fact that the line is to have simple contact with the curve completely determines the line. In order that the tangent should have contact of the second order with the curve we must have $f^{''} (ξ) = ϕ^{''} (ξ)$ , i.e. $f^{''} (ξ) = 0$ . A point at which the tangent to a curve has contact of the second order is called a point of inflexion.

3. Find the points of inflexion on the graphs of the functions $3 x^{4} - 6 x^{3} + 1$ , $2 x / (1 + x^{2})$ , $sin x$ , $a cos^{2} x + b sin^{2} x$ , $tan x$ , $arc tan x$ .

4. Show that the conic $a x^{2} + 2 h x y + b y^{2} + 2 g x + 2 f y + c = 0$ cannot have a point of inflexion. [Here $a x + h y + g + (h x + b y + f) y_{1} = 0$ and

a + 2 h y_{1} + b y_{1}^{2} + (h x + b y + f) y_{2} = 0,

suffixes denoting differentiations. Thus at a point of inflexion

a + 2 h y_{1} + b y_{1}^{2} = 0,

a (h x + b y + f)^{2} - 2 h (a x + h y + g) (h x + b y + f) + b (a x + h y + g)^{2} = 0,

(ab - h^{2}) {a x^{2} + 2 h x y + b y^{2} + 2 g x + 2 f y} + a f^{2} - 2 f g h + b g^{2} = 0.

But this is inconsistent with the equation of the conic unless

a f^{2} - 2 f g h + b g^{2} = c (ab - h^{2})

or $ab c + 2 f g h - a f^{2} - b g^{2} - c h^{2} = 0$ ; and this is the condition that the conic should degenerate into two straight lines.]

5. The curve $y = (a x^{2} + 2 b x + c) / (α x^{2} + 2 β x + γ)$ has one or three points of inflexion according as the roots of $α x^{2} + 2 β x + γ = 0$ are real or complex.

[The equation of the curve can, by a change of origin (cf. Ex. xlvi. 15), be reduced to the form

η = ξ / (A ξ^{2} + 2 B ξ + C) = ξ / {A (ξ - p) (ξ - q)},

where $p$ , $q$ are real or conjugate. The condition for a point of inflexion will be found to be $ξ^{3} - 3 pq ξ + pq (p + q) = 0$ , which has one or three real roots according as ${pq (p - q)}^{2}$ is positive or negative, i.e. according as $p$ and $q$ are real or conjugate.]

6. Discuss in particular the curves $y = (1 - x) / (1 + x^{2})$ , $y = (1 - x^{2}) / (1 + x^{2})$ , $y = (1 + x^{2}) / (1 - x^{2})$ .

7. Show that when the curve of Ex. 5 has three points of inflexion, they lie on a straight line. [The equation $ξ^{3} - 3 pq ξ + pq (p + q) = 0$ can be put in the form $(ξ - p) (ξ - q) (ξ + p + q) + (p - q)^{2} ξ = 0$ , so that the points of inflexion lie on the line $ξ + A (p - q)^{2} η + p + q = 0$ or $A ξ - 4 (A C - B^{2}) η = 2 B$ .]

8. Show that the curves $y = x sin x$ , $y = (sin x) / x$ have each infinitely many points of inflexion.

9. Contact of a circle with a curve. Curvature.¹ The general equation of a circle, viz.

(x - a)^{2} + (y - b)^{2} = r^{2}, (1)

contains three arbitrary constants. Let us attempt to determine them so that the circle has contact of as high an order as possible with the curve $y = f (x)$ at the point $(ξ, η)$ , where $η = f (ξ)$ . We write $η_{1}$ , $η_{2}$ for $f^{'} (ξ)$ , $f^{''} (ξ)$ . Differentiating the equation of the circle twice we obtain

(x - a) + (y - b) y_{1} 1 + y_{1}^{2} + (y - b) y_{2} = 0, = 0. (2) (3)

If the circle touches the curve then the equations (1) and (2) are satisfied when $x = ξ$ , $y = η$ , $y_{1} = η_{1}$ . This gives $(ξ - a) / η_{1} = - (η - b) = r / 1 + η_{1}^{2}$ . If the contact is of the second order then the equation (3) must also be satisfied when $y_{2} = η_{2}$ . Thus $b = η + {(1 + η_{1}^{2}) / η_{2}}$ ; and hence we find

a = ξ - \frac{η _{1} ( 1 + η _{1}^{2} )}{η _{2}}, b = η + \frac{1 + η _{1}^{2}}{η _{2}}, r = \frac{( 1 + η _{1}^{2} ) ^{3/2}}{η _{2}} .

The circle which has contact of the second order with the curve at the point $(ξ, η)$ is called the circle of curvature, and its radius the radius of curvature. The measure of curvature (or simply the curvature) is the reciprocal of the radius: thus the measure of curvature is $f^{''} (ξ) / {1 + [f^{'} (ξ)]^{2}}^{3/2}$ , or

\frac{d ^{2} η}{d ξ ^{2}} / {1 + (\frac{d η}{d ξ})^{2}}^{3/2} .

10. Verify that the curvature of a circle is constant and equal to the reciprocal of the radius; and show that the circle is the only curve whose curvature is constant.

11. Find the centre and radius of curvature at any point of the conics $y^{2} = 4 a x$ , $(x / a)^{2} + (y / b)^{2} = 1$ .

12. In an ellipse the radius of curvature at $P$ is $C D^{3} / ab$ , where $C D$ is the semi-diameter conjugate to $C P$ .

13. Show that in general a conic can be drawn to have contact of the fourth order with the curve $y = f (x)$ at a given point $P$ .

[Take the general equation of a conic, viz.

a x^{2} + 2 h x y + b y^{2} + 2 g x + 2 f y + c = 0,

and differentiate four times with respect to $x$ . Using suffixes to denote differentiation we obtain

a x + h y + g + (h x + b y + f) y_{1} a + 2 h y_{1} + b y_{1}^{2} + (h x + b y + f) y_{2} 3 (h + b y_{1}) y_{2} + (h x + b y + f) y_{3} 4 (h + b y_{1}) y_{3} + 3 b y_{2}^{2} + (h x + b y + f) y_{4} = 0, = 0, = 0, = 0.

If the conic has contact of the fourth order, then these five equations must be satisfied by writing $ξ$ , $η$ , $η_{1}$ , $η_{2}$ , $η_{3}$ , $η_{4}$ , for $x$ , $y$ , $y_{1}$ , $y_{2}$ , $y_{3}$ , $y_{4}$ . We have thus just enough equations to determine the ratios $a : b : c : f : g : h$ .]

14. An infinity of conics can be drawn having contact of the third order with the curve at $P$ . Show that their centres all lie on a straight line.

[Take the tangent and normal as axes. Then the equation of the conic is of the form $2 y = a x^{2} + 2 h x y + b y^{2}$ , and when $x$ is small one value of $y$ may be expressed (Ch. V, Misc. Ex. 22) in the form

y = \frac{1}{2} a x^{2} + (\frac{1}{2} ah + ϵ_{x}) x^{3},

where $ϵ_{x} \to 0$ with $x$ . But this expression must be the same as

y = \frac{1}{2} f^{''} (0) x^{2} + {\frac{1}{6} f^{'''} (0) + ϵ_{x}^{'}} x^{3},

where $ϵ_{x}^{'} \to 0$ with $x$ , and so $a = f^{''} (0)$ , $h = f^{'''} (0) /3 f^{''} (0)$ , in virtue of the result of Ex. lv. 15. But the centre lies on the line $a x + h y = 0$ .]

15. Determine a parabola which has contact of the third order with the ellipse $(x / a)^{2} + (y / b)^{2} = 1$ at the extremity of the major axis.

16. The locus of the centres of conics which have contact of the third order with the ellipse $(x / a)^{2} + (y / b)^{2} = 1$ at the point $(a cos α, b sin α)$ is the diameter $x / (a cos α) = y / (b sin α)$ . [For the ellipse itself is one such conic.]

Footnotes

A much fuller discussion of the theory of curvature will be found in Mr Fowler’s tract referred to on p. \DPchg{272}{266}. ↩

152Differentiation of functions of several variables

So far we have been concerned exclusively with functions of a single variable $x$ , but there is nothing to prevent us applying the notion of differentiation to functions of several variables $x$ , $y$ , ….

Suppose then that $f (x, y)$ is a function of two¹ real variables $x$ and $y$ , and that the limits

h \to 0 lim \frac{f ( x + h , y ) - f ( x , y )}{h}, k \to 0 lim \frac{f ( x , y + k ) - f ( x , y )}{k}

exist for all values of $x$ and $y$ in question, that is to say that $f (x, y)$ possesses a derivative $df / d x$ or $D_{x} f (x, y)$ with respect to $x$ and a derivative $df / d y$ or $D_{y} f (x, y)$ with respect to $y$ . It is usual to call these derivatives the partial differential coefficients of $f$ , and to denote them by

\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}

f_{x}^{'} (x, y), f_{y}^{'} (x, y)

or simply $f_{x}^{'}$ , $f_{y}^{'}$ or $f_{x}$ , $f_{y}$ . The reader must not suppose, however, that these new notations imply any essential novelty of idea: ‘partial differentiation’ with respect to $x$ is exactly the same process as ordinary differentiation, the only novelty lying in the presence in $f$ of a second variable $y$ independent of $x$ .

In what precedes we have supposed $x$ and $y$ to be two real variables entirely independent of one another. If $x$ and $y$ were connected by a relation the state of affairs would be very different. In this case our definition of $f_{x}^{'}$ would fail entirely, as we could not change $x$ into $x + h$ without at the same time changing $y$ . But then $f (x, y)$ would not really be a function of two variables at all. A function of two variables, as we defined it in Ch. II, is essentially a function of two independent variables. If $y$ depends on $x$ , $y$ is a function of $x$ , say $y = ϕ (x)$ ; and then

f (x, y) = f {x, ϕ (x)}

is really a function of the single variable $x$ . Of course we may also represent it as a function of the single variable $y$ . Or, as is often most convenient, we may regard $x$ and $y$ as functions of a third variable $t$ , and then $f (x, y)$ , which is of the form $f {ϕ (t), ψ (t)}$ , is a function of the single variable $t$ .

Examples LX.

1. Prove that if $x = r cos θ$ , $y = r sin θ$ , so that $r = x^{2} + y^{2}$ , $θ = arc tan (y / x)$ , then

\frac{\partial r}{\partial x} \frac{\partial x}{\partial r} = \frac{x}{x ^{2} + y ^{2}}, = cos θ, \frac{\partial r}{\partial y} \frac{\partial y}{\partial r} = \frac{y}{x ^{2} + y ^{2}}, = sin θ, \frac{\partial θ}{\partial x} \frac{\partial x}{\partial θ} = - \frac{y}{x ^{2} + y ^{2}}, = - r sin θ, \frac{\partial θ}{\partial y} \frac{\partial y}{\partial θ} = \frac{x}{x ^{2} + y ^{2}}, = r cos θ .

2. Account for the fact that $\frac{\partial r}{\partial x} \neq = 1 / (\frac{\partial x}{\partial r})$ and $\frac{\partial θ}{\partial x} \neq = 1 / (\frac{\partial x}{\partial θ})$ . [When we were considering a function $y$ of one variable $x$ it followed from the definitions that $d y / d x$ and $d x / d y$ were reciprocals. This is no longer the case when we are dealing with functions of two variables. Let $P$ (Fig. 46) be the point $(x, y)$ or $(r, θ)$ . To find $\partial r / \partial x$ we must increase $x$ , say by an increment $M M_{1} = δ x$ , while keeping $y$ constant. This brings $P$ to $P_{1}$ . If along $O P_{1}$ we take $O P^{'} = O P$ , the increment of $r$ is $P^{'} P_{1} = δ r$ , say; and $\partial r / \partial x = lim (δ r / δ x)$ . If on the other hand we want to calculate $\partial x / \partial r$ , $x$ and $y$

being now regarded as functions of $r$ and $θ$ , we must increase $r$ by $Δ r$ , say, keeping $θ$ constant. This brings $P$ to $P_{2}$ , where $P P_{2} = Δ r$ : the corresponding increment of $x$ is $M M_{1} = Δ x$ , say; and

\partial x / \partial r = lim (Δ x /Δ r) .

Now $Δ x = δ x$ :² but $Δ r \neq = δ r$ . Indeed it is easy to see from the figure that

lim (δ r / δ x) = lim (P^{'} P_{1} / P P_{1}) = cos θ,

but

lim (Δ r /Δ x) = lim (P P_{2} / P P_{1}) = sec θ,

so that

lim (δ r /Δ r) = cos^{2} θ .

The fact is of course that $\partial x / \partial r$ and $\partial r / \partial x$ are not formed upon the same hypothesis as to the variation of $P$ .]

3. Prove that if $z = f (a x + b y)$ then $b (\partial z / \partial x) = a (\partial z / \partial y)$ .

4. Find $\partial X / \partial x$ , $\partial X / \partial y$ , … when $X + Y = x$ , $Y = x y$ . Express $x$ , $y$ as functions of $X$ , $Y$ and find $\partial x / \partial X$ , $\partial x / \partial Y$ , ….

5. Find $\partial X / \partial x$ , … when $X + Y + Z = x$ , $Y + Z = x y$ , $Z = x y z$ ; express $x$ , $y$ , $z$ in terms of $X$ , $Y$ , $Z$ and find $\partial x / \partial X$ , ….

[There is of course no difficulty in extending the ideas of the last section to functions of any number of variables. But the reader must be careful to impress on his mind that the notion of the partial derivative of a function of several variables is only determinate when all the independent variables are specified. Thus if $u = x + y + z$ , $x$ , $y$ , and $z$ being the independent variables, then $\partial u / \partial x = 1$ . But if we regard $u$ as a function of the variables $x$ , $x + y = η$ , and $x + y + z = ζ$ , so that $u = ζ$ , then $\partial u / \partial x = 0$ .]

Footnotes

The new points which arise when we consider functions of several variables are illustrated sufficiently when there are two variables only. The generalisations of our theorems for three or more variables are in general of an obvious character. ↩
Of course the fact that $Δ x = δ x$ is due merely to the particular value of $Δ r$ that we have chosen (viz. $P P_{2}$ ). Any other choice would give us values of $Δ x$ , $Δ r$ proportional to those used here. ↩

153Differentiation of a function of two functions

There is a theorem concerning the differentiation of a function of one variable, known generally as the Theorem of the Total Differential Coefficient, which is of very great importance and depends on the notions explained in the preceding section regarding functions of two variables. This theorem gives us a rule for differentiating

f {ϕ (t), ψ (t)},

with respect to $t$ .

Let us suppose, in the first instance, that $f (x, y)$ is a function of the two variables $x$ and $y$ , and that $f_{x}^{'}$ , $f_{y}^{'}$ are continuous functions of both variables (§ 107) for all of their values which come in question. And now let us suppose that the variation of $x$ and $y$ is restricted in that $(x, y)$ lies on a curve

x = ϕ (t), y = ψ (t),

where $ϕ$ and $ψ$ are functions of $t$ with continuous differential coefficients $ϕ^{'} (t)$ , $ψ^{'} (t)$ . Then $f (x, y)$ reduces to a function of the single variable $t$ , say $F (t)$ . The problem is to determine $F^{'} (t)$ .

Suppose that, when $t$ changes to $t + τ$ , $x$ and $y$ change to $x + ξ$ and $y + η$ . Then by definition

\frac{d F ( t )}{d t} = τ \to 0 lim \frac{1}{τ} [f {ϕ (t + τ), ψ (t + τ)} - f {ϕ (t), ψ (t)}] = lim \frac{1}{τ} {f (x + ξ, y + η) - f (x, y)} = lim [\frac{f ( x + ξ , y + η ) - f ( x , y + η )}{ξ} \frac{ξ}{τ} + \frac{f ( x , y + η ) - f ( x , y )}{η} \frac{η}{τ}] .

But, by the Mean Value Theorem,

{f (x + ξ, y + η) - f (x, y + η)} / ξ {f (x, y + η) - f (x, y)} / η = f_{x}^{'} (x + θ ξ, y + η), = f_{y}^{'} (x, y + θ^{'} η),

where $θ$ and $θ^{'}$ each lie between $0$ and $1$ . As $τ \to 0$ , $ξ \to 0$ and $η \to 0$ , and $ξ / τ \to ϕ^{'} (t)$ , $η / τ \to ψ^{'} (t)$ : also

f_{x}^{'} (x + θ ξ, y + η) \to f_{x}^{'} (x, y), f_{y}^{'} (x, y + θ^{'} η) \to f_{y}^{'} (x, y) .

Hence

F^{'} (t) = D_{t} f {ϕ (t), ψ (t)} = f_{x}^{'} (x, y) ϕ^{'} (t) + f_{y}^{'} (x, y) ψ^{'} (t),

where we are to put $x = ϕ (t)$ , $y = ψ (t)$ after carrying out the differentiations with respect to $x$ and $y$ . This result may also be expressed in the form

\frac{df}{d t} = \frac{\partial f}{\partial x} \frac{d x}{d t} + \frac{\partial f}{\partial y} \frac{d y}{d t}

Examples LXI.

1. Suppose $ϕ (t) = (1 - t^{2}) / (1 + t^{2})$ , $ψ (t) = 2 t / (1 + t^{2})$ , so that the locus of $(x, y)$ is the circle $x^{2} + y^{2} = 1$ . Then

ϕ^{'} (t) F^{'} (t) = - 4 t / (1 + t^{2})^{2}, ψ^{'} (t) = 2 (1 - t^{2}) / (1 + t^{2})^{2}, = {- 4 t / (1 + t^{2})^{2}} f_{x}^{'} + {2 (1 - t^{2}) / (1 + t^{2})^{2}} f_{y}^{'},

where $x$ and $y$ are to be put equal to $(1 - t^{2}) / (1 + t^{2})$ and $2 t / (1 + t^{2})$ after carrying out the differentiations.

We can easily verify this formula in particular cases. Suppose, e.g., that $f (x, y) = x^{2} + y^{2}$ . Then $f_{x}^{'} = 2 x$ , $f_{y}^{'} = 2 y$ , and it is easily verified that $F^{'} (t) = 2 x ϕ^{'} (t) + 2 y ψ^{'} (t) = 0$ , which is obviously correct, since $F (t) = 1$ .

2. Verify the theorem in the same way when (a) $x = t^{m}$ , $y = 1 - t^{m}$ , $f (x, y) = x + y$ ; (b) $x = a cos t$ , $y = a sin t$ , $f (x, y) = x^{2} + y^{2}$ .

3. One of the most important cases is that in which $t$ is $x$ itself. We then obtain

D_{x} f {x, ψ (x)} = D_{x} f (x, y) + D_{y} f (x, y) ψ^{'} (x) .

where $y$ is to be replaced by $ψ (x)$ after differentiation.

It was this case which led to the introduction of the notation $\partial f / \partial x$ , $\partial f / \partial y$ . For it would seem natural to use the notation $df / d x$ for either of the functions $D_{x} f {x, ψ (x)}$ and $D_{x} f (x, y)$ , in one of which $y$ is put equal to $ψ (x)$ before and in the other after differentiation. Suppose for example that $y = 1 - x$ and $f (x, y) = x + y$ . Then $D_{x} f (x, 1 - x) = D_{x} 1 = 0$ , but $D_{x} f (x, y) = 1$ .

The distinction between the two functions is adequately shown by denoting the first by $df / d x$ and the second by $\partial f / \partial x$ , in which case the theorem takes the form

\frac{df}{d x} = \frac{\partial f}{\partial x} + \frac{\partial f}{\partial y} \frac{d y}{d x};

though this notation is also open to objection, in that it is a little misleading to denote the functions $f {x, ψ (x)}$ and $f (x, y)$ , whose forms as functions of $x$ are quite different from one another, by the same letter $f$ in $df / d x$ and $\partial f / \partial x$ .

4. If the result of eliminating $t$ between $x = ϕ (t)$ , $y = ψ (t)$ is $f (x, y) = 0$ , then

\frac{\partial f}{\partial x} \frac{d x}{d t} + \frac{\partial f}{\partial y} \frac{d y}{d t} = 0.

5. If $x$ and $y$ are functions of $t$ , and $r$ and $θ$ are the polar coordinates of $(x, y)$ , then $r^{'} = (x x^{'} + y y^{'}) / r$ , $θ^{'} = (x y^{'} - y x^{'}) / r^{2}$ , dashes denoting differentiations with respect to $t$ .

154The Mean Value Theorem for functions of two variables

Many of the results of the last chapter depended upon the Mean Value Theorem, expressed by the equation

ϕ (x + h) - ϕ (x) = h f^{'} (x + θ h),

or as it may be written, if $y = ϕ (x)$ ,

δ y = f^{'} (x + θ δ x) δ x .

Now suppose that $z = f (x, y)$ is a function of the two independent variables $x$ and $y$ , and that $x$ and $y$ receive increments $h$ , $k$ or $δ x$ , $δ y$ respectively: and let us attempt to express the corresponding increment of $z$ , viz.

δ z = f (x + h, y + k) - f (x, y),

in terms of $h$ , $k$ and the derivatives of $z$ with respect to $x$ and $y$ .

Let $f (x + h t, y + k t) = F (t)$ . Then

f (x + h, y + k) - f (x, y) = F (1) - F (0) = F^{'} (θ),

where $0 < θ < 1$ . But, by § 153,

F^{'} (t) = D_{t} f (x + h t, y + k t) = h f_{x}^{'} (x + h t, y + k t) + k f_{y}^{'} (x + h t, y + k t) .

Hence finally

δ z = f (x + h, y + k) - f (x, y) = h f_{x}^{'} (x + θ h, y + θ k) + k f_{y}^{'} (x + θ h, y + θ k),

which is the formula desired. Since $f_{x}^{'}$ , $f_{y}^{'}$ are supposed to be continuous functions of $x$ and $y$ , we have

f_{x}^{'} (x + θ h, y + θ k) f_{y}^{'} (x + θ h, y + θ k) = f_{x}^{'} (x, y) + ϵ_{h, k}, = f_{y}^{'} (x, y) + η_{h, k},

where $ϵ_{h, k}$ and $η_{h, k}$ tend to zero as $h$ and $k$ tend to zero. Hence the theorem may be written in the form

δ z = (f_{x}^{'} + ϵ) δ x + (f_{y}^{'} + η) δ y, (1)

where $ϵ$ and $η$ are small when $δ x$ and $δ y$ are small.

The result embodied in (1) may be expressed by saying that the equation

δ z = f_{x}^{'} δ x + f_{y}^{'} δ y

is approximately true; i.e. that the difference between the two sides of the equation is small in comparison with the larger of $δ x$ and $δ y$ .¹ We must say ‘the larger of $δ x$ and $δ y$ ’ because one of them might be small in comparison with the other; we might indeed have $δ x = 0$ or $δ y = 0$ .

It should be observed that if any equation of the form $δ z = λ δ x + μ δ y$ is ‘approximately true’ in this sense, we must have $λ = f_{x}^{'}$ , $μ = f_{y}^{'}$ . For we have

δ z - f_{x}^{'} δ x - f_{y}^{'} δ y = ϵ δ x + η δ y, δ z - λ δ x - μ δ y = ϵ^{'} δ x + η^{'} δ y

where $ϵ$ , $η$ , $ϵ^{'}$ , $η^{'}$ all tend to zero as $δ x$ and $δ y$ tend to zero; and so

(λ - f_{x}^{'}) δ x + (μ - f_{y}^{'}) δ y = ρ δ x + ρ^{'} δ y

where $ρ$ and $ρ^{'}$ tend to zero. Hence, if $ζ$ is any assigned positive number, we can choose $σ$ so that

∣ (λ - f_{x}^{'}) δ x + (μ - f_{y}^{'}) δ y ∣ < ζ (∣ δ x ∣ + ∣ δ y ∣)

for all values of $δ x$ and $δ y$ numerically less than $σ$ . Taking $δ y = 0$ we obtain $∣ (λ - f_{x}^{'}) δ x ∣ < ζ ∣ δ x ∣$ , or $∣ λ - f_{x}^{'} ∣ < ζ$ , and, as $ζ$ may be as small as we please, this can only be the case if $λ = f_{x}^{'}$ . Similarly $μ = f_{y}^{'}$ .

Footnotes

Or with $∣ δ x ∣ + ∣ δ y ∣$ or $δ x^{2} + δ y^{2}$ . ↩

155Differentials

In the applications of the Calculus, especially in geometry, it is usually most convenient to work with equations expressed not, like equation (1) of § 154, in terms of the increments $δ x$ , $δ y$ , $δ z$ of the functions $x$ , $y$ , $z$ , but in terms of what are called their differentials $d x$ , $d y$ , $d z$ .

Let us return for a moment to a function $y = f (x)$ of a single variable $x$ . If $f^{'} (x)$ is continuous then

δ y = {f^{'} (x) + ϵ} δ x, (1)

where $ϵ \to 0$ as $δ x \to 0$ : in other words the equation

δ y = f^{'} (x) δ x (2)

is ‘approximately’ true. We have up to the present attributed no meaning of any kind to the symbol $d y$ standing by itself. We now agree to define $d y$ by the equation

d y = f^{'} (x) δ x . (3)

If we choose for $y$ the particular function $x$ , we obtain

d x = δ x, (4)

so that

d y = f^{'} (x) d x . (5)

If we divide both sides of (5) by $d x$ we obtain

\frac{d y}{d x} = f^{'} (x), (6)

where $d y / d x$ denotes not, as heretofore, the differential coefficient of $y$ , but the quotient of the differentials $d y$ , $d x$ . The symbol $d y / d x$ thus acquires a double meaning; but there is no inconvenience in this, since (6) is true whichever meaning we choose.

The equation (5) has two apparent advantages over (2). It is exact and not merely approximate, and its truth does not depend on any assumption as to the continuity of $f^{'} (x)$ . On the other hand it is precisely the fact that we can, under certain conditions, pass from the exact equation (5) to the approximate equation (2), which gives the former its importance. The advantages of the ‘differential’ notation are in reality of a purely technical character. These technical advantages are however so great, especially when we come to deal with functions of several variables, that the use of the notation is almost inevitable.

When $f^{'} (x)$ is continuous, we have

lim \frac{d y}{δ y} = 1

when $δ x \to 0$ . This is sometimes expressed by saying that $d y$ is the principal part of $δ y$ when $δ x$ is small, just as we might say that $a x$ is the ‘principal part’ of $a x + b x^{2}$ when $x$ is small.

We pass now to the corresponding definitions connected with a function $z$ of two independent variables $x$ and $y$ . We define the differential $d z$ by the equation

d z = f_{x}^{'} δ x + f_{y}^{'} δ y . (7)

Putting $z = x$ and $z = y$ in turn, we obtain

d x = δ x, d y = δ y, (8)

so that

d z = f_{x}^{'} d x + f_{y}^{'} d y, (9)

which is the exact equation corresponding to the approximate equation (1) of § 154. Here again it is to be observed that the former is of importance only for reasons of practical convenience in working and because the latter can in certain circumstances be deduced from it.

One property of the equation (9) deserves special remark. We saw in § 153 that if $z = f (x, y)$ , $x$ and $y$ being not independent but functions of a single variable $t$ , so that $z$ is also a function of $t$ alone, then

\frac{d z}{d t} = \frac{\partial f}{\partial x} \frac{d x}{d t} + \frac{\partial f}{\partial y} \frac{d y}{d t} .

Multiplying this equation by $d t$ and observing that

d x = \frac{d x}{d t} d t, d y = \frac{d y}{d t} d t, d z = \frac{d z}{d t} d t,

we obtain

d z = f_{x}^{'} d x + f_{y}^{'} d y,

which is the same in form as (9). Thus the formula which expresses $d z$ in terms of $d x$ and $d y$ is the same whether the variables $x$ and $y$ are independent or not. This remark is of great importance in applications.

It should also be observed that if $z$ is a function of the two independent variables $x$ and $y$ , and

d z = λ d x + μ d y,

then $λ = f_{x}^{'}$ , $μ = f_{y}^{'}$ . This follows at once from the last paragraph of § 154.

It is obvious that the theorems and definitions of the last three sections are capable of immediate extension to functions of any number of variables.

Examples LXII.

1. The area of an ellipse is given by $A = π ab$ , where $a$ , $b$ are the semiaxes. Prove that

\frac{d A}{A} = \frac{d a}{a} + \frac{d b}{b},

and state the corresponding approximate equation connecting the increments of the axes and the area.

2. Express $Δ$ , the area of a triangle $A B C$ , as a function of (i) $a$ , $B$ , $C$ , (ii) $A$ , $b$ , $c$ , and (iii) $a$ , $b$ , $c$ , and establish the formulae

\frac{d Δ}{Δ} = 2 \frac{d a}{a} + \frac{c d B}{a sin B} + \frac{b d C}{a sin C}, \frac{d Δ}{Δ} = cot A d A + \frac{d b}{b} + \frac{d c}{c}, d Δ = R (cos A d a + cos B d b + cos C d c),

where $R$ is the radius of the circumcircle.

3. The sides of a triangle vary in such a way that the area remains constant, so that $a$ may be regarded as a function of $b$ and $c$ . Prove that

\frac{\partial a}{\partial b} = - \frac{cos B}{cos A}, \frac{\partial a}{\partial c} = - \frac{cos C}{cos A} .

[This follows from the equations

d a = \frac{\partial a}{\partial b} d b + \frac{\partial a}{\partial c} d c, cos A d a + cos B d b + cos C d c = 0.

4. If $a$ , $b$ , $c$ vary so that $R$ remains constant, then

\frac{d a}{cos A} + \frac{d b}{cos B} + \frac{d c}{cos C} = 0,

and so

\frac{\partial a}{\partial b} = - \frac{cos A}{cos B}, \frac{\partial a}{\partial c} = - \frac{cos A}{cos C} .

[Use the formulae $a = 2 R sin A$ , …, and the facts that $R$ and $A + B + C$ are constant.]

5. If $z$ is a function of $u$ and $v$ , which are functions of $x$ and $y$ , then

\frac{\partial z}{\partial x} = \frac{\partial z}{\partial u} \frac{\partial u}{\partial x} + \frac{\partial z}{\partial v} \frac{\partial v}{\partial x}, \frac{\partial z}{\partial y} = \frac{\partial z}{\partial u} \frac{\partial u}{\partial y} + \frac{\partial z}{\partial v} \frac{\partial v}{\partial y} .

[We have

d z = \frac{\partial z}{\partial u} d u + \frac{\partial z}{\partial v} d v, d u = \frac{\partial u}{\partial x} d x + \frac{\partial u}{\partial y} d y, d v = \frac{\partial v}{\partial x} d x + \frac{\partial v}{\partial y} d y .

Substitute for $d u$ and $d v$ in the first equation and compare the result with the equation

d z = \frac{\partial z}{\partial x} d x + \frac{\partial z}{\partial y} d y .]

6. Let $z$ be a function of $x$ and $y$ , and let $X$ , $Y$ , $Z$ be defined by the equations

x = a_{1} X + b_{1} Y + c_{1} Z, y = a_{2} X + b_{2} Y + c_{2} Z, z = a_{3} X + b_{3} Y + c_{3} Z .

Then $Z$ may be expressed as a function of $X$ and $Y$ . Express $\partial Z / \partial X$ , $\partial Z / \partial Y$ in terms of $\partial z / \partial x$ , $\partial z / \partial y$ . [Let these differential coefficients be denoted by $P$ , $Q$ and $p$ , $q$ . Then $d z - p d x - q d y = 0$ , or

(c_{1} p + c_{2} q - c_{3}) d Z + (a_{1} p + a_{2} q - a_{3}) d X + (b_{1} p + b_{2} q - b_{3}) d Y = 0.

Comparing this equation with $d Z - P d X - Q d Y = 0$ we see that

P = - \frac{a _{1} p + a _{2} q - a _{3}}{c _{1} p + c _{2} q - c _{3}}, Q = - \frac{b _{1} p + b _{2} q - b _{3}}{c _{1} p + c _{2} q - c _{3}} .]

7. If

(a_{1} x + b_{1} y + c_{1} z) p + (a_{2} x + b_{2} y + c_{2} z) q = a_{3} x + b_{3} y + c_{3} z,

then

(a_{1} X + b_{1} Y + c_{1} Z) P + (a_{2} X + b_{2} Y + c_{2} Z) Q = a_{3} X + b_{3} Y + c_{3} Z .

(Math. Trip. 1899.)

8. Differentiation of implicit functions. Suppose that $f (x, y)$ and its derivative $f_{y}^{'} (x, y)$ are continuous in the neighbourhood of the point $(a, b)$ , and that

f (a, b) = 0, f_{b}^{'} (a, b) \neq = 0.

Then we can find a neighbourhood of $(a, b)$ throughout which $f_{y}^{'} (x, y)$ has always the same sign. Let us suppose, for example, that $f_{y}^{'} (x, y)$ is positive near $(a, b)$ . Then $f (x, y)$ is, for any value of $x$ sufficiently near to $a$ , and for values of $y$ sufficiently near to $b$ , an increasing function of $y$ in the stricter sense of § 95. It follows, by the theorem of § 108, that there is a unique continuous function $y$ which is equal to $b$ when $x = a$ and which satisfies the equation $f (x, y) = 0$ for all values of $x$ sufficiently near to $a$ .

Let us now suppose that $f (x, y)$ possesses a derivative $f_{x}^{'} (x, y)$ which is also continuous near $(a, b)$ . If $f (x, y) = 0$ , $x = a + h$ , $y = b + k$ , we have

0 = f (x, y) - f (a, b) = (f_{a}^{'} + ϵ) h + (f_{b}^{'} + η) k,

where $ϵ$ and $η$ tend to zero with $h$ and $k$ . Thus

\frac{k}{h} = - \frac{f _{a}^{'} + ϵ}{f _{b}^{'} + η} \to - \frac{f _{a}^{'}}{f _{b}^{'}},

\frac{d y}{d x} = - \frac{f _{a}^{'}}{f _{b}^{'}} .

9. The equation of the tangent to the curve $f (x, y) = 0$ , at the point $x_{0}$ , $y_{0}$ , is

(x - x_{0}) f_{x_{0}}^{'} (x_{0}, y_{0}) + (y - y_{0}) f_{y_{0}}^{'} (x_{0}, y_{0}) = 0.

156Definite Integrals and Areas

It will be remembered that, in Ch. VI, § 145, we assumed that, if $f (x)$ is a continuous function of $x$ , and $P Q$ is the

graph of $y = f (x)$ , then the region $P pq Q$ shown in Fig. 47 has associated with it a definite number which we call its area. It is clear that, if we denote $O p$ and $O q$ by $a$ and $x$ , and allow $x$ to vary, this area is a function of $x$ , which we denote by $F (x)$ .

Making this assumption, we proved in § 145 that $F^{'} (x) = f (x)$ , and we showed how this result might be used in the calculation of the areas of particular curves. But we have still to justify the fundamental assumption that there is such a number as the area $F (x)$ .

We know indeed what is meant by the area of a rectangle, and that it is measured by the product of its sides. Also the properties of triangles, parallelograms, and polygons proved by Euclid enable us to attach a definite meaning to the areas of such figures. But nothing which we know so far provides us with a direct definition of the area of a figure bounded by curved lines. We shall now show how to give a definition of $F (x)$ which will enable us to prove its existence.¹

Let us suppose $f (x)$ continuous throughout the interval $(a, b)$ , and let us divide up the interval into a number of sub-intervals by means of the points of division $x_{0}$ , $x_{1}$ , $x_{2}$ , …, $x_{n}$ , where

a = x_{0} < x_{1} < \dots < x_{n - 1} < x_{n} = b .

Further, let us denote by $δ_{ν}$ the interval $(x_{ν}, x_{ν + 1})$ , and by $m_{ν}$ the lower bound (§ 102) of $f (x)$ in $δ_{ν}$ , and let us write

s = m_{0} δ_{0} + m_{1} δ_{1} + \dots + m_{n} δ_{n} = \sum m_{ν} δ_{ν},

say.

It is evident that, if $M$ is the upper bound of $f (x)$ in $(a, b)$ , then $s ≦ M (b - a)$ . The aggregate of values of $s$ is therefore, in the language of § 80, bounded above, and possesses an upper bound which we will denote by $j$ . No value of $s$ exceeds $j$ , but there are values of $s$ which exceed any number less than $j$ .

In the same way, if $M_{ν}$ is the upper bound of $f (x)$ in $δ_{ν}$ , we can define the sum

S = \sum M_{ν} δ_{ν} .

It is evident that, if $m$ is the lower bound of $f (x)$ in $(a, b)$ , then $S ≧ m (b - a)$ . The aggregate of values of $S$ is therefore bounded below, and possesses a lower bound which we will denote by $J$ . No value of $S$ is less than $J$ , but there are values of $S$ less than any number greater than $J$ .

It will help to make clear the significance of the sums $s$ and $S$ if we observe that, in the simple case in which $f (x)$ increases steadily from $x = a$ to $x = b$ , $m_{ν}$ is $f (x_{ν})$ and $M_{ν}$ is $f (x_{ν + 1})$ . In this case $s$ is the total area of the rectangles shaded in Fig. 48, and $S$ is the

area bounded by a thick line. In general $s$ and $S$ will still be areas, composed of rectangles, respectively included in and including the curvilinear region whose area we are trying to define.

We shall now show that no sum such as $s$ can exceed any sum such as $S$ . Let $s$ , $S$ be the sums corresponding to one mode of subdivision, and $s^{'}$ , $S^{'}$ those corresponding to another. We have to show that $s ≦ S^{'}$ and $s^{'} ≦ S$ .

We can form a third mode of subdivision by taking as dividing points all points which are such for either $s$ , $S$ or $s^{'}$ , $S^{'}$ . Let $s$ , $S$ be the sums corresponding to this third mode of subdivision. Then it is easy to see that

s ≧ s, s ≧ s^{'}, S ≦ S, S ≦ S^{'} . (1)

For example, $s$ differs from $s$ in that at least one interval $δ_{ν}$ which occurs in $s$ is divided into a number of smaller intervals

δ_{ν, 1}, δ_{ν, 2}, \dots, δ_{ν, p},

so that a term $m_{ν} δ_{ν}$ of $s$ is replaced in $s$ by a sum

m_{ν, 1} δ_{ν, 1} + m_{ν, 2} δ_{ν, 2} + \dots + m_{ν, p} δ_{ν, p},

where $m_{ν, 1}$ , $m_{ν, 2}$ , … are the lower bounds of $f (x)$ in $δ_{ν, 1}$ , $δ_{ν, 2}$ , …. But evidently $m_{ν, 1} ≧ m_{ν}$ , $m_{ν, 2} ≧ m_{ν}$ , …, so that the sum just written is not less than $m_{ν} δ_{ν}$ . Hence $s ≧ s;$ and the other inequalities (1) can be established in the same way. But, since $s ≦ S$ , it follows that

s ≦ s ≦ S ≦ S^{'},

which is what we wanted to prove.

It also follows that $j ≦ J$ . For we can find an $s$ as near to $j$ as we please and an $S$ as near to $J$ as we please,² and so $j > J$ would involve the existence of an $s$ and an $S$ for which $s > S$ .

So far we have made no use of the fact that $f (x)$ is continuous. We shall now show that $j = J$ , and that the sums $s$ , $S$ tend to the limit $J$ when the points of division $x_{ν}$ are multiplied indefinitely in such a way that all the intervals $δ_{ν}$ tend to zero. More precisely, we shall show that,

given any positive number $ϵ$ , it is possible to find $δ$ so that

0 ≦ J - s < ϵ, 0 ≦ S - J < ϵ

whenever $δ_{ν} < δ$ for all values of $ν$ .

There is, by Theorem II of § 106, a number $δ$ such that

M_{ν} - m_{ν} < ϵ / (b - a),

whenever every $δ_{ν}$ is less than $δ$ . Hence

S - s = \sum (M_{ν} - m_{ν}) δ_{ν} < ϵ .

But

S - s = (S - J) + (J - j) + (j - s);

and all the three terms on the right-hand side are positive, and therefore all less than $ϵ$ . As $J - j$ is a constant, it must be zero. Hence $j = J$ and $0 ≦ j - s < ϵ$ , $0 ≦ S - J < ϵ$ , as was to be proved.

We define the area of $P pq Q$ as being the common limit of $s$ and $S$ , that is to say $J$ . It is easy to give a more general form to this definition. Consider the sum

σ = \sum f_{ν} δ_{ν}

where $f_{ν}$ denotes the value of $f (x)$ at any point in $δ_{ν}$ . Then $σ$ plainly lies between $s$ and $S$ , and so tends to the limit $J$ when the intervals $δ_{ν}$ tend to zero. We may therefore define the area as the limit of $σ$ .

Footnotes

The argument which follows is modelled on that given in Goursat’s Cours d’Analyse (second edition), vol. i, pp. 171 et seq.; but Goursat’s treatment is much more general. ↩
The $s$ and the $S$ do not in general correspond to the same mode of subdivision. ↩

157The definite integral

Let us now suppose that $f (x)$ is a continuous function, so that the region bounded by the curve $y = f (x)$ , the ordinates $x = a$ and $x = b$ , and the axis of $x$ , has a definite area. We proved in Ch. VI, § 145, that if $F (x)$ is an ‘integral function’ of $f (x)$ , i.e. if

F^{'} (x) = f (x), F (x) = \int f (x) d x,

then the area in question is $F (b) - F (a)$ .

As it is not always practicable actually to determine the form of $F (x)$ , it is convenient to have a formula which represents the area $P pq Q$ and contains no explicit reference to $F (x)$ . We shall write

(P pq Q) = \int_{a}^{b} f (x) d x .

The expression on the right-hand side of this equation may then be regarded as being defined in either of two ways. We may regard it as simply an abbreviation for $F (b) - F (a)$ , where $F (x)$ is some integral function of $f (x)$ , whether an actual formula expressing it is known or not; or we may regard it as the value of the area $P pq Q$ , as directly defined in § 156.

The number

\int_{a}^{b} f (x) d x

is called a definite integral; $a$ and $b$ are called its lower and upper limits; $f (x)$ is called the subject of integration or integrand; and the interval $(a, b)$ the range of integration. The definite integral depends on $a$ and $b$ and the form of the function $f (x)$ only, and is not a function of $x$ . On the other hand the integral function

F (x) = \int f (x) d x

is sometimes called the indefinite integral of $f (x)$ .

The distinction between the definite and the indefinite integral is merely one of point of view. The definite integral $\int_{a}^{b} f (x) d x = F (b) - F (a)$ is a function of $b$ , and may be regarded as a particular integral function of $f (b)$ . On the other hand the indefinite integral $F (x)$ can always be expressed by means of a definite integral, since

F (x) = F (a) + \int_{a}^{x} f (t) d t .

But when we are considering ‘indefinite integrals’ or ‘integral functions’ we are usually thinking of a relation between two functions, in virtue of which one is the derivative of the other. And when we are considering a ‘definite integral’ we are not as a rule concerned with any possible variation of the limits. Usually the limits are constants such as $0$ and $1$ ; and

\int_{0}^{1} f (x) d x = F (1) - F (0)

is not a function at all, but a mere number.

It should be observed that the integral $\int_{a}^{x} f (t) d t$ , having a differential coefficient $f (x)$ , is a fortiori a continuous function of $x$ .

Since $1/ x$ is continuous for all positive values of $x$ , the investigations of the preceding paragraphs supply us with a proof of the actual existence of the function $lo g x$ , which we agreed to assume provisionally in § 128.

158Area of a sector of a circle. The circular functions

The theory of the trigonometrical functions $cos x$ , $sin x$ , etc., as usually presented in text-books of elementary trigonometry, rests on an unproved assumption. An angle is the configuration formed by two straight lines $O A$ , $O P$ ; there is no particular difficulty in translating this ‘geometrical’ definition into purely analytical terms. The assumption comes at the next stage, when it is assumed that angles are capable of numerical measurement, that is to say

that there is a real number $x$ associated with the configuration, just as there is a real number associated with the region $P pq Q$ of Fig. 47. This point once admitted, $cos x$ and $sin x$ may be defined in the ordinary way, and there is no further difficulty of principle in the elaboration of the theory. The whole difficulty lies in the question, what is the $x$ which occurs in $cos x$ and $sin x$ ? To answer this question, we must define the measure of an angle, and we are now in a position to do so. The most natural definition would be this: suppose that $A P$ is an arc of a circle whose centre is $O$ and whose radius is unity, so that $O A = O P = 1$ . Then $x$ , the measure of the angle, is the length of the arc $A P$ . This is, in substance, the definition adopted in the text-books, in the accounts which they give of the theory of ‘circular measure’. It has however, for our present purpose, a fatal defect; for we have not proved that the arc of a curve, even of a circle, possesses a length. The notion of the length of a curve is capable of precise mathematical analysis just as much as that of an area; but the analysis, although of the same general character as that of the preceding sections, is decidedly more difficult, and it is impossible that we should give any general treatment of the subject here.

We must therefore found our definition on the notion not of length but of area. We define the measure of the angle $A O P$ as twice the area of the sector $A O P$ of the unit circle.

Suppose, in particular, that $O A$ is $y = 0$ and that $O P$ is $y = m x$ , where $m > 0$ . The area is a function of $m$ , which we may denote by $ϕ (m)$ . If we write $μ$ for $(1 + m^{2})^{- \frac{1}{2}}$ , $P$ is the point $(μ, m μ)$ , and we have

ϕ (m) = \frac{1}{2} m μ^{2} + \int_{μ}^{1} 1 - x^{2} d x .

Differentiating with respect to $m$ , we find

ϕ^{'} (m) = \frac{1}{2 ( 1 + m ^{2} )}, ϕ (m) = \frac{1}{2} \int_{0}^{m} \frac{d t}{1 + t ^{2}} .

Thus the analytical equivalent of our definition would be to define $arc tan m$ by the equation

arc tan m = \int_{0}^{m} \frac{d t}{1 + t ^{2}};

and the whole theory of the circular functions could be worked out from this starting point, just as the theory of the logarithm is worked out from a similar definition in Ch. IX. See Appendix III.

Examples LXIII.

Calculation of the definite from the indefinite integral.

1. Show that

\int_{a}^{b} x^{n} d x = \frac{b ^{n + 1} - a ^{n + 1}}{n + 1},

and in particular that

\int_{0}^{1} x^{n} d x = \frac{1}{n + 1} .

2. $\int_{a}^{b} cos m x d x = \frac{sin mb - sin ma}{m}$ , $\int_{a}^{b} sin m x d x = \frac{cos ma - cos mb}{m}$ .

3. $\int_{a}^{b} \frac{d x}{1 + x ^{2}} = arc tan b - arc tan a$ , $\int_{0}^{1} \frac{d x}{1 + x ^{2}} = \frac{1}{4} π$ .

[There is an apparent difficulty here owing to the fact that $arc tan x$ is a many valued function. The difficulty may be avoided by observing that, in the equation

\int_{0}^{x} \frac{d t}{1 + t ^{2}} = arc tan x,

$arc tan x$ must denote an angle lying between $- \frac{1}{2} π$ and $\frac{1}{2} π$ . For the integral vanishes when $x = 0$ and increases steadily and continuously as $x$ increases. Thus the same is true of $arc tan x$ , which therefore tends to $\frac{1}{2} π$ as $x \to \infty$ . In the same way we can show that $arc tan x \to - \frac{1}{2} π$ as $x \to - \infty$ . Similarly, in the equation

\int_{0}^{x} \frac{d t}{1 - t ^{2}} = arc sin x,

where $- 1 < x < 1$ , $arc sin x$ denotes an angle lying between $- \frac{1}{2} π$ and $\frac{1}{2} π$ . Thus, if $a$ and $b$ are both numerically less than unity, we have

\int_{a}^{b} \frac{d x}{1 - x ^{2}} = arc sin b - arc sin a .]

4. $\int_{0}^{1} \frac{d x}{1 - x + x ^{2}} = \frac{2 π}{3 3}$ , $\int_{0}^{1} \frac{d x}{1 + x + x ^{2}} = \frac{π}{3 3}$

5. $\int_{0}^{1} \frac{d x}{1 + 2 x cos α + x ^{2}} = \frac{α}{2 sin α}$ if $- π < α < π$ , except when $α = 0$ , when the value of the integral is $\frac{1}{2}$ , which is the limit of $\frac{1}{2} α cosec α$ as $α \to 0$ .

6. $\int_{0}^{1} 1 - x^{2} d x = \frac{1}{4} π$ , $\int_{0}^{a} a^{2} - x^{2} d x = \frac{1}{4} π a^{2}$ $(a > 0)$ .

7. $\int_{0}^{π} \frac{d x}{a + b cos x} = \frac{π}{a ^{2} - b ^{2}}$ , if $a > ∣ b ∣$ . [For the form of the indefinite integral see Exs. liii. 3, 4. If $∣ a ∣ < ∣ b ∣$ then the subject of integration has an infinity between $0$ and $π$ . What is the value of the integral when $a$ is negative and $- a > ∣ b ∣$ ?]

8. $\int_{0}^{\frac{1}{2} π} \frac{d x}{a ^{2} cos ^{2} x + b ^{2} sin ^{2} x} = \frac{π}{2 ab}$ , if $a$ and $b$ are positive. What is the value of the integral when $a$ and $b$ have opposite signs, or when both are negative?

9. Fourier’s integrals. Prove that if $m$ and $n$ are positive integers then

\int_{0}^{2 π} cos m x sin n x d x

is always equal to zero, and

\int_{0}^{2 π} cos m x cos n x d x, \int_{0}^{2 π} sin m x sin n x d x

are equal to zero unless $m = n$ , when each is equal to $π$ .

10. Prove that $\int_{0}^{π} cos m x cos n x d x$ and $\int_{0}^{π} sin m x sin n x d x$ are each equal to zero except when $m = n$ , when each is equal to $\frac{1}{2} π$ ; and that

\int_{0}^{π} cos m x sin n x d x = \frac{2 n}{n ^{2} - m ^{2}}, \int_{0}^{π} cos m x sin n x d x = 0,

according as $n - m$ is odd or even.

159Calculation of the definite integral from its definition as the limit of a sum

In a few cases we can evaluate a definite integral by direct calculation, starting from the definitions of §§ 156 and 157. As a rule it is much simpler to use the indefinite integral, but the reader will find it instructive to work through a few examples.

Examples LXIV.

1. Evaluate $\int_{a}^{b} x d x$ by dividing $(a, b)$ into $n$ equal parts by the points of division $a = x_{0}$ , $x_{1}$ , $x_{2}$ , …, $x_{n} = b$ , and calculating the limit as $n \to \infty$ of

(x_{1} - x_{0}) f (x_{0}) + (x_{2} - x_{1}) f (x_{1}) + \dots + (x_{n} - x_{n - 1}) f (x_{n - 1}) .

[This sum is

\frac{b - a}{n} [a + (a + \frac{b - a}{n}) + (a + 2 \frac{b - a}{n}) + \dots + {a + (n - 1) \frac{b - a}{n}}] = \frac{b - a}{n} [na + \frac{b - a}{n} {1 + 2 + \dots + (n - 1)}] = (b - a) {a + (b - a) \frac{n ( n - 1 )}{2 n ^{2}}},

which tends to the limit $\frac{1}{2} (b^{2} - a^{2})$ as $n \to \infty$ . Verify the result by graphical reasoning.]

2. Calculate $\int_{a}^{b} x^{2} d x$ in the same way.

3. Calculate $\int_{a}^{b} x d x$ , where $0 < a < b$ , by dividing $(a, b)$ into $n$ parts by the points of division $a$ , $a r$ , $a r^{2}$ , … $a r^{n - 1}$ , $a r^{n}$ , where $r^{n} = b / a$ . Apply the same method to the more general integral $\int_{a}^{b} x^{m} d x$ .

4. Calculate $\int_{a}^{b} cos m x d x$ and $\int_{a}^{b} sin m x d x$ by the method of Ex. 1.

5. Prove that $n r = 0 \sum n - 1 \frac{1}{n ^{2} + r ^{2}} \to \frac{1}{4} π$ as $n \to \infty$ .

[This follows from the fact that

\frac{n}{n ^{2}} + \frac{n}{n ^{2} + 1 ^{2}} + \dots + \frac{n}{n ^{2} + ( n - 1 ) ^{2}} = r = 0 \sum n - 1 \frac{( 1/ n )}{1 + ( r / n ) ^{2}},

which tends to the limit $\int_{0}^{1} \frac{d x}{1 + x ^{2}}$ as $n \to \infty$ , in virtue of the direct definition of the integral.]

6. Prove that $\frac{1}{n ^{2}} r = 0 \sum n - 1 n^{2} - r^{2} \to \frac{1}{4} π$ . [The limit is $\int_{0}^{1} 1 - x^{2} d x$ .]

160General properties of the definite integral

The definite integral possesses the important properties expressed by the following equations.¹

(1) $\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x$ .

This follows at once from the definition of the integral by means of the integral function $F (x)$ , since $F (b) - F (a) = - {F (a) - F (b)}$ . It should be observed that in the direct definition it was presupposed that the upper limit is greater than the lower; thus this method of definition does not apply to the integral $\int_{b}^{a} f (x) d x$ when $a < b$ . If we adopt this definition as fundamental we must extend it to such cases by regarding the equation (1) as a definition of its right-hand side.

(2) $\int_{a}^{a} f (x) d x = 0$ .

(3) $\int_{a}^{b} f (x) d x + \int_{b}^{c} f (x) d x = \int_{a}^{c} f (x) d x$ .

(4) $\int_{a}^{b} k f (x) d x = k \int_{a}^{b} f (x) d x$ .

(5) $\int_{a}^{b} {f (x) + ϕ (x)} d x = \int_{a}^{b} f (x) d x + \int_{a}^{b} ϕ (x) d x$ .

The reader will find it an instructive exercise to write out formal proofs of these properties, in each case giving a proof starting from ( $α$ ) the definition by means of the integral function and ( $β$ ) the direct definition.

The following theorems are also important.

(6) If $f (x) ≧ 0$ when $a ≦ x ≦ b$ , then $\int_{a}^{b} f (x) d x ≧ 0$ .

We have only to observe that the sum $s$ of § 156 cannot be negative. It will be shown later (Misc. Ex. 41) that the value of the integral cannot be zero unless $f (x)$ is always equal to zero: this may also be deduced from the second corollary of § 121.

(7) If $H ≦ f (x) ≦ K$ when $a ≦ x ≦ b$ , then

H (b - a) ≦ \int_{a}^{b} f (x) d x ≦ K (b - a) .

This follows at once if we apply (6) to $f (x) - H$ and $K - f (x)$ .

(8) $\int_{a}^{b} f (x) d x = (b - a) f (ξ)$ ,

where $ξ$ lies between $a$ and $b$ .

This follows from (7). For we can take $H$ to be the least and $K$ the greatest value of $f (x)$ in $(a, b)$ . Then the integral is equal to $η (b - a)$ , where $η$ lies between $H$ and $K$ . But, since $f (x)$ is continuous, there must be a value of $ξ$ for which $f (ξ) = η$ (§ 100).

If $F (x)$ is the integral function, we can write the result of (8) in the form

F (b) - F (a) = (b - a) F^{'} (ξ),

so that (8) appears now to be only another way of stating the Mean Value Theorem of § 125. We may call (8) the First Mean Value Theorem for Integrals.

(9) The Generalised Mean Value Theorem for integrals. If $ϕ (x)$ is positive, and $H$ and $K$ are defined as in (7), then

H \int_{a}^{b} ϕ (x) d x ≦ \int_{a}^{b} f (x) ϕ (x) d x ≦ K \int_{a}^{b} ϕ (x) d x;

and

\int_{a}^{b} f (x) ϕ (x) d x = f (ξ) \int_{a}^{b} ϕ (x) d x,

where $ξ$ is defined as in (8).

This follows at once by applying Theorem (6) to the integrals

\int_{a}^{b} {f (x) - H} ϕ (x) d x, \int_{a}^{b} {K - f (x)} ϕ (x) d x .

The reader should formulate for himself the corresponding result which holds when $ϕ (x)$ is always negative.

(10) The Fundamental Theorem of the Integral Calculus. The function

F (x) = \int_{a}^{x} f (t) d t

has a derivative equal to $f (x)$ .

This has been proved already in § 145, but it is convenient to restate the result here as a formal theorem. It follows as a corollary, as was pointed out in § 157, that $F (x)$ is a continuous function of $x$ .

Examples LXV.

1. Show, by means of the direct definition of the definite integral, and equations (1)–(5) above, that

(i) $\int_{- a}^{a} ϕ (x^{2}) d x = 2 \int_{0}^{a} ϕ (x^{2}) d x$ , $\int_{- a}^{a} x ϕ (x^{2}) d x = 0$ ;

(ii) $\int_{0}^{\frac{1}{2} π} ϕ (cos x) d x = \int_{0}^{\frac{1}{2} π} ϕ (sin x) d x = \frac{1}{2} \int_{0}^{π} ϕ (sin x) d x$ ;

(iii) $\int_{0}^{mπ} ϕ (cos^{2} x) d x = m \int_{0}^{π} ϕ (cos^{2} x) d x$ ,

$m$ being an integer. [The truth of these equations will appear geometrically intuitive, if the graphs of the functions under the sign of integration are sketched.]

2. Prove that $\int_{0}^{π} \frac{sin n x}{sin x} d x$ is equal to $π$ or to $0$ according as $n$ is odd or or even. [Use the formula $(sin n x) / (sin x) = 2 cos {(n - 1) x} + 2 cos {(n - 3) x} + \dots$ , the last term being $1$ or $2 cos x$ .]

3. Prove that $\int_{0}^{π} sin n x cot x d x$ is equal to $0$ or to $π$ according as $n$ is odd or even.

4. If $ϕ (x) = a_{0} + a_{1} cos x + b_{1} sin x + a_{2} cos 2 x + \dots + a_{n} cos n x + b_{n} sin n x$ , and $k$ is a positive integer not greater than $n$ , then

\int_{0}^{2 π} ϕ (x) d x = 2 π a_{0}, \int_{0}^{2 π} cos k x ϕ (x) d x = π a_{k}, \int_{0}^{2 π} sin k x ϕ (x) d x = π b_{k} .

If $k > n$ then the value of each of the last two integrals is zero. [Use Ex. lxiii. 9.]

5. If $ϕ (x) = a_{0} + a_{1} cos x + a_{2} cos 2 x + \dots + a_{n} cos n x$ , and $k$ is a positive integer not greater than $n$ , then

\int_{0}^{π} ϕ (x) d x = π a_{0}, \int_{0}^{π} cos k x ϕ (x) d x = \frac{1}{2} π a_{k} .

If $k > n$ then the value of the last integral is zero. [Use Ex. lxiii. 10.]

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

\int_{0}^{2 π} \frac{d x}{a ^{2} cos ^{2} x + b ^{2} sin ^{2} x} = \frac{2 π}{ab} .

[Use Ex. lxiii. 8 and Ex. 1 above.]

7. If $f (x) ≦ ϕ (x)$ when $a ≦ x ≦ b$ , then $\int_{a}^{b} f d x ≦ \int_{a}^{b} ϕ d x$ .

8. Prove that

00 < \int_{0}^{\frac{1}{2} π} sin^{n + 1} x d x < \int_{0}^{\frac{1}{4} π} tan^{n + 1} x d x < \int_{0}^{\frac{1}{2} π} sin^{n} x d x, < \int_{0}^{\frac{1}{4} π} tan^{n} x d x .

9. ² If $n > 1$ then

.5 < \int_{0}^{\frac{1}{2}} \frac{d x}{1 - x ^{2 n}} < .524.

[The first inequality follows from the fact that $1 - x^{2 n} < 1$ , the second from the fact that $1 - x^{2 n} > 1 - x^{2}$ .]

10. Prove that

\frac{1}{2} < \int_{0}^{1} \frac{d x}{4 - x ^{2} + x ^{3}} < \frac{1}{6} π .

11. Prove that $(3 x + 8) /16 < 1/ 4 - 3 x + x^{3} < 1/ 4 - 3 x$ if $0 < x < 1$ , and hence that

\frac{19}{32} < \int_{0}^{1} \frac{d x}{4 - 3 x + x ^{3}} < \frac{2}{3} .

12. Prove that

.573 < \int_{1}^{2} \frac{d x}{4 - 3 x + x ^{3}} < .595.

[Put $x = 1 + u$ : then replace $2 + 3 u^{2} + u^{3}$ by $2 + 4 u^{2}$ and by $2 + 3 u^{2}$ .]

13. If $α$ and $ϕ$ are positive acute angles then

ϕ < \int_{0}^{ϕ} \frac{d x}{1 - sin ^{2} α sin ^{2} x} < \frac{ϕ}{1 - sin ^{2} α sin ^{2} ϕ} .

If $α = ϕ = \frac{1}{6} π$ , then the integral lies between $.523$ and $.541$ .

14. Prove that

\int_{a}^{b} f (x) d x ≦ \int_{a}^{b} ∣ f (x) ∣ d x .

[If $σ$ is the sum considered at the end of § 156, and $σ^{'}$ the corresponding sum formed from the function $∣ f (x) ∣$ , then $∣ σ ∣ ≦ σ^{'}$ .]

15. If $∣ f (x) ∣ ≦ M$ , then

\int_{a}^{b} f (x) ϕ (x) d x ≦ M \int_{a}^{b} ∣ ϕ (x) ∣ d x .

Footnotes

All functions mentioned in these equations are of course continuous, as the definite integral has been defined for continuous functions only. ↩
Exs. 9–13 are taken from Prof. Gibson’s Elementary Treatise on the Calculus. ↩

161Integration by parts and by substitution

It follows from § 138 that

\int_{a}^{b} f (x) ϕ^{'} (x) d x = f (b) ϕ (b) - f (a) ϕ (a) - \int_{a}^{b} f^{'} (x) ϕ (x) d x .

This formula is known as the formula for integration of a definite integral by parts.

Again, we know (§ 133) that if $F (t)$ is the integral function of $f (t)$ , then

\int f {ϕ (x)} ϕ^{'} (x) d x = F {ϕ (x)} .

Hence, if $ϕ (a) = c$ , $ϕ (b) = d$ , we have

\int_{c}^{d} f (t) d t = F (d) - F (c) = F {ϕ (b)} - F {ϕ (a)} = \int_{a}^{b} f {ϕ (x)} ϕ^{'} (x) d x;

which is the formula for the transformation of a definite integral by substitution.

The formulae for integration by parts and for transformation often enable us to evaluate a definite integral without the labour of actually finding the integral function of the subject of integration, and sometimes even when the integral function cannot be found. Some instances of this will be found in the following examples. That the value of a definite integral may sometimes be found without a knowledge of the integral function is only to be expected, for the fact that we cannot determine the general form of a function $F (x)$ in no way precludes the possibility that we may be able to determine the difference $F (b) - F (a)$ between two of its particular values. But as a rule this can only be effected by the use of more advanced methods than are at present at our disposal.

Examples LXVI.

1. Prove that

\int_{a}^{b} x f^{''} (x) d x = {b f^{'} (b) - f (b)} - {a f^{'} (a) - f (a)} .

2. More generally,

\int_{a}^{b} x^{m} f^{(m + 1)} (x) d x = F (b) - F (a),

where

F (x) = x^{m} f^{(m)} (x) - m x^{m - 1} f^{(m - 1)} (x) + m (m - 1) x^{m - 2} f^{(m - 2)} (x) - \dots + (- 1)^{m} m! f (x) .

3. Prove that

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

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

\int_{0}^{\frac{1}{2} π} \frac{x cos x sin x d x}{( a ^{2} cos ^{2} x + b ^{2} sin ^{2} x ) ^{2}} = \frac{π}{4 a b ^{2} ( a + b )} .

[Integrate by parts and use Ex. lxiii. 8.]

5. If

f_{1} (x) = \int_{0}^{x} f (t) d t, f_{2} (x) = \int_{0}^{x} f_{1} (t) d t, \dots, f_{k} (x) = \int_{0}^{x} f_{k - 1} (t) d t,

then

f_{k} (x) = \frac{1}{( k - 1 )!} \int_{0}^{x} f (t) (x - t)^{k - 1} d t .

[Integrate repeatedly by parts.]

6. Prove by integration by parts that if

u_{m, n} = \int_{0}^{1} x^{m} (1 - x)^{n} d x,

where $m$ and $n$ are positive integers, then $(m + n + 1) u_{m, n} = n u_{m, n - 1}$ , and deduce that

u_{m, n} = \frac{m ! n !}{( m + n + 1 )!} .

7. Prove that if

u_{n} = \int_{0}^{\frac{1}{4} π} tan^{n} x d x

then $u_{n} + u_{n - 2} = 1/ (n - 1)$ . Hence evaluate the integral for all positive integral values of $n$ .

[Put $tan^{n} x = tan^{n - 2} x (sec^{2} x - 1)$ and integrate by parts.]

8. Deduce from the last example that $u_{n}$ lies between $1/ {2 (n - 1)}$ and $1/ {2 (n + 1)}$ .

9. Prove that if

u_{n} = \int_{0}^{\frac{1}{2} π} sin^{n} x d x

then $u_{n} = {(n - 1) / n} u_{n - 2}$ . [Write $sin^{n - 1} x sin x$ for $sin^{n} x$ and integrate by parts.]

10. Deduce that $u_{n}$ is equal to

\frac{2 \cdot 4 \cdot 6 \dots ( n - 1 )}{3 \cdot 5 \cdot 7 \dots n}, \frac{1}{2} π \frac{1 \cdot 3 \cdot 5 \dots ( n - 1 )}{2 \cdot 4 \cdot 6 \dots n},

according as $n$ is odd or even.

11. The Second Mean Value Theorem. If $f (x)$ is a function of $x$ which has a differential coefficient of constant sign for all values of $x$ from $x = a$ to $x = b$ , then there is a number $ξ$ between $a$ and $b$ such that

\int_{a}^{b} f (x) ϕ (x) d x = f (a) \int_{a}^{ξ} ϕ (x) d x + f (b) \int_{ξ}^{b} ϕ (x) d x .

[Let $\int_{a}^{x} ϕ (t) d t = Φ (x)$ . Then

\int_{a}^{b} f (x) ϕ (x) d x = \int_{a}^{b} f (x) Φ^{'} (x) d x = f (b) Φ (b) - \int_{a}^{b} f^{'} (x) Φ (x) d x = f (b) Φ (b) - Φ (ξ) \int_{a}^{b} f^{'} (x) d x,

by the generalised Mean Value Theorem of § 160: i.e.

\int_{a}^{b} f (x) ϕ (x) d x = f (b) Φ (b) + {f (a) - f (b)} Φ (ξ),

which is equivalent to the result given.]

12. Bonnet’s form of the Second Mean Value Theorem. If $f^{'} (x)$ is of constant sign, and $f (b)$ and $f (a) - f (b)$ have the same sign, then

\int_{a}^{b} f (x) ϕ (x) d x = f (a) \int_{a}^{X} ϕ (x) d x,

where $X$ lies between $a$ and $b$ . [For $f (b) Φ (b) + {f (a) - f (b)} Φ (ξ) = μ f (a)$ , where $μ$ lies between $Φ (ξ)$ and $Φ (b)$ , and so is the value of $Φ (x)$ for a value of $x$ such as $X$ . The important case is that in which $0 ≦ f (b) ≦ f (x) ≦ f (a)$ .]

Prove similarly that if $f (a)$ and $f (b) - f (a)$ have the same sign, then

\int_{a}^{b} f (x) ϕ (x) d x = f (b) \int_{X}^{b} ϕ (x) d x,

where $X$ lies between $a$ and $b$ . [Use the function $Ψ (ξ) = \int_{ξ}^{b} ϕ (x) d x$ . It will be found that the integral can be expressed in the form

f (a) Ψ (a) + {f (b) - f (a)} Ψ (ξ) .

The important case is that in which $0 ≦ f (a) ≦ f (x) ≦ f (b)$ .]

13. Prove that

\int_{X}^{X^{'}} \frac{sin x}{x} d x < \frac{2}{X}

if $X^{'} > X > 0$ . [Apply the first formula of Ex. 12, and note that the integral of $sin x$ over any interval whatever is numerically less than $2$ .]

14. Establish the results of Ex. lxv. 1 by means of the rule for substitution. [In (i) divide the range of integration into the two parts $(- a, 0)$ , $(0, a)$ , and put $x = - y$ in the first. In (ii) use the substitution $x = \frac{1}{2} π - y$ to obtain the first equation: to obtain the second divide the range $(0, π)$ into two equal parts and use the substitution $x = \frac{1}{2} π + y$ . In (iii) divide the range into $m$ equal parts and use the substitutions $x = π + y$ , $x = 2 π + y$ , ….]

15. Prove that

\int_{a}^{b} F (x) d x = \int_{a}^{b} F (a + b - x) d x .

16. Prove that

\int_{0}^{\frac{1}{2} π} cos^{m} x sin^{m} x d x = 2^{- m} \int_{0}^{\frac{1}{2} π} cos^{m} x d x .

17. Prove that

\int_{0}^{π} x ϕ (sin x) d x = \frac{1}{2} π \int_{0}^{π} ϕ (sin x) d x .

[Put $x = π - y$ .]

18. Prove that

\int_{0}^{π} \frac{x sin x}{1 + cos ^{2} x} d x = \frac{1}{4} π^{2} .

19. Show by means of the transformation $x = a cos^{2} θ + b sin^{2} θ$ that

\int_{a}^{b} (x - a) (b - x) d x = \frac{1}{8} π (b - a)^{2} .

20. Show by means of the substitution $(a + b cos x) (a - b cos y) = a^{2} - b^{2}$ that

\int_{0}^{π} (a + b cos x)^{- n} d x = (a^{2} - b^{2})^{- (n - \frac{1}{2})} \int_{0}^{π} (a - b cos y)^{n - 1} d y,

when $n$ is a positive integer and $a > ∣ b ∣$ , and evaluate the integral when $n = 1$ , $2$ , $3$ .

21. If $m$ and $n$ are positive integers then

\int_{a}^{b} (x - a)^{m} (b - x)^{n} d x = (b - a)^{m + n + 1} \frac{m ! n !}{( m + n + 1 )!} .

[Put $x = a + (b - a) y$ , and use Ex. 6.]

162Proof of Taylor’s Theorem by Integration by Parts

We shall now give the alternative form of the proof of Taylor’s Theorem to which we alluded in § 147.

Let $f (x)$ be a function whose first $n$ derivatives are continuous, and let

F_{n} (x) = f (b) - f (x) - (b - x) f^{'} (x) - \dots - \frac{( b - x ) ^{n - 1}}{( n - 1 )!} f^{(n - 1)} (x) .

Then

F_{n}^{'} (x) = - \frac{( b - x ) ^{n - 1}}{( n - 1 )!} f^{(n)} (x),

and so

F_{n} (a) = F_{n} (b) - \int_{a}^{b} F_{n}^{'} (x) d x = \frac{1}{( n - 1 )!} \int_{a}^{b} (b - x)^{n - 1} f^{(n)} (x) d x .

If now we write $a + h$ for $b$ , and transform the integral by putting $x = a + t h$ , we obtain

f (a + h) = f (a) + h f^{'} (a) + \dots + \frac{h ^{n - 1}}{( n - 1 )!} f^{(n - 1)} (a) + R_{n}, (1)

where

R_{n} = \frac{h ^{n}}{( n - 1 )!} \int_{0}^{1} (1 - t)^{n - 1} f^{(n)} (a + t h) d t . (2)

Now, if $p$ is any positive integer not greater than $n$ , we have, by Theorem (9) of § 160,

\int_{0}^{1} (1 - t)^{n - 1} f^{(n)} (a + t h) d t = \int_{0}^{1} (1 - t)^{n - p} (1 - t)^{p - 1} f^{(n)} (a + t h) d t = (1 - θ)^{n - p} f^{(n)} (a + θ h) \int_{0}^{1} (1 - t)^{p - 1} d t,

where $0 < θ < 1$ . Hence

R_{n} = \frac{( 1 - θ ) ^{n - p} f ^{(n)} ( a + θ h ) h ^{n}}{p ( n - 1 )!} . (3)

If we take $p = n$ we obtain Lagrange’s form of $R_{n}$ (§ 148). If on the other hand we take $p = 1$ we obtain Cauchy’s form, viz.¹

R_{n} = \frac{( 1 - θ ) ^{n - 1} f ^{(n)} ( a + θ h ) h ^{n}}{( n - 1 )!} . (4)

Footnotes

The method used in § 147 can also be modified so as to obtain these alternative forms of the remainder. ↩

163Application of Cauchy’s form to the Binomial Series

If $f (x) = (1 + x)^{m}$ , where $m$ is not a positive integer, then Cauchy’s form of the remainder is

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

Now $(1 - θ) / (1 + θ x)$ is less than unity, so long as $- 1 < x < 1$ , whether $x$ is positive or negative; and $(1 + θ x)^{m - 1}$ is less than a constant $K$ for all values of $n$ , being in fact less than $(1 + ∣ x ∣)^{m - 1}$ if $m > 1$ and than $(1 - ∣ x ∣)^{m - 1}$ if $m < 1$ Hence

∣ R_{n} ∣ < K ∣ m ∣ (n - 1 m - 1) ∣ x^{n} ∣ = ρ_{n},

say But $ρ_{n} \to 0$ as $n \to \infty$ , by Ex. xxvii. 13, and so $R_{n} \to 0$ . The truth of the Binomial Theorem is thus established for all rational values of $m$ and all values of $x$ between $- 1$ and $1$ . It will be remembered that the difficulty in using Lagrange’s form, in Ex. lvi. 2, arose in connection with negative values of $x$ .

164Integrals of complex functions of a real variable

So far we have always supposed that the subject of integration in a definite integral is real. We define the integral of a complex function $f (x) = ϕ (x) + i ψ (x)$ of the real variable $x$ , between the limits $a$ and $b$ , by the equations

\int_{a}^{b} f (x) d x = \int_{a}^{b} {ϕ (x) + i ψ (x)} d x = \int_{a}^{b} ϕ (x) d x + i \int_{a}^{b} ψ (x) d x;

and it is evident that the properties of such integrals may be deduced from those of the real integrals already considered.

There is one of these properties that we shall make use of later on. It is expressed by the inequality¹

\int_{a}^{b} f (x) d x ≦ \int_{a}^{b} ∣ f (x) ∣ d x . (1)

This inequality may be deduced without difficulty from the definitions of §§ 156 and 157. If $δ_{ν}$ has the same meaning as in § 156, $ϕ_{ν}$ and $ψ_{ν}$ are the values of $ϕ$ and $ψ$ at a point of $δ_{ν}$ , and $f_{ν} = ϕ_{ν} + i ψ_{ν}$ , then we have

\int_{a}^{b} f d x = \int_{a}^{b} ϕ d x + i \int_{a}^{b} ψ d x = lim \sum ϕ_{ν} δ_{ν} + i lim \sum ψ_{ν} δ_{ν} = lim \sum (ϕ_{ν} + i ψ_{ν}) δ_{ν} = lim \sum f_{ν} δ_{ν},

and so

\int_{a}^{b} f d x = ∣ lim \sum f_{ν} δ_{ν} ∣ = lim ∣ \sum f_{ν} δ_{ν} ∣;

while

\int_{a}^{b} ∣ f ∣ d x = lim \sum ∣ f_{ν} ∣ δ_{ν} .

The result now follows at once from the inequality

∣ \sum f_{ν} δ_{ν} ∣ ≦ \sum ∣ f_{ν} ∣ δ_{ν} .

It is evident that the formulae (1) and (2) of § 162 remain true when $f$ is a complex function $ϕ + i ψ$ .

Footnotes

The corresponding inequality for a real integral was proved in Ex. lxv. 14. ↩

Misc.Miscellaneous Examples on Chapter VII

1. Verify the terms given of the following Taylor’s Series:

(1) (2) (3) (4) tan x sec x x cosec x x cot x = x + \frac{1}{3} x^{3} + \frac{2}{15} x^{5} + \dots, = 1 + \frac{1}{2} x^{2} + \frac{5}{24} x^{4} + \dots, = 1 + \frac{1}{6} x^{2} + \frac{7}{360} x^{4} + \dots, = 1 - \frac{1}{3} x^{2} - \frac{1}{45} x^{4} - \dots .

2. Show that if $f (x)$ and its first $n + 2$ derivatives are continuous, and $f^{(n + 1)} (0) \neq = 0$ , and $θ_{n}$ is the value of $θ$ which occurs in Lagrange’s form of the remainder after $n$ terms of Taylor’s Series, then

θ_{n} = \frac{1}{n + 1} + \frac{n}{2 ( n + 1 ) ^{2} ( n + 2 )} {\frac{f ^{(n + 2)} ( 0 )}{f ^{(n + 1)} ( 0 )} + ϵ_{x}} x,

where $ϵ_{x} \to 0$ as $x \to 0$ . [Follow the method of Ex. lv. 12.]

3. Verify the last result when $f (x) = 1/ (1 + x)$ . [Here $(1 + θ_{n} x)^{n + 1} = 1 + x$ .]

4. Show that if $f (x)$ has derivatives of the first three orders then

f (b) = f (a) + \frac{1}{2} (b - a) {f^{'} (a) + f^{'} (b)} - \frac{1}{12} (b - a)^{3} f^{'''} (α),

where $a < α < b$ . [Apply to the function

f (x) - f (a) - \frac{1}{2} (x - a) {f^{'} (a) + f^{'} (x)} - (\frac{x - a}{b - a})^{3} [f (b) - f (a) - \frac{1}{2} (b - a) {f^{'} (a) + f^{'} (b)}]

arguments similar to those of § 147.]

5. Show that under the same conditions

f (b) = f (a) + (b - a) f^{'} {\frac{1}{2} (a + b)} + \frac{1}{24} (b - a)^{3} f^{'''} (α) .

6. Show that if $f (x)$ has derivatives of the first five orders then

f (b) = f (a) + \frac{1}{6} (b - a) [f^{'} (a) + f^{'} (b) + 4 f^{'} {\frac{1}{2} (a + b)}] - \frac{1}{2880} (b - a)^{5} f^{(5)} (α) .

7. Show that under the same conditions

f (b) = f (a) + \frac{1}{2} (b - a) {f^{'} (a) + f^{'} (b)} - \frac{1}{12} (b - a)^{2} {f^{''} (b) - f^{''} (a)} + \frac{1}{720} (b - a)^{5} f^{(5)} (α) .

8. Establish the formulae

(i) $f (a) g (a) f (b) g (b) = (b - a) f (a) g (a) f^{'} (β) g^{'} (β)$ ,

where $β$ lies between $a$ and $b$ , and

(ii) $f (a) g (a) h (a) f (b) g (b) h (b) f (c) g (c) h (c) = \frac{1}{2} (b - c) (c - a) (a - b) f (a) g (a) h (a) f^{'} (β) g^{'} (β) h^{'} (β) f^{''} (γ) g^{''} (γ) h^{''} (γ)$ ,

where $β$ and $γ$ lie between the least and greatest of $a$ , $b$ , $c$ . [To prove (ii) consider the function

ϕ (x) = f (a) g (a) h (a) f (b) g (b) h (b) f (x) g (x) h (x) - \frac{( x - a ) ( x - b )}{( c - a ) ( c - b )} f (a) g (a) h (a) f (b) g (b) h (b) f (c) g (c) h (c)

which vanishes when $x = a$ , $x = b$ , and $x = c$ . Its first derivative, by Theorem B of § 121, must vanish for two distinct values of $x$ lying between the least and greatest of $a$ , $b$ , $c$ ; and its second derivative must therefore vanish for a value $γ$ of $x$ satisfying the same condition. We thus obtain the formula

f (a) g (a) h (a) f (b) g (b) h (b) f (c) g (c) h (c) = \frac{1}{2} (c - a) (c - b) f (a) g (a) h (a) f (b) g (b) h (b) f^{''} (γ) g^{''} (γ) h^{''} (γ) .

The reader will now complete the proof without difficulty.]

9. If $F (x)$ is a function which has continuous derivatives of the first $n$ orders, of which the first $n - 1$ vanish when $x = 0$ , and $A ≦ F^{(n)} (x) ≦ B$ when $0 ≦ x ≦ h$ , then $A (x^{n} / n!) ≦ F (x) ≦ B (x^{n} / n!)$ when $0 ≦ x ≦ h$ .

Apply this result to

f (x) - f (0) - x f^{'} (0) - \dots - \frac{x ^{n - 1}}{( n - 1 )!} f^{(n - 1)} (0),

and deduce Taylor’s Theorem.

10. If $Δ_{h} ϕ (x) = ϕ (x) - ϕ (x + h)$ , $Δ_{h}^{2} ϕ (x) = Δ_{h} {Δ_{h} ϕ (x)}$ , and so on, and $ϕ (x)$ has derivatives of the first $n$ orders, then

Δ_{h}^{n} ϕ (x) = r = 0 \sum n (- 1)^{r} (r n) ϕ (x + r h) = (- h)^{n} ϕ^{(n)} (ξ),

where $ξ$ lies between $x$ and $x + nh$ . Deduce that if $ϕ^{(n)} (x)$ is continuous then ${Δ_{h}^{n} ϕ (x)} / h^{n} \to (- 1)^{n} ϕ^{(n)} (x)$ as $h \to 0$ . [This result has been stated already when $n = 2$ , in Ex. lv. 13.]

11. Deduce from Ex. 10 that $x^{n - m} Δ_{h}^{n} x^{m} \to m (m - 1) \dots (m - n + 1) h^{n}$ as $x \to \infty$ , $m$ being any rational number and $n$ any positive integer. In particular prove that

x x {x - 2 x + 1 + x + 2} \to - \frac{1}{4} .

12. Suppose that $y = ϕ (x)$ is a function of $x$ with continuous derivatives of at least the first four orders, and that $ϕ (0) = 0$ , $ϕ^{'} (0) = 1$ , so that

y = ϕ (x) = x + a_{2} x^{2} + a_{3} x^{3} + (a_{4} + ϵ_{x}) x^{4},

where $ϵ_{x} \to 0$ as $x \to 0$ . Establish the formula

x = ψ (y) = y - a_{2} y^{2} + (2 a_{2}^{2} - a_{3}) y^{3} - (5 a_{2}^{3} - 5 a_{2} a_{3} + a_{4} + ϵ_{y}) y^{4},

where $ϵ_{y} \to 0$ as $y \to 0$ , for that value of $x$ which vanishes with $y$ ; and prove that

\frac{ϕ ( x ) ψ ( x ) - x ^{2}}{x ^{4}} \to a_{2}^{2}

as $x \to 0$ .

13. The coordinates $(ξ, η)$ of the centre of curvature of the curve $x = f (t)$ , $y = F (t)$ , at the point $(x, y)$ , are given by

- (ξ - x) / y^{'} = (η - y) / x^{'} = (x^{' 2} + y^{' 2}) / (x^{'} y^{''} - x^{''} y^{'});

and the radius of curvature of the curve is

(x^{' 2} + y^{' 2})^{3/2} / (x^{'} y^{''} - x^{''} y^{'}),

dashes denoting differentiations with respect to $t$ .

14. The coordinates $(ξ, η)$ of the centre of curvature of the curve $27 a y^{2} = 4 x^{3}$ , at the point $(x, y)$ , are given by

3 a (ξ + x) + 2 x^{2} = 0, η = 4 y + (9 a y) / x .

(Math. Trip. 1899.)

15. Prove that the circle of curvature at a point $(x, y)$ will have contact of the third order with the curve if $(1 + y_{1}^{2}) y_{3} = 3 y_{1} y_{2}^{2}$ at that point. Prove also that the circle is the only curve which possesses this property at every point; and that the only points on a conic which possess the property are the extremities of the axes. [Cf. Ch. VI, Misc. Ex. 10 (iv).]

16. The conic of closest contact with the curve $y = a x^{2} + b x^{3} + c x^{4} + \dots + k x^{n}$ , at the origin, is $a^{3} y = a^{4} x^{2} + a^{2} b x y + (a c - b^{2}) y^{2}$ . Deduce that the conic of closest contact at the point $(ξ, η)$ of the curve $y = f (x)$ is

18 η_{2}^{3} T = 9 η_{2}^{4} (x - ξ)^{2} + 6 η_{2}^{2} η_{3} (x - ξ) T + (3 η_{2} η_{4} - 4 η_{3}^{2}) T^{2},

where $T = (y - η) - η_{1} (x - ξ)$ . (Math. Trip. 1907.)

17. Homogeneous functions.¹ If $u = x^{n} f (y / x, z / x, \dots)$ then $u$ is unaltered, save for a factor $λ^{n}$ , when $x$ , $y$ , $z$ , … are all increased in the ratio $λ : 1$ . In these circumstances $u$ is called a homogeneous function of degree $n$ in the variables $x$ , $y$ , $z$ , …. Prove that if $u$ is homogeneous and of degree $n$ then

x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} + z \frac{\partial u}{\partial z} + \dots = n u .

This result is known as Euler’s Theorem on homogeneous functions.

18. If $u$ is homogeneous and of degree $n$ then $\partial u / \partial x$ , $\partial u / \partial y$ , … are homogeneous and of degree $n - 1$ .

19. Let $f (x, y) = 0$ be an equation in $x$ and $y$ (e.g. $x^{n} + y^{n} - x = 0$ ), and let $F (x, y, z) = 0$ be the form it assumes when made homogeneous by the introduction of a third variable $z$ in place of unity (e.g. $x^{n} + y^{n} - x z^{n - 1} = 0$ ). Show that the equation of the tangent at the point $(ξ, η)$ of the curve $f (x, y) = 0$ is

x F_{ξ} + y F_{η} + z F_{ζ} = 0,

where $F_{ξ}$ , $F_{η}$ , $F_{ζ}$ denote the values of $F_{x}$ , $F_{y}$ , $F_{z}$ when $x = ξ$ , $y = η$ , $z = ζ = 1$ .

20. Dependent and independent functions. Jacobians or functional determinants. Suppose that $u$ and $v$ are functions of $x$ and $y$ connected by an identical relation

ϕ (u, v) = 0. (1)

Differentiating (1) with respect to $x$ and $y$ , we obtain

\frac{\partial ϕ}{\partial u} \frac{\partial u}{\partial x} + \frac{\partial ϕ}{\partial v} \frac{\partial v}{\partial x} = 0, \frac{\partial ϕ}{\partial u} \frac{\partial u}{\partial y} + \frac{\partial ϕ}{\partial v} \frac{\partial v}{\partial y} = 0, (2)

and, eliminating the derivatives of $ϕ$ ,

J = u_{x} v_{x} u_{y} v_{y} = u_{x} v_{y} - u_{y} v_{x} = 0, (3)

where $u_{x}$ , $u_{y}$ , $v_{x}$ , $v_{y}$ are the derivatives of $u$ and $v$ with respect to $x$ and $y$ . This condition is therefore necessary for the existence of a relation such as (1). It can be proved that the condition is also sufficient; for this we must refer to Goursat’s Cours d’ Analyse, vol. i, pp. 125 et seq.

Two functions $u$ and $v$ are said to be dependent or independent according as they are or are not connected by such a relation as (1). It is usual to call $J$ the Jacobian or functional determinant of $u$ and $v$ with respect to $x$ and $y$ , and to write

J = \frac{\partial ( u , v )}{\partial ( x , y )} .

Similar results hold for functions of any number of variables. Thus three functions $u$ , $v$ , $w$ of three variables $x$ , $y$ , $z$ are or are not connected by a relation $ϕ (u, v, w) = 0$ according as

J = u_{x} v_{x} w_{x} u_{y} v_{y} w_{y} u_{z} v_{z} w_{z} = \frac{\partial ( u , v , w )}{\partial ( x , y , z )}

does or does not vanish for all values of $x$ , $y$ , $z$ .

21. Show that $a x^{2} + 2 h x y + b y_{2}$ and $A x^{2} + 2 H x y + B y^{2}$ are independent unless $a / A = h / H = b / B$ .

22. Show that $a x^{2} + b y^{2} + c z^{2} + 2 f y z + 2 g z x + 2 h x y$ can be expressed as a product of two linear functions of $x$ , $y$ , and $z$ if and only if

ab c + 2 f g h - a f^{2} - b g^{2} - c h^{2} = 0.

[Write down the condition that $p x + q y + r z$ and $p^{'} x + q^{'} y + r^{'} z$ should be connected with the given function by a functional relation.]

23. If $u$ and $v$ are functions of $ξ$ and $η$ , which are themselves functions of $x$ and $y$ , then

\frac{\partial ( u , v )}{\partial ( x , y )} = \frac{\partial ( u , v )}{\partial ( ξ , η )} \frac{\partial ( ξ , η )}{\partial ( x , y )} .

Extend the result to any number of variables.

24. Let $f (x)$ be a function of $x$ whose derivative is $1/ x$ and which vanishes when $x = 1$ . Show that if $u = f (x) + f (y)$ , $v = x y$ , then $u_{x} v_{y} - u_{y} v_{x} = 0$ , and hence that $u$ and $v$ are connected by a functional relation. By putting $y = 1$ , show that this relation must be $f (x) + f (y) = f (x y)$ . Prove in a similar manner that if the derivative of $f (x)$ is $1/ (1 + x^{2})$ , and $f (0) = 0$ , then $f (x)$ must satisfy the equation

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

25. Prove that if $f (x) = \int_{0}^{x} \frac{d t}{1 - t ^{4}}$ then

f (x) + f (y) = f {\frac{x 1 - y ^{4} + y 1 - x ^{4}}{1 + x ^{2} y ^{2}}} .

26. Show that if a functional relation exists between

u = f (x) + f (y) + f (z), v = f (y) f (z) + f (z) f (x) + f (x) f (y), w = f (x) f (y) f (z),

then $f$ must be a constant. [The condition for a functional relation will be found to be

f^{'} (x) f^{'} (y) f^{'} (z) {f (y) - f (z)} {f (z) - f (x)} {f (x) - f (y)} = 0.]

27. If $f (y, z)$ , $f (z, x)$ , and $f (x, y)$ are connected by a functional relation then $f (x, x)$ is independent of $x$ . (Math. Trip. 1909.)

28. If $u = 0$ , $v = 0$ , $w = 0$ are the equations of three circles, rendered homogeneous as in Ex. 19, then the equation

\frac{\partial ( u , v , w )}{\partial ( x , y , z )} = 0

represents the circle which cuts them all orthogonally. (Math. Trip. 1900.)

29. If $A$ , $B$ , $C$ are three functions of $x$ such that

A B C A^{'} B^{'} C^{'} A^{''} B^{''} C^{''}

vanishes identically, then we can find constants $λ$ , $μ$ , $ν$ such that $λ A + μ B + ν C$ vanishes identically; and conversely. [The converse is almost obvious. To prove the direct theorem let $α = B C^{'} - B^{'} C$ , …. Then $α^{'} = B C^{''} - B^{''} C$ , …, and it follows from the vanishing of the determinant that $β γ^{'} - β^{'} γ = 0$ , …; and so that the ratios $α : β : γ$ are constant. But $α A + β B + γ C = 0$ .]

30. Suppose that three variables $x$ , $y$ , $z$ are connected by a relation in virtue of which (i) $z$ is a function of $x$ and $y$ , with derivatives $z_{x}$ $z_{y}$ , and (ii) $x$ is a function of $y$ and $z$ , with derivatives $x_{y}$ , $x_{z}$ . Prove that

x_{y} = - z_{y} / z_{x}, x_{z} = 1/ z_{x} .

[We have

d z = z_{x} d x + z_{y} d y, d x = x_{y} d y + x_{z} d z

The result of substituting for $d x$ in the first equation is

d z = (z_{x} x_{y} + z_{y}) d y + z_{x} x_{z} d z,

which can be true only if $z_{x} x_{y} + z_{y} = 0$ , $z_{x} x_{z} = 1$ .]

31. Four variables $x$ , $y$ , $z$ , $u$ are connected by two relations in virtue of which any two can be expressed as functions of the others. Show that

y_{z}^{u} z_{x}^{u} x_{y}^{u} = - y_{z}^{x} z_{x}^{y} x_{y}^{z} = 1, x_{z}^{u} z_{x}^{y} + y_{z}^{u} z_{y}^{x} = 1,

where $y_{z}^{u}$ denotes the derivative of $y$ , when expressed as a function of $z$ and $u$ , with respect to $z$ . (Math. Trip. 1897.)

32. Find $A$ , $B$ , $C$ , $λ$ so that the first four derivatives of

\int_{a}^{a + x} f (t) d t - x [A f (a) + B f (a + λ x) + C f (a + x)]

vanish when $x = 0$ ; and $A$ , $B$ , $C$ , $D$ , $λ$ , $μ$ so that the first six derivatives of

\int_{a}^{a + x} f (t) d t - x [A f (a) + B f (a + λ x) + C f (a + μx) + D f (a + x)]

vanish when $x = 0$ .

33. If $a > 0$ , $a c - b^{2} > 0$ , and $x_{1} > x_{0}$ , then

\int_{x_{0}}^{x_{1}} \frac{d x}{a x ^{2} + 2 b x + c} = \frac{1}{a c - b ^{2}} arc tan {\frac{( x _{1} - x _{0} ) a c - b ^{2}}{a x _{1} x _{0} + b ( x _{1} + x _{0} ) + c}},

the inverse tangent lying between $0$ and $π$ .²

34. Evaluate the integral $\int_{- 1}^{1} \frac{sin α d x}{1 - 2 x cos α + x ^{2}}$ . For what values of $α$ is the integral a discontinuous function of $α$ ? (Math. Trip. 1904.)

[The value of the integral is $\frac{1}{2} π$ if $2 nπ < α < (2 n + 1) π$ , and $- \frac{1}{2} π$ if $(2 n - 1) π < α < 2 nπ$ , $n$ being any integer; and $0$ if $α$ is a multiple of $π$ .]

35. If $a x^{2} + 2 b x + c > 0$ when $x_{0} ≦ x ≦ x_{1}$ , $f (x) = a x^{2} + 2 b x + c$ , and

y = f (x), y_{0} = f (x_{0}), y_{1} = f (x_{1}), X = (x_{1} - x_{0}) / (y_{1} + y_{0}),

then

\int_{x_{0}}^{x_{1}} \frac{d x}{y} = \frac{1}{a} lo g \frac{1 + X a}{1 - X a}, \frac{- 2}{- a} arc tan {X - a},

according as $a$ is positive or negative. In the latter case the inverse tangent lies between $0$ and $\frac{1}{2} π$ . [It will be found that the substitution $t = \frac{x - x _{0}}{y + y _{0}}$ reduces the integral to the form $2 \int_{0}^{X} \frac{d t}{1 - a t ^{2}}$ .]

36. Prove that

\int_{0}^{a} \frac{d x}{x + a ^{2} - x ^{2}} = \frac{1}{4} π .

(Math. Trip. 1913.)

37. If $a > 1$ then

\int_{- 1}^{1} \frac{1 - x ^{2}}{a - x} d x = π {a - a^{2} - 1} .

38. If $p > 1$ , $0 < q < 1$ , then

\int_{0}^{1} \frac{d x}{{ 1 + ( p ^{2} - 1 ) x } { 1 - ( 1 - q ^{2} ) x }} = \frac{2 ω}{( p + q ) sin ω},

where $ω$ is the positive acute angle whose cosine is $(1 + pq) / (p + q)$ .

39. If $a > b > 0$ , then

\int_{0}^{2 π} \frac{sin ^{2} θ d θ}{a - b cos θ} = \frac{2 π}{b ^{2}} {a - a^{2} - b^{2}} .

(Math. Trip. 1904.)

40. Prove that if $a > b^{2} + c^{2}$ then

\int_{0}^{π} \frac{d θ}{a + b cos θ + c sin θ} = \frac{2}{a ^{2} - b ^{2} - c ^{2}} arc tan {\frac{a ^{2} - b ^{2} - c ^{2}}{c}},

the inverse tangent lying between $0$ and $π$ .

41. If $f (x)$ is continuous and never negative, and $\int_{a}^{b} f (x) d x = 0$ , then $f (x) = 0$ for all values of $x$ between $a$ and $b$ . [If $f (x)$ were equal to a positive number $k$ when $x = ξ$ , say, then we could, in virtue of the continuity of $f (x)$ , find an interval $(ξ - δ, ξ + δ)$ throughout which $f (x) > \frac{1}{2} k$ ; and then the value of the integral would be greater than $δ k$ .]

42. Schwarz’s inequality for integrals. Prove that

(\int_{a}^{b} ϕ ψ d x)^{2} ≦ \int_{a}^{b} ϕ^{2} d x \int_{a}^{b} ψ^{2} d x .

[Use the definitions of §§ 156 and 157, and the inequality

(\sum ϕ_{ν} ψ_{ν} δ_{ν})^{2} ≦ \sum ϕ_{ν}^{2} δ_{ν} \sum ψ_{ν}^{2} δ_{ν}

(Ch. I, Misc. Ex. 10).]

43. If

P_{n} (x) = \frac{1}{( β - α ) ^{n} n !} (\frac{d}{d x})^{n} {(x - α) (β - x)}^{n},

then $P_{n} (x)$ is a polynomial of degree $n$ , which possesses the property that

\int_{α}^{β} P_{n} (x) θ (x) d x = 0

if $θ (x)$ is any polynomial of degree less than $n$ . [Integrate by parts $m + 1$ times, where $m$ is the degree of $θ (x)$ , and observe that $θ^{(m + 1)} (x) = 0$ .]

44. Prove that

\int_{α}^{β} P_{m} (x) P_{n} (x) d x = 0

if $m \neq = n$ , but that if $m = n$ then the value of the integral is $(β - α) / (2 n + 1)$ .

45. If $Q_{n} (x)$ is a polynomial of degree $n$ , which possesses the property that

\int_{α}^{β} Q_{n} (x) θ (x) d x = 0

if $θ (x)$ is any polynomial of degree less than $n$ , then $Q_{n} (x)$ is a constant multiple of $P_{n} (x)$ .

[We can choose $κ$ so that $Q_{n} - κ P_{n}$ is of degree $n - 1$ : then

\int_{α}^{β} Q_{n} (Q_{n} - κ P_{n}) d x = 0, \int_{α}^{β} P_{n} (Q_{n} - κ P_{n}) d x = 0,

and so

\int_{α}^{β} (Q_{n} - κ P_{n})^{2} d x = 0.

Now apply Ex. 41.]

46. Approximate Values of definite integrals. Show that the error in taking $\frac{1}{2} (b - a) {ϕ (a) + ϕ (b)}$ as the value of the integral $\int_{a}^{b} ϕ (x) d x$ is less than $\frac{1}{12} M (b - a)^{3}$ , where $M$ is the maximum of $∣ ϕ^{''} (x) ∣$ in the interval $(a, b)$ ; and that the error in taking $(b - a) ϕ {\frac{1}{2} (a + b)}$ is less than $\frac{1}{24} M (b - a)^{3}$ . [Write $f^{'} (x) = ϕ (x)$ in Exs. 4 and 5.] Show that the error in taking

\frac{1}{6} (b - a) [ϕ (a) + ϕ (b) + 4 ϕ {\frac{1}{2} (a + b)}]

as the value is less than $\frac{1}{2880} M (b - a)^{5}$ , where $M$ is the maximum of $ϕ^{(4)} (x)$ . [Use Ex. 6. This rule, which gives a very good approximation, is known as Simpson’s Rule. It amounts to taking one-third of the first approximation given above and two-thirds of the second.]

Show that the approximation assigned by Simpson’s Rule is the area bounded by the lines $x = a$ , $x = b$ , $y = 0$ , and a parabola with its axis parallel to $O Y$ and passing through the three points on the curve $y = ϕ (x)$ whose abscissae are $a$ , $\frac{1}{2} (a + b)$ , $b$ .

It should be observed that if $ϕ (x)$ is any cubic polynomial then $ϕ^{(4)} (x) = 0$ , and Simpson’s Rule is exact. That is to say, given three points whose abscissae are $a$ , $\frac{1}{2} (a + b)$ , $b$ , we can draw through them an infinity of curves of the type $y = α + β x + γ x^{2} + δ x^{3}$ ; and all such curves give the same area. For one curve $δ = 0$ , and this curve is a parabola.

47. If $ϕ (x)$ is a polynomial of the fifth degree, then

\int_{0}^{1} ϕ (x) d x = \frac{1}{18} {5 ϕ (α) + 8 ϕ (\frac{1}{2}) + 5 ϕ (β)},

$α$ and $β$ being the roots of the equation $x^{2} - x + \frac{1}{10} = 0$ . (Math. Trip. 1909.)

48. Apply Simpson’s Rule to the calculation of $π$ from the formula $\frac{1}{4} π = \int_{0}^{1} \frac{d x}{1 + x ^{2}}$ . [The result is $.7833 \dots$ . If we divide the integral into two, from $0$ to $\frac{1}{2}$ and $\frac{1}{2}$ to $1$ , and apply Simpson’s Rule to the two integrals separately, we obtain $.7853916 \dots$ . The correct value is $.7853981 \dots$ .]

49. Show that

8.9 < \int_{3}^{5} 4 + x^{2} d x < 9.

(Math. Trip. 1903.)

50. Calculate the integrals

\int_{0}^{1} \frac{d x}{1 + x}, \int_{0}^{1} \frac{d x}{1 + x ^{4}}, \int_{0}^{π} sin x d x, \int_{0}^{π} \frac{sin x}{x} d x,

to two places of decimals. [In the last integral the subject of integration is not defined when $x = 0$ : but if we assign to it, when $x = 0$ , the value $1$ , it becomes continuous throughout the range of integration.]

Footnotes

In this and the following examples the reader is to assume the continuity of all the derivatives which occur. ↩
In connection with Exs. 33–35, 38, and 40 see a paper by Dr Bromwich in vol. xxxv of the Messenger of Mathematics. ↩