## 108. Implicit functions.

We have already, in Ch. II, met with the idea of an implicit function. Thus, if $$x$$ and $$y$$ are connected by the relation $\begin{equation*} y^{5} – xy – y – x = 0, \tag{1} \end{equation*}$ then $$y$$ is an ‘implicit function’ of $$x$$.

But it is far from obvious that such an equation as this does really define a function $$y$$ of $$x$$, or several such functions. In Ch. II we were content to take this for granted. We are now in a position to consider whether the assumption we made then was justified.

We shall find the following terminology useful. Suppose that it is possible to surround a point $$(a, b)$$, as in § 107, with a square throughout which a certain condition is satisfied. We shall call such a square a neighbourhood of $$(a, b)$$, and say that the condition in question is satisfied in the neighbourhood of $$(a, b)$$, or near $$(a, b)$$, meaning by this simply that it is possible to find some square throughout which the condition is satisfied. It is obvious that similar language may be used when we are dealing with a single variable, the square being replaced by an interval on a line.

Theorem. If $$f(x, y)$$ is a continuous function of $$x$$ and $$y$$ in the neighbourhood of $$(a, b)$$,

(ii) $$f(a, b) = 0$$,

(iii) $$f(x, y)$$ is, for all values of $$x$$ in the neighbourhood of $$a$$, a steadily increasing function of $$y$$, in the stricter sense of § 95,

then (1) there is a unique function $$y = \phi(x)$$ which, when substituted in the equation $$f(x, y) = 0$$, satisfies it identically for all values of $$x$$ in the neighbourhood of $$a$$,

(2) $$\phi(x)$$ is continuous for all values of $$x$$ in the neighbourhood of $$a$$.

In the figure the square represents a ‘neighbourhood’ of $$(a, b)$$ throughout which the conditions (i) and (iii) are satisfied, and $$P$$ the point $$(a, b)$$. If we take $$Q$$ and $$R$$ as in the figure, it follows from (iii) that $$f(x, y)$$ is positive at $$Q$$ and negative at $$R$$. This being so, and $$f(x, y)$$ being continuous at $$Q$$ and at $$R$$, we can draw lines $$QQ’$$ and $$RR’$$ parallel to $$OX$$, so that $$R’Q’$$ is parallel to $$OY$$ and $$f(x, y)$$ is positive at all points of $$QQ’$$ and negative at all points of $$RR’$$. In particular $$f(x, y)$$ is positive at $$Q’$$ and negative at $$R’$$, and therefore, in virtue of (iii) and § 100, vanishes once and only once at a point $$P’$$ on $$R’Q’$$. The same construction gives us a unique point at which $$f(x, y) = 0$$ on each ordinaten between $$RQ$$ and $$R’Q’$$. It is obvious, moreover, that the same construction can be carried out to the left of $$RQ$$. The aggregate of points such as $$P’$$ gives us the graph of the required function $$y = \phi(x)$$.

It remains to prove that $$\phi(x)$$ is continuous. This is most simply effected by using the idea of the ‘limits of indetermination’ of $$\phi(x)$$ as $$x \to a$$ (§ 96). Suppose that $$x \to a$$, and let $$\lambda$$ and $$\Lambda$$ be the limits of indetermination of $$\phi(x)$$ as $$x \to a$$. It is evident that the points $$(a, \lambda)$$ and $$(a, \Lambda)$$ lie on $$QR$$. Moreover, we can find a sequence of values of $$x$$ such that $$\phi(x) \to \lambda$$ when $$x \to a$$ through the values of the sequence; and since $$f\{x, \phi(x)\} = 0$$, and $$f(x, y)$$ is a continuous function of $$x$$ and $$y$$, we have $f(a, \lambda) = 0.$ Hence $$\lambda = b$$; and similarly $$\Lambda = b$$. Thus $$\phi(x)$$ tends to the limit $$b$$ as $$x \to a$$, and so $$\phi(x)$$ is continuous for $$x = a$$. It is evident that we can show in exactly the same way that $$\phi(x)$$ is continuous for any value of $$x$$ in the neighbourhood of $$a$$.

It is clear that the truth of the theorem would not be affected if we were to change ‘increasing’ to ‘decreasing’ in condition (iii).

As an example, let us consider the equation (1), taking $$a = 0$$, $$b = 0$$. It is evident that the conditions (i) and (ii) are satisfied. Moreover $f(x, y) – f(x, y’) = (y – y’) (y^{4} + y^{3}y’ + y^{2}y’^{2} + yy’^{3} + y’^{4} – x – 1)$ has, when $$x$$, $$y$$, and $$y’$$ are sufficiently small, the sign opposite to that of $$y – y’$$. Hence condition (iii) (with ‘decreasing’ for ‘increasing’) is satisfied. It follows that there is one and only one continuous function $$y$$ which satisfies the equation (1) identically and vanishes with $$x$$.

The same conclusion would follow if the equation were $y^{2} – xy – y – x = 0.$ The function in question is in this case $y = \tfrac{1}{2}\{1 + x – \sqrt{1 + 6 x + x^{2}}\},$ where the square root is positive. The second root, in which the sign of the square root is changed, does not satisfy the condition of vanishing with $$x$$.

There is one point in the proof which the reader should be careful to observe. We supposed that the hypotheses of the theorem were satisfied ‘in the neighbourhood of $$(a, b)$$’, that is to say throughout a certain square $$\xi – \delta \leq x \leq \xi + \delta$$, $$\eta – \delta \leq y \leq \eta + \delta$$. The conclusion holds ‘in the neighbourhood of $$x = a$$’, that is to say throughout a certain interval $$\xi – \delta_{1} \leq x \leq \xi + \delta_{1}$$. There is nothing to show that the $$\delta_{1}$$ of the conclusion is the $$\delta$$ of the hypotheses, and indeed this is generally untrue.

## 109. Inverse Functions.

Suppose in particular that $$f(x, y)$$ is of the form $$F(y) – x$$. We then obtain the following theorem.

If $$F(y)$$ is a function of $$y$$, continuous and steadily increasing $$or decreasing$$, in the stricter sense of § 95, in the neighbourhood of $$y = b$$, and $$F(b) = a$$, then there is a unique continuous function $$y = \phi(x)$$ which is equal to $$b$$ when $$x = a$$ and satisfies the equation $$F(y) = x$$ identically in the neighbourhood of $$x = a$$.

The function thus defined is called the inverse function of $$F(y)$$.

Suppose for example that $$y^{3} = x$$, $$a = 0$$, $$b = 0$$. Then all the conditions of the theorem are satisfied. The inverse function is $$x = \sqrt[3]{y}$$.

If we had supposed that $$y^{2} = x$$ then the conditions of the theorem would not have been satisfied, for $$y^{2}$$ is not a steadily increasing function of $$y$$ in any interval which includes $$y = 0$$: it decreases when $$y$$ is negative and increases when $$y$$ is positive. And in this case the conclusion of the theorem does not hold, for $$y^{2} = x$$ defines two functions of $$x$$, viz. $$y = \sqrt{x}$$ and $$y = -\sqrt{x}$$, both of which vanish when $$x = 0$$, and each of which is defined only for positive values of $$x$$, so that the equation has sometimes two solutions and sometimes none. The reader should consider the more general equations $y^{2n} = x, \quad y^{2n+1} = x,$ in the same way. Another interesting example is given by the equation $y^{5} – y – x = 0,$ already considered in Ex. XIV. 7.

Similarly the equation $\sin y = x$ has just one solution which vanishes with $$x$$, viz. the value of $$\arcsin x$$ which vanishes with $$x$$. There are of course an infinity of solutions, given by the other values of $$\arcsin x$$ (cf. Ex. XV. 10), which do not satisfy this condition.

So far we have considered only what happens in the neighbourhood of a particular value of $$x$$. Let us suppose now that $$F(y)$$ is positive and steadily increasing (or decreasing) throughout an interval $${[a, b]}$$. Given any point $$\xi$$ of $${[a, b]}$$, we can determine an interval $$i$$ including $$\xi$$, and a unique and continuous inverse function $$\phi_{i} (x)$$ defined throughout $$i$$.

From the set $$I$$ of intervals $$i$$ we can, in virtue of the Heine-Borel Theorem, pick out a finite sub-set covering up the whole interval $${[a, b]}$$; and it is plain that the finite set of functions $$\phi_{i} (x)$$, corresponding to the sub-set of intervals $$i$$ thus selected, define together a unique inverse function $$\phi(x)$$ continuous throughout $${[a, b]}$$.

We thus obtain the theorem:

if $$x = F(y)$$, where $$F(y)$$ is continuous and increases steadily and strictly from $$A$$ to $$B$$ as $$x$$ increases from $$a$$ to $$b$$, then there is a unique inverse function $$y = \phi(x)$$ which is continuous and increases steadily and strictly from $$a$$ to $$b$$ as $$x$$ increases from $$A$$ to $$B$$.

It is worth while to show how this theorem can be obtained directly without the help of the more difficult theorem of § 108. Suppose that $$A < \xi < B$$, and consider the class of values of $$y$$ such that (i) $$a <y < b$$ and (ii) $$F(y) \leq \xi$$. This class has an upper bound $$\eta$$, and plainly $$F(\eta) \leq \xi$$. If $$F(\eta)$$ were less than $$\xi$$, we could find a value of $$y$$ such that $$y > \eta$$ and $$F(y) < \xi$$, and $$\eta$$ would not be the upper bound of the class considered. Hence $$F(\eta) = \xi$$. The equation $$F(y) = \xi$$ has therefore a unique solution $$y = \eta = \phi(\xi)$$, say; and plainly $$\eta$$ increases steadily and continuously with $$\xi$$, which proves the theorem.