## 26. C. Explicit Algebraical Functions.

The next important class of functions is that of explicit algebraical functions. These are functions which can be generated from $$x$$ by a finite number of operations such as those used in generating rational functions, together with a finite number of operations of root extraction. Thus $\frac{\sqrt{1 + x} – \sqrt[3]{1 – x}} {\sqrt{1 + x} + \sqrt[3]{1 – x}},\quad \sqrt{x} + \sqrt{x +\sqrt{x}},\quad \left(\frac{x^{2} + x + \sqrt{3}}{x\sqrt[3]{2} – \pi}\right)^{\frac{2}{3}}$ are explicit algebraical functions, and so is $$x^{m/n}$$$$\sqrt[n]{x^{m}}$$), where $$m$$ and $$n$$ are any integers.

It should be noticed that there is an ambiguity of notation involved in such an equation as $$y = \sqrt{x}$$. We have, up to the present, regarded () $$\sqrt{2}$$ as denoting the positive square root of $$2$$, and it would be natural to denote by $$\sqrt{x}$$, where $$x$$ is any positive number, the positive square root of $$x$$, in which case $$y = \sqrt{x}$$ would be a one-valued function of $$x$$. It is however often more convenient to regard $$\sqrt{x}$$ as standing for the two-valued function whose two values are the positive and negative square roots of $$x$$.

The reader will observe that, when this course is adopted, the function $$\sqrt{x}$$ differs fundamentally from rational functions in two respects. In the first place a rational function is always defined for all values of $$x$$ with a certain number of isolated exceptions. But $$\sqrt{x}$$ is undefined for a whole range of values of $$x$$ ( all negative values). Secondly the function, when $$x$$ has a value for which it is defined, has generally two values of opposite signs.

The function $$\sqrt[3]{x}$$, on the other hand, is one-valued and defined for all values of $$x$$.

Example XIII

1.  $$\sqrt{(x – a)(b – x)}$$, where $$a < b$$, is defined only for $$a \leq x \leq b$$. If $$a < x < b$$ it has two values: if $$x = a$$ or $$b$$ only one, viz. $$0$$.

2. Consider similarly $\begin{gathered} \sqrt{(x – a)(x – b)(x – c)} \quad (a < b < c), \\ \sqrt{x(x^{2} – a^{2})},\quad \sqrt[3]{(x – a)^{2}(b – x)}\quad (a < b), \\ \frac{\sqrt{1 + x} – \sqrt{1 – x}} {\sqrt{1 + x} + \sqrt{1 – x}},\quad \sqrt{x + \sqrt{x}}.\end{gathered}$

3. Trace the curves $$y^{2} = x$$, $$y^{3} = x$$, $$y^{2} = x^{3}$$.

4. Draw the graphs of the functions $y = \sqrt{a^{2} – x^{2}},\quad y = b\sqrt{1 – (x^{2}/a^{2})}.$

## 27. D. Implicit Algebraical Functions.

It is easy to verify that if $y = \frac{\sqrt{1 + x} – \sqrt[3]{1 – x}} {\sqrt{1 + x} + \sqrt[3]{1 – x}},$ then $\left(\frac{1 + y}{1 – y}\right)^{6} = \frac{(1 + x)^{3}}{(1 – x)^{2}};$ or if $y = \sqrt{x} + \sqrt{x + \sqrt{x}},$ then $y^{4} – (4y^{2} + 4y + 1)x = 0.$ Each of these equations may be expressed in the form $\begin{equation*} y^{m} + R_{1}y^{m-1} + \dots + R_{m} = 0, \tag {1}\end{equation*}$ where $$R_{1}$$, $$R_{2}$$, …, $$R_{m}$$ are rational functions of $$x$$: and the reader will easily verify that, if $$y$$ is any one of the functions considered in the last set of examples, $$y$$ satisfies an equation of this form. It is naturally suggested that the same is true of any explicit algebraic function. And this is in fact true, and indeed not difficult to prove, though we shall not delay to write out a formal proof here. An example should make clear to the reader the lines on which such a proof would proceed. Let $y = \frac{x + \sqrt{x} + \sqrt{x + \sqrt{x}} + \sqrt[3]{1 + x}} {x – \sqrt{x} + \sqrt{x + \sqrt{x}} – \sqrt[3]{1 + x}}.$ Then we have the equations $\begin{gathered} y = \frac{x + u + v + w} {x – u + v – w}, \\ u^{2} = x,\quad v^{2} = x + u,\quad w^{3} = 1 + x,\end{gathered}$ and we have only to eliminate $$u$$$$v$$$$w$$ between these equations in order to obtain an equation of the form desired.

We are therefore led to give the following definition: a function $$y = f(x)$$ will be said to be an algebraical function of $$x$$ if it is the root of an equation such as ,  the root of an equation of the $$m$$th degree in $$y$$, whose coefficients are rational functions of $$x$$. There is plainly no loss of generality in supposing the first coefficient to be unity.

This class of functions includes all the explicit algebraical functions considered in § 26. But it also includes other functions which cannot be expressed as explicit algebraical functions. For it is known that in general such an equation as  cannot be solved explicitly for $$y$$ in terms of $$x$$, when $$m$$ is greater than $$4$$, though such a solution is always possible if $$m = 1$$, $$2$$$$3$$, or $$4$$ and in special cases for higher values of $$m$$.

The definition of an algebraical function should be compared with that of an algebraical number given in the last chapter (Misc. Exs. 32).

Example XIV

1. If $$m = 1$$, $$y$$ is a rational function.

2. If $$m = 2$$, the equation is $$y^{2} + R_{1}y + R_{2} = 0$$, so that $y = \tfrac{1}{2}\{-R_{1} \pm \sqrt{R_{1}^{2} – 4R_{2}}\}.$ This function is defined for all values of $$x$$ for which $$R_{1}^{2} \geq 4R_{2}$$. It has two values if $$R_{1}^{2} > 4R_{2}$$ and one if $$R_{1}^{2} = 4R_{2}$$.

If $$m = 3$$ or $$4$$, we can use the methods explained in treatises on Algebra for the solution of cubic and biquadratic equations. But as a rule the process is complicated and the results inconvenient in form, and we can generally study the properties of the function better by means of the original equation.

3. Consider the functions defined by the equations $y^{2} – 2y – x^{2} = 0,\quad y^{2} – 2y + x^{2} = 0,\quad y^{4} – 2y^{2} + x^{2} = 0,$ in each case obtaining $$y$$ as an explicit function of $$x$$, and stating for what values of $$x$$ it is defined.

4. Find algebraical equations, with coefficients rational in $$x$$, satisfied by each of the functions $\sqrt{x} + \sqrt{1/x},\quad \sqrt[3]{x} + \sqrt[3]{1/x},\quad \sqrt{x + \sqrt{x}},\quad \sqrt{x + \sqrt{x + \sqrt{x}}}.$

5. Consider the equation $$y^{4} = x^{2}$$.

[Here $$y^{2} = \pm x$$. If $$x$$ is positive, $$y = \sqrt{x}$$: if negative, $$y = \sqrt{-x}$$. Thus the function has two values for all values of $$x$$ save $$x = 0$$.]

6. An algebraical function of an algebraical function of $$x$$ is itself an algebraical function of $$x$$.

[For we have \begin{aligned} {4} y^{m} &+ R_{1}(z)y^{m-1} &&+ \dots &&+ R_{m}(z) &&= 0, \end{aligned}where \begin{aligned} z^{n} &+ S_{1}(x)z^{n-1} &&+ \dots &&+ S_{n}(x) &&= 0. \end{aligned} Eliminating $z$ we find an equation of the form \begin{aligned} y^{p} &+ T_{1}(x)y^{p-1} &&+ \dots &&+ T_{p}(x) &&= 0.\end{aligned} Here all the capital letters denote rational functions.]

7. An example should perhaps be given of an algebraical function which cannot be expressed in an explicit algebraical form. Such an example is the function $$y$$ defined by the equation $y^{5} – y – x = 0.$ But the proof that we cannot find an explicit algebraical expression for $$y$$ in terms of $$x$$ is difficult, and cannot be attempted here.