28-29. Transcendental functions

1. Draw the graphs of \(\cos x\), \(\sin x\), and \(a\cos x + b\sin x\).

[Since \(a\cos x + b\sin x = \beta\cos(x – \alpha)\), where \(\beta = \sqrt{a^{2} + b^{2}}\), and \(\alpha\) is an angle whose cosine and sine are \(a/\sqrt{a^{2} + b^{2}}\) and \(b/\sqrt{a^{2} + b^{2}}\), the graphs of these three functions are similar in character.]

2. Draw the graphs of \(\cos^{2} x\), \(\sin^{2} x\), \(a\cos^{2} x + b\sin^{2} x\).

3. Suppose the graphs of \(f(x)\) and \(F(x)\) drawn. Then the graph of \[f(x)\cos^{2} x + F(x)\sin^{2} x\] is a wavy curve which oscillates between the curves \(y = f(x)\), \(y = F(x)\). Draw the graph when \(f(x) = x\), \(F(x) = x^{2}\).

4. Show that the graph of \(\cos px + \cos qx\) lies between those of \(2\cos\frac{1}{2}(p – q)x\) and \(-2\cos\frac{1}{2}(p + q)x\), touching each in turn. Sketch the graph when \((p – q)/(p + q)\) is small.

5. Draw the graphs of \(x + \sin x\), \((1/x) + \sin x\), \(x\sin x\), \((\sin x)/x\).

6. Draw the graph of \(\sin(1/x)\).

[If \(y = \sin(1/x)\), then \(y = 0\) when \(x = 1/m\pi\), where \(m\) is any integer. Similarly \(y = 1\) when \(x = 1/(2m + \frac{1}{2})\pi\) and \(y = -1\) when \(x = 1/(2m – \frac{1}{2})\pi\). The curve is entirely comprised between the lines \(y = -1\) and \(y = 1\) (fig. 13). It oscillates up and down, the rapidity of the oscillations becoming greater and greater as \(x\) approaches \(0\). For \(x = 0\) the function is undefined. When \(x\) is large \(y\) is small.² The negative half of the curve is similar in character to the positive half.]

7. Draw the graph of \(x\sin(1/x)\).

[This curve is comprised between the lines \(y = -x\) and \(y = x\) just as the last curve is comprised between the lines \(y = -1\) and \(y = 1\) (fig. 14).]

8. Draw the graphs of \(x^{2}\sin(1/x)\), \((1/x)\sin(1/x)\), \(\sin^{2}(1/x)\), \(\{x\sin(1/x)\}^{2}\), \(a\cos^{2}(1/x) + b\sin^{2}(1/x)\), \(\sin x + \sin(1/x)\), \(\sin x\sin(1/x)\).

9. Draw the graphs of \(\cos x^{2}\), \(\sin x^{2}\), \(a\cos x^{2} + b\sin x^{2}\).

10. Draw the graphs of \(\arccos x\) and \(\arcsin x\).

[If \(y = \arccos x\), \(x = \cos y\). This enables us to draw the graph of \(x\), considered as a function of \(y\), and the same curve shows \(y\) as a function of \(x\). It is clear that \(y\) is only defined for \(-1 \leq x \leq 1\), and is infinitely many-valued for these values of \(x\). As the reader no doubt remembers, there is, when \(-1 < x < 1\), a value of \(y\) between \(0\) and \(\pi\), say \(\alpha\), and the other values of \(y\) are given by the formula \(2n\pi \pm \alpha\), where \(n\) is any integer, positive or negative.]

11. Draw the graphs of \[\tan x,\quad \cot x,\quad \sec x,\quad \csc x,\quad \tan^{2} x,\quad \cot^{2} x,\quad \sec^{2} x,\quad \csc^{2} x.\]

12. Draw the graphs of \(\arctan x\), \(\operatorname{arccot} x\), \(\operatorname{arcsec} x\), \(\operatorname{arccsc} x\). Give formulae (as in Ex. 10) expressing all the values of each of these functions in terms of any particular value.

13. Draw the graphs of \(\tan(1/x)\), \(\cot(1/x)\), \(\sec(1/x)\), \(\csc(1/x)\).

14. Show that \(\cos x\) and \(\sin x\) are not rational functions of \(x\).

[A function is said to be periodic, with period \(a\), if \(f(x) = f(x + a)\) for all values of \(x\) for which \(f(x)\) is defined. Thus \(\cos x\) and \(\sin x\) have the period \(2\pi\). It is easy to see that no periodic function can be a rational function, unless it is a constant. For suppose that \[f(x) = P(x)/Q(x),\] where \(P\) and \(Q\) are polynomials, and that \(f(x) = f(x + a)\), each of these equations holding for all values of \(x\). Let \(f(0) = k\). Then the equation \(P(x) – kQ(x) = 0\) is satisfied by an infinite number of values of \(x\), viz. \(x = 0\), \(a\), \(2a\), etc., and therefore for all values of \(x\). Thus \(f(x) = k\) for all values of \(x\), \(f(x)\) is a constant.]

15. Show, more generally, that no function with a period can be an algebraical function of \(x\).

[Let the equation which defines the algebraical function be \[\begin{equation*}y^{m} + R_{1}y^{m-1} + \dots + R_{m} = 0 \tag{1}\end{equation*}\] where \(R_{1}\), … are rational functions of \(x\). This may be put in the form \[P_{0}y^{m} + P_{1}y^{m-1} + \dots + P_{m} = 0,\] where \(P_{0}\), \(P_{1}\), … are polynomials in \(x\). Arguing as above, we see that \[P_{0}k^{m} + P_{1}k^{m-1} + \dots + P_{m} = 0\] for all values of \(x\). Hence \(y = k\) satisfies the equation for all values of \(x\), and one set of values of our algebraical function reduces to a constant.

Now divide (1) by \(y – k\) and repeat the argument. Our final conclusion is that our algebraical function has, for any value of \(x\), the same set of values \(k\), \(k’\), …; it is composed of a certain number of constants.]

16. The inverse sine and inverse cosine are not rational or algebraical functions. [This follows from the fact that, for any value of \(x\) between \(-1\) and \(+1\), \(\arcsin x\) and \(\arccos x\) have infinitely many values.]

1. Let \(y = [x]\), where \([x]\) denotes the greatest integer not greater than \(x\). The graph is shown in Fig. 15a. The left-hand end points of the thick lines, but not the right-hand ones, belong to the graph.

2. \(y = x – [x]\). (Fig. 15b.)

3. \(y = \sqrt{x – [x]}\). (Fig. 15c.)

4. \(y = [x] + \sqrt{x – [x]}\). (Fig. 15d.)

5. \(y = (x – [x])^{2}\), \([x] + (x – [x])^{2}\).

6. \(y = [\sqrt{x}]\), \([x^{2}]\), \(\sqrt{x} – [\sqrt{x}]\), \(x^{2} – [x^{2}]\), \([1 – x^{2}]\).

7. Let \(y\) be defined as the largest prime factor of \(x\) (cf. . 6). Then \(y\) is defined only for integral values of \(x\). If \[\begin{aligned} {3} x &= 1,\ 2,\ 3,\ 4,\ 5,\ 6,\ 7,\ 8,\ 9,\ &10,&\ 11,\ &12,&\ 13,\ \dots, \\ \end{aligned}\] then\[\begin{aligned} y &= 1,\ 2,\ 3,\ 2,\ 5,\ 3,\ 7,\ 2,\ 3,\ & 5,&\ 11,\ & 3,&\ 13,\ \dots.\end{aligned}\] The graph consists of a number of isolated points.

8. Let \(y\) be the denominator of \(x\) (Exs. X. 7). In this case \(y\) is defined only for rational values of \(x\). We can mark off as many points on the graph as we please, but the result is not in any ordinary sense of the word a curve, and there are no points corresponding to any irrational values of \(x\).

Draw the straight line joining the points \((N – 1, N)\), \((N, N)\), where \(N\) is a positive integer. Show that the number of points of the locus which lie on this line is equal to the number of positive integers less than and prime to \(N\).

9. Let \(y = 0\) when \(x\) is an integer, \(y = x\) when \(x\) is not an integer. The graph is derived from the straight line \(y = x\) by taking out the points \[\dots\ (-1, -1),\quad (0, 0),\quad (1, 1),\quad (2, 2),\ \dots\] and adding the points \((-1, 0)\), \((0, 0)\), \((1, 0)\), … on the axis of \(x\).

The reader may possibly regard this as an unreasonable function. Why, he may ask, if \(y\) is equal to \(x\) for all values of \(x\) save integral values, should it not be equal to \(x\) for integral values too? The answer is simply, why should it? The function \(y\) does in point of fact answer to the definition of a function: there is a relation between \(x\) and \(y\) such that when \(x\) is known \(y\) is known. We are perfectly at liberty to take this relation to be what we please, however arbitrary and apparently futile. This function \(y\) is, of course, a quite different function from that one which is always equal to \(x\), whatever value, integral or otherwise, \(x\) may have.

10. Let \(y = 1\) when \(x\) is rational, but \(y = 0\) when \(x\) is irrational. The graph consists of two series of points arranged upon the lines \(y = 1\) and \(y = 0\). To the eye it is not distinguishable from two continuous straight lines, but in reality an infinite number of points are missing from each line.

11. Let \(y = x\) when \(x\) is irrational and \(y = \sqrt{(1 + p^{2})/(1 + q^{2})}\) when \(x\) is a rational fraction \(p/q\).

The irrational values of \(x\) contribute to the graph a curve in reality discontinuous, but apparently not to be distinguished from the straight line \(y = x\).

Now consider the rational values of \(x\). First let \(x\) be positive. Then \(\sqrt{(1 + p^{2})/(1 + q^{2})}\) cannot be equal to \(p/q\) unless \(p = q\), \(x = 1\). Thus all the points which correspond to rational values of \(x\) lie off the line, except the one point \((1, 1)\). Again, if \(p < q\), \(\sqrt{(1 + p^{2})/(1 + q^{2})} > p/q\); if \(p > q\), \(\sqrt{(1 + p^{2})/(1 + q^{2})} < p/q\). Thus the points lie above the line \(y = x\) if \(0 < x < 1\), below if \(x > 1\). If \(p\) and \(q\) are large, \(\sqrt{(1 + p^{2})/(1 + q^{2})}\) is nearly equal to \(p/q\). Near any value of \(x\) we can find any number of rational fractions with large numerators and denominators. Hence the graph contains a large number of points which crowd round the line \(y = x\). Its general appearance (for positive values of \(x\)) is that of a line surrounded by a swarm of isolated points which gets denser and denser as the points approach the line.

The part of the graph which corresponds to negative values of \(x\) consists of the rest of the discontinuous line together with the reflections of all these isolated points in the axis of \(y\). Thus to the left of the axis of \(y\) the swarm of points is not round \(y = x\) but round \(y = -x\), which is not itself part of the graph. See Fig. 16.