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 $-1\le x\le 1$, 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 $$\tan x,\cot x,\sec x,\csc x,{\tan }^{2}x,{\cot }^{2}x,{\sec }^{2}x,{\csc }^{2}x.$$

12. Draw the graphs of $\arctan x$, $\mathrm{arccot}x$, $\mathrm{arcsec}x$, $\mathrm{arccsc}x$. 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 $$\begin{array}{}\text{(1)}& y^m+R_1{y}^{m-1}+\cdots +R_m=0\end{array}$$ where $R_1$, … are rational functions of $x$. This may be put in the form $$P_0{y}^m+P_1{y}^{m-1}+\cdots +P_m=0,$$ where $P_0$, $P_1$, … are polynomials in $x$. Arguing as above, we see that $$P_0{k}^m+P_1{k}^{m-1}+\cdots +P_m=0$$ 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 $k$, $k^{\prime }$, …; 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 $$\begin{array}{rlrlrl}3x& =1,2,3,4,5,6,7,8,9,& 10,& 11,& 12,& 13,\dots ,\end{array}$$ then$$\begin{array}{rlrlrl}y& =1,2,3,2,5,3,7,2,3,& 5,& 11,& 3,& 13,\dots .\end{array}$$ 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 $$\dots (-1,-1),(0,0),(1,1),(2,2),\dots$$ 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.