62. Oscillating Functions

Consider the behaviour as $n$ tends to $\mathrm{\infty }$ of the following functions:

1. $(-1{)}^n$, $5+3(-1{)}^n$, $(1,000,000/n)+(-1{)}^n$, $1,000,000(-1{)}^n+(1/n)$.

2. $(-1{)}^{n}n$, $1,000,000+(-1{)}^{n}n$.

3. $1,000,000–n$, $(-1{)}^n(1,000,000–n)$.

4. $n\{1+(-1{)}^n\}$. In this case the values of $\varphi (n)$ are $$0,4,0,8,0,12,0,16,\dots .$$ The odd terms are all zero and the even terms tend to $+\mathrm{\infty }$: $\varphi (n)$ oscillates infinitely.

5. $n^2+(-1{)}^{n}2n$. The second term oscillates infinitely, but the first is very much larger than the second when $n$ is large. In fact $\varphi (n)\ge n^2–2n$ and $n^2–2n=(n–1{)}^2–1$ is greater than any assigned value $\mathrm{\Delta }$ if $n>1+\sqrt{\mathrm{\Delta }+1}$. Thus $\varphi (n)\to +\mathrm{\infty }$. It should be observed that in this case $\varphi (2k+1)$ is always less than $\varphi (2k)$, so that the function progresses to infinity by a continual series of steps forwards and backwards. It does not however ‘oscillate’ according to our definition of the term.

6. $n^2\{1+(-1{)}^n\}$, $(-1{)}^n{n}^2+n$, $n^3+(-1{)}^n{n}^2$.

7. $\sin n\theta \pi$. We have already seen (Exs. XXIII. 9) that $\varphi (n)$ oscillates finitely when $\theta$ is rational, unless $\theta$ is an integer, when $\varphi (n)=0$, $\varphi (n)\to 0$.

The case in which $\theta$ is irrational is a little more difficult. But it is not difficult to see that $\varphi (n)$ still oscillates finitely. We can without loss of generality suppose $0<\theta <1$. In the first place $|\varphi (n)|<1$. Hence $\varphi (n)$ must oscillate finitely or tend to a limit. We shall consider whether the second alternative is really possible. Let us suppose that $$lim\sin n\theta \pi =l.$$ Then, however small $ϵ$ may be, we can choose $n_0$ so that $\sin n\theta \pi$ lies between $l-ϵ$ and $l+ϵ$ for all values of $n$ greater than or equal to $n_0$. Hence $\sin (n+1)\theta \pi –\sin n\theta \pi$ is numerically less than $2ϵ$ for all such values of $n$, and so $|\sin \frac{1}{2}\theta \pi \cos (n+\frac{1}{2})\theta \pi |<ϵ$.

Hence $$\cos (n+\frac{1}{2})\theta \pi =\cos n\theta \pi \cos \frac{1}{2}\theta \pi –\sin n\theta \pi \sin \frac{1}{2}\theta \pi$$ must be numerically less than $ϵ/|\sin \frac{1}{2}\theta \pi |$. Similarly $$\cos (n–\frac{1}{2})\theta \pi =\cos n\theta \pi \cos \frac{1}{2}\theta \pi +\sin n\theta \pi \sin \frac{1}{2}\theta \pi$$ must be numerically less than $ϵ/|\sin \frac{1}{2}\theta \pi |$; and so each of $\cos n\theta \pi \cos \frac{1}{2}\theta \pi$, $\sin n\theta \pi \sin \frac{1}{2}\theta \pi$ must be numerically less than $ϵ/|\sin \frac{1}{2}\theta \pi |$. That is to say, $\cos n\theta \pi \cos \frac{1}{2}\theta \pi$ is very small if $n$ is large, and this can only be the case if $\cos n\theta \pi$ is very small. Similarly $\sin n\theta \pi$ must be very small, so that $l$ must be zero. But it is impossible that $\cos n\theta \pi$ and $\sin n\theta \pi$ can both be very small, as the sum of their squares is unity. Thus the hypothesis that $\sin n\theta \pi$ tends to a limit $l$ is impossible, and therefore $\sin n\theta \pi$ oscillates as $n$ tends to $\mathrm{\infty }$.

The reader should consider with particular care the argument ‘$\cos n\theta \pi \cos \frac{1}{2}\theta \pi$ is very small, and this can only be the case if $\cos n\theta \pi$ is very small’. Why, he may ask, should it not be the other factor $\cos \frac{1}{2}\theta \pi$ which is ‘very small’? The answer is to be found, of course, in the meaning of the phrase ‘very small’ as used in this connection. When we say ‘$\varphi (n)$ is very small’ for large values of $n$, we mean that we can choose $n_0$ so that $\varphi (n)$ is numerically smaller than any assigned number, if . Such an assertion is palpably absurd when made of a fixed number such as $\cos \frac{1}{2}\theta \pi$, which is not zero.

Prove similarly that $\cos n\theta \pi$ oscillates finitely, unless $\theta$ is an even integer.

8. $\sin n\theta \pi +(1/n)$, $\sin n\theta \pi +1$, $\sin n\theta \pi +n$, $(-1{)}^n\sin n\theta \pi$.

9. $a\cos n\theta \pi +b\sin n\theta \pi$, ${\sin }^{2}n\theta \pi$, $a{\cos }^{2}n\theta \pi +b{\sin }^{2}n\theta \pi$.

10. $a+bn+(-1{)}^n(c+dn)+e\cos n\theta \pi +f\sin n\theta \pi$.

11. $n\sin n\theta \pi$. If $\theta$ is integral, then $\varphi (n)=0$, $\varphi (n)\to 0$. If $\theta$ is rational but not integral, or irrational, then $\varphi (n)$ oscillates infinitely.

12. $n(a{\cos }^{2}n\theta \pi +b{\sin }^{2}n\theta \pi )$. In this case $\varphi (n)$ tends to $+\mathrm{\infty }$ if $a$ and $b$ are both positive, but to $-\mathrm{\infty }$ if both are negative. Consider the special cases in which $a=0$, $b>0$, or $a>0$, $b=0$, or $a=0$, $b=0$. If $a$ and $b$ have opposite signs $\varphi (n)$ generally oscillates infinitely. Consider any exceptional cases.

13. $\sin (n^2\theta \pi )$. If $\theta$ is integral, then $\varphi (n)\to 0$. Otherwise $\varphi (n)$ oscillates finitely, as may be shown by arguments similar to though more complex than those used in Exs. XXIII. 9 and Exs. XXIV. 7.¹

14. $\sin (n!\theta \pi )$. If $\theta$ has a rational value $p/q$, then $n!\theta$ is certainly integral for all values of $n$ greater than or equal to $q$. Hence $\varphi (n)\to 0$. The case in which $\theta$ is irrational cannot be dealt with without the aid of considerations of a much more difficult character.

15. $\cos (n!\theta \pi )$, $a{\cos }^2(n!\theta \pi )+b{\sin }^2(n!\theta \pi )$, where $\theta$ is rational.

16. $an–[bn]$, $(-1{)}^n(an–[bn])$.

17. $[\sqrt{n}]$, $(-1{)}^n[\sqrt{n}]$, $\sqrt{n}–[\sqrt{n}]$.

18. The smallest prime factor of $n$. When $n$ is a prime, $\varphi (n)=n$. When $n$ is even, $\varphi (n)=2$. Thus $\varphi (n)$ oscillates infinitely.

19. The largest prime factor of $n$.

20. The number of days in the year $n$ A.D.

1. If $\varphi (n)\to +\mathrm{\infty }$ and $\psi (n)\ge \varphi (n)$ for all values of $n$, then $\psi (n)\to +\mathrm{\infty }$.

2. If $\varphi (n)\to 0$, and $|\psi (n)|\le |\varphi (n)|$ for all values of $n$, then $\psi (n)\to 0$.

3. If $lim|\varphi (n)|=0$, then $lim\varphi (n)=0$.

4. If $\varphi (n)$ tends to a limit or oscillates finitely, and $|\psi (n)|\le |\varphi (n)|$ when $n\ge n_0$, then $\psi (n)$ tends to a limit or oscillates finitely.

5. If $\varphi (n)$ tends to $+\mathrm{\infty }$, or to $-\mathrm{\infty }$, or oscillates infinitely, and $$|\psi (n)|\ge |\varphi (n)|$$ when $n\ge n_0$, then $\psi (n)$ tends to $+\mathrm{\infty }$ or to $-\mathrm{\infty }$ or oscillates infinitely.

6. ‘If $\varphi (n)$ oscillates and, however great be $n_0$, we can find values of $n$ greater than $n_0$ for which $\psi (n)>\varphi (n)$, and values of $n$ greater than $n_0$ for which $\psi (n)<\varphi (n)$, then $\psi (n)$ oscillates’. Is this true? If not give an example to the contrary.

7. If $\varphi (n)\to l$ as $n\to \mathrm{\infty }$, then also $\varphi (n+p)\to l$, $p$ being any fixed integer. [This follows at once from the definition. Similarly we see that if $\varphi (n)$ tends to $+\mathrm{\infty }$ or $-\mathrm{\infty }$ or oscillates so also does $\varphi (n+p)$.]

8. The same conclusions hold (except in the case of oscillation) if $p$ varies with $n$ but is always numerically less than a fixed positive integer $N$; or if $p$ varies with $n$ in any way, so long as it is always positive.

9. Determine the least value of $n_0$ for which it is true that $$(a)n^2+2n>999,999(n\ge n_0),(b)n^2+2n>1,000,000(n\ge n_0).$$

10. Determine the least value of $n_0$ for which it is true that $$(a)n+(-1{)}^n>1000(n\ge n_0),(b)n+(-1{)}^n>1,000,000(n\ge n_0).$$

11. Determine the least value of $n_0$ for which it is true that $$(a)n^2+2n>\mathrm{\Delta }(n\ge n_0),(b)n+(-1{)}^n>\mathrm{\Delta }(n\ge n_0),$$ $\mathrm{\Delta }$ being any positive number.

[(a) $n_0=[\sqrt{\mathrm{\Delta }+1}]$: () $n_0=1+[\mathrm{\Delta }]$ or $2+[\mathrm{\Delta }]$, according as $[\mathrm{\Delta }]$ is odd or even, i.e. $n_0=1+[\mathrm{\Delta }]+\frac{1}{2}\{1+(-1{)}^{[\mathrm{\Delta }]}\}$.]

12. Determine the least value of $n_0$ such that $$(a)n/(n^2+1)<.0001,(b)(1/n)+\{(-1{)}^n/n^2\}<.00001,$$ when $n\ge n_0$. [Let us take the latter case. In the first place $$(1/n)+\{(-1{)}^n/n^2\}\le (n+1)/n^2,$$ and it is easy to see that the least value of $n_0$, such that $(n+1)/n^2<.000001$ when $n\ge n_0$, is $1,000,002$. But the inequality given is satisfied by $n=1,000,001$, and this is the value of $n_0$ required.]