For the sake of periodicity- Part 1

Algebra Level 5

If $f(x)$ and $g(x)$ have the fundamental periods as $a$ and $b$ respectively, then the fundamental period of $f(x)\cdot g(x)$ is the lowest common multiple of $a$ and $b$ .

Inspired by this problem. .

True, if

a=b

Always False True, if L.C.M.

(a,b)

exists None of the given choices Always true

2 solutions

Jequil Hartz
Aug 19, 2015

$\sin{(\pi x)}$ and $\cos{(\pi x)}$ have a period of $2$ , but their product is $\sin{(\pi x)}\cos{(\pi x)} = \frac{1}{2} \sin{(2 \pi x)}$ which has a period of $1$ .

Thus, the fundamental period of a product of periodic functions being the LCM of the fundamental periods is not always true if a = b, not always true in general, and not true if the LCM exists. Three choices have been eliminated just with this example. Now the answer is either always false or none of the given choices.

To prove against the always false choice, we must find a case where this is true.

Consider $\cos^{2}{(\pi x)}$ which has a fundamental period of $1$ .

Now $\cos^{4}{(\pi x)}$ also has a fundamental period of $1$ , but $\cos^{4}{(\pi x)}$ is the product of $\cos^{2}{(\pi x)}$ with itself.

$LCM(1, 1) = 1$ which implies that this is a case where the fundamental period of the product is the lowest common multiple of the fundamental periods.

Thus, the answer must be none of the given choices which is also paradoxical as it is one of the given choices.

Moderator note:

Great! You eliminated all the other possibilities.

What, if anything, can we say about the fundamental period of $f(x)g(x)$ ?

For example, can the period be 0 if $ab \neq 0$ ?

Great! You eliminated all the other possibilities.

What, if anything, can we say about the fundamental period of $f(x)g(x)$ ?

For example, can the period be 0 if $ab \neq 0$ ?

Calvin Lin Staff - 5 years, 9 months ago

Just a heads up that I don't really have much of a proof for any of what I am about to write. Determination of the below can be determined empirically.

From using functions of the form $\sin{(n \pi x)}$ and $\cos{(m \pi x)}$ it would appear that the product of two functions of this form

$\sin{(n \pi x)}\cos{(m \pi x)}, \sin{(n \pi x)}\sin{(m \pi x)}, or \cos{(n \pi x)}\cos{(m \pi x)}$ for $m \neq n$ for any numbers $m$ and $n$ .

will not be periodic if m is an irrational multiple of n and will be periodic if m is a rational multiple of n.

As such it is possible to have a product with a period of zero without the either of the functions having a period of zero.

As an example, consider $\sin{(2 \pi x)}\sin{x}$ which is the product of two functions with periods $1$ and $2 \pi$ .

Note that $m$ can be taken as $2 \pi$ which would mean that $n = \frac{1}{\pi}$ which would imply that $m$ is an irrational multiple of $n$ .

It can be checked that $\sin{(2 \pi x)}\sin{x}$ is not periodic.

I would be happy if anyone could add any insight and rigor to this.

Jequil Hartz - 5 years, 9 months ago

Great analysis!

What is one simple way of showing that $\sin ( 2 \pi x ) \sin x$ is not periodic?

Hint: What can we say about the set of zeros? Why is this set not periodic?

Calvin Lin Staff - 5 years, 9 months ago

@Calvin Lin – That is a really great hint. I never considered the possibility for using the set of zeroes to find candidates for periods of a function (hopefully this is what you intended with the hint.).

Let $f(x) = \sin{(2\pi x)}\sin{x}$ and let us assume that $f(x + c) = f(x)$ for all x for some nonzero number c.

Seeing that this is true for all x, choose x to be zero such that $f(c) = f(0) = 0$ .

This implies that if the period exists, it will be in the set of numbers that are mapped to zero by $f(x)$ . Let us call this set $S$ .

In order for $f(x)$ to be zero, either of its components must be zero.

Thus, $S = \left\{\frac{n}{2}, m \pi \: | \: m, n \in \mathbb{Z}, mn \neq 0 \right\}$ .

Let us assume that c is of the form $\frac{n}{2}$ such that

$f(x + \frac{n}{2}) = \sin{(2\pi x + n \pi)}\sin{(x + \frac{n}{2})} = \sin{(2 \pi x)}\sin{x}$

This must be true for all values of x for some value of n, so let us choose $x = \pi$ .

Thus, $f( \pi + \frac{n}{2}) = \sin{(2{\pi}^{2} + n \pi)}\sin{(\pi + \frac{n}{2})} = 0$ .

Given the possible values of n, it is impossible for $\sin{(2{\pi}^{2} + n \pi)}$ to equal zero for any value n, thus $\sin{(\pi + \frac{n}{2})}$ must be equal to zero for some n which implies that $\pi + \frac{n}{2} = k \pi$ for some $k \in \mathbb{Z}$ , but this is only possible if $n = 0$ which is not possible given the set condition of $c \neq 0$ .

Thus, c is not of the form of $\frac{n}{2}$ .

Then if c exists, it must be of the form $m \pi$ such that

$f(x + m\pi) = \sin{(2\pi x + 2 m {\pi}^{2})}\sin{(x + m \pi)} = \sin{(2 \pi x)}\sin{x}$

This must be true for all values of x for some value of m, so let us choose $x = \frac{1}{2}$ .

Thus, $f(\frac{1}{2} + m\pi) = \sin{(\pi + 2 m {\pi}^{2})}\sin{(\frac{1}{2} + m \pi)} = 0$ .

Given the possible values of m, it is impossible for $\sin{ (\pi + 2m{\pi}^{2} )}$ to equal zero for any value m, thus $\sin{(\frac{1}{2} + m \pi)}$ must be equal to zero for some m which implies that $\frac{1}{2} + m \pi = k \pi$ for some $k \in \mathbb{Z}$ , but this is never possible with the possible values of m.

Thus, c is not of the form of $m \pi$ .

Thus, $c \notin S$ which contradicts that assumption that $c \in S$ . Therefore, c does not exist. This means that $f(x) = \sin{(2\pi x)}\sin{x}$ is not periodic.

Jequil Hartz - 5 years, 9 months ago

@Jequil Hartz – Yup! There are lots of ways to show that something is not periodic, even though it seems like it could be. For example, if we were looking at the function $\cos \theta \cos ( \pi \theta)$ , then we can argue that the maximum of 1 is achieved only when $\theta = 0$ , and hence it is not a periodic function.

You have the right idea, though there is a much easier way to express it. Basically, what we are after, is that the $S$ cannot be translated onto itself (other than the trivial translation),

Calvin Lin Staff - 5 years, 9 months ago

@Calvin Lin – Ah. Good to know. Learning to produce far more clever arguments is the whole reason for continued practice I suppose. At least I was on the right track and learned new things about periodic functions today.

Jequil Hartz - 5 years, 9 months ago

@Jequil Hartz – For the easier argument that $S = \{ n, n\pi \}$ is not periodic:

All that we use is $\pi$ is irrational .
Assume that the set is periodic with period $k$ . Clearly, we must either have $k = N$ or $k = N \pi$ , since 0 is mapped to some other value.
In the event that the period $k = N$ , we will show that $\pi + N \not \in S$ . Suppose it is, then $\pi + N = M$ or $M \pi$ . In the first case, we have $\pi = M - N$ which is an integer. In case, we have $\pi = \frac{N}{M-1}$ which is a rational number. Hence, we have a contradiction.
In the event that the period is $k = N \pi$ , in a similar manner, $1 + N \pi \not \in S$ .
Hence we are done.

Calvin Lin Staff - 3 years, 3 months ago

Tanishq Varshney
Jun 27, 2015

For example

$f(x)=\sin x$ , its period is $2\pi$

$g(x)=\cos x$ , its period is $2\pi$

$f(x).g(x)=\sin x \cos x$ its period is $pi$ .

which is not the LCM of the fundamental periods of the given functions.

The solution is incomplete.You would have to prove that the answer cannot be "Always False" maybe by giving an example where the property holds.

Sudeep Salgia - 5 years, 11 months ago

For the sake of periodicity- Part 1

Inspired by this problem. .

2 solutions

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