# A fundamental period claim

This arose from Sandeep's problem Do you know its property - 9

Starting question. If two functions $f$ and $g$ are periodic, is $f+g$ also periodic?

Be warned that this note requires significant mathematical maturity, which is why I set it at level 5.

Note: The LCM of 2 real numbers is defined in the same way. Find the smallest positive number (if it exists) such that for some integers $n,m$, we have $n \alpha = m \beta$.

Claim 1: If the fundamental period of $f$ is $\alpha$ and the fundamental period of $g$ is $\beta$, then the fundamental period of $f + g$ is the LCM of $\alpha, \beta$?

Is this claim true? No, not necessarily. We only know that their LCM must be a period, but it need not be the fundamental period. There could be a smaller period due to cancellation (see Krishna's comment). What is true, is the following:

Fact: If the period of $f$ is $\alpha$ and the period of $g$ is $\beta$, and $\frac{\alpha} { \beta }$ is rational, then a period of $f + g$ is the LCM of $\alpha, \beta$.

That leaves us to consider the irrational case.

Claim 2: If the period of $f$ is $\alpha$ and the period of $g$ is $\beta$, and $\frac{\alpha} { \beta }$ is irrational, then $f+g$ is never periodic.

It is easy to find examples where $f + g$ is not periodic, like $\sin x + \sin \pi x$. However, must this statement be always true?

We know that there is no obvious candidate for a period, but can't there be some "weird" functions which cancel out, like in the rational case? The following 4 hints will guide us through the rest of this investigation. You should try working on these hints first, before reading further

Hint 1: Prove that a continuous periodic function is bounded.

Hint 2: Prove that if $f$ and $g$ are both continuous, periodic, non-constant functions with irrational ratios, then $f+g$ is not periodic.

Hint 3: Find a function that is not constant, whose periods is dense in $\mathbb{R}$ but the set of periods is not $\mathbb{R}$. Note: The function is not continuous.

Hint 4: Now find $f$ and $g$ with irrational ratio of periods, such that $f+g$ is periodic (and non-constant).

Note: I used the Axiom of Choice (which I believe in).

Proof of hint 1. Let $f$ be a continuous, periodic function with period $\alpha$. Suppose that $f$ is not bounded. That means that for any $N$, there exists an $x_N$ such that $| f ( x_N )| > N$. WLOG, we may assume that $0 \leq x_N \leq \alpha$, as we translate it to the first period.

Since the points $x_N$ lie in the closed interval $[0, \alpha ]$, there exists an accumulation point, which means that there is a certain subsequence $x_{N_k}$ which converges to $x^*$. Since $f$ is a continuous function, we have $f( x^*) = \lim_{k\rightarrow \infty} f(x_{N_k} )$. But the RHS tends to infinity, which contradicts the assumption that $f(x^*)$ is finite. Hence, the assumption is false, and $f$ is bounded. $_\square$

Proof of hint 2. We first prove the following claim:

Claim If a function $f$ is continuous, periodic, and it's periods are dense in $\mathbb{R}$, then it is the constant function.

This follows immediately by considering $f( \alpha )$, where $\alpha$ is a period of $f$. It is dense in $\mathbb{R}$, and their values are all equal. By continuity, it is the constant function. $_\square$

Now, on to the proof. Suppose that $f$ and $g$ are continuous, periodic, non-constant functions with irrational period ratio, and $f+ g$ is periodic. If $f+g$ is the constant function, then we can show that $f$ and $g$ must have the same period, which contradicts the assumption. Hence, we may assume that $f + g$ has a smallest period $\gamma > 0$.

We are given that $f( x+ \gamma) + g( x + \gamma ) = f(x) + g(x)$. Define $h(x) = f( x + \gamma) - f(x) = g(x) - g( x + \gamma)$, which is a continuous function (since it is the difference of 2 continuous functions).

For any pair of integers $n, m$, observe that \begin{aligned} & h ( x + n \alpha + m \beta) \\ = & f( x + n \alpha + m\beta + \gamma) - f( x + n \alpha + m \beta ) \\ = & f ( x + m \beta + \gamma) - f ( x + m \beta ) = h ( x + m \beta ) \\ = & g ( x + m \beta ) - g ( x + m\beta + \gamma)\\ =& g ( x ) - g ( x + \gamma ) \\ =& h ( x ) \end{aligned}

This implies that the function $h (x)$ is periodic with period $n \alpha + m \beta$. But since the ratio $\frac{ \alpha } { \beta }$ is irrational, this implies that $\mathbb{Z} \alpha + \mathbb{Z} \beta$ is dense in $\mathbb{R}$, and thus $h(x)$ must be a constant, which we will denote by $h$.

If $h \neq 0$, then $f( x + n \gamma) = f(x) + n h$ which is an unbounded sequence, contradicting Hint 1.

If $h = 0$, then $f$ and $g$ are both periodic with period $gamma$. Now, at least one of $\frac{ \alpha } { \gamma }$ or $\frac { \beta } { \gamma}$ is irrational (WLOG, let it be the first ). Once again, this implies that $\mathbb{Z} \alpha + \mathbb{Z} \gamma$ is dense in $\mathbb{R}$, and thus that $f$ is a constant function. However, this contradicts the assumption. $_\square$

Hence, what we can conclude at this point is

Fact: If $f$ is a continuous periodic function, then either $f$ is constant, or $f$ has a fundamental period (smallest).

Fact: If $f$ and $g$ are continuous, periodic, non-constant functions with an irrational ratio of fundamental periods, then $f + g$ is not periodic.

The claim is true, if we restrict out attention to only continuous functions.

Now, we move on to the fun and interesting part. The world is extremely well-behaved (relatively) when functions are continuous. Let's observe what happens when things go crazy.

Proof of Hint 3. Consider the function $f(x) = \begin{cases} 0 & x \in \mathbb{Q} \\ 1 & x \not \in \mathbb{Q} \\ \end{cases}.$

This function is periodic, and the set of periods is all rational numbers. This is dense in $\mathbb{R}$, but is not $\mathbb{R}$. $_\square$

In particular, this shows us that a non-continuous, non-constant periodic function need not have a fundamental period.

Proof of Hint 4. We know that $1, e, \pi$ are rational-linearly independent. What this means is that if $q_1 \times 1 + q_2 \times e + q_3 \times \pi = 0$, where $q_i$ are rational numbers, then $q _ i = 0$.

Define the rational span of $(a,b)$ to be the set of all numbers of the form $\mathbb{Q} a + \mathbb{Q} b$.

Let $\chi_{1, e }$ be the characteristic function of the rational span of $(1,e)$, and $\chi_{e, \pi}$ and $\chi_{1, \pi}$ similarly defined.

Then, set $f = \chi_{ 1, e } + \chi_{e, \pi}$ and $g = - \chi{ e, \pi} + \chi { 1 , \pi }$. Then, we have $e$ is a period of $f$, $\pi$ is a period of $g$ and $1$ is a period of $f + g$. $_ \square$

Thus, we have 2 functions with irrational ratio of periods, but sum to give a periodic function.

Note by Calvin Lin
4 years, 8 months ago

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $ ... $ or $ ... $ to ensure proper formatting.
2 \times 3 $2 \times 3$
2^{34} $2^{34}$
a_{i-1} $a_{i-1}$
\frac{2}{3} $\frac{2}{3}$
\sqrt{2} $\sqrt{2}$
\sum_{i=1}^3 $\sum_{i=1}^3$
\sin \theta $\sin \theta$
\boxed{123} $\boxed{123}$

Sort by:

What an amazing observation you've just shared!! I'm astonished!! Let me think..

- 4 years, 8 months ago

I've added the details on how to proceed. It would still be worthwhile to consider and work through the hints yourself.

Staff - 4 years, 8 months ago

This claim is not always true.

If both the functions are even and complimentary to each other(differ by phase difference of $\displaystyle \frac{\pi}{2}$) then the period is

L.C.M of $\frac{1}{2}$( $\alpha$ & $\beta$)

Example:-

f = $|\sin x|$

g = $|\cos x|$

f + g = $|\sin x| + |\cos x|$

Here period of 'f + g' is $\displaystyle \frac{\pi}{2}$

- 4 years, 8 months ago

For rational ratios, what would be the correct claim?

Staff - 4 years, 8 months ago

On further investigation I found only 2 exceptions of the claim, one I discussed above and the other when function converts into a constant value or constant function(LCM Method fails here)

Ex:-

f = $\sin^{2} x$

g = $\cos^{2} x$

f + g = $\sin^{2} x$ + $\cos^{2} x$ = 1

Function is periodic but fundamental period is not defined

Similarly

f + g = $\sec^{2} x$ - $\tan^{2} x$ = 1

- 4 years, 8 months ago

Nice example!

As mentioned in "Fact", all that we know is "a period of $f+g$ is the LCM". As you brought up, we are not guaranteed that $f+g$ must have a fundamental period.

Staff - 4 years, 8 months ago

But Calvin sir here has brought a very interesting question. That the sum function is still periodic even when their LCM doesn't exist.

- 4 years, 8 months ago

'Time' Period???? :-)

- 4 years, 8 months ago

Aah! Fixed it

- 4 years, 8 months ago

Let LCM of $T_1$ & $T_2$ be $T$ then $T$ will be the period of $(f+g)$, provided there does not exist a positive number $K( for which $f(x+K)+g(x+K)=f(x)+g(x)$ else $K$ will be the period. Generally the situation arises due to the interchanging of $f(x)$ and $g(x)$ by addition of the positive number $K$. In case of $(|\sin x|+|\cos x|)$ $|\sin x|$ & $|\cos x|$ interchanged by adding $\frac{π}{2}$ & the result being$|\cos x| +|\sin x|=|\sin x|+|\cos x|$ hence the period is $\frac{π}{2}$ instead of $π$.

- 4 years, 8 months ago

So, the interesting case occurs when $\frac { \alpha } { \beta }$ is irrational.

Question: Does there exist 2 functions $f$ and $g$, with fundamental periods $\alpha , \beta$ and $\frac{ \alpha } { \beta } \neq \mathbb{Q}$, such that $f + g$ is periodic?

Hint: Prove that if $f$ and $g$ are bounded, then $f+g$ is not periodic.

Hint: Prove that if $f$ and $g$ are continuous and periodic, then they are bounded.

Hint: Consider what happens if $f$ and $g$ are not continuous.

Staff - 4 years, 8 months ago

• L.C.M of rational with rational is defined

• L.C.M of irrational with similar irrational is defined

• But L.C.M of rational with irrational is not defined(called as aperiodic or not periodic)

Example :-

f + g = $\sin x$ + {x} (fractional part of 'x')

Is aperiodic

- 4 years, 8 months ago

The point of the above comment is to get people to think about proving, or disproving, the following:

If $\frac{ \alpha} { \beta }$ is irrational, must $f + g$ be aperiodic?

It is true that in a lot of cases, $f + g$ is aperiodic. But, can we create a scenario where $f + g$ is periodic?? This is interesting, because I currently believe that the answer is "There is such a function!"

Do not simply repeat what you believe, or what you recall someone else saying is true. Instead, find a justification of your statements.

Staff - 4 years, 8 months ago

If only period is concerned one such example is $f(x)=\sin x$ & $g(x)=0$. But if you talk about fundamental period i can't think of any example. Do you have any?? @Calvin Lin

- 4 years, 8 months ago

Find the period of Y=tanx/sin (2/3) x and y=sinx*cos3x

- 4 years, 4 months ago

Verrryyy nice mr calvin

- 4 years, 1 month ago

Is this a high level of mathematical maturity?? Cause all these things were taught to me by my mathematics teacher recently. I'm currently in class 12th.

- 4 years, 8 months ago

The start is easy to understand. Claim 1 and claim 2 are often commonly stated, and often without any proof. However, neither claims are true. Claim 1 is easily fixed, but Claim 2 is much more complicated. It is not true, and it can be hard to come up with a counter-example by yourself.

In order to work through and appreciate the hints, you need a certain level of mathematical maturity. See the proof of Hint 1 as an example.

Staff - 4 years, 8 months ago

Yes! I still have a really long way to go... I got a little hint of what you were doing but I couldn't understand everything. I can imagine graphically that those hints are true, but would be unable to prove them mathematically.

- 4 years, 8 months ago

Keep at it!

I started this note thinking that it was easy / straightforward, given how a lot of people state it as if it was fact. But now you can tell your teacher that it's not true! You can even show him the counter examples!

As I explored further, I found that I had difficulty accounting for the "cancellation", and it slowly began to cross my mind that the statement might not be true. I tried to detail how I explored this topic, starting with the simplification of continuous functions which is well loved, and then moving on to discontinuous functions which are not well-understood. I tried to motivate / justify the counter-example, but this can be hard to do.

Staff - 4 years, 8 months ago

@Calvin Lin sir in claim 2 can you please explain how can the ratio of two numbers be an irrational number.........according to defination an irrational number can't be expressed as the ratio of two numbers

- 4 years, 8 months ago

$\frac{π}{1}=π$

- 4 years, 8 months ago

Ohhh okk thanks

- 4 years, 8 months ago

An irrational number cannot be expressed as the ratio of 2 integers.

Note that every number can be expressed as the ratio of two numbers, namely $x = \frac{x}{1}$.

Staff - 4 years, 8 months ago