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 \[ \begin{align} & h ( x + n \alpha + m \beta) \\ = & f( x + n \alpha + m\beta + \gamma) - f( x + n \alpha + m \beta ) \\ = & f ( x + m \beta + \gamma) - f ( x + m \beta ) = h ( x + m \beta ) \\ = & g ( x + m \beta ) - g ( x + m\beta + \gamma)\\ =& g ( x ) - g ( x + \gamma ) \\ =& h ( x ) \end{align} \]

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.

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## Comments

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

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

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This claim is

notalways true.If both the functions are

evenand complimentary to each other(differ by phase difference of \(\displaystyle \frac{\pi}{2}\)) then the period isL.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 \(\displaystyle \frac{\pi}{2}\)

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For rational ratios, what would be the correct claim?

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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

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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.

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

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'Time' Period???? :-)

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Aah! Fixed it

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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 \(K(<T)\) 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 \(π\).

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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.

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L.C.M of irrational with

similarirrational is definedBut L.C.M of rational with irrational is

notdefined(called as aperiodic or not periodic)Example :-

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

Is aperiodic

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

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.

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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

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

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Verrryyy nice mr calvin

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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.

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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.

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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.

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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.

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@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

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\(\frac{π}{1}=π\)

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Ohhh okk thanks

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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 \( x = \frac{x}{1} \).

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