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

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.

Note by Calvin Lin
2 years, 2 months ago

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What an amazing observation you've just shared!! I'm astonished!! Let me think.. Sanjeet Raria · 2 years, 2 months ago

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@Sanjeet Raria I've added the details on how to proceed. It would still be worthwhile to consider and work through the hints yourself. Calvin Lin Staff · 2 years, 2 months ago

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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 \(\displaystyle \frac{\pi}{2}\) Krishna Sharma · 2 years, 2 months ago

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@Krishna Sharma 'Time' Period???? :-) Danny Kills · 2 years, 2 months ago

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@Danny Kills Aah! Fixed it Krishna Sharma · 2 years, 2 months ago

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@Krishna Sharma But Calvin sir here has brought a very interesting question. That the sum function is still periodic even when their LCM doesn't exist. Sanjeet Raria · 2 years, 2 months ago

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@Krishna Sharma For rational ratios, what would be the correct claim? Calvin Lin Staff · 2 years, 2 months ago

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@Calvin Lin 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 Krishna Sharma · 2 years, 2 months ago

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@Krishna Sharma 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. Calvin Lin Staff · 2 years, 2 months ago

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@Krishna Sharma 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 \(π\). Sanjeet Raria · 2 years, 2 months ago

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Verrryyy nice mr calvin Ishak Tahir · 1 year, 8 months ago

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Find the period of Y=tanx/sin (2/3) x and y=sinx*cos3x Lengheng Nim · 1 year, 10 months ago

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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 Sanjeet Raria · 2 years, 2 months ago

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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. Calvin Lin Staff · 2 years, 2 months ago

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@Calvin Lin

  • 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 Krishna Sharma · 2 years, 2 months ago

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@Krishna Sharma 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. Calvin Lin Staff · 2 years, 2 months ago

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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. Danny Kills · 2 years, 2 months ago

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@Danny Kills 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. Calvin Lin Staff · 2 years, 2 months ago

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@Calvin Lin 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. Danny Kills · 2 years, 2 months ago

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@Danny Kills 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. Calvin Lin Staff · 2 years, 2 months ago

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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 Aman Sharma · 2 years, 2 months ago

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@Aman Sharma 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 \( x = \frac{x}{1} \). Calvin Lin Staff · 2 years, 2 months ago

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@Aman Sharma \(\frac{π}{1}=π\) Sanjeet Raria · 2 years, 2 months ago

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@Sanjeet Raria Ohhh okk thanks Aman Sharma · 2 years, 2 months ago

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