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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, 12 months ago

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

- 2 years, 12 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 - 2 years, 12 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 $$\displaystyle \frac{\pi}{2}$$

- 2 years, 12 months ago

'Time' Period???? :-)

- 2 years, 12 months ago

Aah! Fixed it

- 2 years, 12 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.

- 2 years, 12 months ago

For rational ratios, what would be the correct claim?

Staff - 2 years, 12 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

- 2 years, 11 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 - 2 years, 11 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 $$K(<T)$$ 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 $$π$$.

- 2 years, 12 months ago

Verrryyy nice mr calvin

- 2 years, 5 months ago

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

- 2 years, 7 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

- 2 years, 12 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 - 2 years, 12 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

- 2 years, 12 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 - 2 years, 12 months 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.

- 2 years, 12 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 - 2 years, 12 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.

- 2 years, 11 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 - 2 years, 11 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

- 2 years, 12 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 - 2 years, 12 months ago

$$\frac{π}{1}=π$$

- 2 years, 12 months ago

Ohhh okk thanks

- 2 years, 12 months ago