A fundamental period claim

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

Starting question. If two functions ff and gg are periodic, is f+g 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,mn,m , we have nα=mβ n \alpha = m \beta .


Claim 1: If the fundamental period of ff is α \alpha and the fundamental period of g g is β \beta , then the fundamental period of f+g f + g is the LCM of α,β\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 ff is α \alpha and the period of g g is β \beta , and αβ \frac{\alpha} { \beta } is rational, then a period of f+g f + g is the LCM of α,β\alpha, \beta .

That leaves us to consider the irrational case.

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

It is easy to find examples where f+g f + g is not periodic, like sinx+sinπx \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 ff and gg are both continuous, periodic, non-constant functions with irrational ratios, then f+gf+g is not periodic.

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

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

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


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

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


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

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

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

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

We are given that f(x+γ)+g(x+γ)=f(x)+g(x) f( x+ \gamma) + g( x + \gamma ) = f(x) + g(x) . Define h(x)=f(x+γ)f(x)=g(x)g(x+γ) 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,mn, m , observe that h(x+nα+mβ)=f(x+nα+mβ+γ)f(x+nα+mβ)=f(x+mβ+γ)f(x+mβ)=h(x+mβ)=g(x+mβ)g(x+mβ+γ)=g(x)g(x+γ)=h(x) \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) h (x) is periodic with period nα+mβ n \alpha + m \beta . But since the ratio αβ \frac{ \alpha } { \beta } is irrational, this implies that Zα+Zβ \mathbb{Z} \alpha + \mathbb{Z} \beta is dense in R \mathbb{R} , and thus h(x) h(x) must be a constant, which we will denote by hh .

If h0 h \neq 0 , then f(x+nγ)=f(x)+nh f( x + n \gamma) = f(x) + n h which is an unbounded sequence, contradicting Hint 1.

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


Hence, what we can conclude at this point is

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

Fact: If ff and gg are continuous, periodic, non-constant functions with an irrational ratio of fundamental periods, then f+g 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)={0xQ1x∉Q. 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 R \mathbb{R} , but is not R \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,π 1, e, \pi are rational-linearly independent. What this means is that if q1×1+q2×e+q3×π=0 q_1 \times 1 + q_2 \times e + q_3 \times \pi = 0 , where qi q_i are rational numbers, then qi=0 q _ i = 0 .

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

Let χ1,e \chi_{1, e } be the characteristic function of the rational span of (1,e) (1,e) , and χe,π \chi_{e, \pi} and χ1,π \chi_{1, \pi} similarly defined.

Then, set f=χ1,e+χe,π f = \chi_{ 1, e } + \chi_{e, \pi} and g=χe,π+χ1,π g = - \chi{ e, \pi} + \chi { 1 , \pi } . Then, we have e e is a period of ff, π \pi is a period of gg and 11 is a period of f+g 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, 10 months ago

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

Sanjeet Raria - 4 years, 10 months ago

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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.

Calvin Lin Staff - 4 years, 10 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 π2\displaystyle \frac{\pi}{2}) then the period is

L.C.M of 12\frac{1}{2}( α\alpha & β\beta)

Example:-

f = sinx|\sin x|

g = cosx|\cos x|

f + g = sinx+cosx|\sin x| + |\cos x|

Here period of 'f + g' is π2\displaystyle \frac{\pi}{2}

Krishna Sharma - 4 years, 10 months ago

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For rational ratios, what would be the correct claim?

Calvin Lin Staff - 4 years, 10 months ago

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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 = sin2x\sin^{2} x

g = cos2x\cos^{2} x

f + g = sin2x\sin^{2} x + cos2x\cos^{2} x = 1

Function is periodic but fundamental period is not defined

Similarly

f + g = sec2x\sec^{2} x - tan2x\tan^{2} x = 1

Krishna Sharma - 4 years, 9 months ago

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@Krishna Sharma Nice example!

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

Calvin Lin Staff - 4 years, 9 months ago

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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.

Sanjeet Raria - 4 years, 10 months ago

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'Time' Period???? :-)

Danny Kills - 4 years, 10 months ago

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Aah! Fixed it

Krishna Sharma - 4 years, 10 months ago

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Let LCM of T1T_1 & T2T_2 be TT then TT will be the period of (f+g)(f+g), provided there does not exist a positive number K(<T)K(<T) for which f(x+K)+g(x+K)=f(x)+g(x)f(x+K)+g(x+K)=f(x)+g(x) else KK will be the period. Generally the situation arises due to the interchanging of f(x)f(x) and g(x)g(x) by addition of the positive number KK. In case of (sinx+cosx)(|\sin x|+|\cos x|) sinx|\sin x| & cosx|\cos x| interchanged by adding π2\frac{π}{2} & the result beingcosx+sinx=sinx+cosx|\cos x| +|\sin x|=|\sin x|+|\cos x| hence the period is π2\frac{π}{2} instead of ππ.

Sanjeet Raria - 4 years, 10 months ago

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

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

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

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

Hint: Consider what happens if ff and gg are not continuous.

Calvin Lin Staff - 4 years, 10 months ago

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  • 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 = sinx\sin x + {x} (fractional part of 'x')

Is aperiodic

Krishna Sharma - 4 years, 10 months ago

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

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

It is true that in a lot of cases, f+g f + g is aperiodic. But, can we create a scenario where f+g 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 - 4 years, 10 months ago

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

Sanjeet Raria - 4 years, 10 months ago

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

lengheng nim - 4 years, 5 months ago

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Verrryyy nice mr calvin

ishak tahir - 4 years, 3 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 - 4 years, 10 months ago

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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.

Calvin Lin Staff - 4 years, 10 months ago

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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.

Danny Kills - 4 years, 9 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 - 4 years, 9 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 - 4 years, 10 months ago

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π1=π\frac{π}{1}=π

Sanjeet Raria - 4 years, 10 months ago

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Ohhh okk thanks

Aman Sharma - 4 years, 10 months ago

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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=x1 x = \frac{x}{1} .

Calvin Lin Staff - 4 years, 10 months ago

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