Conditions on f such that the anti-derivative F is periodic?

[MxwllsPersuasns]

Homework Statement

Let f : R → Rⁿ be a smooth function. Give necessary and sufficient conditions on f so that the antiderivative F(x) = ∫f(t)dt (from 0 to x) is periodic with period p ≠ 0

Homework Equations

The Attempt at a Solution

My initial thought is that as long as f is periodic then F will be periodic as well -- the oscillatory nature of the differentiation operators on Cos and Sin comes to mind. But I imagine it can't be that simple, perhaps there's extra conditions which are stipulated to ensure F is periodic. Or maybe my thought that f need be periodic is not strictly true.

Any guidance?

 

[FactChecker]

What can you say about the derivative of a periodic function? I can think of two things. I probably shouldn't say more now on a homework problem.

 

[MxwllsPersuasns]

Hmm well I would say that the derivative would also be a periodic function... (I think this is a theorem?) and hmm... my intuition says that perhaps something to do with the slope at the point of periodicity? As in if you approach from the left you get the opposite of approaching the point from the right? Am I on the right track?

 

[ehild]

Homework Statement

Let f : R → Rⁿ be a smooth function. Give necessary and sufficient conditions on f so that the antiderivative F(x) = ∫f(t)dt (from 0 to x) is periodic with period p ≠ 0

Homework Equations

The Attempt at a Solution

My initial thought is that as long as f is periodic then F will be periodic as well -- the oscillatory nature of the differentiation operators on Cos and Sin comes to mind. But I imagine it can't be that simple, perhaps there's extra conditions which are stipulated to ensure F is periodic. Or maybe my thought that f need be periodic is not strictly true.

Any guidance?

Is this function periodic? And its primitive function?

 

[MxwllsPersuasns]

I would say.. yes? I mean the values of f(x) oscillates between 0 and, say Y. If we take the distance between each change of value (first from Y to 0 then 0 to Y and so on) to be X then we'd say we have values of Y at x = (0, X), (2X, 3X), (4X, 5X) and thus we have beginning points at every 2nX and endpoints at every 2nX + 1. The same can be said for the interval where f(x) is zero and since we have the same f(x) values for increasing values of x, separated with regularity, then I would say it's periodic.

When you say primitive I'm assuming anti-derivative? I would guess based on my assertion above that the derivative of a periodic function is itself periodic that the converse is also true (Yes I'm aware this doesn't always happen and I shouldn't assume that, but I'm not -- just guessing)

 

[ehild]

I would say.. yes?
When you say primitive I'm assuming anti-derivative? I would guess based on my assertion above that the derivative of a periodic function is itself periodic that the converse is also true (Yes I'm aware this doesn't always happen and I shouldn't assume that, but I'm not -- just guessing)

Yes, the function is periodic, but what is its antiderivative as function of x? Try to sketch! ( It increases in the first half period, then stays constant in the next half period, then increases again...

 

[FactChecker]

Think about what you can say about the values of a continuous periodic function at the beginning and end of a period. Translate that into a property of its derivative.

 

[MxwllsPersuasns]

@ehild - Well since the integral measures the "accumulation" -- the area under the graph -- it seems like the graph would be increasing linearly (with the coefficient of x being whatever constant value (Y) is obtained in the original function. Then just be constant since the derivative is zero then increase again then constant, etc.. so it would look kind of like a hill then plateau then hill then plateau? Sorry if I'm not getting this..

@FactChecker - Let's take the periodic function to be sin(x) and so the period begins at x = 0 and ends at x = pi. In this case we start at f(0) = sin(0) = 0 but begin increasing until we reach 1 at x = pi/2 then start decreasing until we end, again, at f(pi) = 0. In both these cases though the derivative would be opposite? Since moving forward in the period from x = 0 our function is increasing thus our derivative is positive and then moving forward after pi/2 our derivative would be negative since our function is decreasing. So could we say something like f'(0) = -f'(pi)?

 

[FactChecker]

In both these cases though the derivative would be opposite?

No. A continuous differentiable periodic function has the same values and the same slope at the beginning and end of a period. That is because the end of one period is the start of the next period.

Suppose f(x) is periodic, with x₁ and x₂ the beginning and end of one period. Then f(x₁) = f(x₂). State that in terms of the integral of its derivative, f'(x).

 

[MxwllsPersuasns]

since f(x₁) = f(x₂) that translates to something like ∫f'(x)dx from 0 to x₁ = ∫f'(x)dx from 0 to x₂, yes?

 

[ehild]

@ehild - Well since the integral measures the "accumulation" -- the area under the graph -- it seems like the graph would be increasing linearly (with the coefficient of x being whatever constant value (Y) is obtained in the original function. Then just be constant since the derivative is zero then increase again then constant, etc.. so it would look kind of like a hill then plateau then hill then plateau? Sorry if I'm not getting this..

Yes, and is F(x) periodic ? So, if a function is periodic, its anti-derivative can be non-periodic.

 

[MxwllsPersuasns]

Ah yes I see.. so then f being periodic is not a necessary or sufficient condition. I currently only have that the derivatives need agree at the beginning and end of the period, though not much else.

 

[ehild]

Ah yes I see.. so then f being periodic is not a necessary or sufficient condition. I currently only have that the derivatives need agree at the beginning and end of the period, though not much else.

You see that F(x) starts the second period with the area under f(x) from 0 to p (period). In order to be periodic, F(p) should be zero. What does it mean for $\int_0^p{f(x) dx}$?
What would be F(x) of the function shown in the figure?

 

[MxwllsPersuasns]

So this means the integral from to the period would then be 0? It seems you're suggesting that given the picture, we can clearly see that F(x) would indeed be 0 as every bit above the x-axis is canceled out by an equal bit under the x axis. Correct?

 

[ehild]

So this means the integral from to the period would then be 0? It seems you're suggesting that given the picture, we can clearly see that F(x) would indeed be 0 as every bit above the x-axis is canceled out by an equal bit under the x axis. Correct?

F(p) is zero, but F(x) is not. F(p) must be the same as F(0), and it is true for any multiple of the period.

 

[MxwllsPersuasns]

So wait, are we still talking about conditions on the derivative such that the antiderivative is periodic? This seems like you're explaining properties of the antiderivative of a periodic function. But we aren't necessarily assuming that f is periodic? Sorry if I'm seeming dense.. it's been a long day.

 

[ehild]

So wait, are we still talking about conditions on the derivative such that the antiderivative is periodic?

Yes.

This seems like you're explaining properties of the antiderivative of a periodic function. But we aren't necessarily assuming that f is periodic? Sorry if I'm seeming dense.. it's been a long day.

If we assume that F(x) periodic, should be F'(x)=f(x) periodic too?
If f(x) is periodic, F(x) to be periodic means that F(kp)=F(0) should hold. (k is any integer). What does it mean for f(x)?

 

[MxwllsPersuasns]

Well we showed that the primitive of a periodic function isn't necessarily periodic but I'm pretty sure the derivative of a periodic function is itself periodic, in fact I believe there's a theorem about that. And what does F(kp) = F(0) mean for f(x)? Well didn't you just say f(x) is periodic and so then couldn't we say the same about f? That it equals the same value at the beginning of the period (x = 0) that it does at every integer multiple of the end of the period (kp), yes?

 

[MxwllsPersuasns]

or would it be something like f(np) = nf(p)?

 

[ehild]

The problem asked to give necessary and sufficient conditions on f so that the antiderivative F(x) = ∫f(t)dt (from 0 to x) is periodic with period p ≠ 0. (f was a smooth real function).
You have to know what is necessary and sufficient condition. Read :
http://philosophy.wisc.edu/hausman/341/Skill/nec-suf.htm
Necessary condition is what follows from a statement. From F(x) periodic, it follows that f(x) is also periodic (assuming it is smooth) So f(x) must be periodic so as its antiderivative is periodic.
But it is not sufficient. If f(x) is periodic, F(x) can be non-periodic, as we saw from the figures. What else should be true for f(x) that F(x) = ∫f(t)dt (from 0 to x) is periodic?