Prove a series is periodic

[physics1000]

$\sum _{n=0}^{\infty }\:\frac{sin\left(2^nx\right)}{2^n}$
I have to show the series is periodic, $2\pi$ ( I think ), it is related to fourier - analyze fourrier course at academics, so I might be right, or just periodic, we learned only periodic of $2\pi$
( of course also continuous, but that is easy, no need ).
My problem is, we have not learned how to show a series is perodic.
I know in order to show a function is perodic, I have to show $f(x) = f(x+t)$, but in series, I dont know how it will be... and also how can a series be periodic... it doesnt sound right to me...
I dont need a solution, just the beginning, what should I Do
should I do $f(x)=f(x+T)$ on the series? or something else?

 

[hilbert2]

If every term in the series has period $2\pi$, then also every partial sum does. So if it's a converging series, it should be impossible for the limiting function to not be periodic.

 

[physics1000]

If every term in the series has period $2\pi$, then also every partial sum does. So if it's a converging series, it should be impossible for the limiting function to not be periodic.

sorry, I dont understand what you mean?
It is a converging series, it is the geometric series, proved them both are continuous. but I dont understand how to prove periodic, I have to prove it, I cant tell by a sentence

 

[topsquark]

$\sum _{n=0}^{\infty }\:\frac{sin\left(2^nx\right)}{2^n}$
I have to show the series is periodic, $2\pi$ ( I think ), it is related to fourier - analyze fourrier course at academics, so I might be right, or just periodic, we learned only periodic of $2\pi$
( of course also continuous, but that is easy, no need ).
My problem is, we have not learned how to show a series is perodic.
I know in order to show a function is perodic, I have to show $f(x) = f(x+t)$, but in series, I dont know how it will be... and also how can a series be periodic... it doesnt sound right to me...
I dont need a solution, just the beginning, what should I Do
should I do $f(x)=f(x+T)$ on the series? or something else?

Hint: Take a look at $sin(2^n(x))$. Translate by $2 \pi$: $sin(2^n (x - 2 \pi )) = sin(2^n (x) - 2^n (2 \pi ))$. What do we know about $2^n \cdot 2$ for all n in $[0, \infty )$? What does that say about the sine?

-Dan

 

[physics1000]

Hint: Take a look at $sin(2^n(x))$. Translate by $2 \pi$: $sin(2^n (x - 2 \pi )) = sin(2^n (x) - 2^n (2 \pi ))$. What do we know about $2^n \cdot 2$ for all n in $[0, \infty )$? What does that say about the sine?

-Dan

wait, why are you assuming T=$2\pi$? to show periodic is $f(x)= f(x+T)$, not T=$2\pi$.
I can assume T= 2pi?

 

[topsquark]

wait, why are you assuming T=$2\pi$? to show periodic is $f(x)= f(x+T)$, not T=$2\pi$.
I can assume T= 2pi?

Does $T = 2\pi$ work?

-Dan

 

[physics1000]

Does $T = 2\pi$ work?

-Dan

Hi, I didnt try yet, because I dont want to delete if its wrong.
I mean, I know trigonometry identitys can help with such stuff, but I mean.
You chose T=-$2\pi$, but what reason? should'nt I just do x=x+T on the series? the question, how do I continue from here

 

[physics1000]

wait, why are you assuming T=$2\pi$? to show periodic is $f(x)= f(x+T)$, not T=$2\pi$.
I can assume T= 2pi?

about the $2^n * 2$, no idea actually, never heard something special, all i know it is equal to $2^{n+1}$ and of course that it does not converge, because sinus does not have a permament value, it changes 2pi

 

[PeroK]

I know in order to show a function is perodic, I have to show $f(x) = f(x+t)$, but in series, I dont know how it will be... and also how can a series be periodic... it doesnt sound right to me...

What you have is a function, expressed as a series. For example, the sine function (like many common functions) can be expressed as a power series:$$\sin x = \sum_{n = 0}^{\infty} (-1)^n \frac{x^{2n+1}}{(2n+1)!}$$Although it's not quite so easy to prove that is periodic.

 

[physics1000]

What you have is a function, expressed as a series. For example, the sine function (like many common functions) can be expressed as a power series:$$\sin x = \sum_{n = 0}^{\infty} (-1)^n \frac{x^{2n+1}}{(2n+1)!}$$Although it's not quite so easy to prove that is periodic.

Yea, that I know.
I did alot of it at calculus 2 ( forgot already how to build series from such, was good at it, but a few months passed, so.. yea:\ ).
But anyway, just to understand, because its not that I dont understand what you did.
in order to show a series if periodic, my function.
I just have to show $f(x) = f(x+T)$? and I must demand $T = 2\pi$? if yes to the second thing, why? because its on fourrier course?

 

[PeroK]

I just have to show $f(x) = f(x+T)$? and I must demand $T = 2\pi$? if yes to the second thing, why? because its on fourrier course?

A function has period $2\pi$ if it has period $2\pi$. That's the case whatever the course you're on.

Although the answer "the function $f$ has period $2\pi$ because this is a Fourier course" has a certain appeal. Note the spelling of Fourier.

 

[physics1000]

A function has period $2\pi$ if it has period $2\pi$. That's the case whatever the course you're on.

Although the answer "the function $f$ has period $2\pi$ because this is a Fourier course" has a certain appeal. Note the spelling of Fourier.

I know, the problem is there aint any clue for other perdiocs.
So I guess they really mean Fourier ( sorry, bad english ).
I guess I will just say T=$2\pi$, I will try to prove and see if I am getting any hard there, thanks!

 

[PeroK]

There is a big clue in post #2. Also, with a series you can always write out the terms to give you ideas:
$$\sum_{n = 0}^{\infty} \frac{\sin(2^nx)}{2^n} = \sin x + \frac{\sin(2x)}{2} + \dots $$

 

[physics1000]

Ahh, to be honest, even when I did $T = 2\pi$ could not solve it...
I tried using $sin(a+b) = sin(a)cos(b) + sin(b)cos(a)$.
regarding what PeroK just said ( what about to post, you suddenly replied, so answering it ).
I dont see thought how to show it is periodic, it is periodic logically, it is sinus after all.
the problem is the proof.
I tried it myself also, writing the terms, but it can not be a proof, proof has to be formal, that is my problem.

 

[PeroK]

I tried it myself also, writing the terms, but it can not be a proof, proof has to be formal, that is my problem.

I'm not sure what you've tried. A formal proof is only an informal proof with some technical details added. If you have an informal proof you are half way there. Anyway, if we call our function $f$, we have:
$$f(x + 2\pi) = \sum_{n = 0}^{\infty} \frac{\sin(2^n(x+2\pi))}{2^n} = \sin( x+2\pi) + \frac{\sin(2(x+2\pi))}{2} + \frac{\sin(4(x+2\pi))}{4}+ \dots = ?$$

 

[physics1000]

I'm not sure what you've tried. A formal proof is only an informal proof with some technical details added. If you have an informal proof you are half way there. Anyway, if we call our function $f$, we have:
$$f(x + 2\pi) = \sum_{n = 0}^{\infty} \frac{\sin(2^n(x+2\pi))}{2^n} = \sin( x+2\pi) + \frac{\sin(2(x+2\pi))}{2} + \frac{\sin(4(x+2\pi))}{4}+ \dots = ?$$

You are saying to use the fact that if the fraction of sum are periodic, then the sum itself also?
it just sounds to me like an awfully lazy proof, it is true, but there has to be a nicer way to prove it. if not I will just do this. Althought I dont have any other better way, so I can not say something bad about the proof :)
But thanks for the idea, it is a new way of proof to me, like that. never thought of that

 

[PeroK]

it just sounds to me like an awfully lazy proof,

I'd get a warning from the mentors if I said what I feel about this post. Suffice to say, it's the last help you're getting from me.

 

[physics1000]

I'd get a warning from the mentors if I said what I feel about this post. Suffice to say, it's the last help you're getting from me.

Sorry if you got insulted, did not mean it.
Its just that I wanted something more formal.
I didnt mean you being lazy, but the proof.
Regardless, thank for the help here, really appreciated ( And again, sorry ).

 

[topsquark]

about the $2^n * 2$, no idea actually, never heard something special, all i know it is equal to $2^{n+1}$ and of course that it does not converge, because sinus does not have a permament value, it changes 2pi

Notice that $2^{n+1}$ is even for all n in $[0, \infty )$. So $sin(2^n x - 2^n \cdot (2 \pi ) ) = sin( 2^n x - \text{even integer} \cdot \pi )$. Can you simplify this?

And as to how I found $2 \pi$, I simply graphed it. That would be the first tactic I'd recommend.

Sit down and play with it for a while.

-Dan

 

[WWGD]

I know, the problem is there aint any clue for other perdiocs.
So I guess they really mean Fourier ( sorry, bad english ).
I guess I will just say T=$2\pi$, I will try to prove and see if I am getting any hard there, thanks!

Or you can experiment on your own to help you make an educated guess, like topsquark suggested. Set f(x)=f(x+T), expand, see what it gets you.

 

[jaumzaum]

Consider your function $$f(x)=\sum_{n=0}^\infty a_n(x)$$ where $$a_n(x)=\frac {sin(2^n x)}{2^n}$$
Take $$g(x)=\sum_{n=0}^\infty b_n(x)$$$$b_n(x)=a_n(x+2\pi)=\frac {sin(2^n x +2^n 2\pi)}{2^n}$$
We know that the sine function is periodic with period 2 pi $$sin(y+2k\pi)= sin(y)$$ So:
$$b_n(x)=\frac {sin(2^n x)}{2^n}=a_n(x)$$
And if all the terms of the sums are equal and the sum converges, they must be equal

 

[DaveE]

OK, back to basics, I think. A function f is periodic, with a period T, iff f(x) = f(x+T).
Your series is simply a sum of functions. So, let's assume you have two functions f₁ and f₂, both of which are periodic with period T. Show that the function f(x) = f₁(x) + f₂(x) is also periodic with period T.
Now assume you have functions g and g₁ which are periodic with period T, and that g(x) = g₁(x) + g₂(x). Show that the function g₂ also has a period of T.
Try these again with three functions. Can you see that if each term in the sum is periodic with period T, then the sum is also?
This understanding will be the basis of your proof, I think. Maybe a proof by induction?
Finally, notice that sin(x) has a period of 2πk for k=1,2,3,... We tend to assume the fundamental period of 2π, but it is has a period of 4π, 6π, 8π, etc. Also if a function f(x) is periodic the function a⋅f(x) is also periodic for any constant a.

 

[pasmith]

You should know that $\sin(kx)$ has period $2\pi/k$. So $\sin(2^n x)$ has period $2\pi/2^n$.

Now the first partial sum $\sin(x)$ has period $2\pi$, so you can show by induction that each partial sum has period $2\pi$.

The result then follows from the following proposition, the proof of which is left as an exercise:

Proposition
If $f_n$ are a sequence of real functions each of which has period $T$ and such that the pointwise limit $f(x) = \lim_{n \to \infty} f_n(x)$ exists, then $f$ also has period $T$.

 

[PeroK]

Proposition
If $f_n$ are a sequence of real functions each of which has period $T$ and such that the pointwise limit $f(x) = \lim_{n \to \infty} f_n(x)$ exists, then $f$ also has period $T$.

I don't see that we need that in this case, as every term in the two series ($f(x)$ and $f(x+2\pi)$) are equal.

 

[physics1000]

Hi guys.
Thank you very much for the help.
I used all the information from here and managed to solve it.
I actually wrote two solutions ( one with the partial sums ) and other that I got information here.

 

[vela]

A function f is periodic, with a period T, iff f(x) = f(x+T).

To be pedantic, it shouldn't be an if-and-only-if there. The period is defined as the smallest $T$ for which $f(x)=f(x+T)$ for all $x$. Even though $\sin x = \sin(x+4\pi)$, we don't say $\sin x$ is periodic with a period of $4\pi$.

 

[DaveE]

To be pedantic, it shouldn't be an if-and-only-if there. The period is defined as the smallest $T$ for which $f(x)=f(x+T)$ for all $x$. Even though $\sin x = \sin(x+4\pi)$, we don't say $\sin x$ is periodic with a period of $4\pi$.

OK, I wouldn't argue with that. I would have (actually did) call that the fundamental period myself. But I'm an engineer, not a mathematician. We may have different semantics. You may want to edit this wikipedia page.

 

[PeroK]

OK, I wouldn't argue with that. I would have (actually did) call that the fundamental period myself. But I'm an engineer, not a mathematician. We may have different semantics. You may want to edit this wikipedia page.

I agree. The function $\sin x$ definitely has a period of $4\pi$ as well as $2\pi$.

 

[pasmith]

I don't see that we need that in this case, as every term in the two series ($f(x)$ and $f(x+2\pi)$) are equal.

It's not a difficult proof.

 

[DaveE]

To be pedantic, it shouldn't be an if-and-only-if there. The period is defined as the smallest $T$ for which $f(x)=f(x+T)$ for all $x$. Even though $\sin x = \sin(x+4\pi)$, we don't say $\sin x$ is periodic with a period of $4\pi$.

Anyway, now I'm confused. Can you explain, regardless of how many values T can take (because it's value wasn't specified), why it isn't iff in either interpretation? Is it the "if" part you don't like, or the "only if" part?

 

[vela]

Anyway, now I'm confused. Can you explain, regardless of how many values T can take (because it's value wasn't specified), why it isn't iff in either interpretation? Is it the "if" part you don't like, or the "only if" part?

If you define T as the smallest value, then if T is "the" period, we know f(x)=f(x+T) for all x, but if some value T satisfies f(x)=f(x+T) for all x, that doesn't imply T is "the" period but rather just "a" period. So it comes down to how one defines T. I've always seen it defined as the smallest value that satisfies f(x)=f(x+T) for all x, but apparently, not everyone defines it that way.

 

[PeroK]

If you define T as the smallest value, then if T is "the" period, we know f(x)=f(x+T) for all x, but if some value T satisfies f(x)=f(x+T) for all x, that doesn't imply T is "the" period but rather just "a" period. So it comes down to how one defines T. I've always seen it defined as the smallest value that satisfies f(x)=f(x+T) for all x, but apparently, not everyone defines it that way.

In this case, can you prove that $2\pi$ is the fundamental period of the function in question? It's easy to see that it does not have period $\pi$. But, it might be tricky to show that it can't be anything else less than $2\pi$.

 

[vela]

In this case, can you prove that $2\pi$ is the fundamental period of the function in question? It's easy to see that it does not have period $\pi$. But, it might be tricky to show that it can't be anything else less than $2\pi$.

I'd offer the physicist's proof: Look at the graph. ;)

 

[Office_Shredder]

In this case, can you prove that $2\pi$ is the fundamental period of the function in question? It's easy to see that it does not have period $\pi$. But, it might be tricky to show that it can't be anything else less than $2\pi$.

I believe it's a theorem that either the fundamental period is $2\pi/n$ for some $n$, or the function has arbitrarily small periods. The second one can't be true because the function is continuous and not constant (probably? Maybe this is a trick question haha). The first one doesn't feel impossible to manage but I agree isn't easy.

Edit to add: it is not constant, since $f(0)=0$ and $f(\pi/2)=1$