Proving period of f/g could be less than period of f and g

[a1b2c3zzz]

Homework Statement

Prove that if f and g are periodic with period p , then f/g is also periodic, but its period could be smaller than p.

The Attempt at a Solution

So, the first part seems simple enough.

$\frac{f(x+p)}{g(x+p)}=\frac{f}{g}(x + p)=\frac{f}{g}(x)$

But how exactly do I show that the period could be smaller? Here's what I'm thinking:

$\frac{-f}{-g}(x)=\frac{f}{g}(x)$

and what I get from this is that the values of the function would repeat at another interval (where they are both negative) as well as the interval np. However, I could be looking at this entirely the wrong way. Any hints you guys can give me?

 

[micromass]

You need to find a concrete example.

I guess taking $f=g$ would be cheating?

What are some periodic functions you know?

 

[AlephZero]

I guess taking $f=g$ would be cheating?

I don't see why it's cheating. If the question meant that $p$ was the smallest period of $f$ and $g$, it should have said so!

 

[a1b2c3zzz]

Yes, I'm taking trig so I know the periods of the trigonometric functions (on the unit circle, that is - we are fairly early into the course). And also, I can easily see this with the tangent function

$\sin (x+2\pi)=\sin x$ and $\cos (x+2\pi)=\cos x$

$\sin (x+\frac{2\pi}{2})=\sin (x+\pi)$ and $\cos (x+\frac{2\pi}{2})=\cos (x+\pi)$

$\sin (x+\pi)=-\sin x$ and $\cos (x+\pi)=\cos x$

$\frac{\sin (x+\pi)}{\cos (x+\pi)}=\frac{-\sin x}{-\cos x}=\frac{\sin x}{\cos x}$

But would it work for absolutely any? I suppose that I should ask, do all periodic functions behave this way? Forgive me, this is my first encounter with them haha.

 

[Ray Vickson]

Yes, I'm taking trig so I know the periods of the trigonometric functions (on the unit circle, that is - we are fairly early into the course). And also, I can easily see this with the tangent function

$\sin (x+2\pi)=\sin x$ and $\cos (x+2\pi)=\cos x$

$\sin (x+\frac{2\pi}{2})=\sin (x+\pi)$ and $\cos (x+\frac{2\pi}{2})=\cos (x+\pi)$

$\sin (x+\pi)=-\sin x$ and $\cos (x+\pi)=\cos x$

$\frac{\sin (x+\pi)}{\cos (x+\pi)}=\frac{-\sin x}{-\cos x}=\frac{\sin x}{\cos x}$

But would it work for absolutely any? I suppose that I should ask, do all periodic functions behave this way? Forgive me, this is my first encounter with them haha.

If $a(x)$ has period $p$, what are the periods of $f_1(x) = a(x) + a(x + p/2)$ and $f_2(x) = a(x)\, a(x + p/2)$?

 

[epenguin]

f could have g as a factor and another factor that has a frequency that is a multiple of 1/ρ .

In fact to think about whether the given conditions could be satisfied any other way?

 

[a1b2c3zzz]

Hey guys, thanks for the help so far! However I'd like to ask for no more hints. I've had other coursework to do so I haven't had much time to look at this one, but I will come back to it when I get the chance and post my solution.

EDIT: I came up with this using $f(x)=g(x)$

If $f(x)=g(x)$ and $f$ and $g$ both have period $p$

then $\frac{f}{g}(x+p)=1=\frac{f}{g}(x)$

and $\frac{f}{g}(x+\frac{p}{2})=1=\frac{f}{g}(x)$

but $f(x+\frac{p}{2})\neq f(x)$ and $g(x+\frac{p}{2})\neq g(x)$

Does this look correct to you guys?

 

[epenguin]

If f = g then f/g = 1 . You could say that repeats with any period like to mention, question of definition, rather trivial and surely not what they are looking for. Think about my suggestion and question.

 

[a1b2c3zzz]

If f = g then f/g = 1 . You could say that repeats with any period like to mention, question of definition, rather trivial and surely not what they are looking for. Think about my suggestion and question.

I seem to be missing some fundamental information on combining functions.

f could have g as a factor and another factor that has a frequency that is a multiple of 1/ρ .

In fact to think about whether the given conditions could be satisfied any other way?

I am not sure how to go about this, but here is my best shot.

If $g$ is a factor of $f$ and both have period $p$

then $\frac{f(x+p)}{g(x+p)}$ will equal an integer, correct?

and if $b$ is a factor of $f$ and has period $1/p$

then $\frac{f(x+p)}{b(x+\frac{1}{p})}$... well, this will also equal an integer but how does this affect the period?

Here is my try with your suggestion, Ray.

If a(x) has period p, what are the periods of f1(x)=a(x)+a(x+p/2) and f2(x)=a(x)a(x+p/2)?

So if $a(x)$ has a period of $p$...

The period of $f1(x)=a(x)+a(x+p/2)$ is, maybe, $\frac{p}{4}$? Would that total the values to $p$ by adding it to each $a$?

The second one is even more confusing to me, and I feel like a big dummy. However, in my book, in the "combining functions" section, there isn't a single problem that I cannot do. Is there something I am missing?

 

[a1b2c3zzz]

Sorry to double post guys. The only thing I can figure out, with the given knowledge in the textbook, is setting $f(x)=sin x$ and $g(x)=cos x$ or setting $f=g$.

The suggestions you guys have given me seem to go over my head.

I suppose I should point out that I am in a precalculus class. The only mention of periodic functions in the text is the definition

"A function $f$ is periodic if there is a positive number $p$ such that $f(t+p) = f(t)$ for every $t$."

I don't believe that they expect us to know more than this. The focus of the book is on plug-n-chug problems, but even the other more theoretical problems haven't given me this much trouble. I was thinking that perhaps it was possible that problems like this show up later, in a much more general sense, and require knowledge I have not yet attained.

 

[epenguin]

I seem to be missing some fundamental information on combining functions.
I am not sure how to go about this, but here is my best shot.

If $g$ is a factor of $f$ and both have period $p$

then $\frac{f(x+p)}{g(x+p)}$ will equal an integer, correct?

and if $b$ is a factor of $f$ and has period $1/p$

then $\frac{f(x+p)}{b(x+\frac{1}{p})}$... well, this will also equal an integer but how does this affect the period?

then $\frac{f(x+p)}{g(x+p)}$ will equal an integer, correct?

Er no $\frac{f(x+p)}{g(x+p)}$ merely equals $\frac{f(x)}{g(x)}$ which is what you showed in #1 , or what I think you meant to say (your notation of it is dodgy) .

You are making too heavy weather of this problem which is not one of deep mathematical knowledge or calculation of anything you don't know, perhaps you are overthinking.

We are not supposed to give answers to the problems but I have!

 

[Redbelly98]

Sorry to double post guys. The only thing I can figure out, with the given knowledge in the textbook, is setting $f(x)=sin x$ and $g(x)=cos x$ ...

And that is enough to answer the given problem.

...or setting $f=g$.

Whenever you find a trivially true solution, you might want to check with your teacher/professor about accepting that as an answer. Or, submit it along with the more interesting solution. Yes, technically any constant function is periodic, but if you are then asked to give the period of a constant function...

 

[epenguin]

Sorry, I don't think that's a trivial answer, I think it's not an answer at all. The question asked for the period of f/g not to be equal or less than that of f, g but to be less. Your ratio (which inessentially is tan(x) ) is just equal period to the two functions you give.

My point in a semi-concrete example was: suppose both f and g repeat themselves every 10 seconds. Then so does f/g, I think we arrived at that.

Suppose f(x) is a product f(x) = g(x)h(x) and h has periodicity 1 sec. What is the periodicity of f/g in that case?

Nothing calling on any special knowledge.

 

[micromass]

Sorry, I don't think that's a trivial answer, I think it's not an answer at all. The question asked for the period of f/g not to be equal or less than that of f, g but to be less. Your ratio (which inessentially is tan(x) ) is just equal period to the two functions you give.

My point in a semi-concrete example was: suppose both f and g repeat themselves every 10 seconds. Then so does f/g, I think we arrived at that.

Suppose f(x) is a product f(x) = g(x)p(x) and p has periodicity 1 sec. What is the periodicity of f/g in that case?

Nothing calling on any special knowledge.

The period of the tangent function is less than the period of sine and cosine, so it's a valid answer.

 

[epenguin]

The period of the tangent function is less than the period of sine and cosine, so it's a valid answer.

oops yes. But an infinity of examples I gave are right.

 

[micromass]

oops yes. But an infinity of examples I gave are right.

Certainly. Your post is much deeper than just suggesting the tan(x) counterexample!

 

[a1b2c3zzz]

Hey guys. I asked my instructor for a solution a few days ago (it was not an assigned problem) and he wrote this, almost verbatim (this is from my notebook):

Since $f$ and $g$ are periodic with $p$,

$f(x+p)=f(x)$ for all $x$ and $g(x+p)=g(x)$ for all $x$.

Thus $\frac{f(x+p)}{g(x+p)} = \frac{f}{g}(x)$ for all $x$ unless $g$ is undefined in which case both are undefined.

But consider $f(x)=\sin x$ (period $2\pi$) and $g(x)=\cos x$ (period $2\pi$). Their quotient $\frac{f(x)}{g(x)} = \frac{\sin x}{\cos x} = tan x$ has a period of $\pi$

So I suppose I already had the answer they were looking for.

Sorry I didn't understand, epenguin, but I appreciate the help!

 

[Redbelly98]

Sorry, I don't think that's a trivial answer, I think it's not an answer at all.

Sorry if I wasn't clear, I meant that f=g (so f/g=1) is a trivial answer, not the sin/cos solution.

 

[Ray Vickson]

Homework Statement

Prove that if f and g are periodic with period p , then f/g is also periodic, but its period could be smaller than p.

The Attempt at a Solution

So, the first part seems simple enough.

$\frac{f(x+p)}{g(x+p)}=\frac{f}{g}(x + p)=\frac{f}{g}(x)$

But how exactly do I show that the period could be smaller? Here's what I'm thinking:

$\frac{-f}{-g}(x)=\frac{f}{g}(x)$

and what I get from this is that the values of the function would repeat at another interval (where they are both negative) as well as the interval np. However, I could be looking at this entirely the wrong way. Any hints you guys can give me?

I hope enough time has passed that I will not violate PF rules by posting a solution. Say $a(x), b(x)$ are two (different) functions with period $p$. Then, of course, $a(x)/b(x)$ is periodic, and has period $p$ or less; assume its period is not less than $p$. Then the function
F(x) = \frac{a(x) \, a\left(x+\frac{p}{2}\right)}{b(x)\, b\left(x+\frac{p}{2}\right)}
has period $p/2$. It can be written in the form
F(x) = \frac{f(x)}{g(x)}, \text{ with } f(x) = \frac{a(x)}{b(x)}\text{ and }\:<br /> g(x) = \frac{b\left(x+\frac{p}{2}\right)}{a\left(x+\frac{p}{2}\right)}.
Here the functions $f(x), g(x)$ have period $p$.

Using a similar construction you can cook up $f,g$ of period $p$ but with $f/g$ of period $p/n$, for any fixed integer $n > 2$.

 

[Redbelly98]

I hope enough time has passed that I will not violate PF rules by posting a solution.

Since the OP has posted with a valid solution already, I think you are on safe ground.

.

Ray, I may be missing something obvious but it is not evident to me how your method extends to n>2.

If n=3, then I presume we are to consider the function

F(x) = \frac{a(x) \ a(x+p/3) \ a(x+2p/3)}{b(x) \ b(x+p/3) \ b(x+2p/3)},

which has period p/3. But then I am having trouble identifying suitable functions for f(x) and g(x).

 

[Ray Vickson]

Since the OP has posted with a valid solution already, I think you are on safe ground.

.

Ray, I may be missing something obvious but it is not evident to me how your method extends to n>2.

If n=3, then I presume we are to consider the function

F(x) = \frac{a(x) \ a(x+p/3) \ a(x+2p/3)}{b(x) \ b(x+p/3) \ b(x+2p/3)},

which has period p/3. But then I am having trouble identifying suitable functions for f(x) and g(x).

You could take
f(x) = \frac{a(x)}{b(x) \, b(x+p/3)}, \: g(x) = \frac{b(x + 2p/3)}{a(x+p/3)\, a(x+2p/3)}
or various other similar combinations.