Period of sin(e*x) + sin(pi*x)

[besulzbach]

Consider f(x) = \sin{(ex)}
and g(x) = \sin{(\pi x)}
Determine the period of f(x) + g(x)?

Is it possible?

Would LCM[\frac{2\pi}{e}, 2] = \frac{2\pi}{e}?

Homework Equations

e = \lim_{h \to 0} (1 + h)^\frac{1}{h} \approx 2.718281828
The period of the sum of the functions would be the LCM of their periods.

The Attempt at a Solution

I just do not know the LCM of \frac{2\pi}{e} and 2

 

[UltrafastPED]

e and pi are incommensurate; they have no LCM.

And the sum f(x) + g(x) is not periodic.

 

[besulzbach]

e and pi are incommensurate; they have no LCM.

And the sum f(x) + g(x) is not periodic.

Thanks, but, would the LCM of 2e and e exist? They are incommensurate, but, nonetheless, it seems really straightforward to stipulate that LCM.

If that is possible, sin(ex) + sin(2ex) would be considered periodic with a period of 2pi/e?

Thanks in anticipate, I have no background whatsoever in this area and this week I just can't pick up any math book.

 

[Mark44]

e and pi are incommensurate; they have no LCM.

I don't know what you mean about incommensurate - e and $\pi$ certainly have no factors in common, but their LCM is e*$\pi$.

And the sum f(x) + g(x) is not periodic.

 

[UltrafastPED]

Incommensurate: have no common factors.

LCM is usually defined as an integer, though it can be extended to rationals.

 

[UltrafastPED]

The LCM of (2x, x) is 2 as the x divides out.

Edit: Sorry - that would be GCD, not LCM. The two are related:
http://www.cut-the-knot.org/arithmetic/GcdLcmProperties.shtml

As to the ratio of e/pi, the continued fraction expansion (Thanks, Wolfram|Alpha!) is:
[0; 1, 6, 2, 2, 1, 2, 6, 8, 2, 1, 1, 1, 4, 3, 1, 1, 66, 2, 1, 1, 2, 4, 2, 7, 46, 10, 2,
1, 18, 1, 23, 10, 14, 1, 1, 4, 3, 1, 6, 2, 5, 1, 1, 1, 18, 355, 1, 1, 5, 3, 4, 1, ...]

e has a structure to its continued fraction expansion; pi does not. This looks more like pi.
The CF expansion provides a "best estimate" rational number for each expansion, getting better with each additional "term". See http://mathworld.wolfram.com/ContinuedFraction.html

 

[Mark44]

I had forgotten that LCM normally applies to integers.

 

[besulzbach]

The LCM of (2x, x) is 2 as the x divides out.

What? So:
LCM[222, 444]
= LCM[222, 222 * 2]
= 2? (as the 222 would cancel out?)
LCM of 222 and 444 is clearly 444.
As the GCD is 222.

 

[D H]

e and pi are incommensurate; they have no LCM.

As far as I know, nobody knows whether pi/e is rational or irrational. It's one of many outstanding problems concerning combinations pi and e.

 

[ainster31]

h(x)=f(x)+g(x) appears to be periodical with a period of approximately 16.08358.

 

[besulzbach]

h(x)=f(x)+g(x) appears to be periodical with a period of approximately 16.08358.

That is the plot of sin(ex) + sin(πx)
period of sin(ex) = 2π/e
period of sin(πx) = 2π/π = 2
Therefore the period of sin(ex) + sin(πx) would be the LCM of 2π/e and 2.
I think that the this LCM is 2π/e. But the graphics show that I'm absolutely wrong.

 

[D H]

<image elided>

Please don't post wide images like that!

h(x)=f(x)+g(x) appears to be periodical with a period of approximately 16.08358.

How did you get that figure? I see ten beats between -96 and +52, or an apparent period of about 14.8. What you are seeing in that graph is the beat frequency, 2*pi/(pi-e) ≈ 14.84. It has *nothing* to do with whether pi/e is rational. It has *everything* to do with the fact that pi and e are somewhat close together. You get beats.

 

[ainster31]

It appears to be periodic with $T=\frac { 30π }{ e+π }$.

I'm not sure. Hopefully someone can clarify.

Edit: It looks like D H has clarified.

 

[besulzbach]

What I got from this topic (up to this moment):
\sin{(\mathrm{e} x)} + \sin{(\pi x)} is aperiodic.

\sin{(2\pi x)} + \sin{(\pi x)} has a definable period.

What I would still like to know:

How do I get the period of \sin{(2\pi x)} + \sin{(\pi x)} with arbitrary precision?

I don't want to blankly stare at a graph and guess values, I want a formula, method, something.

I thank you all again for your answers.

 

[Dick]

What I got from this topic (up to this moment):
\sin{(\mathrm{e} x)} + \sin{(\pi x)} is aperiodic.

\sin{(2\pi x)} + \sin{(\pi x)} has a definable period.

What I would still like to know:

How do I get the period of \sin{(2\pi x)} + \sin{(\pi x)} with arbitrary precision?

I don't want to blankly stare at a graph and guess values, I want a formula, method, something.

I thank you all again for your answers.

What's wrong with your LCM method? In this case the two periods are integers.

 

[Dick]

And in the case where they are not integers, if you call the periods p1 and p2, you want to find the smallest number n such n=k1*p1 and n=k2*p2 where k1 and k2 are integers. Just think about what periodic means. And if you want an algorithm in the case where p1 and p2 are rational, then express them over the least common denominator, so p1=q1/d and p2=q2/d. Then seems to me like the LCM should be defined as LCM(q1,q2)/d.

 

[besulzbach]

What's wrong with your LCM method? In this case the two periods are integers.

I made a mistake there, I know the period of \sin{2\pi x} + \sin{\pi x}

Which two periods you were talking about?

f(x) = \sin{2\pi x}
P_{f}=\frac{2\pi} {2\pi}=1

g(x) = \sin{\pi x}
P_{g}=\frac{2\pi} {\pi}=2

h(x) = f(x) + g(x) = \sin{2\pi x} + \sin{\pi x}
P_{h}=LCM[2, 1]=2

What looks right.

Last doubt is:
\sin{\mathrm{e} x} + \sin{\pi x} is really aperiodic?
Is there a LCM to \frac{2\pi} {\mathrm{e}} and 2?

If I need two INTEGERS k₁ and k₂ such that \frac{2\pi} <br /> <br /> {\mathrm{e}} \cdot k_{1} equals 2 \cdot k_{2}, then
\frac{2\pi} {\mathrm{e}} \cdot k_{1} = 2 \cdot k_{2}
\frac{2\pi} {\mathrm{e}} \cdot \frac{k_{1}}{k_{2}} = 2
\frac{k_{1}}{k_{2}} = \frac{2\mathrm{e}}{{2\pi}}
\frac{k_{1}}{k_{2}} = \frac{\mathrm{e}}{{\pi}}

Supposing that \frac{\mathrm{e}}{{\pi}} \notin \mathbb{Q}
And that there are no k_{1} \in \mathbb{Z} and k_{2} \in \mathbb{Z} such that \frac{k_{1}}{k_{2}} \notin \mathbb{Q}
We can assume that \sin{\mathrm{e} x} + \sin{\pi x} is aperiodic.

Was this proof any right?

 

[Dick]

I made a mistake there, I know the period of \sin{2\pi x} + \sin{\pi x}

Which two periods you were talking about?

f(x) = \sin{2\pi x}
P_{f}=\frac{2\pi} {2\pi}=1

g(x) = \sin{\pi x}
P_{g}=\frac{2\pi} {\pi}=2

h(x) = f(x) + g(x) = \sin{2\pi x} + \sin{\pi x}
P_{h}=LCM[2, 1]=2

What looks right.

Last doubt is:
\sin{\mathrm{e} x} + \sin{\pi x} is really aperiodic?
Is there a LCM to \frac{2\pi} {\mathrm{e}} and 2?

If I need two INTEGERS k₁ and k₂ such that \frac{2\pi} <br /> <br /> {\mathrm{e}} \cdot k_{1} equals 2 \cdot k_{2}, then
\frac{2\pi} {\mathrm{e}} \cdot k_{1} = 2 \cdot k_{2}
\frac{2\pi} {\mathrm{e}} \cdot \frac{k_{1}}{k_{2}} = 2
\frac{k_{1}}{k_{2}} = \frac{2\mathrm{e}}{{2\pi}}
\frac{k_{1}}{k_{2}} = \frac{\mathrm{e}}{{\pi}}

Supposing that \frac{\mathrm{e}}{{\pi}} \notin \mathbb{Q}
And that there are no k_{1} \in \mathbb{Z} and k_{2} \in \mathbb{Z} such that \frac{k_{1}}{k_{2}} \notin \mathbb{Q}
We can assume that \sin{\mathrm{e} x} + \sin{\pi x} is aperiodic.

Was this proof any right?

As DH pointed out, nobody REALLY knows whether e/pi is rational or not. Googling around seems to confirm this. So it's a pretty bad question in that regard. But if e/pi were proved to be irrational that would show it.

 

[besulzbach]

As DH pointed out, nobody REALLY knows whether e/pi is rational or not. Googling around seems to confirm this. So it's a pretty bad question in that regard. But if e/pi were proved to be irrational that would show it.

I understand it.

So, if we suppose they are irrational*, then we can say that sin(ex) + sin(πx) IS aperiodic, right?

And, of course, that e/π is also irrational.

 

[Dick]

I understand it.

So, if we suppose they are irrational*, then we can say that sin(ex) + sin(πx) IS aperiodic, right?

And, of course, that e/π is also irrational.

Yes, if you suppose that the ratio is irrational (and I think almost everybody would suspect that it is - it would be really surprising if e were a rational multiple of pi), then the function is aperiodic. And e and pi are irrational. That at least has been proved. In fact, they are transcendental. But that they are individually irrational doesn't matter, it's the ratio that matters.

 

[besulzbach]

Got everything clarified, thanks to everyone who answered my questions.