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

  1. Oct 17, 2013 #1
    Consider [itex]f(x) = \sin{(ex)}[/itex]
    and [itex]g(x) = \sin{(\pi x)}[/itex]
    Determine the period of [itex]f(x) + g(x)[/itex]?

    Is it possible?

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

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

    3. The attempt at a solution
    I just do not know the LCM of [itex]\frac{2\pi}{e}[/itex] and [itex]2[/itex]
  2. jcsd
  3. Oct 17, 2013 #2


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    e and pi are incommensurate; they have no LCM.

    And the sum f(x) + g(x) is not periodic.
  4. Oct 17, 2013 #3
    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.
  5. Oct 17, 2013 #4


    Staff: Mentor

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

  6. Oct 17, 2013 #5


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    Incommensurate: have no common factors.

    LCM is usually defined as an integer, though it can be extended to rationals.
  7. Oct 17, 2013 #6


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

    Edit: Sorry - that would be GCD, not LCM. The two are related:

    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
    Last edited: Oct 17, 2013
  8. Oct 17, 2013 #7


    Staff: Mentor

    I had forgotten that LCM normally applies to integers.
  9. Oct 17, 2013 #8
    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.
  10. Oct 17, 2013 #9

    D H

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    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.
  11. Oct 17, 2013 #10

    h(x)=f(x)+g(x) appears to be periodical with a period of approximately 16.08358.
  12. Oct 17, 2013 #11
    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.
    Last edited: Oct 17, 2013
  13. Oct 17, 2013 #12

    D H

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    Please don't post wide images like that!

    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.
  14. Oct 17, 2013 #13
    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.
  15. Oct 17, 2013 #14
    What I got from this topic (up to this moment):
    [itex]\sin{(\mathrm{e} x)} + \sin{(\pi x)}[/itex] is aperiodic.

    [itex]\sin{(2\pi x)} + \sin{(\pi x)}[/itex] has a definable period.

    What I would still like to know:

    How do I get the period of [itex]\sin{(2\pi x)} + \sin{(\pi x)}[/itex] 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.
  16. Oct 17, 2013 #15


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    What's wrong with your LCM method? In this case the two periods are integers.
  17. Oct 17, 2013 #16


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    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.
    Last edited: Oct 17, 2013
  18. Oct 17, 2013 #17
    I made a mistake there, I know the period of [itex]\sin{2\pi x} + \sin{\pi x}[/itex]

    Which two periods you were talking about?

    [itex]f(x) = \sin{2\pi x}[/itex]
    [itex]P_{f}=\frac{2\pi} {2\pi}=1[/itex]

    [itex]g(x) = \sin{\pi x}[/itex]
    [itex]P_{g}=\frac{2\pi} {\pi}=2[/itex]

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

    What looks right.

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

    If I need two INTEGERS k1 and k2 such that [itex]\frac{2\pi}

    {\mathrm{e}} \cdot k_{1}[/itex] equals [itex]2 \cdot k_{2}[/itex], then
    [itex]\frac{2\pi} {\mathrm{e}} \cdot k_{1} = 2 \cdot k_{2}[/itex]
    [itex]\frac{2\pi} {\mathrm{e}} \cdot \frac{k_{1}}{k_{2}} = 2[/itex]
    [itex]\frac{k_{1}}{k_{2}} = \frac{2\mathrm{e}}{{2\pi}}[/itex]
    [itex]\frac{k_{1}}{k_{2}} = \frac{\mathrm{e}}{{\pi}}[/itex]

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

    Was this proof any right?
  19. Oct 17, 2013 #18


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    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.
  20. Oct 17, 2013 #19
    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.
  21. Oct 17, 2013 #20


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    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.
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