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Periodicity of function

  1. Oct 30, 2008 #1
    1. The problem statement, all variables and given/known data

    Check the periodicity of the function f(x)=sinx*cosx, and if it is periodic, find the period of the function.

    2. Relevant equations

    The function f(x), [itex]x \in D[/itex], it is periodic if there is real number [itex]T \neq 0[/itex], so that:

    1o [itex]x \in D[/itex] and [itex]x \pm T \in D[/itex];

    2o f(x+T)=f(x)

    3. The attempt at a solution

    [tex]f(x)=sinx*cosx=sin(x+2\pi)*cos(x+2\pi)[/tex]

    How will I continue now?
     
  2. jcsd
  3. Oct 30, 2008 #2

    HallsofIvy

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    Good, that shows that f is periodic. Further it shows that [itex]2\pi[/itex] is a period. It is not necessarily the period, the fundamental period- that is, the smallest period. You do know that every period is a multiple of the the fundamental period. So, if the fundamental period is not [itex]2\pi[/itex], it must be [itex]\pi[/itex], or [itex]\pi/2[/itex], or [itex]\pi/3[/itex] (very unlikely I would think) or [itex]\pi/4[/tex], ..., Try those and see if any is the period.
     
  4. Oct 30, 2008 #3

    Mark44

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    Dyavol,
    Do you know any trig identies that involve sin(x)cos(x)?
     
  5. Oct 30, 2008 #4
    Thanks for the reply. I tried with [itex]\pi[/itex] and it works.

    Mark44, yes I do.

    Maybe I should use sin2x=sinx*cosx

    So if I try:

    [tex]sin(2x+2\pi)=sin(2(x+\pi))[/tex]

    So, the [itex]\pi[/itex] is the period.

    HallsofIvy and Mark44 thanks for the help.
     
  6. Oct 30, 2008 #5

    Mark44

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    The identity is actually sin2x = 2*sinx * cosx.
     
  7. Oct 30, 2008 #6
    Yes, sorry, it should be [itex]\frac{sin2x}{2}=sinx*cosx[/itex]. But anyway, the period will be same.
     
  8. Dec 7, 2008 #7
    Another problem:
    [tex]f(x)=sinx+\frac{1}{2}sin2x+\frac{1}{3}sin3x[/tex]
    My way of solving:
    [tex]\frac{6sinx+3sin2x+2sin3x}{6}=\frac{6sinx+3sin2x+2sin(2x+x)}{6}=\frac{6sinx+3sin2x+2(sin2x*cosx+cos2x*sinx)}[/tex][tex]{6}=\frac{6sinx+3sin2x+4sinx*cos^2x+2cos^2x*sinx-2sin^3x}{6}[/tex][tex]
    =\frac{6sinx+3sin2x+4sinx-4sin^3x+2sinx-2sin^3x-2sin^3x}{6}[/tex]
    [tex]=\frac{12sinx-12sin^3x+6sinx*cosx}{6}=
    2sinx-2sin^3x+sinxcosx=sinx(2-2sin^2x+cosx)=[/tex][tex]
    sinx(2-2+2cos^2x+cosx)=sinx(2cos^2x+cosx)=
    sinx*cosx(2cosx+1)[/tex]
    I am stuck in here. Aren't any simpler way out?

    Thanks in advance.
     
    Last edited: Dec 7, 2008
  9. Dec 7, 2008 #8
    Since that method doesn't seem to get you anywhere, first find the periods of sin x, sin(2x), and sin(3x) individually. You want the lowest number that is a multiple of all 3 periods.
     
  10. Dec 8, 2008 #9
    sin(x+2п)+sin(2x+2п)+sin(3x+2п)=sin(x+2п)+sin2(x+п)+sin3(x+2п/3).
    The periods are 2п, п and 2п/3. How this method works? Could you please give me some link to see how this works?
    Thanks.
    Regards.
     
  11. Dec 8, 2008 #10

    Redbelly98

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    If you have three periods A, B, and C, then you need to find the least common multiple of those 3 numbers.
     
  12. Dec 8, 2008 #11
    Is that some theorem? Could you please tell me where that theorem comes from?
     
  13. Dec 8, 2008 #12

    Redbelly98

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    It is probably a theorem, but my education and experience is more science and engineering rather than math. So I don't know where the theorem comes from, I just know that it will give the (fundamental) period of the function.

    Once you have the least common multiple of 2п, п and 2п/3, you can at least verify that it fits the definition of a period (Equation #2 from your post #1).
     
  14. Dec 8, 2008 #13
    How to find it? Sorry for not understanding I got translation problem. Should I multiple all of them?
     
  15. Dec 8, 2008 #14

    HallsofIvy

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    You should have learned how to find the least common multiple of three numbers in arithmetic: the least common denominator used to add or subtract fractions is the least common multiple of the denominators. Multiply the numbers together will give a common multiple but not necessarily the least common multiple.

    And I don't know if we should characterize the fact that if f has period fundamental p1 and g has fundamental period p2[/sup], then f+ g has fundamental period equal to the least common multiple of p1[/sup] and p2[/sup] as a "theorem". It should be clear that any period of f+ g must be a period of both f and g: and so a multiple of the fundamental period of each.
     
  16. Dec 8, 2008 #15
    Oh, do you mean like [tex]\frac{1}{2}+\frac{1}{3}+1[/tex]. The least common multiple here is 6, right? In my country we call it НЗС.

    I got the periods 2п, п and 2п/3 Now, probably [tex]\frac{6\pi+3\pi+2\pi}{3}=\frac{11\pi}{3}[/tex] is the least common multiple.
     
    Last edited: Dec 9, 2008
  17. Dec 8, 2008 #16

    Redbelly98

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    FYI, when using LaTex you'll need to use \pi for the Greek letter pi. The п symbol does not work withing LaTex for some reason.

    That being said: for 11п/3 to be a common multiple of 2п, п and 2п/3 the following must all be true:

    1. 11п/3 is an integer multiple of 2п,
    2. 11п/3 is an integer multiple of п,
    AND
    3. 11п/3 is an integer multiple of 2п/3

    Are these true statements? If yes, you have found a common multiple. If not, can you find a number that would make those statements true?

    (And finally, you'll need to find the smallest positive number that makes those 3 statements true.)
     
  18. Dec 8, 2008 #17
    Once again as mentioned, find individual periods of sin(x), sin(2x) and sin(3x) and then find the LCM. If you experience difficulty in understanding why this is so, try graphing the individual components of the function on the same graph.
     
  19. Dec 10, 2008 #18
    Thanks for the replies.
    @Redbelly I got problem with translation, but I forget to look on Google, so I now sow on some random sites what is LCM. In Russia we call it НОК.

    п, 2п , 2п/3 | п
    1, 2, 2/3 | :2
    1, 1, 1/3

    As far as I know LCM is only about integers.

    As I suppose 2п is the least common multiple of all of them.

    @arunbg I made graph and sow that all of them intercept in 2п.
     
  20. Dec 10, 2008 #19

    Redbelly98

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    Yes, that's right.

    2п is an integer multiple of all 3 periods:
    1 x 2п
    2 x п
    3 x 2п/3
     
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