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Period of trig functions

  1. Apr 18, 2008 #1
    Period of trig functions!!

    This is a rather easy-silly quiestion but i just don't know how to show it.

    I know how to find the period of trig functions of the form f(x)=Asin(wx+c) etc. i mean i know how to show that the period of this is [tex]\frac{2\pi}{w}[/tex].

    However, how would one find the period of let's say the following function, is there any analitycal method of showing it:

    [tex]f(x)=asin(wx+c)+kcos(f \pi x+d)-rtan(zx)[/tex]

    Is there any way to prove it analiytically?

    letters in front of x are constants.
  2. jcsd
  3. Apr 18, 2008 #2


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    Hi sutupidmath! :smile:

    There almost certainly isn't a period.

    The three individual parts have their own periods. Unless those periods have rational-number ratios, there won't be a common period!

    For example, the orbit of the moon is a combination of three periodic functions (one is called synodic … I forget the names of the other two), and they nearly coincide every 18 years, but not quite, and so eclipses aren't regular, but come in groups which only last a few hundred years. :cry:
  4. Apr 18, 2008 #3
    THIS is probably a more difficult question than i'm cabable of answering but here are my thoughts for what its worth

    i guess if you got the roots of the equation that would give you a starting point
    i could call the first root the start of the first period

    then i'm guessing at the start of each subsequent period the values of
    d^nf(x)/dx^n [thats the nth derivitive], should be the same

    for sin(x) for eg the roots all differ by pi
    then the first deriv is cos(x)
    evaluated at the roots gives you 1,-1, 1,-1, 1,-1
    so obviously every second root has the same value so the period is 2*pi

    i'm pretty darn sure this method is 99% unreliable
    and god knows how many derivites you'd have to go to
    but as tiny tim says there almost certainly isnt a period
    [i actually think there almost certainly should be if all terms are trig?]

    anyway i doubt this is of any use apart from food for thought
    Last edited: Apr 18, 2008
  5. Apr 19, 2008 #4


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    I thought it was well known that the period a sum of periodic functions was the least common multiple of their periods. If f has period p1 and g has period p2, P= np1, P= mp2, then (f+g)(P)= f(np1)+ g(np2)= f(0)+ g(0)= (f+g)(0).
  6. Apr 19, 2008 #5
    that works fine if there IS a common multiple.
  7. Apr 19, 2008 #6


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    ??Any finite collection of numbers has a common multiple: multiply them together!
  8. Apr 19, 2008 #7
    Halls, pardone my ignorance, but can you just elaborate a little bit more this. MOreover, my question should have read this way: How do we show that the least common multiple of a sum of periodic functions is their period?

    P.S. I was so ashamed to post this question here, i thought it was too obvious, but i am glad it is not that obvious by the way!!!
  9. Apr 19, 2008 #8
    Ok, I should have said if there is least common multiple. There is no such thing in the real numbers, since everything is a multiple of everything else.
  10. Apr 19, 2008 #9


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    We require that the LCM be an integer multiple of all the numbers involved. So we really want the Least Common INTEGER Multiple.
  11. Apr 20, 2008 #10
    Ok let's suppose that the constants before "x" are all integers, then how do we show that the LCM is the period of the sum of trig functions?
  12. Apr 20, 2008 #11
    I would think that you could evaluate it at a point, say x=0. Then evaluate it at x = common integer multiple (CIM) + 0 and show that the function has the same value in this case. You could then use induction to show that it is periodic with n*(CIM), using trig. identities and what not :smile:

    Find CIM.
    Show f(x+CIM)=f(x) for some arbitrary x.
    Assume, f(x+n*CIM)=f(x).
    Show that f(x+(n+1)*CIM)=f(x+n*CIM)

    That should work, please correct me if I'm wrong.
    Last edited: Apr 20, 2008
  13. Apr 20, 2008 #12

    I think this suggestion works, lets see how i went about it, correct me if i am missing sth.

    Let :

    [tex]f(x)=sin(ax+b)+cos(cx+d)[/tex] where a,c are integers.



    Then from here i can show that [tex] T_1=\frac{2\pi}{a},T_2=\frac{2\pi}{c}[/tex]

    Now, let the Leas Common Multiple of T_1,T_2 be :

    [tex] n\pi[/tex] such that :

    [tex]r_1 *\frac{2\pi}{a}=n\pi,and,r_2 * \frac{2\pi}{c}=n\pi=>2r_1\pi=an\pi,2r_2\pi=cn\pi[/tex]

    Where r_1,r_2 are also integers.

    Now, let's try to show that



    Now since [tex] 2r_1\pi, 2r_2\pi[/tex] will always be even, it means that the above expretion is equal to

    [texf(x)=sin(ax+b)+cos(cx+d)[/tex] What we needed to show.

    Now i think i could extend it to n+1 also, by just supposing first that its period is npi, and then trying to show that also its period is (n+1)pi.

    Would this be approximately correct??
    Last edited: Apr 20, 2008
  14. Apr 21, 2008 #13
    In performing the above step you have shown that the inductive hypothesis
    f(x)=f(x+n*LCM) is true, so you're done. :smile: {Assuming its correct if course, I could only skim through the working}
    The normal approach is to assume this statement and show that f(x+(n+1)*LCM)=f(x+n*LCM).
    Last edited: Apr 21, 2008
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