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A very weird problem

  1. Sep 24, 2006 #1
    what would be the period of cos(2pi*x)*sin(2pi*x)??
    calculator is telling me it's 2, but i somehow dont wanna believe that



    also, what would be the period of cos(2pi*m*x)*sin(2pi*m*x)??
     
  2. jcsd
  3. Sep 24, 2006 #2

    SGT

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    [tex]cos 2\pi x sin 2\pi x = \frac{1}{2}sin 4\pi x[/tex],
    so the period is [tex]\frac{4\pi}{2\pi} = 2[/tex]
     
  4. Sep 24, 2006 #3
    That's almost correct, but the period is 1/2: if x increases from 0 to 1/2, then [tex] 4 \pi x[/tex] increases from 0 to [tex]2 \pi[/tex].
     
  5. Sep 24, 2006 #4
    As a rule, the period of sin(n*x) or cos(n*x) is 2*pi/n. More generally, if f(x) is a periodic function with period T, than f(n*x) has period T/n. You can deduce this simply by thinking that when you multiply the argument of a periodic function by a number you actually multiply its speed of oscillation by that number, meaning that you divide its period by that number.
    So the period of cos(2pi*x) is 1. The same for sin(2pi*x). To calculate the period of cos(2pi*x)*sin(2pi*x) you can write these two functions in the complex form and see that their product has the same period as cos(4pi*x) which is 0.5. Do the same for cos(2pi*m*x)*sin(2pi*m*x).
     
    Last edited: Sep 24, 2006
  6. Sep 24, 2006 #5

    SGT

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    You're right
     
  7. Sep 24, 2006 #6
    i agree with you people that the period should be 1/2.
    but when i look at the graph, it seems like it's 2!
    can anybody plot for me and see if they are getting the period 1/2 or 2, just based on the plot?
    thanks!
     
  8. Sep 25, 2006 #7

    HallsofIvy

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    Staff Emeritus
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    I graphed it on a TI 93 and it looks to me like the period is clearly 1/2. Of course, since 2 is a multiple of 1/2, 2 is a period.
     
  9. Sep 25, 2006 #8

    SGT

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    Here is a plot.
     

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