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Function's period

  1. Feb 13, 2004 #1
    [more on the same] function's period

    i have this function(taken from fourier analysis):

    n - 1,2,3...

    f(x) = cos( (2*pi*n)/L * x )

    the literature says this function has L period (n parameter):

    f(x + L) = cos( (2*pi*n)/L * (x + L) ) =
    = cos( 2*pi*n/L * x + 2*pi*n) = cos( 2*pi*n/L * x)

    so it's true that L f's period for any n.....
    but L/n is also f's period because:

    f(x + L/n) = cos( (2*pi*n)/L * (x + L/n) ) =
    = cos( 2*pi*n/L * x + 2*pi) = cos( 2*pi*n/L * x)


    so what happens here??
     
    Last edited: Feb 13, 2004
  2. jcsd
  3. Feb 13, 2004 #2

    matt grime

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    n varies, and as there is no summation there, one only conclude that when you treat f as unique, rather than a family of functions, one for each n, that you might have misread the literature
     
  4. Feb 13, 2004 #3
    actually there is a summation over n
    it is just one term from fourier series...
     
  5. Feb 13, 2004 #4

    matt grime

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    So, you need to know why

    [tex]\sum_n b_n\cos(2\pi nx/L)[/tex]

    has period L and not L/n? Erm, is because n varies not an acceptable anwer?

    It doesnt't matter that each individual term may have period less than L, only that the sum has period L.

    And assuming b_n are such that that sum makes sense obviously
     
    Last edited: Feb 13, 2004
  6. Feb 13, 2004 #5
    you are right...:smile: (too much reading)

    thank you very much
     
  7. Feb 13, 2004 #6
    i want to rephrase what i asked before

    lets take this function,

    f(x)=cos(2*pi*3/L*x)

    n is parameter and lets say n=3 so f(x) becomes

    f(x)=cos(2*pi*3/L*x)

    as i said earlier both numbers L/3 and L
    are sutisfying the following condition

    (#) f(x + T)=f(x): T f's period

    or to be more specific

    f(x + L)=f(x) and f(x + L/3)=f(x)

    I asked how it can be that both equilities are true...
    Meanwhile it occured to me that this condition (#) alone doesn't says
    that T is a period and if we want T to be a period we must demand
    that T is also minimal number for which (#) holds

    so the answer to my original question is that f's period is L/n

    Is what i wrote is correct?
     
  8. Feb 13, 2004 #7

    matt grime

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    cos(x) has period 2pi

    cos(kx) has period 2pi/k for every k in R

    that enough?
     
  9. Feb 13, 2004 #8
    you mean that cos(kx) has a period 2*p/k?

    but what about what i said before that?
    ...that period is defined by:

    1) f(x+L)=f(x)
    2) L is minimal among all other numbers

    is that correct?
     
  10. Feb 13, 2004 #9

    matt grime

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    yep that looks about right. wolfram's mathworld is your friend for these things
     
  11. Feb 13, 2004 #10
    Hi wormhole,

    Let us say that L or L/n are circle's radius.

    So, in both cases you have f(x+circle)=f(x)
     
  12. Feb 13, 2004 #11
    thanks guys for your help:smile:
    no more question...
     
  13. Feb 14, 2004 #12

    matt grime

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    Erm, we could say that, but it would be wrong.
     
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