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Show cos²(x) is periodic

  1. Jun 4, 2007 #1
    1. The problem statement, all variables and given/known data
    Show cos²(x) is periodic

    3. The attempt at a solution
    I don't know how :confused:, could somebody please show me step by step?
    Thanks in advance!

    edit: I've got another show/justify question.
    If f is an odd function, then show f of f is an odd function.

    Whilst I know odd of odd results in an odd function, it seems such an explanation is not suffucient.
    I have been shown this example, but can't really make any sense of it
    f o f(-x) = f(f(-x))
    = f(-f(x))
    = -f(f(x))
    = -f o f(x)
     
    Last edited: Jun 4, 2007
  2. jcsd
  3. Jun 5, 2007 #2

    D H

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    A function f(x) is periodic if there exists some P>0 such that f(x+P)=f(x) over the domain of f. Try rewriting cos2(x) in terms of cos(2x).
     
  4. Jun 5, 2007 #3

    Mute

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    For the first one, a function f(x) is periodic, with a period of T if f(x) = f(x + T). You can use this fact to solve your first problem:

    [tex]\cos^2(x) = \cos^2(x + T)[/tex]

    A useful identity is [tex]\cos^2x = \frac{1}{2}\left(1 + \cos{2x})[/tex]. Use that identity on both sides of the above equation, then group terms in x. Since T is a constant, it must be independent of x, so something must kill all of the x terms. Find the condition on T such that this happens. If you can find such a T, then f(x) is periodic with a period of T.

    For the second one, the properties of odd functions should be enough.

    An odd function is a function such that [tex]f(-x) = -f(x)[/tex], so

    [tex]f(f(-x))=f(-f(x)) = f(-y) = -f(y) = -f(f(x))[/tex]

    Using [itex]f(x) = y[/itex] to show a little more explicitly how it works. Is that not a sufficient demonstration of what the question asks for?
     
  5. Jun 5, 2007 #4
    in regards to the 2nd question, I don't understand how the negative can move positions from inside the brackets to outside and then outside again.
     
  6. Jun 5, 2007 #5

    D H

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    You are told f(x) is an odd function: f(-x)=-f(x). You are asked to show that f(f(x)) is an odd function. Your "example" is a lot more than just an example. It is a proof of this conjecture.

    Define y=f(x). Then f(f(x)) = f(y). You want to evaluate f(f(-x)). What is f(-x)? It is -f(x)=-y, since f is an odd function. (Note: This is not true for all functions, but it is true for all odd functions by definition.) Thus f(f(-x))=f(-f(x))=f(-y). Once again using that f is odd, f(-y)=-f(y). Using y=f(x), -f(y)=-f(f(x)). Thus f(f(x))=-f(f(x)), so f(f(x)) is odd.
     
  7. Jun 5, 2007 #6

    Dick

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    Why would you need an identity? It just says to show that it is periodic, not to find the smallest period. cos(x+2pi)=cos(x) -> cos^2(x+2pi)=cos^(x).
     
    Last edited: Jun 5, 2007
  8. Jun 5, 2007 #7
    I'm really quite confused.
    I have been shown this alternative method

    cos²(x+2pi)
    =[cos(x+2pi)]²
    =[cos(x)+cos(2pi)]²
    =[cos(x)]²

    However I dont see how the cos(2pi) part dissapears as cos(2pi)=1 :confused:
     
  9. Jun 5, 2007 #8

    Gib Z

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    Umm Cos (a+b) doesnt equal cos A + cos B by the way, who ever showed you that method. But the general method was correct, show that cos x is periodic to show cos^2 x is.
     
  10. Jun 5, 2007 #9
    My bad, it should be
    =[cos(x+2pi)]²
    =[cos(x)]²

    without that step, I thought something was fishy...

    But my question still remains, how do you get from [cos(x+2pi)]² to [cos(x)]²?
     
  11. Jun 5, 2007 #10

    Gib Z

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    umm...Think of a unit circle?
     
  12. Jun 5, 2007 #11
    silly me....

    still struggling to comprehend how the negative can move through the brackets in the 2nd question though..
     
  13. Jun 5, 2007 #12

    Gib Z

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    Ok f(x)= - f(x) is given to us.

    We want to show f( f(x) ) = - f( f(x) )

    Ok so on the LHS replace the f(x) inside, with -f(x) since we know we can.

    Now we can f( f(x) ) = f ( -f(x) )

    Now lets replace with the f(x) with u for a second,to make things less confusing. So it becomes f (-u). f is an odd function though, so f(-u)=-f(u)

    Hence f( f(x) ) = - f( f(x) ). Done!
     
  14. Jun 5, 2007 #13
    I don't get this part. Why do we replace with -f(x)?
     
  15. Jun 5, 2007 #14

    Gib Z

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    f(x) = - f(x) is given to us?
     
  16. Jun 5, 2007 #15

    NateTG

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    Your notation...could use some help...or something.

    f(x)=-f(x)
    so
    2f(x)=0
    so
    f(x)=0
    ...
     
  17. Jun 5, 2007 #16

    VietDao29

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    Ok, if f is an odd function, then the following property should hold: f(-x) = -f(x), or, some also may write f(x) = -(f(-x)).

    Now, f o f(-x) = f(f(-x)), you can get this step, right?
    Since f(x) is odd, so f(-x) = -f(x). Right? So we have:
    f o f(-x) = f(f(-x)) = f(-f(x))

    Now, again, since f(x) is odd, we have: f(-f(x)) = -f(f(x)) (Notice the movement of the minus sign)

    If you still cannot see why this is true, let k = f(x), we have:
    f o f(-x) = f(f(-x)) = f(-f(x)) = f(-k) = -f(k) = -f(f(x))

    Now, the final step -f(f(x)) is actually, -f o f(x).

    So, we have shown that:
    f o f(-x) = - f o f(x), hence, f o f(x) is an odd function.

    Can you get it?

    Ok, here's some other similar problems, you can try to see if you can do it.

    ---------------------------

    Problem 1:
    f(x) is an odd function, and g(x) is an even function.
    a. Is f o g(x) odd, or even?
    b. Is g o f(x) odd, or even?

    Problem 2:
    If f o g o h(x) is an even function, and we know that f(x), and g(x) are all odd functions, what can we say about h(x)?
     
    Last edited: Jun 5, 2007
  18. Jun 7, 2007 #17

    Gib Z

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    I have a severe mental retardation, i can see that now >.< ahh i can't believe i didn't notice that...its f(-x) = -f(x) btw :(
     
  19. Jun 7, 2007 #18
    After many minutes thinking over it, I finally got it. Thanks to all!
     
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