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Math Help - Periodic Functions

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    Periodic Functions

    Can someone delete this thread? Can't find the delete thread button.
    Last edited by chrozer; February 10th 2009 at 05:29 AM. Reason: delete
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    Quote Originally Posted by chrozer View Post
    I think that both (c) is equal, but I do not know why. And I'm not to sure if (a) or (b) is equal or not? Any ideas or explanations?
    (a) f(t-2c) = f(t)
    Let t'=t-2c
    so f(t)=f(t'+2c)=f((t'+c)+c)=f(t'+c)=f(t')=f(t-2c)

    (b) f(t+ \frac {1}{2}c) = f(\frac {1}{2}t)

    (c) f(\frac {1}{2}(t+c)) = f(\frac {1}{2}t)
    b) is false, it makes nonsense...

    c) is false, because it would mean it has a period \frac c2. But c is the smallest number such that f(t+c)=f(t)
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    Let t'=t-2c
    so f(t)=f(t'+2c)=f((t'+c)+c)=f(t'+c)=f(t')=f(t-2c)
    Ok...part (a)...I just don't seek how you got that. Where did you get f(t'+2c) and how does f((t'+c)+c)=f(t'+c)?

    (b) is false, it makes nonsense...
    Why is it nonsense? What makes it false?
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    Quote Originally Posted by chrozer View Post
    Ok...part (a)...I just don't seek how you got that. Where did you get f(t'+2c)
    Because I let t=t'+2c
    It's easier if you make a substitution, as you'll report to the formula you know
    and how does f((t'+c)+c)=f(t'+c)?
    Because c is the period !
    Again, if you can't see it, let T=t'+c, you'd have f(T+c)=f(T), which is f((t'+c)+c)=f(t'+c)
    Whatever you have, if you add c, it will give the same !

    Why is it nonsense? What makes it false?
    Hmmm I've been looking for an explanation... I think I found it

    f\left(\tfrac t2\right)=f \left(\tfrac t2+c\right)
    =f \left(\tfrac 12 (t+2c)\right)=f \left(\tfrac 12 (t+c)+\tfrac c2\right)
    Can it possibly equal f \left(\tfrac 12(t+c)\right) ?
    Only if :
    - c=0 (which is not possible otherwise it would not make sense for the periodicity)
    - f is c/2 periodic. But as I explained for c), it's not possible.

    Thus b) is false
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    Quote Originally Posted by Moo View Post
    Because I let t=t'+2c
    It's easier if you make a substitution, as you'll report to the formula you know

    Because c is the period !
    Again, if you can't see it, let T=t'+c, you'd have f(T+c)=f(T), which is f((t'+c)+c)=f(t'+c)
    Whatever you have, if you add c, it will give the same !


    Hmmm I've been looking for an explanation... I think I found it

    f\left(\tfrac t2\right)=f \left(\tfrac t2+c\right)
    =f \left(\tfrac 12 (t+2c)\right)=f \left(\tfrac 12 (t+c)+\tfrac c2\right)
    Can it possibly equal f \left(\tfrac 12(t+c)\right) ?
    Only if :
    - c=0 (which is not possible otherwise it would not make sense for the periodicity)
    - f is c/2 periodic. But as I explained for c), it's not possible.

    Thus b) is false
    Ok I see now...thanx for the explanation.
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