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Math Help - Prove that function is periodic?

  1. #1
    Super Member fardeen_gen's Avatar
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    Prove that function is periodic?

    Prove that if the graph of the function y = f(x), defined throughout the number scale, is symmetrical about the two lines x = a\ \mbox{and}\ x = b,\ (a<b), then the function is a periodic one.
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    Proof by Induction

    (i) Suppose f(a-x)=f(a+x) for all x
    (ii) Suppose f(b-x)=f(b+x) for all x
    (iii) Define \Delta=b-a

    Base Case (n=1)
    f(a)=f(b-\Delta)=f(b+\Delta)=f(a+2\Delta)

    General Case
    Suppose f(a)=f(a+2n\Delta) for some n. Then f(a)=f(b+(2n-1)\Delta)=f(b+(1-2n)\Delta)=f(a+(2-2n)\Delta)=f(a+2(n+1)\Delta) . So if f(a)=f(a+2n\Delta) then f(a)=f(a+2(n+1)\Delta).

    Induction
    Therefore, f(a)=f(a+2n\Delta) for all n. So the function has a period of 2\Delta.

    Similarly, the argument can be made in the reverse direction.
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    Quote Originally Posted by Media_Man View Post
    Therefore, f(a)=f(a+2n\Delta) for all n. So the function has a period of 2\Delta.

    Similarly, the argument can be made in the reverse direction.
    Well generally the period must be a constant... here \Delta depends on 'a'

    Something is wrong
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  4. #4
    Moo
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    Quote Originally Posted by Isomorphism View Post
    Well generally the period must be a constant... here \Delta depends on 'a'

    Something is wrong
    But you fix a and b before proving that it is periodic.
    So it's not really a problem

    It is a constant with respect to the variable. But a and b act like constants.
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    Lord of certain Rings
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    Quote Originally Posted by Moo View Post
    But you fix a and b before proving that it is periodic.
    So it's not really a problem

    It is a constant with respect to the variable. But a and b act like constants.
    a and b are already fixed . It has been shown that f(\text{\LARGE{a}}) = f(\text{\LARGE{a}}+2n\Delta) for a specific value of \text{\LARGE{a}}.

    One has to show f(x) = f(x+\Delta) for any value of x, which does not depend on \Delta. Here x is a and \Delta = b - a
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    Moo
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    Quote Originally Posted by Isomorphism View Post
    a and b are already fixed . It has been shown that f(\text{\LARGE{a}}) = f(\text{\LARGE{a}}+2n\Delta) for a specific value of \text{\LARGE{a}}.

    One has to show f(x) = f(x+\Delta) for any value of x, which does not depend on \Delta. Here x is a and \Delta = b - a
    Oh sorry, thought you were pointing out something else
    The problem is not the period, it's the fact that he set x=a, isn't it ?
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    Correction

    Isomorphism is absolutely right. The definition of periodic is not simply "there exists a \Delta for which f(x)=f(x+n\Delta) for some x." This has to be true for all x. The proof is the same though.


    (i) Suppose f(a-x)=f(a+x) for all x
    (ii) Suppose f(b-x)=f(b+x) for all x
    (iii) Define \Delta=b-a

    Base Case (n=1)
    f(a+x)=f(a-x)=f(b-\Delta-x)=f(b+\Delta+x)=f(a+x+2\Delta)

    General Case
    Suppose f(a+x)=f(a+x+2n\Delta) for some n. Then f(a+x)=f(b+x+(2n-1)\Delta)= f(b-x+(1-2n)\Delta)=f(a-x+(2-2n)\Delta)=f(a+x+2(n+1)\Delta) . So if f(a+x)=f(a+x+2n\Delta) then f(a+x)=f(a+x+2(n+1)\Delta).

    Induction
    Therefore, f(a+x)=f(a+x+2n\Delta) for all n\geq 1. Letting x'=a+x, f(x')=f(x'+2n\Delta), so the function has a period of 2\Delta.

    Again, a separate, albeit identical proof is necessary to prove the function is periodic in the negative direction.
    Last edited by Media_Man; May 24th 2009 at 12:20 PM. Reason: misspelled a word
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  8. #8
    Senior Member pankaj's Avatar
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    It is not difficult;
    f(a-x)=f(a+x)
    Replace x by a-x
    f(x)=f(2a-x)

    Similarly
    f(x)=f(2b-x)

    Therefore,
    f(2a-x)=f(2b-x)

    Replace x by 2a-x
    f(x)=f(2b-2a+x)

    Therefore,period is T=2(b-a) though it still remains to be proved that 2(b-a) is the least value of T for which f(x+T)=f(x).

    Quote Originally Posted by Media_Man View Post

    Again, a separate, albeit identical proof is necessary to prove the function is periodic in the negative direction.
    Consider f:[0,\infty)\rightarrow R
    f(x)=\sin x
    Is f(x) not periodic
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  9. #9
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    Quote Originally Posted by pankaj View Post
    Therefore,period is T=2(b-a) though it still remains to be proved that 2(b-a) is the least value of T for which f(x+T)=f(x).
    Thats not required. The question merely demands the function to be periodic. It does not ask for a certain period period
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    Quote Originally Posted by pankaj View Post
    Therefore,period is T=2(b-a) though it still remains to be proved that 2(b-a) is the least value of T for which f(x+T)=f(x).
    This is also not always the case. Consider f(x)=sin(x). This has two lines of symmetry at x=\frac{\pi}{2} and x=\frac{3\pi}{2} giving it a period of 2\pi . But applying your algorithm with a=\frac{\pi}{2} and b=\frac{17\pi}{2} would certainly not reveal the "lowest" possible period.

    You could say that the lowest value of \Delta satisfying f(x)=f(x+\Delta) for all x also satisfies \Delta |2(b-a) (or \frac{2(b-a)}{\Delta} is a whole number).
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  11. #11
    Senior Member pankaj's Avatar
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    Quote Originally Posted by Isomorphism View Post
    Thats not required. The question merely demands the function to be periodic. It does not ask for a certain period period
    Well,I wanted to know that does a function periodic function need to satisfy the following criterion:
    f(x+T)=f(x), and
    f(x-T)=f(x).

    Actually I had read that for f to be periodic the condition f(x+T)=f(x) is to be satisfied where T is a positive constant independent of x.
    But I had not read about f(x-T)=f(x) that is why I stated the example in particular.
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