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Math Help - Find the period of Cosine of Quadratic function

  1. #1
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    Find the period of Cosine of Quadratic function

    Hi all,

    Hope some here can help me with this math problem.

    Given,
    y1 = ax^2 + b.
    y2 = cos (y1).
    where a and b are constants. Is y2 periodic with respect to x.? Visually using example, the graph, seems to be periodic. How do u find the exact period of such a function?

    Thanks in advance.

    regards,
    cybershakith
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  2. #2
    MHF Contributor Also sprach Zarathustra's Avatar
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    cos(ax^2+b+T)=cos(ax^2+b)

    Find T.


    Edit:
    Rough mistake!

    Thank you @topsquark!
    Last edited by Also sprach Zarathustra; March 19th 2011 at 07:53 PM.
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Also sprach Zarathustra View Post
    cos(ax^2+b+T)=cos(ax^2+b)

    Find T.
    Actually I would say to solve cos(a(x - T)^2 + b) = cos(ax^2 + b). Then show that the period T depends on x.

    -Dan
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  4. #4
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    I tried something along those lines.

    cos (a*x^2 + b) = cos (a*(x+T)^2 + b)

    Hence,
    a*x^2 + b + 2* PI*k = a*(x+T)^2 + b, where is k is an integer.

    Which reduces to,

    2*PI*k = a*2*x*T + a * T^2

    So T is function of x, is it really then periodic?
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  5. #5
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    Okay. So the function is not periodic.

    But let's take an example,
    y = Cos (2*PI*ax^2 + 2*PI*b)
    where is a =0.01277777778 and b = 255.5555556;

    From plotting this graph, it seems like the y values are peridoc over x = 900.
    So how does it happen?

    for x =0;
    y = Cos (2*PI*ax^2 + 2*PI*b) = Cos 2*PI * 255.5555556;

    for x = 900;
    y = Cos (2*PI*ax^2 + 2*PI*b) = Cos 2*PI*( 255.5555556 + 0.01277777778*900^2 ) = Cos 2*PI*(10605.5555556);


    for x =11;
    y = Cos (2*PI*ax^2 + 2*PI*b) = Cos 2*PI*( 255.5555556 + 0.01277777778*11^2 ) = Cos 2*PI*(257.10166671138);

    for x = 911;
    y = Cos (2*PI*ax^2 + 2*PI*b) = Cos 2*PI*( 255.5555556 + 0.01277777778*911^2 ) = Cos 2*PI*(10860.10166855538); //slight difference due to lack of precision.

    This is true for all x, it seems.

    this is because fractional part of ax^2 and a(x+T)^2 terms are the same.

    So is it periodic?
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  6. #6
    Grand Panjandrum
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    Quote Originally Posted by cybershakith View Post
    Hi all,

    Hope some here can help me with this math problem.

    Given,
    y1 = ax^2 + b.
    y2 = cos (y1).
    where a and b are constants. Is y2 periodic with respect to x.? Visually using example, the graph, seems to be periodic. How do u find the exact period of such a function?

    Thanks in advance.

    regards,
    cybershakith
    It is not periodic.

    To convince yourself look at the zeros of y2.

    CB
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  7. #7
    Grand Panjandrum
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    Quote Originally Posted by cybershakith View Post
    Okay. So the function is not periodic.

    But let's take an example,
    y = Cos (2*PI*ax^2 + 2*PI*b)
    where is a =0.01277777778 and b = 255.5555556;

    From plotting this graph, it seems like the y values are peridoc over x = 900.
    So how does it happen?
    If you are plotting this on a computer with a fixed plotting interval, you are probably seeing aliasing between the function and the sampling function

    CB
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