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Math Help - Power series

  1. #1
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    Power series

    Hi, can anyone help with this question, I can't get me head around power series.

    A sound wave is given by the function:

    f(t) = 0.5e^{-2t} \cos 4t


    Write down the first four terms of the power series expansions of e^{2t}and cos 4t

    Determine the cubic (up to and including the third power of t) approximation of f(t) and calculate the accurate and approximate values of f(0.02) giving your results correct to 3 decimal places.


    Thanks
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  2. #2
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    Quote Originally Posted by srobrien View Post
    Hi, can anyone help with this question, I can't get me head around power series.

    A sound wave is given by the function:

    f(t) = 0.5e^{-2t} \cos 4t


    Write down the first four terms of the power series expansions of e^{2t}and cos 4t

    Determine the cubic (up to and including the third power of t) approximation of f(t) and calculate the accurate and approximate values of f(0.02) giving your results correct to 3 decimal places.


    Thanks
    Are you allowed to take some things for granted?

    Like e^x = \sum_{k = 0}^{\infty}\frac{x^k}{k!}

    and \cos{x} = \sum_{k = 0}^{\infty}\frac{(-1)^kx^{2k}}{(2k)!}?


    Now replace x with -2t and 4t as required...
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  3. #3
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    Hi thanks for the help.

    R
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  4. #4
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    Ok, gave the first part a shot today, don't know if im right so feedback would be gratefully received.

    First 4 terms of e^{-2t}:

    e^{-2t} = 1 + (-2t) + \frac{(-2t)^2}{2!} + \frac{(-2t)^3}{3!} + ...


    e^{-2t} = 1 -2t + \frac{(4t)^2}{2!} - \frac{(8t)^3}{3!} + ...

    e^{-2t} = 1 -2t + 2t^2 - \frac{(8t)^3}{3!} + ...


    First 4 terms of \cos4t:

    \cos4t = 1 - \frac{(4t)^2}{2!} + \frac{(4t)^4}{4!} - \frac{(4t)^6}{6!} + ...

    \cos4t = 1 - \frac{(16t)^2}{2!} + \frac{(256t)^4}{4!} - \frac{(4096t)^6}{6!} + ...

    \cos4t = 1 - (8t)^2 + \frac{(64t)^4}{3!} - \frac{(4096t)^6}{6!} + ...

    Am I anywhere close? If so where is next from here?
    Sorry about the basic questions, this subject was briefly covered by a module and didn't really go into much depth.

    Thanks again
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