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Thread: 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:

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


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

    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:

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


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

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

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


    Now replace $\displaystyle x$ with $\displaystyle -2t$ and $\displaystyle 4t$ 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 $\displaystyle e^{-2t}$:

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


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

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


    First 4 terms of $\displaystyle \cos4t$:

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

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

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

    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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