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Math Help - Obtain a MacLaurin Series for a function with given conditions for x

  1. #1
    Junior Member Coop's Avatar
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    Obtain a MacLaurin Series for a function with given conditions for x

    Hi,

    The question asks to find a Maclaurin series for the function

    f(x) = (x-sinx)/x^3 when x does not = 0

    and

    f(x) = 1/6 when x = 0

    Since sinx = x - x^3/3! + x^5/5! - ...

    x-sinx = x^3/3! - x^5/5! + x^7/7! - ...

    and (x-sinx)/x^3 = 1/3! - x^2/5! + x^4/7! - ... = 1/6 - x^2/5! + x^4/7! - ...

    I noticed that I can make a series out of the second term onward, and in doing so would make f(x) = 1/6 when x = 0. This gives me 1/6 - series(n=0,infinity) ((-1)^(x+1) * x^(2n+2))/(2n+3)!

    Would that be my answer? Just curious, because I've never done a problem like this before.
    Last edited by Coop; April 23rd 2013 at 05:05 PM.
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  2. #2
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    Re: Obtain a MacLaurin Series for a function with given conditions for x

    Your series would just be \displaystyle \begin{align*} \frac{1}{3!} - \frac{x^2}{5!} + \frac{x^4}{7!} - \dots + \dots \end{align*}. I don't know why you're subtracting it from \displaystyle \begin{align*}  \frac{1}{6}\end{align*}.
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    Junior Member Coop's Avatar
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    Re: Obtain a MacLaurin Series for a function with given conditions for x

    Quote Originally Posted by Prove It View Post
    Your series would just be \displaystyle \begin{align*} \frac{1}{3!} - \frac{x^2}{5!} + \frac{x^4}{7!} - \dots + \dots \end{align*}. I don't know why you're subtracting it from \displaystyle \begin{align*}  \frac{1}{6}\end{align*}.
    But how would I write that as a series in sigma notation without it being equal to 0 when x = 0? I was assuming the answer had to be in sigma notation.
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    Re: Obtain a MacLaurin Series for a function with given conditions for x

    It wouldn't be equal to 0 when x = 0, the constant term is \displaystyle \frac{1}{3!} = \frac{1}{6}, like we require.

    Anyway, in sigma notation, we could write this series as \displaystyle \sum_{n = 0}^{\infty}\frac{(-1)^{n}x^{2n}}{(2n + 3)!}.
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    Junior Member Coop's Avatar
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    Re: Obtain a MacLaurin Series for a function with given conditions for x

    Quote Originally Posted by Prove It View Post
    It wouldn't be equal to 0 when x = 0, the constant term is \displaystyle \frac{1}{3!} = \frac{1}{6}, like we require.

    Anyway, in sigma notation, we could write this series as \displaystyle \sum_{n = 0}^{\infty}\frac{(-1)^{n}x^{2n}}{(2n + 3)!}.
    Oh thanks, I was thinking 0^0 = 0 or undefined; I couldn't really find a definite answer online, does 0^0 = 1? That's why I was confused.

    0^0 - Wolfram|Alpha
    Last edited by Coop; April 23rd 2013 at 06:10 PM.
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    Re: Obtain a MacLaurin Series for a function with given conditions for x

    No, \displaystyle 0^0 is undefined.
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    Re: Obtain a MacLaurin Series for a function with given conditions for x

    Would that be my answerReplica Handbags,Replica Wallets For Sale
    ? Just curious, because I've never www.louisvuittonend.com
    done a problem like this before.
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