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Math Help - 2 questions on Maclaurin series

  1. #1
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    Talking 2 questions on Maclaurin series

    Hello! I'm stuck on these questions... Can someone help?

    First: how would you write f(x) = x / (1 +
    x3) as a series?

    I mean like

    Σ = .....
    k=0

    I know that 1/(1-x) =

    Σ
    xk
    k=0

    But how do i convert this into f(x)?

    Second: h
    ow would you find the radius of convergence of the following binomial series:


    Σ (3k k)
    x2k+1
    k=0

    with (3k k) is like (n k) = n! / k!(n-k)!

    I know how to do it when there's
    xk in the sommation but with x2k+1?

    Thanks for the help, I've been spending hours on these two questions!

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  2. #2
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    Re: 2 questions on Maclaurin series

    Quote Originally Posted by jones123 View Post
    Hello! I'm stuck on these questions... Can someone help?

    First: how would you write f(x) = x / (1 +
    x3) as a series?

    I mean like

    Σ = .....
    k=0

    I know that 1/(1-x) =

    Σ
    xk
    k=0

    But how do i convert this into f(x)?

    Second: h
    ow would you find the radius of convergence of the following binomial series:


    Σ (3k k)
    x2k+1
    k=0

    with (3k k) is like (n k) = n! / k!(n-k)!

    I know how to do it when there's
    xk in the sommation but with x2k+1?

    Thanks for the help, I've been spending hours on these two questions!

    1. You could write \displaystyle \frac{x}{1 + x^3} = x \left[ \frac{1}{1 - \left( -x^3 \right) } \right] , which you can then write as a geometric series.

    2. Use the Ratio Test.
    Thanks from topsquark
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  3. #3
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    Re: 2 questions on Maclaurin series

    Hmm I don't quite understand the second one. What difference makes the x^k in stead of x^(2k+1) in the ratio test ?

    The radius of convergence of the sum with x^k should be 4/27, the sum of x^(2k+1): 2/sqrt(27)
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  4. #4
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    Re: 2 questions on Maclaurin series

    You have \sum_{k= 0}^\infty\begin{pmatrix}3k \\ k\end{pmatrix}x^{2k+1}= \tex]\sum_{k= 0}^\infty \frac{(2k+ 1)!}{k! (k+1)!} x^{2k+1}
    To use the "ratio test" to determine whether the series \sum_{k=0}^\infty} a_n converges, look at \lim_{k\to\infty}\frac{a_{k+1}}{a_k}

    Here, a_k= \frac{(2k+1)!}{k!(k+1)!}x^{2k+1} so a_{k+1}= \frac{(2(k+1)+1)!}{(k+1)!((k+1)+1)!}x^{2(k+1)+1}= \frac{(2k+3)!}{(k+1)!(k+ 2)!}x^{2k+ 3}

    and so \frac{a_{k+1}}{a_k}= \frac{(2k+3)!}{(k+1)!(k+2)!}\frac{k!(k+1)!}{(2k+1)  !}\frac{x^{2k+3}}{x^{2k+1}} = \frac{(2k+ 3)!}{(2k+1)!}\frac{k!}{(k+ 1)!}\frac{(k+1)!}{(k+2)!}x^2

    Can you complete it from there?
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