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Math Help - power series

  1. #1
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    power series

    The function   f(x) = \frac{5}{(1 - 3 x)^2} is represented as a power series  f(x) =<br />
\sum_{n=0}^\infty c_n x^n .

    Find the first few coefficients in the power series.
    C_0 =
    C_1 =
    C_2 =
    C_3 =
    C_4 =

    Find the radius of convergence R of the series.
     R =
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by viet View Post
    The function   f(x) = \frac{5}{(1 - 3 x)^2} is represented as a power series  f(x) =<br />
\sum_{n=0}^\infty c_n x^n .

    Find the first few coefficients in the power series.
    C_0 =
    C_1 =
    C_2 =
    C_3 =
    C_4 =

    Find the radius of convergence R of the series.
     R =
    To the first question: Recall that the power series expansion for  \frac {1}{(1 - x)^2} is the series \sum_{n = 1}^{ \infty}n x^{n - 1} = \sum_{n = 0}^{ \infty} (n + 1) x^n (this is derived by taking the derivative of the well know power series for \frac {1}{1 - x})

    So we have that:

    \frac {5}{(1 - 3x)^2} = \sum_{n = 0}^{ \infty} 5 \cdot (n + 1)(3x)^n = \sum_{n = 0}^{ \infty} 5 \cdot 3^n (n + 1) x^n

    therefore, we have

    C_n = 5 \cdot 3^n (n + 1)

    use that to find the first few coefficients


    For the second question:

    use the root test (or if you prefer, the ratio test, either one works nicely in this case) to find the radius of convergence. do you need help with that part?
    Last edited by Jhevon; July 8th 2007 at 12:18 PM.
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  3. #3
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    yea looks like i need help on the second question
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by viet View Post
    yea looks like i need help on the second question
    Ok, we wish to find the radius of convergence for the series \sum_{n = 0}^{ \infty} 5 \cdot (n + 1)(3x)^n

    Since you didn't specify what test you wanted me to use, I will pick one myself. I choose the root test, it's less typing than the ration (why do i always do that?! type "ration" when i mean to type "ratio"!) test in this case.

    We have \lim_{n \to \infty} \left| 5 (n + 1) \cdot (3x)^n \right|^{ \frac {1}{n}} = \lim_{n \to \infty} \left| 5^{ \frac {1}{n}} (n + 1)^{ \frac {1}{n}} \cdot 3x \right| = | 1 \cdot 1 \cdot 3x | = 3 |x|

    we have convergence if 3|x| < 1 \implies |x| < \frac {1}{3}

    Thus the radius of convergence is \frac {1}{3}
    Last edited by Jhevon; July 9th 2007 at 10:26 AM. Reason: expressing a peeve
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