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Math Help - General term

  1. #1
    No one in Particular VonNemo19's Avatar
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    General term

    I'm having trouble finding the coefficients for the maclaurin series for f(x)=\sqrt{1-x}. i have

    \frac{\Pi_{k=2}^n(2k-3)}{2^nn!}. Is this right?
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  2. #2
    Super Member
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    By differentiation ,

     f(x) = \sqrt{1-x} = (1-x)^{ \frac{1}{2} }

     f'(x) = -\frac{1}{2} \cdot (1-x)^{- \frac{1}{2} }

     f''(x) = - \frac{1}{2} \cdot \frac{1}{2} \cdot (1-x)^{- \frac{3}{2} }

     f'''(x) = - \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{3}{2} \cdot (1-x)^{- \frac{5}{2} }


    Observation is a good tool to conclude that

     f^{(n)}(x) = - \frac{1}{2} \cdot \frac{1 \cdot 3 ... \cdot (2n-3) }{ 2^{n-1} } \cdot (1-x)^{- \frac{2n-1}{2} } ( Isolate the first  \frac{1}{2} )  n \geq 2

    Therefore ,

     f^{(n}(0) = - \frac{1}{2} \cdot \frac{1 \cdot 3 ... \cdot (2n-3) }{ 2^{n-1} }

    But if we " fill up " the spaces between the odd numbers ,

     1 \cdot 3 \cdot 5 \cdot ... \cdot (2n-3)

     = 1 \cdot 2 \cdot 3 \cdot 4 ... \cdot (2n-3) \cdot (2n-2)  \cdot \frac{1}{2 \cdot 4 \cdot ... \cdot (2n-2)}

     = \frac{ (2n-2)!}{ 2^{n-1} \cdot (n-1)!}

    we will find that

     f^{(n)}(0) is also equal to :

     - \frac{ (2n-2)! }{ 2^{2n-1} (n-1)! }

    When accompanying with  \frac{1}{n!} to form the coefficient for the Maclaurin series , we can obtain an elegant expression .

     - \frac{ (2n-2)! }{ 2^{2n-1} (n-1)! n! }

     = - \frac{ (2n-2)! }{  2^{2n-1} (n-1)! n! } \cdot \frac{ (2n-1)(2n) }{ n } \cdot \frac{n}{2n(2n-1)}

     = - \frac{ \binom{2n}{n} }{ 2^{2n} (2n-1) }

     = - \frac{1}{4^n} \frac{ \binom{2n}{n} }{ 2n-1}
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