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Math Help - Maclaurin series

  1. #1
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    Maclaurin series

    Find the Maclaurin series (don't show that Rn(x)-->0). Also find the radius of convergence:
    (1+x)^{-3}

    So I got the first few terms: (1)^{-3}-\frac{3(2)^{-4}}{2!}x+\frac{12(3)^{-5}}{3!}x^2-\frac{60(4)^{-6}}{4!}x^3+\frac{360(5)^{-7}}{5!}x^4-...

    I can't account for the coefficient
    \sum_{n=0}^\infty \frac{(n+1)^{-3-n}(-x)^n}{n+1}
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  2. #2
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    Try this: If [tex]f(u)= \frac{1}{u}= u^{-3}, then f'= -3u^{-4}, f"= 12u^{-5}, f"'= -60u^{-6}, f""= 360u^{-7} and, in general, f^{(n)}= (-1)^n \frac{(n+2)!}{2} u^{-n-2}.

    In your problem, \frac{1}{(1+x)^{-3}}, let u= 1+ x. x= 0 when u= 1.
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  3. #3
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    Quote Originally Posted by superdude View Post
    Find the Maclaurin series (don't show that Rn(x)-->0). Also find the radius of convergence:
    (1+x)^{-3}

    So I got the first few terms: (1)^{-3}-\frac{3(2)^{-4}}{2!}x+\frac{12(3)^{-5}}{3!}x^2-\frac{60(4)^{-6}}{4!}x^3+\frac{360(5)^{-7}}{5!}x^4-... ???

    I can't account for the coefficient
    \sum_{n=0}^\infty \frac{(n+1)^{-3-n}(-x)^n}{n+1}

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