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Thread: 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:
    $\displaystyle (1+x)^{-3}$

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

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

    In your problem, $\displaystyle \frac{1}{(1+x)^{-3}}$, 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:
    $\displaystyle (1+x)^{-3}$

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

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

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