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Thread: Don't get this taylor series question?

  1. #1
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    Don't get this taylor series question?

    Find the Taylor series for f(x) centered at the given value of a. [Assume that f has a power series expansion. Do not show that Rn(x) → 0.] f(x) = 2/x , a = −3
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    Your power series will be

    $\displaystyle \displaystyle f(x) = \sum_{n = 0}^{\infty}\frac{f^{(n)}(a)}{n!}(x-a)$, where $\displaystyle \displaystyle f^{(n)}(x)$ represents the $\displaystyle \displaystyle n^{\textrm{th}}$ derivative of $\displaystyle \displaystyle f(x)$.
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    this is the answer i got(after deriving over and over)

    this is the pattern



    shouldnt this be the pattern of the taylor series about a=-3?
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    Why have you put $\displaystyle \displaystyle (x + 3)^n$ as the denominator? A polynomial does not have algebraic terms in denominators...

    I suggest you reread what I posted above. You need to evaluate a number of derivatives at the point $\displaystyle \displaystyle x = -3$ and substitute these and $\displaystyle \displaystyle x = a = -3$ into the Taylor series formula.
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  5. #5
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    The nth derivative of 2/x is $\displaystyle \frac{2(-1)^n (n!)}{x^{n+1}}$ but the coefficient of the Taylor's series is that evaluated at x= -3 divided by n!. You would then multiply by $\displaystyle (x+ 3)^n$.
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