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Math Help - 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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  2. #2
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    Your power series will be

    \displaystyle f(x) = \sum_{n = 0}^{\infty}\frac{f^{(n)}(a)}{n!}(x-a), where \displaystyle f^{(n)}(x) represents the \displaystyle n^{\textrm{th}} derivative of \displaystyle f(x).
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  3. #3
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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 (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 x = -3 and substitute these and \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 \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 (x+ 3)^n.
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