# Thread: Don't get this taylor series question?

1. ## 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

2. 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 $\displaystyle \displaystyle f(x)$.

3. 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?

4. 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.

5. 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 $\displaystyle (x+ 3)^n$.