So i have a function: $f: \mathbb{R} \to \mathbb{R}$ $$f(x)=\frac{1}{\sqrt{(x^2+x+1)^5}}$$

I want to expand it into a taylor series centered at $x=-\frac{1}{2}$ I normally did expansions in Maclaurin series, also i don't know how to do this one. Any help would be appreciated.


As you must obtain a power series in $(x+1/2)$, it is better to set $t=x+1/2$ and substitute that into your expression to obtain $$ f(t)={1\over\sqrt{(3/4+t^2)^5}}= \left({4\over3}\right)^{5/2}\left(1+{4\over3}t^2\right)^{-5/2}. $$ Now expand $f(t)$ in Maclaurin series (easy, if you remember the expansion of $(1+x)^\alpha$) and then substitute back $t=(x+1/2)$.

  • $\begingroup$ same problem, but i need to have help figuring out why this: $(\frac{3}{4}^{-\frac{5}{2}}\sum_{k=0}^{\infty}\binom{-\frac{5}{2}}{k})(\frac{4y^2}{3})^k = (\frac{3}{4})^{-\frac{5}{2}}\sum_{k=0}^{\infty}\frac{(-1)^k(2k+3)!!}{2k!!}(\frac{4y^2}{3})^k$ I am wondering why can i do it like this? If you could help, i would appreciate it. $\endgroup$ – MathIsTheWayOfLife Aug 30 '16 at 9:40
  • $\begingroup$ never mind, i figured it out. $\endgroup$ – MathIsTheWayOfLife Aug 30 '16 at 9:41

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