I want to prove that $ \frac{1}{x^{k}} $ is analytic in $ \mathbb{R} \setminus [0] $ for |x|<1 its obvious. i want to prove for the rest of $ \mathbb{R} $ so i tried that:
I proved by induction that $ \left(\frac{1}{x^{k}}\right)^{(n)}=\frac{\left(k+n-1\right)!}{\left(k-1\right)!}\frac{\left(-1\right)^{n}}{x^{\left(k+n\right)}} $
then, I wrote lagrange reminder of the function
$ |R_{n}\left(x\right)|=|\frac{\frac{\left(n+k\right)!}{\left(k-1\right)!}\frac{\left(-1\right)^{n+1}}{c^{\left(k+n+1\right)}}}{\left(n+1\right)!}\left(x-x_{0}\right)^{n+1} $ = $ |\frac{\left(n+k\right)!}{\left(n+1\right)!\left(k-1\right)!c^{\left(k+n+1\right)}}\left(x-x_{0}\right)^{n+1}| $
so all i have to do in order to prove it, is to find $ \delta $ such for any $ x\in\left(x_{0}-\delta,x_{0}+\delta\right) $ the reminder goes to $ 0 $. i dont know how to calculate this limit and find delta, any help will be great. Thanks
It's enough to prove that $f(x)=\frac{1}{x}$ is analytic; the general case follows by an easy induction, because products of analytic functions are analytic. Now, if $a\neq 0,$ we have $\frac{1}{x}=\frac{1}{a+(x-a)}=\frac{1}{a}\frac{1}{1+(x-a)/a)}=\frac{1}{a}\sum^\infty_{n=0}\left(\frac{x-a}{a}\right)^n$ and by uniqueness of power series, this is the Taylor series of $f$ about $x=a$ and it has radius of convergence $|a|.$We just need to check that the remainder $R_n(x)\to 0$ as $n\to \infty.$ We have $R_n(x)=\frac{f^{n+1}(\xi)}{(n+1)!}(x-a)^{n+1}$ for $a<\xi<x$ and $|f^{n+1}(\xi)|=\left|\frac{(-1)^{(n+1)}(n+1)!}{\xi^{k+1}}\right|\le \frac{(n+1)!}{a^{n+1}}$ so $|R_n(x)|\le \left(\frac{|x-a|}{|a|}\right)^{n+1}\to 0$ as $n\to \infty$ within its radius of convergence.