# Taylor polynomials: remainder formula for expansion around $\infty$.

By definition of Taylor polynomials, we have $$f(x)=f(x_0)+f'(x_0)(x-x_0)+...+\frac{f^{(n)}(x_0)}{n!}(x-x_0)^n+R_n(x,x_0)$$ where $R_n$ is the $n-$th remainder . Let $U(x_0)$ the neighborhood of $x_0$. If $f\in \mathcal C^{n+1}\left(U(x_0)\right)$, for $x\in U(x_0)$ it has: $$R_n(x,x_0)=\frac{f^{(n+1)}(\xi)}{(n+1)!}(x-x_0)^{n+1}, \ \ \ \xi\in[x_0,x]$$ We assume now to expand $f(x)$ (bounded) in a Taylor series around $+\infty$. How can we modify the ramainder formula? I had thought $$R_n(x,\infty)=\frac{f^{(n+1)}(\xi)}{x^n (n+1)!}, \ \ \ \xi\in[0,1/x]$$ Suggestions are welcome.

Then we let $x=\frac{1}{z}$ and notice that $f(\frac{1}{z}) =g(z)$ can be expanded in powers of z .
We expand then we substitute back $z=\frac{1}{x}$ thereby expanding f near infinity.