I am working on a project, where I make the 4. degree Taylorpolynomaial for the function $$g = \frac{1}{1-x}$$ Which is:
$$P_4(x)=1+x+x^2+x^3+x^4$$ I understand that to make a Taylorpolynomial, the function we are approximating needs to be differentiable in an interval $I$ where $x_0 \in I$. And that the domain of $g$ is $$\mathbb{R}\setminus \{ 1 \}$$
$I$ cannot contain $x=1$ since $g$ isn't differentiable here, therefore our approximation of $g$ becomes very bad, since our requirements for the Taylorpolynomial don't hold here and afterwards.
My question then is, why doesn't our approximation hold before $x=-1$
This is because, in this case, we have the exact value for the remainder in Taylor's formula:
From a well-known high-school factorisation formula, we have $$\frac1{1-x}=1+x+x^2+\dots+x^n+\frac{x^{n+1}}{1-x},$$ so that the remainder $\:\dfrac{x^{n+1}}{1-x}\:$ tends to $0$ as $\:n\to \infty\:$ if & only if $\:|x|<1$.