Taylor polynomial of $\frac{1}{2-x}$ Can someone show how to find the Taylor polynomial of $\frac{1}{2-x}$?
I tried this: $\frac{1}{2-x}=\frac{1}{1-(x-1)}$ and then use that $ \ T_n(\frac{1}{1-x})=1+x+\dots +x^n.$ But this gives $1+(x-1)+\dots +(x-1)^n$ which is the wrong answer. Why?

Here is how you find it about the point $x=0$.
$$ \frac{1}{2-x} = \frac{1}{2(1-x/2)} = \frac{1}{2}( 1+x/2+x^2/2^2+\dots\,. ). $$
Added: If you want to derive the Taylor series at $x=1$ then you can advance like this
$$  \frac{1}{2-x} = \frac{1}{1-(x-1)}. $$
Now let $t=x-1$ which gives
$$ T_n(\frac{1}{1-t}) = 1+t+\dots+t^n. $$
Substitute back $t=x-1$ gives what you have

$$T_n(\frac{1}{1-t}) = = 1+(x-1)+\dots+(x-1)^n. $$

