Correct representation of $f(x)=(1-x)^2$ as Taylor series

Question

This is my first problem on Taylor series, so I would appreciate if someone could point out any errors. Thanks.

I want to $i$) represent $f(x)=(1-x)^2$ as a Taylor series centered at $0$, and $ii)$ show for which values of $x$ is the function properly represented.

$i$) Notice that $(1+x)^k=\sum_{n=0}^\infty {k\choose n} x^n$.

Then

$$f(x)=\sum_{n=0}^\infty {2\choose n}(-x)^n$$

$ii$) Because this is a binomial series, the convergence radius is $R=1$. Now see that

$f'(x)=-2(1-x), f''(x)=2, f'''(x)=0$ and then $f^{(n+1)}(x)=0, x>2$.

Therefore $f^{(n+1)}(x)\leq1$. Then according to Taylor's inequality

$$|R_n(x)|\leq\frac{x^{n+1}}{(n+1)!}=0$$

Then, because of the squeeze theorem, $\lim_{n\to\infty} R_n(x)=0$, and this proves that $f(x)$ is the sum of its Taylor series for $|x|<R \iff x\in(-1, 1)$.

Answer 1

The required Taylor expansion is: $$(1-x)^2=1-2x+x^2$$

Answer 2

Any polynomial is its own Taylor series centred at $0$.

If you want its Taylor series centred at another value $a$, just determine its expansion along powers of $x-a$ with the same algorithm as the algorithm used to deduce the expansion of a number in base $b$ from its expansion in base $10$.