Find the Taylor series about 0 for a function $p(x)$

I have to find the Taylor series for the function of a arbitrary poylnomium given by $$p(x) = a_0 + a_1x + ... + a_{k-1}x^{k-1}+a_kx^k$$ By a few calculations I have realized that the Taylor series is just \begin{align*} \sum_{n = 0}^\infty \frac{p^{(n)}(a)}{n!}(x-a)^n & = p(a) + p'(a)(x-a) + \frac{p''(a)}{2!}(x-a)^2 + ... + \frac{p^{(n)}}{n!}(x-a)^n \\ & = a_0 + a_1(x-0)^1 + 2!\frac{a_2}{2!}(x-0)^2 + ... + n!\frac{a_k}{n!}(x-0)^n \\ & = a_0 + a_1x + a_2x^2+...+a_kx^n \end{align*} if I have not made any mistakes. Is this correct? Furthermore I now have to argue/prove that if $$x_0 \in \mathbb{R}$$ and $$T_n(x)$$ is a Taylor polynomium for p of grade $$n$$ about $$x_0$$ that $$T_n(x) = p(x)$$ for all $$x \in \mathbb{R}$$ when $$n \geq k$$. I have divided this into two scenarios. When $$n > k$$ and when $$n = k$$.

When $$n = k$$ I have just proven this in a) so I have said this is obvious. But I am not sure how to handle the last part when $$n > k$$. Can you help me in the right direction?

Lastly I have to find a function $$f: ]-1,1[ \rightarrow \mathbb{R}$$ which Taylor series about $$0$$ is equal to the series given by $$2 + 2x + 2x^2 + 2x^3 +...+2x^n$$ Is this just $$f(x) = 2 + 2x^2 + 2x^3 + ... + 2x^n$$ in continuation of the the part I have to prove? Otherwise I am not sure what to do.

Regards

Mathias

• Remember that the idea behind Taylor series is to find the polynomial which is the best approximation for a given function. When the function is not a polynomial, the series gives better and better approximations as you take more and more terms. So the best approximation to a polynomial is itself. Mar 22 '20 at 14:06

Let $$k\in\mathbb{N}$$, and $$p\left(x\right)=\sum\limits_{j=0}^{k}{a_{j}x^{j}}$$, we know that for any $$i\leq k$$, we have $$p^{\left(i\right)}\left(x\right)=\sum\limits_{j=i}^{k}{a_{j}\frac{j!}{\left(j-i\right)!}x^{j-i}}\cdot$$
Thus, for any real $$a$$, we get :\begin{aligned} p\left(x\right)&=\sum\limits_{j=0}^{k}{a_{j}\left(a+x-a\right)^{j}}\\&=\sum\limits_{j=0}^{k}{a_{j}\sum\limits_{i=0}^{j}{\binom{j}{i}a^{j-i}\left(x-a\right)^{i}}}\\&=\sum\limits_{i=0}^{k}{\sum\limits_{j=i}^{k}{a_{j}\binom{j}{i}a^{j-i}\left(x-a\right)^{i}}} \\&=\sum\limits_{i=0}^{k}{\frac{1}{i!}\left(\sum\limits_{j=i}^{k}{a_{j}\frac{j!}{\left(j-i\right)!}a^{j-i}}\right)\left(x-a\right)^{i}}\\p\left(x\right)&=\sum\limits_{i=0}^{k}{\frac{p^{\left(i\right)}\left(a\right)}{i!}\left(x-a\right)^{i}} \end{aligned}