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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.

Thanks in advance for your help.

Regards

Mathias

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    $\begingroup$ 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. $\endgroup$
    – MasB
    Mar 22, 2020 at 14:06

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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}

(In the third line, we switched the order of the double summation)

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