Let $x_0\in\mathbb{R}$ and let $T_n(x)$ be a Taylorpolynomial for $p$ of degree $n$ by $x_0$. Explain why $T_n(x)=p(x) \ \forall \ x \in\mathbb{R}$ when $n\geq k$. I know that the idea behind the Taylor Polynomial is to find the polynomial which best approximates a given function. Therefore, the best approximation to a polynomial must be the polynomial itself. However, i cannot see why it is only for $n\geq k$.
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What is $k$ here? – mathcounterexamples.net Mar 29 '20 at 12:57
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Ah sorry- our $p(x)$ is given by $p(x)=a_kx^k+a_{k-1}x^{k-1}+...+a_1x+a_0$ – Benjamin Bech Mar 29 '20 at 12:59
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In term of pure logic, the sentence Explain why $T_n(x)=p(x) \ \forall \ x \in\mathbb{R}$ when $n\geq k$. doesn't mean that $T_n(x)$ is not the best approximation when $n <k$. Also you'll need to precise what best approximation means. – mathcounterexamples.net Mar 29 '20 at 13:26