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If $f : \mathbb R \to\mathbb R $ is a polynomial function of degree $n$ with $a \in\mathbb R$. Show that the $n$-th Taylor polynomial $P_{f,a,n}$ of $f$ at $a$ is equal to $f$.

I know that I need to show that $P_{f,a,n}(x) = f(x)$ for all $x \in\mathbb R$, and I think that I need to do this by using induction and the Taylor series, any help would be appreciated.

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    $\begingroup$ Write down the exact definition of a taylor polynomial for $f$ around $a$. notice that $(x-a)^k=0$ whenever $x=a$. $\endgroup$ – Alex R. Dec 22 '15 at 19:54
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Outline: We do an induction on degree. Suppose the result is true at $n=k$. We show the result is true at $n=k+1$.

Let $f(x)$ be a polynomial of degree $k+1$, and let $P(x)$ be its $(k+1)$-th Taylor polynomial about $x=a$. Then differentiating $P(x)$ term by term we find that $P'(x)$ is the $k$-th Taylor polynomial of $f'(x)$ about $x=a$. Since $f'(x)$ has degree $k$, by the induction hypothesis we have $f'(x)=P'(x)$ for all $x$.

It follows (calculus) that $f(x)$ and $P(x)$ differ by a constant. Evaluation at $x=a$ shows that that constant is $0$.

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