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There is an exercise in an exam which my professor gave to us saying:

If $p=\Bbb{R}\rightarrow\Bbb{R}$ a polynomial of degree $n$. Proof that for any $a,x\in\Bbb{R}$ we have $p(x)=p(a)+p'(a)\cdot(x-a)+...+\frac{p^{(n)}(a)}{n!}\cdot(x-a)^n$.

Using Taylor's Theorem with Lagrange remainder, we have that: As $p:\Bbb{R}\rightarrow\Bbb{R}$ is a function of class $C^{n-1}$, $n$ times derivable in an open interval $(a,x)$. Than exist $c\in (a,x)$ such that $$p(x)=p(a)+p'(a)\cdot(x-a)+...+\frac{p^{(n-1)}(a)}{(n-1)!}\cdot(x-a)^{n-1}+\frac{p^{(n)}(c)}{n!}\cdot(x-a)^{n}$$ But, by this theorem, $a\notin(a,x)$, so how can I do this proof?

Edit: we have to use Taylor's Theorem with Lagrange remainder and we cannot use integral.

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    $\begingroup$ Hint: Take the Taylor polynomial of $p$ up to the $n$th degree term plus the remainder (which has degree $n+1$). $\endgroup$ – Gary Feb 14 at 17:28
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    $\begingroup$ It's not necessary to use Taylor's Theorem, this is a statement about polynomials. Clearly LHS and RHS are polynomials of the same degree which have the same value at $a$. Subtract $p(a)$ from both sides, divide by $x-a$ and induction will get you home. $\endgroup$ – ancientmathematician Feb 14 at 17:38
  • $\begingroup$ Or, as a slight alternative to ancientmathematician's comment - by inspection, the identity works for $p(t) = (t-a)^m$ for $m = 0, \ldots, n$, and both sides are linear in $p$. So then all that's left is to show that $1, t-a, (t-a)^2, \ldots, (t-a)^n$ span the space of polynomials of degree $n$ or less. $\endgroup$ – Daniel Schepler Feb 14 at 18:24
  • $\begingroup$ If you have the absurd requirement of using Taylor with Lagrange remainder, then use it with the Taylor to order $n+1$ and there is nothing to prove: $p^{(n+1)}(c)=0$. So, the remainder is zero. $\endgroup$ – user748968 Feb 14 at 19:23
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Since $p$ is also absolutely continuous on $[a,\,x]$, we can use an exact form of the Lagrange remainder, $\int_a^x\color{blue}{p^{(n+1)}(t)}\frac{(x-t)^n}{n!}dt$. The blue factor is $0$ for a degree-$n$ polynomial $p$.

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  • $\begingroup$ As we didn't see integral yet, we are cannot use it. $\endgroup$ – Rebeca Lie Yatsuzuka Silva Feb 14 at 19:21
  • $\begingroup$ @RebecaLieYatsuzukaSilva Prove its validity by induction, then, if only for polynomials. $\endgroup$ – J.G. Feb 14 at 19:41
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As @Gary 's and @Tora's answers, just to confirm, the proof, using Taylor with Lagrange remainder and not using integral would be this?

Using Taylor's Theorem with Lagrange remainder, we have that: As $p:\Bbb{R}→\Bbb{R}$ is a function of class $C^n$, $n+1$ times derivable in an open interval $(a,x)$. Than exist $c∈(a,x)$ such that $p(x)=p(a)+p′(a)⋅(x−a)+...+\frac{p^{(n)}(a)}{n!}⋅(x−a)^n+\frac{p^{(n+1)}(c)}{(n+1)!}⋅(x−a)^{n+1}$.

As $p(x)$ is a polynomial of degree $n$, $p^{(n+1)}(c)=0$ (prove this by induction?), so $p(x)=p(a)+p′(a)⋅(x−a)+...+\frac{p^{(n)}(a)}{n!}⋅(x−a)^n$

Comment: Doing this just for this question be placed as answered.

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  • $\begingroup$ Yes, this is it. $\endgroup$ – Gary Feb 14 at 20:02

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