# A special polynomial expansion

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.

• Hint: Take the Taylor polynomial of $p$ up to the $n$th degree term plus the remainder (which has degree $n+1$). – Gary Feb 14 at 17:28
• 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. – ancientmathematician Feb 14 at 17:38
• 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. – Daniel Schepler Feb 14 at 18:24
• 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. – user748968 Feb 14 at 19:23

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 $$p$$.

• As we didn't see integral yet, we are cannot use it. – Rebeca Lie Yatsuzuka Silva Feb 14 at 19:21
• @RebecaLieYatsuzukaSilva Prove its validity by induction, then, if only for polynomials. – J.G. Feb 14 at 19:41

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.

• Yes, this is it. – Gary Feb 14 at 20:02