1. ## Taylor Polynomial Question

Hi everyone,

I was wondering if someone could give me a hand with the following problem.

Suppose that p is a polynomial of degree d. Given any x0 inR, show that the (d+1)th Taylor polynomial for p at x0 is equal to p.

What I've tried so far is basically the brute method. I tried setting up a general polynomial p of degree d and expanding and seeing what I came up with. As you can imagine, this hasn't been going so well . Any suggestions?

2. Originally Posted by seven.j
Hi everyone,

I was wondering if someone could give me a hand with the following problem.

Suppose that p is a polynomial of degree d. Given any x0 inR, show that the (d+1)th Taylor polynomial for p at x0 is equal to p.

What I've tried so far is basically the brute method. I tried setting up a general polynomial p of degree d and expanding and seeing what I came up with. As you can imagine, this hasn't been going so well . Any suggestions?

Tell me if this is too heavy duty:

Spoiler:

Let $\displaystyle \text{deg}(p)=n$ fix $\displaystyle x_0\in\mathbb{R}$ (where $\displaystyle \mathbb{R}_n[x]$ is just the set of all $\displaystyle n^{\text{th}}$-degree polynomials of degree $\displaystyle n$) and let $\displaystyle D_{x_0}^j:\mathbb{R}_n[x]\to\mathbb{R}_n[x],\quad j\in\{0,\cdots,n\}$ be the mapping $\displaystyle \left(D_{x_0}^j(p)\right)(x)=(x-x_0)^j p^{(j)}(x_0)$. This mapping is linear since

\displaystyle \begin{aligned}\left(D_{x_0}^j(\alpha p+\beta q)\right)(x) &= (x-x_0)^j\left(\alpha p+\beta q)^{(j)}(x)\\ &= (x-x_0)^j (\alpha p^{(j)}+\beta q^{(j)})(x_0)\\ &= (x-x_0)^j\left(\alpha p^{(j)}(x_0)+\beta q^{(j)}(x_0)\right)\\ &= \alpha (x-x_0)^j p^{(j)}(x_0)+\beta (x-x_0)^j q^{(j)}(x_0)\\ &= \alpha\left(D_{x_0}^j(p)\right)(x)+\beta\left(D_{x _0}^j(q)\right)(x)\end{aligned}

and thus the function

$\displaystyle T_n(p):\mathbb{R}_n[x]\to\mathbb{R}_n[x]$

where

$\displaystyle \displaystyle \left(T_n(p)\right)(x)=\sum_{j=0}^{n}\frac{1}{j!}D _{x_0}^j(p)$

is linear. But, since $\displaystyle \{1,\cdots,x^n\}$ is a basis for $\displaystyle \mathbb{R}_n[x]$; to prove that $\displaystyle T_n(p)=p$ for all $\displaystyle p\in\mathbb{R}_n[x]$ it suffices to check it for $\displaystyle p(x)=x^j,\text{ }j=0,\cdots,n$. To do this we merely note that

\displaystyle \displaystyle \begin{aligned}T_n(x^m) &= \sum_{j=0}^{n}\frac{1}{j!}D^j_{x_0}(x^m)\\ &= \sum_{j=0}^{m}\frac{m(m-1)\cdots(m-j+1)}{j!}(x-x_0)^j x_0^{m-j}\\ &= \sum_{j=0}^{m}{m\choose j}(x-x_0)^j x_0^{m-j}\\ &= (x-x_0+x_0)^m\\ &=x^m\end{aligned}

where we've used the fact that after the $\displaystyle m^{\text{th}}$ term the terms vanish (that's how we can account for the change of the upper index) and at the last step we employed the Binomial theorem. The conclusion follows by noticing that $\displaystyle T_n(p)$ is the $\displaystyle n^{\text{th}}$ degree Taylor polynomial at $\displaystyle x_0$, where of course we realize that the remainder term is zero since the $\displaystyle n+1^{\text{th}}$ derivative of $\displaystyle p$ is identically zero.