# Proof involving Taylor Polynomials / Lagrange Error Bound

## Homework Statement

I'm given that the function $f(x)$ is n times differentiable over an interval $I$ and that there exists a polynomial $Q(x)$ of degree less than or equal to $n$ s.t.
$$\left|f(x) - Q(x)\right| \leq K\left|x - a\right|^{n+1}$$
for a constant $K$ and for $a \in I$

I am to show that $Q(x)$ is the Taylor polynomial for $x$ at $a$

## Homework Equations

For $Q(x)$ to be the Taylor polynomial for $x$ at $a$:

$$Q(x) = \sum_{k = 0}^{n}\frac{f^{(k)}(a)}{k!} (x - a)^k$$

The $n^{th}$ error/remainder term of $Q(x)$ is:
$$R_{n,a}(x) = \left|f(x) - Q(x)\right|$$

The Lagrange error bound is:
$$R_{n,a}(x) = \left|f(x) - Q(x)\right| \leq \frac{M}{(n+1)!}|x - a|^{n+1}$$
where $M = sup\left|f^{n+1}(x)\right|$

## The Attempt at a Solution

I think I could easily go from knowing that $Q(x)$ was the Taylor polynomial of $f$ around $a$ and prove that there exists a constant $K = \frac{M}{(n+1)!}$ s.t. $\left|f(x) - Q(x)\right| \leq K\left|x - a\right|^{n+1}$, since this would just be proving the Lagrange error bound theorem, and would just involve some integration/induction. However, perhaps because I keep thinking about how to prove the converse, I'm completely stuck on how to go from knowing that Q and K exist to make the initial inequality true, to proving that Q is the Taylor polynomial. I'm positive that I'm over-complicating the problem, but I just cannot figure out where to start. Does anyone have any suggestions or jumping off points for this proof?