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## Homework Statement

I'm given that the function [itex]f(x)[/itex] is n times differentiable over an interval [itex]I[/itex] and that there exists a polynomial [itex]Q(x)[/itex] of degree less than or equal to [itex]n[/itex] s.t.

[tex]\left|f(x) - Q(x)\right| \leq K\left|x - a\right|^{n+1}[/tex]

for a constant [itex]K[/itex] and for [itex]a \in I[/itex]

I am to show that [itex]Q(x)[/itex] is the Taylor polynomial for [itex]x[/itex] at [itex]a[/itex]

## Homework Equations

For [itex]Q(x)[/itex] to be the Taylor polynomial for [itex]x[/itex] at [itex]a[/itex]:

[tex]Q(x) = \sum_{k = 0}^{n}\frac{f^{(k)}(a)}{k!} (x - a)^k[/tex]

The [itex]n^{th}[/itex] error/remainder term of [itex]Q(x)[/itex] is:

[tex]R_{n,a}(x) = \left|f(x) - Q(x)\right|[/tex]

The Lagrange error bound is:

[tex]R_{n,a}(x) = \left|f(x) - Q(x)\right| \leq \frac{M}{(n+1)!}|x - a|^{n+1}[/tex]

where [itex]M = sup\left|f^{n+1}(x)\right|[/itex]

## The Attempt at a Solution

I think I could easily go from knowing that [itex]Q(x)[/itex] was the Taylor polynomial of [itex]f[/itex] around [itex]a[/itex] and prove that there exists a constant [itex]K = \frac{M}{(n+1)!}[/itex] s.t. [itex]\left|f(x) - Q(x)\right| \leq K\left|x - a\right|^{n+1}[/itex], 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?