moxy
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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?