Why Taylor Expansion (1st order, multilvariable) is the best approximation of f(x) by polynomials?

Question

I don't how to start the question here, to use "ball" to solve problem seems very hard for me... Could anyone please help?

Answer 1

The problem, as stated, doesn't seem to be correct. Consider $f:\mathbb{R}^n\rightarrow \mathbb{R}$ given by $f(x) = a_0+b_0\cdot x$ for some $a_0 \in \mathbb{R}$ and $b_0 \in \mathbb{R}^n$. Then we see the first approximation of $f$ at some point $x^\ast$ is itself which we could write as $f(x)= a_0+b_0\cdot x^\ast+b_0(x-x^\ast)$. Now, consider $p(x)$ a first order polynomial $p(x) = f(x^\ast) + b_1\cdot (x-x^\ast)$ where $b_1 \neq b_0$, i.e. $p(x^\ast) = f(x^\ast)$ but $p(x)\neq f(x)$.

Now, observe \begin{align} |f(x)-p(x)|= |(b_0-b_1)\cdot(x-x^\ast)|. \end{align} If we choose $x$ such that $(x-x^\ast)\perp (b_0-b_1)$, which is always possible for any neighborhood of $x^\ast$, we see $|f(x)-p(x)| = 0$ but $|x-x^\ast| \neq 0$. Thus there can't exists $\eta>0$ such that $|f(x)-p(x)|\geq\eta|x-x^\ast|$ for any neighborhood of $x^\ast$.