Suppose $f\in C^2$ and using the second taylor expansion at $q$. We have
$f(x)=f(q)+<\bigtriangledown f(q),x-q>+R_2(x)$ for $x\in \overline {B\epsilon (p)}$
How to prove that $|R_2(x)|\le M||x-q||^2$?
Any material related to the proof is appreciated! Thanks!
Can anyone help me? Thanks!
Assume $q=f(q)=\nabla f(q)=0$, and let an $x$ be given. Consider the auxiliary function $$\phi(t):=f(t\>x)\qquad(0\leq t\leq1)\ .$$ Then $$\phi'(t)=\sum_{k=1}^n f_{.k}(t\>x) x_k,\qquad \phi''(t)=\sum_{i,\>k} f_{.ik}(t\>x)x_i x_k\ .$$ From $f\in C^2$ it follows that there is a neighborhood $U$ of $0$ and an $M>0$ such that $$|\phi''(t)|\leq M\>|x|^2\qquad(0\leq t\leq 1)\ ,$$ whatever $x\in U$. From $$f(x)=\phi(1)=\int_0^1\phi'(t)=(t-1)\phi'(t)\biggr|_{t=0}^1+\int_0^1(1-t)\phi''(t)\>dt\tag{1}$$ we now obtain $$|f(x)|\leq {1\over2}M\>|x|^2\qquad(x\in U)\ ,$$ since the first term on the right hand side of $(1)$ evaluates to $0$.