I would like to show the validity of the multivariable version of Taylor series expansion up to second-order terms (if possible without using one of the explicit forms for the remainder term):
Let $f : \mathbb{R}^n \to \mathbb{R}$ be twice differentiable at $\vec{a} = (a_1, \ldots, a_n)$ and let $\vec{x} = (x_1, \ldots, x_n)$. Then
$f(x) = f(a) + \sum_{i=1}^{n} \frac{\partial f(a)}{\partial x_i} (x_i-a_i) + \frac{1}{2}\sum_{i,j=1}^n \frac{\partial^2 f(a)}{\partial x_i \partial x_j} (x_i-a_i)(x_j-a_j) + R(x)$
where $\lim_{x \to a} \frac{R(x)}{||x-a||^2} = 0$.
I let $g(t) = f(a + t(x-a))$. Then I write $g$ in terms of its Taylor Series about $0$, evaluated at $t=1$ to get $g(1) = g(0) + g'(0) + \frac{1}{2} g''(0) + R(1)$. Then, using the chain rule \begin{align*} g(1) &= f(x) \\ g(0) &= f(a) \\ g'(0) &= \sum_{i=1}^n \frac{\partial f(a)}{\partial x_i} (x_i-a_i) \\ g''(0) &= \sum_{i,j=1}^n \frac{\partial^2 f(a)}{\partial x_i \partial x_j} (x_i-a_i)(x_j-a_j) \end{align*} Then letting $R(x) = R(1)$, how do I show that $\lim_{x \to a} \frac{R(x)}{||x-a||^2} = 0$ using the fact that $g(t) = g(0) + g'(0)t + \frac{1}{2} g''(0)t^2 + R(t)$ where $\lim_{x \to 0} \frac{R(t)}{t^2} = 0$? Or perhaps another approach is necessary?
$$R(x)=R(1)$$
$$R(1) =\frac{g'''(c)}{3!} = \frac16\sum_{i,j,k=1}^n \frac{\partial^3 f(a+c(x-a))}{\partial x_i \partial x_j \partial x_k} (x_i-a_i)(x_j-a_j)(x_k-a_k)$$
If there is upbound M for all the third derectives, then we get,
$$|R(x)| \lt |\frac M6|~|\sum_{i,j,k=1}^n (x_i-a_i)(x_j-a_j)(x_k-a_k)|$$
$$\lt |\frac M6|~(\sum_{i}^n (x_i-a_i)^2)^{\frac 32}$$