Doubt in multivariate Taylor's theorem The way wikipedia presents it, I don't understand how many functions $h_\alpha$ do we have in the remainder term?
Using matrix notation, $$f({\boldsymbol {x}})=f({\boldsymbol {a}})+\nabla f(\boldsymbol a)^\top({\boldsymbol {x}}-{\boldsymbol {a}})+h({\boldsymbol {x}})\lVert {\boldsymbol {x}}-{\boldsymbol {a}}\rVert ,\qquad \lim _{{\boldsymbol {x}}\to {\boldsymbol {a}}}h({\boldsymbol {x}})=0.$$
is the expression we get for Taylor's theorem for linear term. Can we say something analogous about quadratic term in terms of Hessian?
In other words, can someone tell me how to write the remainder of multivariable Taylor's theorem in matrix notation for two terms (i.e. when the function is assumed to be twice differentiable at the point)?
 A: In general, you have
$$
\begin{align*}
f(x + h) &= f(x) + f'(x) \cdot h + \ldots + \dfrac{1}{(p-1)!} f^{(p-1)}(x) \cdot h^{(p-1)} \\
&+ \int\limits_0^1 dt\ \dfrac{(1-t)^{p-1}}{(p-1)!} f^{(p)}(x+th) \cdot h^{(p)},
\end{align*}
$$
where the integral is well defined,
$$
\int\limits_0^1 dt\ \dfrac{(1-t)^{p-1}}{(p-1)!} \Big(f^{(p)}(x+th) \cdot h^{(p)}\Big) = \left( \int\limits_0^1 dt\  \dfrac{(1-t)^{p-1}}{(p-1)!} f^{(p)}(x+th) \right) \cdot h^{(p)}
$$
Where $f^{(k)}$ is the $k$th derivative of $f$ (which is a $k$-linear function) and $h^{(k)} = (h, \ldots, h) = h \otimes \mathbf{1}_k$ (and $h \in \mathbf{R}^d,$ assuming $f:\mathbf{R}^d \to \mathbf{R}^c$).
See, (8.14.3) of Foundations of Modern Analysis by Jean Dieudonné. Note that when $p = 2,$ $f^{(2)}(x)$ is a symmetric bilinear function whose matrix representation is the Hessian, in other words, $f^{(2)}(x) \cdot (h_1, h_2) = h_1^\intercal \mathbf{H}_f(x) h_2,$ where $\mathbf{H}_f(x)$ is the Hessian of $f$ at $x.$
Ammend. The previous formula assumes $f$ to be $p$ times differentiable with continuity in a ball centred at $x.$ If we only assume $f$ to be $p$ times differentiable at $x$ (so that $f$ is $p-1$ times differentiable in a ball around $x$ and thr $(p-1)$th derivative is assumed differentiable at $x$), we obtain the weaker form of the previous result:
$$
\|f(x+h) - T_f^p(x; h)\| = o(\|h\|^p),
$$
where $T_f^p(x; h)$ is the Taylor polynomial of $f$ of degree (at most) $p$ evaluated at $h,$ namely
$$
T_f^p(x; h) = f(x) + f'(x) \cdot h + \ldots + \dfrac{1}{p!} f^{(p)}(x) \cdot h^{(p)}.
$$
The proof of this second result is given in Cartan's Differential Calculus (unfortunately out of print for decades now), Theorem 5.6.3.
