Parsiad Azimzadeh

Multivariate Taylor's Theorem


This short post derives Taylor’s Theorem for multivariate functions by employing Taylor’s Theorem in a single variable.


Let $f : \mathbb{R}^d \rightarrow \mathbb{R}$. For vectors $x$ and $v$ in $\mathbb{R}^d$, define $g : \mathbb{R} \rightarrow \mathbb{R}$ by $g(t) = f(x + tv)$. If $g$ is $K$ times differentiable at zero, Taylor’s theorem in 1d tells us

\[\label{eq:1d}\tag{1} f(x + tv) = g(t) = \sum_{k = 0}^K \frac{t^k}{k!} g^{(k)}(0) + o(t^K) \text{ as } t \rightarrow 0.\]


\[\label{eq:derivative}\tag{2} g^{(k)}(t) = \sum_{i_1, \ldots, i_k} v_{i_1} \cdots v_{i_k} \frac{\partial^k f}{\partial x_{i_1} \cdots x_{i_k}}(x + tv).\]

By chain rule,

\[g^{(k + 1)}(t) = \sum_{i_1, \ldots, i_k} v_{i_1} \cdots v_{i_k} \left \langle v, \nabla \left[ \frac{\partial^k f}{\partial x_{i_1} \cdots x_{i_k}} \right] (x + tv) \right \rangle.\]

Simplifying, we arrive at \eqref{eq:derivative} with $k$ replaced by $k + 1$. Since \eqref{eq:derivative} is trivially satisfied at $k = 1$, it follows by induction that \eqref{eq:derivative} holds for all positive integers $k$.

The form \eqref{eq:derivative} is redundant since, assuming the conditions of Clairaut’s theorem, partial derivatives commute (e.g., $f_{x_1 x_2} = f_{x_2 x_1}$). For a multi-index $\alpha = (\alpha_1, \ldots, \alpha_d)$ in $\mathbb{Z}^d_{\geq 0}$, define $|\alpha| = \alpha_1 + \cdots + \alpha_d$ and

\[D^\alpha f = \frac{\partial^{|\alpha|} f}{\partial x_1^{\alpha_1} \cdots \partial x_d^{\alpha_d}}.\]

With this notation we can write \eqref{eq:derivative} as

\[g^{(k)}(t) = \sum_{|\alpha| = k} \frac{k!}{\alpha_1! \cdots \alpha_d!} v_1^{\alpha_1} \cdots v_d^{\alpha_d} D^\alpha f(x + tv).\]

Substituting this into \eqref{eq:1d}, we obtain the desired Taylor polynomial:

\[f(x + tv) = \sum_{k = 0}^K t^k \sum_{|\alpha| = k} \frac{1}{\alpha_1! \cdots \alpha_d!} v_1^{\alpha_1} \cdots v_d^{\alpha_d} D^\alpha f(x) + o(t^K) \text{ as } t \rightarrow 0\]


If, in addition, $g$ is $K + 1$ times differentiable, we can extend the Cauchy or Lagrange form of the remainder term to the multivariate setting. For example, the Lagrange form is

\[o(t^K) = \frac{t^{K + 1}}{\left( K + 1 \right)!} g^{(K + 1)}(\theta)\]

where $\theta$ is some number between zero and $t$. Substituting \eqref{eq:derivative} into the above, we can obtain by triangle inequality, the (loose) bound

\[o(t^K) \leq \frac{ \left( d \left| t \right| \left \Vert v \right \Vert_\infty \right)^{K + 1} }{\left( K + 1 \right)!} \max_{|\alpha| = K + 1} \left |D^\alpha f(x + \theta v) \right|.\]