# Multi-Dimensional Taylor Series¶

date: 2012/10/8

Everyone knows how to write the Taylor series. Given a scalar function $$f(\xi)$$, its Taylor series can be written as

(1)$f(\xi; \xi_r) = \sum_{k=0}^{\infty} \frac{1}{k!} \frac{\mathrm{d}^k f}{\mathrm{d} \xi^k} (\xi-\xi_r)^k$

where $$\xi_r$$ is the centered point. There is few challenge in obtaining Eq. (1). However, if the function of interest has more than one variable, e.g., $$u(x_1, x_2, \ldots, x_N)$$, Eq. (1) can no longer be used.

To obtain a formula for the Taylor series of a multi-dimensional function, it is obvious that we should use partial derivatives $$\partial/\partial x_1, \partial/\partial x_2, \ldots, \partial/\partial x_N$$ rather than simple derivative $$\mathrm{d}/\mathrm{d}\xi$$. To insert the partial derivatives, we need to use directional derivative. To proceed, consider a scalar function

$u(x_1, x_2, \ldots, x_N)$

in an $$\mathbb{E}^N$$-space, in which the coordinate axes are denoted by

$\mathbf{x} = (x_1, x_2, \ldots, x_N)$

Let

$\newcommand\defeq{\stackrel{\mathrm{def}}{=}} \mathbf{y} \defeq \mathbf{x} - \mathbf{x}_r$

denote a vector pointing from a referencing point $$\mathbf{x}_r$$ to $$\mathbf{x}$$. The directional derivative of $$u$$ is then written as

$\nabla_yu = \vec{\nabla}u\cdot\left(\frac{\mathbf{y}}{y}\right)y = \left[\sum_{i=1}^N(x_i - {x_r}_i)\frac{\partial}{\partial x_i}\right] u$

Similarly, higher-order directional derivatives can be written as

(2)$\nabla_y^k u = \left[\sum_{i=1}^N(x_i-{x_r}_i) \frac{\partial}{\partial x_i}\right]^k u$

Aided by replacing the simple derivative in Eq. (1) with the directional derivative, the Taylor series of $$u$$ about $$\mathbf{x}_r$$ can be expressed as

(3)$u(\mathbf{x}; \mathbf{x}_r) = \sum_{k=0}^{\infty} \frac{1}{k!}\nabla_y^k u$

Substituting Eq. (2) into Eq. (3) gives

(4)$u(\mathbf{x}; \mathbf{x}_r) = \sum_{k=0}^{\infty} \frac{1}{k!} \left[\sum_{i=1}^N(x_i-{x_r}_i) \frac{\partial}{\partial x_i}\right]^k u$

Equation is the Taylor series of a multi-dimensional function $$u(x_1, x_2, \ldots, x_N)$$.

It should be noted that symmetry of the higher-order partial derivatives is not assumed. That is, the mixed partial derivatives are not commutative.