Taylor Series Expansion of a Vector Cross product

Question

This is confusing me a little. What is the Taylor series expansion of:

$$f(u,v) = u \times v$$

Around points (i.e., vectors) $u_0$ and $v_0$ up until the first derivative (i.e., no higher order terms than order 2)? Thank you!

Here's what I have so far: so we pick a small values near $(u_0, v_0)$ i.e.,: $$u = \bar{u} + u_0 $$ $$v = \bar{v} + v_0 $$

we want to express $f(u,v)$ as follows: $$f(u,v) \approx f(u_0, v_0) + \frac{\partial f (u,v)}{\partial u}\Big|_{(u_0,v_0)}\,\bar{u} + \frac{\partial f (u,v)}{\partial v}\Big|_{(u_0,v_0)}\,\bar{v}$$.

Which means the Taylor series expansion is given as follows: $$ % f(u,v) \approx u_0 \times v_0 + \frac{\partial}{ \partial u}(u \times v)\Big |_{(u_0,v_0)}\cdot \bar{u} \ \ + \frac{\partial}{ \partial v}(u \times v)\Big |_{(u_0,v_0)}\cdot \bar{v} % $$

$$ = u_0 \times v_0 + (\mathbb{1} \times v_0) \cdot \bar{u} \ \ + (u_0 \times \mathbb{1}) \cdot \bar{v} $$

where $\mathbb{1}$ is simply a column vector of ones i.e., $\frac{\partial }{\partial u}\begin{bmatrix}u_1 \\u_2 \\u_3 \end{bmatrix} = \begin{bmatrix}1 \\1 \\1 \end{bmatrix}$.

Is this correct? Or should $\mathbb{1}$ be the idenity matrix $I_{3\times3}$ instead?

Answer 1

I guess @TedShifrin's comment is the direct approach, where you substitute $u = \bar{u} + u_0, v = \bar{v} + v_0$ in the function, expand, and drop terms that are second order and higher (i.e. involve multiplication, or cross product in this case, of at least two of $\bar{u}, \bar{v}$.

Here's my approach, kind of working backwards from the definition. We ultimately want something of the form $$f(u_0, v_0) + \frac{\partial f (u,v)}{\partial u}\Big|_{(u_0,v_0)}\,\bar{u} + \frac{\partial f (u,v)}{\partial v}\Big|_{(u_0,v_0)}\,\bar{v}, \quad \bar{u} \equiv u - u_0, \; \bar{v} \equiv v - v_0, \; \tag{1}\label{eq1}$$

Since $u$ is a vector, I don't know how to confidently calculate $\frac{\partial f (u,v)}{\partial u}$. So I start with the fundamental definition of the derivative for scalar variables, $$\frac{dy}{dx}\Big|_{x_0} = \lim_{x \to x_0} \frac{y(x) - y(x_0)}{x - x_0} = \lim_{\bar{x} \to 0} \frac{y(x_0 + \bar{x}) - y(x_0)}{\bar{x}}$$

Now if we try to directly use this formula for our vector situation, we have a problem as we'll be dividing by a vector, which is undefined. $$\frac{\partial f (u,v)}{\partial u}\Big|_{(u_0,v_0)} = \lim_{\bar{u} \to 0} \frac{f(u_0 + \bar{u},v_0) - f(u_0,v_0)}{\bar{u}}$$

But if we ignore the limit for now and rearrange, $$\frac{\partial f (u,v)}{\partial u}\Big|_{(u_0,v_0)} \bar{u} = f(u_0 + \bar{u},v_0) - f(u_0,v_0)$$

This we can use in our desired expansion, Eq. \eqref{eq1}, \begin{align} f(u_0, v_0) + &\left(f(u_0 + \bar{u},v_0) - f(u_0,v_0)\right)+ \left(f(u_0, v_0 + \bar{v}) - f(u_0,v_0)\right) \\&= f(u_0 + \bar{u},v_0) + f(u_0, v_0 + \bar{v}) - f(u_0,v_0) \end{align}

Applying the function $f(u,v) = u \times v$, \begin{align} (u_0 + \bar{u})\times v_0 &+ u_0 \times (v_0 + \bar{v}) - u_0 \times v_0 \\ &= u_0 \times v_0 + \bar{u}\times v_0 + u_0 \times \bar{v} \end{align}

So the Taylor series expansion is, $$f(u,v) \approx u_0 \times v_0 + \bar{u}\times v_0 + u_0 \times \bar{v} $$

*I did drop the limit from the derivative definition earlier, so there could be an error in this derivation, but I at least think the final result is correct

Answer 2

$ \def\LR#1{\left(#1\right)} \def\qiq{\quad\implies\quad} \def\p{\partial} \def\grad#1#2{\frac{\p #1}{\p #2}} \def\m#1{\left[\begin{array}{r}#1\end{array}\right]} \def\gradLR#1#2{\LR{\grad{#1}{#2}}} $Based on this answer the relevant derivatives for a cross-product are $$\eqalign{ f &= f(a,b) \;=\; a\times b \;=\; Ab \;=\; B^Ta \quad \\ }$$ $$\eqalign{ \grad fb &= A &= \m{ 0\; & -a_3 & a_2\; \\ a_3 & 0\; & -a_1\; \\ -a_2 & a_1 & 0\;\; \\ } \\ \grad fa &= B^T &= \m{ 0\; & b_3 & \:-b_2\; \\ -b_3 & 0\; & b_1\; \\ b_2 & \:-b_1 & 0\;\; \\ } \;=\; -B \\ \\ }$$ So the Taylor series centered at $(a_0,b_0)$ yields $$\eqalign{ f(a,b) &= f(a_0,b_0) \;+\; A_0(b-b_0) \;+\; B^T_0(a-a_0) \\ &= f(a_0,b_0) \;+\; a_0\times(b-b_0) \;-\; b_0\times(a-a_0) \\ &= f(a_0,b_0) \;+\; a_0\times b \;+\; a\times b_0 \\ }$$