# I Linearizing vectors using Taylor Series

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1. Aug 9, 2016

### SPFF

I am linearizing a vector equation using the first order taylor series expansion. I would like to linearize the equation with respect to both the magnitude of the vector and the direction of the vector.

Does that mean I will have to treat it as a Taylor expansion about two variables, x(direction) and x(magnitude)? Or would linearizing with respect to the vector itself make up for both magnitude and direction?

2. Aug 9, 2016

### Krylov

The magnitude of a vector is not independent of the vector itself. I don't know your precise expression, but note that if you linearize a function of the vector $(x,y)$ with magnitude $r = \sqrt{x^2 + y^2}$ around the vector $(x_0,y_0)$ with magnitude $r_0 = \sqrt{x_0^2 + y_0^2}$ then you will automatically linearize any occurence of $r$ as well, since
\begin{align*} r &\approx r_0 + \frac{\partial r}{\partial x}\Bigr|_{(x,y) = (x_0,y_0)}(x - x_0) + \frac{\partial r}{\partial y}\Bigr|_{(x,y) = (x_0, y_0)}(y - y_0)\\ &= r_0 + \frac{x_0}{r_0}(x - x_0) + \frac{y_0}{r_0}(y - y_0) \end{align*}
neglecting terms of quadratic and higher order. It then also follows, for example, that
$$r^3 \approx r_0^3 + 3 r_0 x_0 (x - x_0) + 3 r_0 y_0 (y - y_0) + \text{h.o.t.}$$

Last edited: Aug 9, 2016
3. Aug 9, 2016

### SPFF

I guess what I'm trying to say is, given some vector r which represents say the position of an object in space, if I wanted to linearize its function with respect to its direction only (i.e. perturb the position with respect to its direction not magnitude), then I could do a Taylor expansion where the derivative is with respect to the unit vector that gives the direction of R but not its magnitude. Now if the function F(r) looks something like GMr/r^3 and I want to linearize with respect to direction and magnitude, would a taylor expansion wrt just the vector r be enough, or would I have to do the taylor expansion with respect to two variable, r and r.
Im leaning towards the latter.

4. Aug 9, 2016

### Krylov

Yes, so am I, but I would prefer to think of it as only doing a Taylor expansion of the function $\mathbf{r} \mapsto \tfrac{\mathbf{r}}{r^3}$ w.r.t. $\mathbf{r}$ where I would take into account that $r$ itself is a function of $\mathbf{r}$ as well. So, to make it (overly, admittedly) explicit, assuming you are in three-dimensional space, you are expanding the function
$$\begin{bmatrix} x\\ y\\ z \end{bmatrix} \mapsto (x^2 + y^2 + z^2)^{-\frac{3}{2}} \begin{bmatrix} x\\ y\\ z \end{bmatrix}$$
which depends on three variables and has three components. (You can of course just Taylor each component separately.) Once you have found the linear approximation, you may then restrict yourself to certain subclasses of perturbations, such as those for which the length or the direction stay constant.

Of course, instead of writing everything out in components, you can also write directly (calling the above function $F$),
$$F({\mathbf{r}) = F(\mathbf{r}_0}) +DF(\mathbf{r}_0)\cdot(\mathbf{r} - \mathbf{r}_0) + \text{h.o.t.}$$
where the second term on the right is a matrix-vector product. The $i$th row of the $3 \times 3$ derivative matrix $DF(\mathbf{r}_0)$ is just the gradient of the $i$th component of $F$ evaluated at $\mathbf{r} = \mathbf{r}_0$. (It is sometimes called the "vector gradient", but I'm not sure whether I find that terminology helpful.) This way you get a vector-valued expansion that looks just like that of a scalar-valued function of one variable.

Last edited: Aug 10, 2016