Multivariable taylor polynomial of a transformation

Question

I asked a similar question yesterday but I realized it was worded very poorly so I deleted it and I am going to try and rephrase my question. I want to find the taylor approximation (taylor polynomial) at the point $(1,0,0)$ of the following transformation:

$$f(r, \phi,\theta)=\begin{pmatrix}br \sin{\phi}\cos{\theta}\\ar\sin{\phi} \sin{\theta}\\cr \cos{\phi}\end{pmatrix}$$

where $a,b,c \in \mathbb R$.

Since the taylor approximation that I linked and want to use can only be used for scalar valued functions, I have to approximate every component seperately. The formula for the expansion is:

$$T_{2,f}=f(\vec{x}_0)+\nabla^Tf(\vec{x}_0)(\vec{x}-\vec{x}_0)+\frac{1}{2}(\vec{x}-\vec{x}_0)^TH_f(\vec{x}_0)(\vec{x}-\vec{x}_0)$$

where $H_f$ is the Hessian matrix and $\vec{x}_0=(1,0,0)^T$. Now I am facing the following problem:

1. What does $\vec{x}$ in this formula refer to?

Up until this point I had only used this formula with cartesian coordinates where $\vec{x}=(x,y,z)^T$ but in this case I am not sure what this is referring to. Do I use $\vec{x}=(r,\phi,\theta)$ or do I use the spherical basis coordinates?

2. In case I have to use spherical coordinates for my $\vec{x}$, vector, do I also have to somehow transform my $\vec{x}_0$ vector or is this vector already in spherical coordinates?

EDIT 1: based on the suggestion from Duca_Conte I tried to do it by converting to cartesian coordinates $f(r,\phi, \theta)=f(x,y,z)=(bx,ay,cz)^T$ but the problem becomes "too easy" (almost trivial) and the answer doesn't make much sense so I am not sure that change of coordinates works. If anyone as any idea how to do this I would be very grateful.

EDIT 2: This is the solution I have so far.

1. Write the transformation in cartesian coordinates:

$$f(r,\phi, \theta)=f(x,y,z)=\begin{pmatrix}bx \\ay \\cz\end{pmatrix}$$

2. Calculate the second-order taylor approximation for each component:

$$T_{2,f_1}=b+b(\frac{\partial f_1}{\partial x},\frac{\partial f_1}{\partial y},\frac{\partial f_1}{\partial z})\begin{pmatrix}x-1\\y\\z\end{pmatrix}+0=2b$$

$$T_{2,f_2}=0 \\ T_{2,f_1}=0$$

$$\implies T_{2,f}= \begin{pmatrix}2b \\ 0\\0\end{pmatrix}$$

This doesn't look right and I am wondering if I am making a big mistake somewhere.

Can anyone tell me if this is the correct way to do this and if not, suggest a solution?

Answer 1

Your problem is mainly a problem of notation; e.g., the names of the variables in your source about the Taylor expansion conflict with the names of the variables in your concrete problem.

You are given a vector-valued map $${\bf f}:\quad {\mathbb R}^3\to{\mathbb R}^3,\qquad(r,\phi,\theta)\mapsto (x,y,z)\ .$$ This map is of a certain practical use, and has a geometric interpretation in terms of affinely scaled spherical coordinates. But this should not disturb us from mechanically applying the rules for finding the Taylor coefficients of ${\bf f}$. In a figure you maybe won't draw the $(r,\phi,\theta)$-space ${\mathbb R}^3$ on the left; but it is there, and has three ordinary axes. You will only draw the $(x,y,z)$-space, and if $a=b=c=1$ you will show the geometric meaning of $r$, $\phi$, $\theta$.

Now $${\bf f}(r,\phi,\theta)=\bigl(x(r,\phi,\theta),y(r,\phi,\theta),z(r,\phi,\theta)\bigr)\ .$$ You have decided to treat the three functions $x(\cdot)$, $y(\cdot)$, $z(\cdot)$ separately. This seems reasonable. Let's consider the function$$x(\cdot):\quad(r,\phi,\theta)\mapsto x(r,\phi,\theta)=b\,r\,\sin\phi\,\cos\theta\ .$$ At this point we need your formula for the Taylor expansion. In terms of the variables at stake it comes as follows: Write $(r,\phi,\theta)=:{\bf r}$ and $(1,0,0)=:{\bf r}_0$.$\ {}^*)\ $ Then $$T_{2,{\bf r}_0} x({\bf r})=x({\bf r}_0)+\nabla x({\bf r}_0)\cdot({\bf r}-{\bf r}_0)+{1\over2}({\bf r}-{\bf r}_0)H_x({\bf r}_0)({\bf r}-{\bf r}_0)\ .\tag{1}$$ Here $T_{2,{\bf r}_0} x$ means "second order Taylor polynomial of the function $x(\cdot)$, centered at ${\bf r}_0$", and I have omitted the ${}^T$ symbols. Now we have to evaluate the things appearing on the RHS of $(1)$: $$\eqalign{x({\bf r}_0)&=x(1,0,0)=0,\cr {\bf r}-{\bf r}_0&=(r-1,\phi,\theta),\cr \nabla x(r,\phi,\theta)&=\bigl(b\sin\phi\cos\theta,b\,r\cos\phi\cos\theta,-b\,r\sin\phi\sin\theta\bigr),\cr \nabla x({\bf r}_0)&=(0,b\,r,0)\ .\cr}$$ I leave it to you to compute the Hessian $H_x({\bf r}_0)$. This will in the end be a $3\times3$-matrix of real numbers.

${}^*)\ $ If $(1,0,0)$ means the point with $(x,y,z)$-coordinates $(1,0,0)$ then ${\bf r}_0=(r_0,\phi_0,\theta_0)$ has to be determined numerically from the equations.

Answer 2

First choose in which coordinates you want to work. Looking at your function $f$, I suppose that the best choice is the spherical.

Of corse you can write your vector $\mathbf{x}$ in polar coordinates, but you have to be careful, since also the gradient and the Hessian must be in the same coordinate system. As long you are coherent with your choice it should work. And yes, also $\mathbf{x}_0$ must be written in spherical coordinates, if not you are subtracting lengths to angles, and it does not make sense.

Here you can find all the transformations you need: https://en.wikipedia.org/wiki/Del_in_cylindrical_and_spherical_coordinates