I have a function $\ f: \mathbb{R}^n \rightarrow \mathbb{R}^m$. I can evaluate the Jacobian of the function cheaply for a single variable $\ f: \mathbb{R} \rightarrow \mathbb{R}^m$, (e.g. a single value parameterizing a homogeneous space). Can I use this information to build a full Jacobian?
1 Answers
Yes, you can use this information to build the full Jacobian one column at a time.
Say the domain $\mathbb{R}^n$ has coordinates $x_1,\dotsc,x_n$ and the function $f$ has components $f_1,\dotsc,f_m$. And say $P=(p_1,\dotsc,p_n)$ is a point in $\mathbb{R}^n$. Our goal is to compute the Jacobian of $f$ at $P$, under the assumption that we can compute Jacobians of single-variable functions (functions $\mathbb{R} \to \mathbb{R}^m$).
The $j$th column of the Jacobian of $f$ is $$ \begin{pmatrix} \frac{\partial f_1}{\partial x_j} \\ \frac{\partial f_2}{\partial x_j} \\ \vdots \\ \frac{\partial f_m}{\partial x_j} \end{pmatrix} . $$ You can compute this by introducing $g(x) = f(p_1,p_2,\dotsc,p_{j-1},x,p_{j+1},\dotsc,x_n)$ ($x$ in the $j$th spot). Here $g : \mathbb{R} \to \mathbb{R}^m$. For each $i$, $\frac{d g_i}{d x} = \frac{\partial f_i}{\partial x_j}$. So the Jacobian (i.e., derivative vector) of $g$ gives you the $j$th column of the Jacobian of $f$. You can compute the Jacobian of $f$ by using $n$ of these "easy" calculations for single-variable functions.
If you'd like to think of this geometrically, you could view this as restricting $f$ to the line through $P$ parallel to the $j$th coordinate axis. (But if you don't like this geometric viewpoint, you don't necessarily need it to carry out the computation of the Jacobian.)
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I think a bit of a typo: "You can compute the Jacobian of $f$ by using $m$ of these..." to "You can compute the Jacobian of $f$ by using $n$ of these...". Correct? – bfletch Jul 07 '17 at 03:14
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Can the Jacobian be filled in less than $n$ (or $m$ if I'm wrong in my above comment) evaluations in the special case that $p_1 = p_2 = ... p_n$? – bfletch Jul 07 '17 at 03:19
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1Thanks for catching the typo! No, I don't think $p_1=p_2=\dotsb=p_n$ gives a shortcut, because the functions $f_i$ might be "asymmetric". If each $f_i$ were a symmetric function (meaning that $f_i(x_1,\dotsc,x_n)=f_i(x_{\sigma(1)},\dotsc,x_{\sigma(n)})$ for every permutation $\sigma$ of ${1,\dotsc,n}$), and if $P$ satisfied $p_1=p_2=\dotsb=p_n$, then every column of the Jacobian would be identical. I think you could cook up conditions where some of the columns would be identical. – Zach Teitler Jul 07 '17 at 04:29