0
$\begingroup$

Suppose you're given a set of $n$ functions of $k$ variables\begin{align}f_1(x_1, x_2,\ldots, x_k)& =0\\f_2(x_1, x_2,\ldots, x_k)& =0\\ & {}\ \vdots \\f_n(x_1, x_2,\ldots, x_k)& =0 \end{align}

And you want to eliminate $n-1$ $x$s to give \begin{align}g_1(x_1, x_2,\ldots, x_{k-n-1})& =0 \end{align} The standard way to keep out of trouble would be to eliminate the same variable from one function paired off with each function in turn to give $n-1$ functions of $k-1$ variables. The same procedure is carried on this new set of functions and so on until you end up with the requred function at the end.

What other strategies work, in particular those that may carry some advantages?

  • 0
    @John Well, I don't think so. It is not even clear that a solution exist. Multiple can exist as well...2011-07-27

1 Answers 1

3

As noted in the comments, the answer depends strongly on the nature of the $f_i$. If they are linear, linear algebra does the trick. If they are perfectly general, chances are you're out of luck - how are you going to eliminate a variable from $\sin x+\log x+e^y+\cos y=17$? An intermediate case which is of great practical importaqnce and has not been mentioned in the comments is the case where the functions are (multivariate) polynomials. Here the keyphrases are Grobner basis and Buchberger's algorithm. It's a long story, but Buchberger's algorithm will get you started.

  • 0
    @Gerry yes, thinking about it, numerical methods don't actually come into it. Thinking about it more, it's jibberish since I've reduced the $dx$ terms by one, but still have the same number of $x$ terms. Still, I can't help thinking use can be made of it...2011-07-28