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Suppose we have a function $f\colon \mathbb{R}^n\to\mathbb{R}$, which is analytic almost everywhere. Can one say that there exists a finite sequence of operations, which will evaluate $f$ for any argument, if the operations are limited to:

  • arbitrary (piecewise-) analytic single-variable functions $g_i\colon \mathbb{R}\to\mathbb{R}$
  • arithmetic

? How can this be (dis)proved?

Ruslan
  • 6,775
  • I suppose you have considered the case $n=2$ with $f$ multiplication; which solution would you propose? You can do some things with logarithms, but you must also cater for multiplication by$~0$. – Marc van Leeuwen Nov 23 '13 at 10:48
  • @MarcvanLeeuwen you're right. Additions are too restrictive. – Ruslan Nov 23 '13 at 11:16

1 Answers1

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This question (especially version 1) appears to be almost exactly what is proved in Kolmogorov–Arnold representation theorem if $f$ is continuous. The theorem states that any continuous function of any number of variables can be represented as a superposition of finite number of one-dimensional functions and additions, i.e. exist such finite families of functions $\Phi_q$ and $\psi_{q,p}$ that

$$f(x_1,...,x_n)=\sum_{q=0}^{2n}\Phi_q\left(\sum_{p=1}^n\psi_{q,p}(x_p)\right).$$

Ruslan
  • 6,775