# What is this property of functions called: $f(a,b,c,d)=g(x(a,b),y(c,d))$

In general (I didn't want to put this long formula in the title): $f(a_1,a_2,\dots,a_k)=g(x_1(a_{i_{1,1}},\dots,a_{i_{1,n_1}}),x_2(a_{i_{2,1}},\dots,a_{i_{2,n_2}}),\dots,x_l(a_{i_{l,1}},\dots,a_{i_{l,n_l}}))$, where $l<k$ and $a_{i_{1,1}}$ through $a_{i_{l,n_l}}$ are a permutation of $a_1$ through $a_l$.

Basically, $f$ can be "separated" into a function $g$ applied to the result of $l$ functions which take disjoint subsequences that form a partition of $f$'s arguments.

A concrete example: $f(a,b,c,d)=ab+cd$; here $g(x,y)=x+y$, $x(a,b)=ab$ and $y(c,d)=cd$. A function that wouldn't be separable like that would be $ab/c+cd$.

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I don't think it really has a name. – Qiaochu Yuan Aug 16 '12 at 2:39
What about some restricted forms? I.e., with $g$ being of a certain form, or for some specific $l$ and $k$? – valtron Aug 16 '12 at 2:41
You may be interested in Hilbert's 13th – user31373 Aug 16 '12 at 4:46
The phrase "nomographic construction" has been associated with this type of formulation, but it would not be understood as such by most readers. See the Wikipedia link given by @LeonidKovalev. The narrow case of (summing) products of functions of one variable is of course the "separation of variables" technique taught, e.g. in introductory differential equations. In the work by Kolomogorov, Arnold, and Shimura, the phrase "superposition of functions" is used but is broader than the construction outlined here. – hardmath Aug 16 '12 at 13:40

You need to be more specific about what properties you want to require of the functions $g, x_i$. If they can be totally generic (so we are working at the level of set theory), then every function has this property. For instance, if we are working on $\mathbb{R}$, you could take $x_1(a,b) = (a,b) \in \mathbb{R}^2$, $x_2$ similarly, and then think of $g$ as a function taking two inputs which are elements of $\mathbb{R}^2$ (which it can then pick apart and apply $f$ to). In fact this is nothing but the associativity of the Cartesian product.
Even if you want all the functions to map back into $\mathbb{R}$, no generality is lost. There exists a bijection $\phi$ from $\mathbb{R}^2$ to $\mathbb{R}$. Then just take $x_i = \phi$, and $g(p,q) = f(\phi^{-1}(p), \phi^{-1}(q))$.