Separation of inverse function

Question

Let's consider $$z=\frac{f(x)+f(y)}{K-f(x)+f(y)}$$ where $K$ is a constant.

Is there any formal method to approximate $z$ as $(g(x)+h(y))$ ?

Answer 1

Here is one method (for simplicity I assume the function is defined over the unit square):

Approximation problem

Suppose we have a continuous function $z : [0,1]\times [0,1] \rightarrow\mathbb{R}$. We want to find functions $g:[0,1]\rightarrow\mathbb{R}$, $h:[0,1]\rightarrow\mathbb{R}$ so that $z(x,y) \approx g(x) + h(y)$. Define the error as: $$ Error = \int_0^1 \int_0^1 (g(x)+h(y)-z(x,y))^2dxdy $$ Since we can always subtract a constant from function $g(x)$ and add it to $h(y)$ without affecting separability or changing the error, without loss of generality we assume that $\int_0^1 g(x)dx = 0$.

Derivation

For each $x$ and $y$, we can imagine $g(x)$ and $g(y)$ as an optimization variable. Fix $x \in [0,1]$. Taking derivatives with respect to $g(x)$ gives: $$ \frac{\partial Error}{\partial g(x)} = \int_0^1 2(g(x)+h(y)-z(x,y))dy = 0 \quad \forall x \in [0,1] $$ Similarly, $$ \frac{\partial Error}{\partial g(y)} = \int_0^1 2(g(x)+h(y)-z(x,y))dx = 0 \quad \forall y \in [0,1] $$ The above equalities reduce to: \begin{align} &g(x) + \int_0^1 h(y)dy - \int_0^1 z(x,y)dy = 0 \quad \forall x \in [0,1] \\ &h(y) + \int_0^1 g(x)dx - \int_0^1 z(x,y)dx = 0 \quad \forall y \in [0,1] \end{align} Define $c = \int_0^1 h(y)dy$ and recall that $\int_0^1 g(x)dx =0$. The value $c$ is not yet known since the function $h(y)$ is not yet determined. However, the above two equalities imply: \begin{align} &g(x) = \int_0^1 z(x,y)dy - c \quad \forall x \in [0,1] \\ &h(y) = \int_0^1 z(x,y)dx \quad \forall y \in [0,1] \end{align} Integrating both sides of the above two equalities gives: \begin{align} &0 = \int_0^1\int_0^1 z(x,y)dxdy - c \\ &c = \int_0^1 \int_0^1 z(x,y)dxdy \end{align} These last two equations are the same. So we need:
$$ c = \int_0^1\int_0^1 z(x,y)dxdy $$ We now have our solution:

Solution:

Define $c = \int_0^1\int_0^1 z(x,y)dxdy$. The optimal separable functions $g^*$ and $h^*$ are: \begin{align} g^*(x) &= \int_0^1z(x,y)dy - c \quad \forall x \in [0,1]\\ h^*(y) &= \int_0^1 z(x,y)dx \quad \forall y \in [0,1] \end{align}

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The above derivation involves an "informal" differentiation with respect to each variable $g(x)$ and $h(y)$. A formal argument is as follows: Let $g(x)$ and $h(y)$ be continuous functions defined over the unit interval such that $\int_0^1 g(x)dx=0$. We want to show these give an approximation error greater than or equal to that of the above functions $g^*(x), h^*(y)$. We have: \begin{align} Error &= \int_0^1 \int_0^1 (g(x)+h(y)-z(x,y))^2 dxdy\\ &= \int_0^1\int_0^1 [(g(x)-g^*(x)+h(y)-h^*(y)) + (g^*(x)+h^*(y)-z(x,y))]^2dxdy\\ &=\int_0^1\int_0^1 (g(x)-g^*(x))^2dxdy\\ &+ \int_0^1\int_0^1 (h(y)-h^*(y))^2dxdy\\ &+\underbrace{2\int_0^1\int_0^1 (g(x)-g^*(x))(h(y)-h^*(y))dxdy}_{0}\\ &+\underbrace{2\int_0^1\int_0^1(g(x)+h(y)-g^*(x)-h^*(y))(g^*(x)+h^*(y)-z(x,y))dxdy}_{0}\\ &+\int_0^1\int_0^1(g^*(x)+h^*(y)-z(x,y))^2dxdy \\ &\geq \int_0^1\int_0^1(g^*(x)+h^*(y)-z(x,y))^2dxdy \end{align} where the first underbrace uses $\int_0^1 g^*(x)dx=\int_0^1 g(x)dx=0$, while the second underbrace uses these together with $g^*(x)=\int_0^1z(x,y)dy - c$ and $h^*(y) = \int_0^1 z(x,y)dx$. $\Box$