Approximating $\log(X-Y)$

Question

Is there a way to approximate $\log(X-Y)$ as $f(X)+f(Y)$?

Answer 1

For exact value the simple answer is no. The reason is a general result. Given fixed $\,n>0,\,$ the general two variable sum of products function $$ h(x,y) := \sum_{i=1}^n f_i(x)\,g_i(y)$$ has the property that the matrix $\,M\,$ with entries $\,M_{j,k} := h(x_j,y_k)\,$ for any values $\,x_1,x_2,\dots,m\,$ and $\,y_1,y_2,\dots,m\,$ has matrix rank at most $\,n.\,$ This implies that the determinant is zero if $\,m>n.$

In your example, $\,h(x,y) = f(x) + f(y)\,$ where $\,n=2,\,$ and with $$f_1(x)=f(x),\quad g_1(y)=1,\quad f_2(x)=1,\quad g_2(y)=f(y).$$

Thus, a single case where the $\,3\times 3\,$ matrix $\,M\,$ has determinant non-zero is a counter-example.

For approximate value, you might as well let $\,f(x)=x\,$ because any number can approximate any other number. You have not specified the range of values of the variables and your definition of "approximation".

The most obvious problem is that if $\,X=Y\,$ then your equation $\,\log(X-Y) = f(X)+f(Y)\,$ has left side $\,\log(0)\,$ which is undefined and right side $\,2f(X),\,$ and thus $\,f(X)\,$ is undefined.