Consider an algebraic equation, whose sides are constituted only by products of functions of single variables.

$$f(x)g(y)h(z) = i(x)j(y)k(z)$$

$f,g,h,i,j,k$ are complex functions of real variable and so $x,y,z \in \mathbb{R}$. If $y$ is the variable to be considered, it is possible to isolate in both sides a factor which is function of just $y$ by hypothesis.

In this kind of problems, what can be a general condition for the equation to be independent from $y$? That is: the equation must be the same for all the real values of $y$, $\forall y \in \mathbb{R}$.

I roughly would answer that the condition is $g(y) = j(y)$, so that they simplify in the above expression: but is it rigorous enough? For example, when $y$ is such that $g(y) = h(y) = 0$, if I simplify $g(y)$ and $h(y)$ I am dividing by $0$.

A function is said independent from a certain variable if its (partial) derivative with respect to that variable is $0$. But what about a whole equation? When it can be said independent from a certain variable?

Note: even if slightly different, this question is related to my previous one.


If $g(y)=j(y)=0$, then your equation $$\operatorname{eqn}:\:\mathbb{R}^3\mapsto\{\bot,\top\},\;\;\operatorname{eqn}(x,y,z):=f(x)g(y)h(z)=i(x)j(y)k(z)\tag1\label{1}$$ becomes $0=0$, i.e. $$\operatorname{eqn}:\:\mathbb{R}^3\mapsto\{\top\},\;\;\operatorname{eqn}(x,y,z):=\top,\tag2\label{2}$$ and it is independent of $y$, because for each $(x,z)$ in $\mathbb{R}^2$ for which we have $\eqref{1}$ (that is, any $(x,z)$ in $\mathbb{R}^2)$, we also have $\eqref{2}$ (because we always have $\eqref{2}$), so $g(y)=j(y)=0$ is not really a problem, as in that case you do not need to divide by anything in the first place (notice that \eqref{2} is actually independent of any variables).

To answer the question, $g(y)=j(y)$ certainly works, but is an overkill. You actually just only need both $g$, $j$ to be constant, and not necessarily equal to each other.

  • $\begingroup$ ok, thank you for your observations. $g,h$ constants may be a sufficient condition, but not a necessary one I think. $g(y) = k_1 y$ and $ j(y) = k_2 y $ make also the equation independent of $y$. Maybe the general form should be $g(y) = kh(y), k \in \mathbb{C}$? $\endgroup$ – BowPark Oct 29 '15 at 16:55
  • $\begingroup$ @BowPark yeah, that looks right. $\endgroup$ – dbanet Oct 29 '15 at 16:59

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