Rewriting $\frac{1-x}{1-y}$ in terms of $\frac{x}{y}$

Question

Is it possible to rewrite $\dfrac{1-x}{1-y}$ in terms of $\dfrac{x}{y}$? That is, the rewritten form includes $\dfrac{x}{y}$?

Answer 1

We can factor a $y$ out of $(1-x)$ as follow. $y(1/y-x/y)$. And so

$$\frac{y}{x}\frac{1/y-x/y}{1/x-y/x}=(\frac{x}{y})^{-1}\frac{1/y-x/y}{1/x-(x/y)^{-1}}$$

is about the best I can see at a glance. You could try and work with that and see if it is possible to rewrite it in terms of $x/y$. Hope this helps.

Answer 2

$\boxed{\text{No.}}$ It's really quite rare for a function of two variables to in fact be a function in one variable of some specific other function of two variables.

Pedantic Answer

If we're talking strictly about functions on the reals and to the reals, no function of the form $g\left(\frac xy\right)$ is defined for any point where $y=0$. However, your equation on the left is always defined when $y=0$. Functions with differing domains are not equal, so no such $g$ exists. Even without being super pedantic, the existence of such a problem should be throwing off alarms. Sanity checks, special cases, and small examples are the bread and butter of figuring out of something is possible or not.

General advice: Make sure function domains are equal before verifying their outputs are the same. The fact that the LHS has a problem at $y=1$ where the RHS has a problem at $y=0$ is a pretty big issue.

Elementary Proof by Contradiction

Keep $\frac xy$ constant while varying $x$ and $y$. If your function really depends only on the ratio, then for all non-zero values of $c$ the expression $\frac{1-cx}{1-cy}$ should have the same output since the ratio $\frac{cx}{cy}$ is constant with respect to $c$. Substitute $x=1$, $y=2$, and $c=1,2$ for a trivial counter-example.

General advice: Try some small examples. Even if they don't yield your counter-example, they'll often give you insight into what you need to try next.

PDE voodoo (not really complete)

I usually see this kind of problem when looking at PDE's, so I'll sketch out the first kind of thing I might try in that domain. This has some finicky details I'm glossing over, but the insight is broader in scope since you're covering nearly the entire domain of your function at once. The insight I'm using is that the property you're describing places extremely tight constraints on the partial derivatives of $f$.

Leaving out some of the gritty details, suppose that $$f(x,y)=\frac{1-x}{1-y}$$ is equal to $g\left(\frac xy\right)$ for some differentiable $g$ for all $x$ and $y$ where $f$ is defined. Taking the partial derivatives of $g$ it isn't hard to see that $$\frac{\partial f}{\partial y}=-\frac{x}{y^2}g'\left(\frac xy\right)=-\frac xy\frac1yg'\left(\frac xy\right)=-\frac xy\frac{\partial f}{\partial x}.$$

Actually computing the partial derivatives of $f$ we find $$\frac{\partial f}{\partial y}=-\frac{1-x}{1-y}\frac{\partial f}{\partial x}.$$

Our conclusion then must be that $$\frac{1-x}{1-y}=\frac xy$$ for all $x$ and $y$ if we want to conclude that such a $g$ exists. This is nonsense, so by contradiction no such $g$ can exist.

General advice: This one's pretty fun I think. If $f(x,y)=g(h(x,y))$ then a similar trick lets us conclude $\frac{\partial f}{\partial x}\frac{\partial h}{\partial y}=\frac{\partial f}{\partial y}\frac{\partial h}{\partial x}$. Note that the $g$ completely drops out of the picture, so instead of worrying about all possible functions that $g$ might be we can just check to see if the equality is satisfied. This is another form of sanity check since even if the equality is satisfied there might not exist such a $g$ (at least I don't think that reverse implication is true).

Isoclines

Since we're having so much fun, let's do it again. For every real $c$, the equation $\frac xy=c$ describes a curve in the $xy$-plane where your function would be constant if your desired property held. In other words, the directional derivative in any direction preserving $\frac xy=c$ (or just $x=cy$ if $y\neq0$) must be $0$.

Well, what direction is that? It's the direction $\langle c,1\rangle$. The gradient of your function is $$\frac1{y-1}\left\langle1,-\frac{x-1}{y-1}\right\rangle.$$

Dotting these two things together to get your directional derivative (ignoring the messy scale factor since we're just checking to see if it's $0$ or not), we get $$\frac c{y-1}-\frac{x-1}{(y-1)^2}.$$

Well then, what is $c$? We defined it to be $\frac xy$. Plugging that into our scaled directional derivative we get the simplified expression $$\frac{y-x}{y(y-1)^2}.$$ This is most assuredly not $0$ for most of the $xy$-plane.

General advice: Check your isoclines. Constant regions give you something concrete to work with.

More general advice: Check your iso-[whatever]. Constant anything in any field of math makes your life easier. Constant topology? Homeomorphisms. Constant Group? Isomorphism. Constant [any other algebraic object]? Isomorphism. The math of when things don't change is surprisingly fruitful.