# Specific problem

So here is my problem. I have a function $$f(x,y\,|\,\alpha) = \frac{x(\alpha y^2 - 1)}{y(\alpha x^2 - 1)}$$ were x,y and $\alpha$ correspond to some physical parameters and thus I know that $\{x,y,\alpha\}$ are all $\in [0,1]$. I dont care what happens to $\alpha$. It is more or less a constant for my purpouse so I'll just drop it in further notation.

I can measure $f(x,y)$ directly (complicated physics experiment, not important here), but the quantity I am really interested in is the ratio $x/y$.

I would have no issue approaching similar problems for any function of single variable (series expand -> try to truncate the terms -> end up with a simple polynomial approximation), but I am a bit stuck here. Basically, I want to know of a systematic (or, at least, "systematic in this case", or even "borderline sane") way of finding $g(x/y)$ which approximates my $f(x,y)$. I want $g$ to have only $x/y$ terms - have $x/y$ ratio appears as a single variable. I don't, however, care if $g$ is not a polynomial in $x/y$

# More general problem

Find a systematic nontrivial way of approximating $f(x)$ as $g(h(x))$ for any $h(x)$ and $f(x)$. The trivial solution is $x = h^{-1}(h(x))$ and so $g(x) = f(h^{-1}(h(x)))$.

I know that my specific problem is multivariate and this one isn't, but I feel like this is a good springboard to start thinking from.

## Example 1

What is the $g$ which best approximates $f(x) = x^2$ with $h(x) = \cos(x)$ about the point $x = 1$?

The trivial solution is $x = \cos^{1}(\cos(x))$, and $g(x) = \left(\cos^{1}(\cos(x))\right)^2$. That's not the one I want as it does not help me with the specific problem above. In the lack of a good starting point lets procede in the spirit on a power series - higer and higer powers of $h(x)$. By throwing a numerical solver at the probelm, I know that And, to a better approximation, .

This is hopeful, but I want an analytical rather than a numerical approach.

## Example 2

What is the $g$ which best approximates $f(x) = \sin(x)$ with $h(x) = \cos(x)$ about the point $x = 1$?

By doing the same numerical approach I can get better and better approximations to $\sin(x)$ using successive powers of $\cos(x)$. If I stop and think for a while before I start, however, I can quite easily get $f(x) = \sin(x) = \sqrt{1-\cos^2{x}} = g(\cos(x))$ and so I'm done (within a $\pm$ sign error anyway).

Hopefully, that illustrates the problem. In example 1 I can't think of what more to do than to throw more and more exotic terms (higher powers / other functions) at the problem while in example 2 I can find the solution with a little thought. Can something like that be done for my specific problem? If not, how do I prove it? Also; , in reality, $\alpha$ is very close to 1 (0.9999) and so maybe there is an approximation which can be made to solve my problem?

• Dear Admins 1: Please move this question to other stack exchange if you deem it more applicable there. 2: I dont know what other tags are applicable here(?). – MarcinKonowalczyk Aug 6 '18 at 12:43

I think that's a tall order. Let's look at the case $\alpha = 1$ to simplify. Then your function can be decomposed as a product of two variables $Z$ and $W$:

$$Z=\frac{x}{y} \text{ and } W = \frac{y^2-1}{x^2-1} \; .$$

In the following picture, I have represented some level curves of the variables $Z$ and $W$, that is values of $(x,y)$ for which the variables $Z$ and $W$ are constant. Blue is when $W$ is constant, orange is when $Z$ is constant.

For your request to work, the orange and blue lines would have to coincide, but only the case $x=y$ satisfies this condition.

The best you can hope to do is therefore to work with two variables $Z$ and $W$. But I suppose you only have experimental control on $Z$ and not on $W$. For near-constant values of $W$, your function $f$ can be approximated as

$$f(x,y) = W \frac{x}{y} \; .$$

If the value of $W$ was not varying much over the range of possible $(x,y)$ pairs, you could use this as a good approximation overall, but the problem is that $W$ can be anything inbetween $0$ and $+\infty$.