# Is there a simple non-degenerate parametrization of rational functions of first order?

Consider the simplest subclass of rational functions which are formed as ratios of two polynomials of first order $$R(x) = \frac{ax + b}{cx + d}$$ Of course, each choice of $a,b,c,d \in \mathbb{R}$ does not correspond to a different rational function. Namely, if we choose a set of values for the constants $a=a_0,b=b_0,c=c_0,d=d_0$, the choice of constants $a=\alpha a_0,b=\alpha b_0,c=\alpha c_0,d=\alpha d_0$ with any $\alpha \neq 0$ will yield the same rational function. Without loss of generality, we fix $d=1$ so that we can be sure we are picking only one of these equivalent parameter choices.

However, this still does not eliminate all the ambiguities. For instance, all choices of constants where $a=bc$ are the same rational function $R(x)=b$. So my question is

1. Is there a one-to-one parametrization of all rational functions of first order?
2. If not, what is the most convenient parametrization and what are all its degeneracies?
• Your answer is probably in Moebius transformation – Somos Nov 4 '17 at 0:15
• @Somos Yes, it came to my mind, but I actually do not know what is the best way to parametrize those anyways. Plus this space of functions is extended by the constant functions which are not allowed in Möbius transforms due to the non-degeneracy requirement. – Void Nov 4 '17 at 0:23
• Have you tried $ad-bc=1$? What do you mean by "best way"? – Somos Nov 4 '17 at 1:46

We want the denominator not to be $0$.

A simple way to write this is $c^2+d^2\neq 0$ or similarly $\begin{cases}c=r\cos(\theta)\\d=r\sin(\theta)\end{cases}\quad$ with $r\neq 0$

Using polar form, we do not have to care about $\cos(\theta)$ and $\sin(\theta)$ being zero simultaneously because this never happen.

Thus let's transcribe the whole expression to polar form:

$R(x)=\dfrac{r_1}{r_2}\times\dfrac{\cos(u)x+\sin(u)}{\cos(t)x+\sin(t)}\quad$ with $(r_1,r_2)\in[0,+\infty[^2\quad$ and $\quad(u,t)\in[0,2\pi[^2$

Since $r_2\neq 0$ and since $r_1$ is arbitrary then we can replace this by a single parameter $r\ge 0$.

Now let's have a look to injectivity:

$r\times\dfrac{\cos(u)x+\sin(u)}{\cos(t)x+\sin(t)}=R\times\dfrac{\cos(U)x+\sin(U)}{\cos(T)x+\sin(T)}$

$\iff (r\cos(u)\cos(T)-R\cos(U)\cos(t))\,x^2+\\\phantom{\iff(}(r\cos(u)\sin(T)+r\sin(u)\cos(T)-R\cos(U)\sin(t)-R\sin(U)\cos(t))\,x+\\\phantom{\iff(}r\sin(u)\sin(T)-R\sin(U)\sin(t)=0\quad\forall x$

We can already notice that having trigonometric functions here, helps simplifying products like $ac$, $bd$ or $ad-bc$.

$\iff\begin{cases} r\cos(u)\cos(T)=R\cos(U)\cos(t)\\ r\sin(u)\sin(T)=R\sin(U)\sin(t)\\ r\cos(u)\sin(T)+r\sin(u)\cos(T)=R\cos(U)\sin(t)-R\sin(U)\cos(t)\\ \end{cases}$

$\overset{(1)-(2)}{\iff}\begin{cases} r\cos(u)\cos(T)=R\cos(U)\cos(t)\\ r\cos(u+T)=R\cos(U+t)\\ r\sin(u+T)=R\cos(U+t)\\ \end{cases}$

$\overset{(*)}{\iff}\begin{cases} r\cos(u-T)=R\cos(U-t)\\ r\cos(u+T)=R\cos(U+t)\\ r\sin(u+T)=R\cos(U+t)\\ \end{cases}\qquad$ $(*): \cos(u)\cos(T)=\frac 12\cos(u+T)+\frac 12\cos(u-T)$

$(2)^2+(3)^2\iff r^2=R^2$ and since both are positive then $r=R$.

And the system reduces (when $r>0$) to $\begin{cases} (u-T)=\pm(U-t)\pmod{2\pi}\\ (u+T)=(U+t)\pmod{2\pi}\\ \end{cases}$

There are four possibilites:

$\begin{cases} U=u\pmod{2\pi}\\T=t\pmod{2\pi}\end{cases}$ or $\begin{cases} U=u+\pi\pmod{2\pi}\\T=t+\pi\pmod{2\pi}\end{cases}\iff\begin{cases} T=t\pmod{\pi}\\U-T=u-t\pmod{2\pi}\end{cases}$

$\begin{cases} u=t\pmod{2\pi}\\U=T\pmod{2\pi}\end{cases}$ or $\begin{cases} u=t+\pi\pmod{2\pi}\\U=T+\pi\pmod{2\pi}\end{cases}\iff\begin{cases} u=t\pmod{\pi}\\U-T=u-t\pmod{2\pi}\end{cases}$

This suggest to set $u-t=t_0$ (resp. $U-T=T_0$).

We get $\begin{cases} T=t\pmod{\pi}\\T_0=t_0\pmod{2\pi}\end{cases}\quad$ which is injective

$\phantom{We i}$or $\begin{cases} t_0=0\pmod{\pi}\\T_0=t_0\pmod{2\pi}\end{cases}\quad$ which is not, since $T,t$ are free.

Thus we have injectivity for : $\quad r>0\ \ ;\ \ t\in[0,\pi[\ \ ;\ \ t_0\in[0,2\pi[\setminus\{0,\pi\}$

$$\boxed{R(x)=r\;\dfrac{\cos(t+t_0)\,x+\sin(t+t_0)}{\cos(t)\,x+\sin(t)}}$$

We lack injectivity for $r=0$ in which case any $(t,t_0)$ give the same $R(x)=0$.

We also lack injectivity for $t_0=0,\pi$ in which case any $t$ give the same $R(x)=\pm r$.

But this is not a major issue, the condition $r=0$ has to be compared to $z=re^{i\theta}$, this is still a great parametrization even though it is not injective for the single value $r=0$. We should also be happy to have transformed the condition $d=1,\ a=bc$ which has an infinity of possibilities to a finite $t_0=0\text{ or }\pi$.

Note that we can now go back to a non-trigonometric parametrization if needed using the $\tan(\frac{\theta}2)$ transformation.

Without details, for $a=\tan(\frac{t_0}2)$ and $b=\tan(\frac t2)$ we get:

Note: these are not the same $a,b$ than previously, I just run short of good variable names.

$R(x)=r\;\dfrac{(4ab-(a^2-1)(b^2-1))x+2a(b^2-1)+2b(a^2-1)}{(a^2+1)((b^2-1)x-2b)}$