I'm designing a transformation that I think needs to be non-affine by definition. Technically it's a chain of two transformations:
The first transforms from the frequency space to a logarithmic chromatic music note space, in units of cents, via
$$ x = 100 \cdot 12 \log_2 \frac f {55} - t $$
where $t$ is some known, desired tuning note.
The second is an antisymmetric non-affine transformation $y(x)$ for the purposes of graphing with the following requirements:
$$y(0) = 0$$ $$y(600) = 600$$ $$y(x) = -y(-x)$$ $$ \text{For all } 0 \le x \le 600, \frac {\partial y}{\partial x} > 0$$ $$ \text{For all } 0 < x < 600, y(x) < x $$
The purpose is to zoom into $x=0$ while still showing $x = -600, x=600$. The function must be invertible and the inverse will be used about as frequently as the function itself, due to the way that matplotlib is written.
The two candidates I thought of are a cubic and a tangent. The cubic implementation is
$$ y = a x^3 + b x , b > 0$$ $$ a = \frac {s - 100} {s(600^2 - s^2)} $$ $$ b = 1 - 600^2 a $$
where $s$ is a scaling parameter, interpreted as the value of $x$ where $y = 100$.
This works well enough; here is a $1 - \left| \frac x {600} \right|$ triangle shown with this horizontal zoom for $s = 320$:
This formulation has two big disadvantages:
- Values of $s > 330$ are not possible without making $b < 0$, in turn giving the curve undesired inflection points
- The inverse $x(y)$ is truly monstrous and I won't include it here unless requested
Is there a different function that will meet these requirements but is more easily invertible?
An easy way to get what you're looking for is using the equation for the motion of a particle in a velocity field, $v$:
$\dot x(t) = v(x(t))$
In your case, $x(t)$ is $y$, $x(0)$ is $x$, and $t$ is $s$. Rewriting with these variable names,
$\dot y(s) = v(y(s))$
$y(0) = x$
This will produce smooth uniform zooming, moving the view of the original function at the same rate, no matter what the value of $s$ is. It is a first order ODE, so $y(x)$ is guaranteed to be invertible. Where $v=0$, the velocity is zero and the action is stationary. Where $\frac{dv}{dx} > 0$, the effect will be a zoom inward, and vice versa with $\frac{dv}{dx} < 0$. So, it is very easy to control the zoom using $v$. This also is a very easy ODE to solve, with a judicious choice of $v$.
First, let's just scale the problem from $\pm600$ to $\pm1$. It is trivial to scale the problem back again afterward. If you choose $v(x) = \frac 1 2 x(1 - x)(1 + x)$, you get the desired zooming in at $x=0$, while keeping $x=0$, $x=1$, and $x=-1$ stationary, as required. (The factor of 1/2 is there to simplify the math.) Math, math, math, and
$$y(x, s) = \frac x {\sqrt{e^{-s} + (1 - e^{-s}) x^2}}$$
Well, that's a pretty function! In $x$, it's just a straight line divided by a hyperbola!
There are other choices of $v$ that give similar results. $v(x) = \sin(\frac \tau 2 x)$ should work, though the integral may be messier. $v(x) = x(1 - |x|)$ will give even simpler results than the cubic I just did—a tanh function will result from it—but the zooming may be more unnatural because of the singularity in $v$ at $x=0$.