Is there a Way to mirror a function across another function

278 Views Asked by At

Let's say I have the function $y=f(x)$. Is there a set of transformations I can apply such that I can get a new equation (call it $f_1(x)$) by mirroring $f(x)$ across some other function $g(x)$ [assuming $g(x)$ is not in the form $y=mx+b$]?

The closest I personally got to answering this, would be to mirror every point on $f(x)$ across the tangent of the closest point on $g(x)$ - as $g(x)$ is not linear, the tangent would allow for us to mirror the point - then somehow extrapolate an equation that would be the transformed equation. While this method only applies to finding individual points, it does not satisfy the need to create an adequate transformation. Thoughts? Suggestions?

1

There are 1 best solutions below

1
On BEST ANSWER

I doubt it. The concepts of reflection (or "mirroring") and projection are somewhat fuzzy, with different meanings based on context, but are usually linked. The most common idea of projection is metric projection, whereby we take a point $x$ in the space, and find the point(s) in a set $C$ that are of minimum distance to $x$. But, there are other forms of projection that don't fit into this mould: e.g. Schwarzian projection, stereographic projection, etc.

I know less about reflection, but I know that it's commonly (but probably not exclusively) defined to be $2P - I$, where $P$ is an associated projection operator and $I$ is the identity operator. If $P$ is the metric projection, then this agrees with your intuition: reflecting about the tangent plane to the nearest point. Indeed, assuming Euclidean distance, the vector from the point to the nearest point on the graph ought to be normal to the tangent plane, if there is one.

The problem of computing the metric projections and reflections are pretty immediately equivalent in terms of difficulty: we have $R = 2P - I$ or equivalently, $P = (R + I) / 2$. However, computing metric projections is hard, in general. If your set is closed and affine, then the problem is easy. If your set is convex, then the projection and reflection have nice properties (single-valued, non-expansive, projection is maximally monotone, etc). But, even with certain convex sets, working out analytically the projection points is tough. For example, projecting/reflecting in the convex set $$\{(x, y, z) : x^6 + y^6 + z^6 \le 1\}$$ would be impossible to do with any closed form, due to insolubility of sextic polynomials. The above set is not especially complicated to describe, but its solution is impossible to describe using elementary symbols.

Without convexity, we still get all the same problems of possible insolubility, but with the added wrinkle that we will also get multiple solutions! It has been known since the 30s that, in Euclidean space, every non-convex set admits points with multiple nearest points. Further, the graph of a function is convex if and only if the function is affine (i.e. a degree $1$ polynomial). So, unless you're reflecting in some (hyper)plane, line, or point, you will get some points that have multiple reflections! It's also worth pointing out that functions reflected even in convex sets may not be functions.

All this said, you may have some success with very simple non-affine functions. If I remember correctly, reflecting in a paraboloid is tractable, for example. I would advise following the following method (which is more or less what you suggested):

  1. Start with a given point $\vec{p}$ and a function $f$.
  2. Compute $\nabla f(\vec{x})$ at each $\vec{x}$ in the space. This direction vector is normal to the graph of $f$.
  3. For each $\vec{x}$ in the space, find the normal lines $t \mapsto \vec{x} + t\nabla f(\vec{x})$ which pass through $\vec{p}$.
  4. For each possible $\vec{x}$, compute $\|\vec{x} - \vec{p}\|$, and find which $\vec{x}$ is closest. Be prepared that there could be whole surfaces of such $\vec{x}$ that need to be compared, so there may be an optimisation problem to be performed here.
  5. The closest such $\vec{x}$ is the projection. To find the reflection, simply compute $2\vec{x} - \vec{p}$.

That, of course, doesn't transform one function into another in any neat way, but it gives you an analytic approach to finding projections and reflections in a given function.