[danbruc]

From school you are used to think of function in their explicit form y = f(x) but you can easily turn that into the implicit form f(x) - y = 0 or more generally f(x, y) = 0. With that you can plot the graph of f(x, y) either as a 3D surface with f(x, y) being the height at point (x, y) or encode the function value at (x, y) into some color at (x, y). Where that surface is equal to zero, i.e. where it intersects the z = 0 plane, that are the points of y = f(x). Points (x, y) at which the value of f(x, y) has small non-zero magnitude are what the article calls low error points or regions, points or regions that almost satisfy y = f(x).

[abtinf]

> f(x, y) = 0. With that you can plot the graph of f(x, y) either as a 3D surface with f(x, y) being the height at point (x, y)

If f(x, y) = 0, wouldn’t using f(x, y) for the height just result in a flat graph?

[layer8]

f(x, y) = 0 is true only for some combinations of x and y. It’s an equation to be solved, not a universal statement like ∀ x, y : f(x, y) = 0, nor a definition like f(x, y) ≔ 0 (or “≝”). The solutions to the equation are the points (x, y) where the graph has height 0. Which points these are depends on how f is defined.

For example, f might be defined as f(x, y) ≔ x² + y² – 1. Then the points (x, y) for which f(x, y) = 0 are those on the unit circle (those for which x² + y² = 1). The graph will have height 0 only for those points.

[roywiggins]

They're really two different types of equal signs.

f(x,y) = x+y might be better written as f(x,y) := x+y where := means "is defined as". Then f(x,y) = 0 is an equation that expands to x+y = 0, or in familiar intro algebra form, y=-x.

g(x,y) := 0 really is a flat plane.

[mitthrowaway2]

When we say "f(x, y) = 0" in this context, we also usually have a separate definition for f(x, y) provided, where that f(x, y) is not necessarily 0 at for all x,y. And so this constraint "f(x, y) = 0" means "find pairs of x and y such that it makes f(x, y) become 0".

If "f(x, y) = 0" is actually the definition of f(x, y), then yes, it would be a pretty boring graph.