The existing implementation compares each component to zero with
an epsilon; effectively `glm::all(glm::is_comp_null(v, epsilon))`.
This probably isn't the desired semantics when calling `glm::is_null`;
rather, we want to determine if the magnitude of the vector is within
`epsilon` units of zero. It's the question of circle versus square.
This behavior matches that of OpenGL Mathematics.
The existing comparison bound of $\epsilon^2$ is improperly scaled for
testing an epsilon of the squared vector magnitude. Let $\epsilon$ be
our specified epsilon and $\delta$ be the permissible delta of the
squared magnitude. Thus, for a nearly-normalized vector, we have
$$\begin{align}
\sqrt{1 + \delta} &= 1 + \epsilon \\
\delta &= (1 + \epsilon)^2 - 1 \\
\delta &= \epsilon^2 + 2\epsilon
\text{ .}\end{align}$$
Since we only care about small epsilon, we can assume
that $\epsilon^2$ is small and just use $\delta = 2\epsilon$. And in
fact, [this is the bound used by GLM][GLM#isNormalized] (MIT license)
... except they're using `length` and not `length2` for some reason.
[GLM#isNormalized]: b06b775c1c/glm/gtx/vector_query.inl (L102)
If we stick an epsilon of `1.0e-6` into the current implementation,
this gives us a computed delta of `1.0e-12`: smaller than the `f32`
machine epsilon, and thus no different than direct float comparison
without epsilon. This also gives an effetive epsilon of `5.0e-13`;
*much* less than the intended `1.0e-6` of intended permitted slack!
By doing a bit more algebra, we can find the effective epsilon is
$\sqrt{\texttt{epsilon}^2 + 1} - 1$. This patch makes the effective
epsilon $\sqrt{2\times\texttt{epsilon} + 1} - 1$ which still isn't
*perfect*, but it's effectively linear in the domain we care about,
only really making a practical difference above an epsilon of 10%.
TL;DR: the existing `is_normalized` considers a vector normalized if
the squared magnitude is within `epsilon*epsilon` of `1`. This is wrong
and it should be testing if it's within `2*epsilon`. This PR fixes it.
For absence of doubt, a comparison epsilon of $\texttt{epsilon}^2$ is
correct when comparing squared magnitude against zero, such as when
testing if a displacement vector is nearly zero.
