Read the equation
Because both vectors in the centered bend statistic point away from the center, a straight trajectory places them in opposite directions and gives cosine −1. This is the reference for an unbent path under that convention. It does not mean the successive motion directions are opposed.
Terms & definitions
| Term | Meaning in this source |
|---|---|
| C_Z | Centered path cosine |
How it works
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Original context
Read the surrounding source passage
CZ(t;k)=⟨Zt−k−Zt,Zt+k−Zt⟩∥Zt−k−Zt∥ ∥Zt+k−Zt∥.
(4.5)
Matrix comparisons use Frobenius quantities. A straight continuing path gives CZ=−1, because its backward and forward displacements point in opposite directions. A larger value indicates a larger local bend. This is a cosine between displacements, not between the states themselves and not a measure of movement amplitude. A zero displacement makes the corresponding angle undefined.
I evaluate lags of five, ten, and 15 training steps on the regular grid. Each lag uses every center with both required neighbors. This preserves the training step offsets and does not impose the longest lag’s edge trimming on shorter lags.
A sharp angle can accompany a tiny movement. I therefore also apply the amplitude eligibility rule in the source plotting code. Let MA(t;k) be the larger of the norms ∥at−k−at∥2 and ∥at+k−at∥2, with MB(t;k) defined analogously. A center is eligible when
Et(k,η)=1{max(MA(t;k),MB(t;k))≥η}.Exact original notation
{C}_{Z}=-1