One-sided derivatives at an optimal Raman frame-count switch

ID: one-sided-derivatives-at-an-optimal-raman-frame-count-switch

One-sided derivatives at an optimal Raman frame-count switch describe the change in the best acquisition variance when several integer designs tie. Let as in frame-count switching in Raman acquisition-budget sensitivity, and let be the active designs with . Suppose the relevant branches are differentiable near the tie and nonactive designs have a positive value gap.
Writing their slopes as gives
Indeed, . Positive selects the smallest slope; negative selects the largest because multiplication reverses their ordering. A single derivative exists only when the selected left and right slopes agree. If an active branch itself has a kink, use its appropriate one-sided slope on each side.
For an original balanced-noise example, take , exposure cap and per-frame readout . Near the three-pair/four-pair switch,
The branches tie at , with variance about . The three-pair branch has slope zero; the four-pair branch has slope . Thus these are the left and right optimum slopes. Smaller pair counts have less capacity and larger counts have more readout near this tie, so they do not change the local optimum.
Joint time-and-readout perturbations at a Raman design tie require following the direction in both resource coordinates; the scalar slopes above hold when only changes. They also quantify a local sensitivity, not a universal prescription to change exposure count whenever a small budget fluctuation occurs.
The primary stimulated-Raman apparatus limited exposures for saturation and switched pump state during shuttered readout.[ref] The rounded timings here are a separate mathematical example. Read noise, drift or switching costs can change the competing branches and their derivatives.

New to topics? Read the docs here!