code4#

class pyhf.interpolators.code4(histogramssets, subscribe=True, alpha0=1)[source]#

Bases: object

The polynomial interpolation and exponential extrapolation strategy.

$${\sigma }_{sb}(\overrightarrow{\alpha })={\sigma }_{sb}^0(\overrightarrow{\alpha })\underset{\text{factors to calculate}}{\underbrace{\prod _{p\in \text{Syst}}{I}_{\text{poly|exp.}}({\alpha }_p;{\sigma }_{sb}^0,{\sigma }_{psb}^{+},{\sigma }_{psb}^{-},{\alpha }_0)}}$$

with

$$\begin{array}{r}{I}_{\text{poly|exp.}}(\alpha ;I^0,I^{+},I^{-},{\alpha }_0)=\{\begin{array}{l}{\left(\frac{I^{+}}{I^0}\right)}^{\alpha }\alpha \ge {\alpha }_0\\ 1+\sum _{i=1}^6{a}_i{\alpha }^i|\alpha |<{\alpha }_0\\ {\left(\frac{I^{-}}{I^0}\right)}^{-\alpha }\alpha <-{\alpha }_0\end{array}\end{array}$$

and the $a_i$ are fixed by the boundary conditions

$${\sigma }_{sb}(\alpha =\pm {\alpha }_0),{\frac{\mathrm{d}{\sigma }_{sb}}{\mathrm{d}\alpha }|}_{\alpha =\pm {\alpha }_0},\mathrm{and}{\frac{{\mathrm{d}}^2{\sigma }_{sb}}{\mathrm{d}{\alpha }^2}|}_{\alpha =\pm {\alpha }_0}.$$

Namely that ${\sigma }_{sb}(\overrightarrow{\alpha })$ is continuous, and its first- and second-order derivatives are continuous as well.

__init__(histogramssets, subscribe=True, alpha0=1)[source]#

Polynomial Interpolation.

Methods

_precompute()[source]#

_precompute_alphasets(alphasets_shape)[source]#