tfp.math.log1psquare

Numerically stable calculation of log(1 + x**2) for small or large |x|.

tfp.math.log1psquare(
    x, name=None
)

For sufficiently large x we use the following observation:

log(1 + x**2) =   2 log(|x|) + log(1 + 1 / x**2)
              --> 2 log(|x|)  as x --> inf

Numerically, log(1 + 1 / x**2) is 0 when 1 / x**2 is small relative to machine epsilon.

Args
x	Float Tensor input.
name	Python string indicating the name of the TensorFlow operation. Default value: 'log1psquare'.

Returns
log1psq	Float Tensor representing log(1. + x**2.).