![]() |
Numerically stable calculation of log(1 + x**2)
for small or large |x|
.
Aliases:
tfp.experimental.substrates.jax.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
: FloatTensor
input.name
: Python string indicating the name of the TensorFlow operation. Default value:'log1psquare'
.
Returns:
log1psq
: FloatTensor
representinglog(1. + x**2.)
.