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.)`.

Except as otherwise noted, the content of this page is licensed under the Creative Commons Attribution 4.0 License, and code samples are licensed under the Apache 2.0 License. For details, see the Google Developers Site Policies. Java is a registered trademark of Oracle and/or its affiliates.

Last updated 2023-11-21 UTC.

Args