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

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.

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

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