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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 | |
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x
|
Float Tensor input.
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name
|
Python string indicating the name of the TensorFlow operation.
Default value: 'log1psquare'.
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Returns | |
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log1psq
|
Float Tensor representing log(1. + x**2.).
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