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# numpy.log1p

`numpy.``log1p`(x, /, out=None, *, where=True, casting='same_kind', order='K', dtype=None, subok=True[, signature, extobj]) = <ufunc 'log1p'>

  Return the natural logarithm of one plus the input array, element-wise.

  Calculates `log(1 + x)`.

  Parameters: | x : array_like Input values. out : ndarray, None, or tuple of ndarray and None, optional A location into which the result is stored. If provided, it must have a shape that the inputs broadcast to. If not provided or None, a freshly-allocated array is returned. A tuple (possible only as a keyword argument) must have length equal to the number of outputs. where : array_like, optional Values of True indicate to calculate the ufunc at that position, values of False indicate to leave the value in the output alone. **kwargs For other keyword-only arguments, see the ufunc docs.
  Returns: | y : ndarray Natural logarithm of 1 + x, element-wise.

  See also

  `expm1`
    `exp(x) - 1`, the inverse of `log1p`.

  Notes

  For real-valued input, `log1p` is accurate also for x so small that 1 + x == 1 in floating-point accuracy.

  Logarithm is a multivalued function: for each x there is an infinite number of z such that exp(z) = 1 + x. The convention is to return the z whose imaginary part lies in [-pi, pi].

  For real-valued input data types, `log1p` always returns real output. For each value that cannot be expressed as a real number or infinity, it yields `nan` and sets the invalid floating point error flag.

  For complex-valued input, `log1p` is a complex analytical function that has a branch cut [-inf, -1] and is continuous from above on it. `log1p` handles the floating-point negative zero as an infinitesimal negative number, conforming to the C99 standard.

  References

  [R53] | M. Abramowitz and I.A. Stegun, “Handbook of Mathematical Functions”, 10th printing, 1964, pp. 67. http://www.math.sfu.ca/~cbm/aands/

  [R54] | Wikipedia, “Logarithm”. http://en.wikipedia.org/wiki/Logarithm

  Examples

      >>> np.log1p(1e-99)
      1e-99
      >>> np.log(1 + 1e-99)
      0.0