__device__ double log1p ( double  x  ) 

Calculate the value of $log_{e}(1+x)$ of the input argument x.

Returns:
  • log1p($\pm 0$) returns $-\infty$.
  • log1p(-1) returns +0.
  • log1p(x) returns NaN for x < -1.
  • log1p($+\infty$) returns $+\infty$.
Note:
For accuracy information for this function see the CUDA C Programming Guide, Appendix C, Table C-2.


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