numpy.expm1¶

numpy.
expm1
(x, /, out=None, *, where=True, casting='same_kind', order='K', dtype=None, subok=True[, signature, extobj]) = <ufunc 'expm1'>¶ Calculate
exp(x)  1
for all elements in the array.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 freshlyallocated 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 keywordonly arguments, see the ufunc docs.
Returns: out : ndarray
Elementwise exponential minus one:
out = exp(x)  1
.See also
log1p
log(1 + x)
, the inverse of expm1.
Notes
This function provides greater precision than
exp(x)  1
for small values ofx
.Examples
The true value of
exp(1e10)  1
is1.00000000005e10
to about 32 significant digits. This example shows the superiority of expm1 in this case.>>> np.expm1(1e10) 1.00000000005e10 >>> np.exp(1e10)  1 1.000000082740371e10