scipy.special.bernoulli#

scipy.special.bernoulli(n)[source]#

Bernoulli numbers B0..Bn (inclusive).

Parameters:

nint: Indicated the number of terms in the Bernoulli series to generate.

Returns:

ndarray: The Bernoulli numbers [B(0), B(1), ..., B(n)].

References

[1]

Zhang, Shanjie and Jin, Jianming. “Computation of Special Functions”, John Wiley and Sons, 1996. https://people.sc.fsu.edu/~jburkardt/f77_src/special_functions/special_functions.html

[2]

“Bernoulli number”, Wikipedia, https://en.wikipedia.org/wiki/Bernoulli_number

Examples

>>> import numpy as np
>>> from scipy.special import bernoulli, zeta
>>> bernoulli(4)
array([ 1.        , -0.5       ,  0.16666667,  0.        , -0.03333333])

The Wikipedia article ([2]) points out the relationship between the Bernoulli numbers and the zeta function, B_n^+ = -n * zeta(1 - n) for n > 0:

>>> n = np.arange(1, 5)
>>> -n * zeta(1 - n)
array([ 0.5       ,  0.16666667, -0.        , -0.03333333])

Note that, in the notation used in the wikipedia article, bernoulli computes B_n^- (i.e. it used the convention that B_1 is -1/2). The relation given above is for B_n^+, so the sign of 0.5 does not match the output of bernoulli(4).