Vous êtes un professionnel et vous avez besoin d'une formation ? RAG (Retrieval-Augmented Generation)
et Fine Tuning d'un LLM Voir le programme détaillé

Module « scipy.special »

Fonction fresnel - module scipy.special

Signature de la fonction fresnel

def fresnel(*args, **kwargs) 

Description

help(scipy.special.fresnel)

fresnel(x[, out1, out2], / [, out=(None, None)], *, where=True, casting='same_kind', order='K', dtype=None, subok=True[, signature])


    fresnel(z, out=None)

    Fresnel integrals.

    The Fresnel integrals are defined as

    .. math::

       S(z) &= \int_0^z \sin(\pi t^2 /2) dt \\
       C(z) &= \int_0^z \cos(\pi t^2 /2) dt.

    See [dlmf]_ for details.

    Parameters
    ----------
    z : array_like
        Real or complex valued argument
    out : 2-tuple of ndarrays, optional
        Optional output arrays for the function results

    Returns
    -------
    S, C : 2-tuple of scalar or ndarray
        Values of the Fresnel integrals

    See Also
    --------
    fresnel_zeros : zeros of the Fresnel integrals

    References
    ----------
    .. [dlmf] NIST Digital Library of Mathematical Functions
              https://dlmf.nist.gov/7.2#iii

    Examples
    --------
    >>> import numpy as np
    >>> import scipy.special as sc

    As z goes to infinity along the real axis, S and C converge to 0.5.

    >>> S, C = sc.fresnel([0.1, 1, 10, 100, np.inf])
    >>> S
    array([0.00052359, 0.43825915, 0.46816998, 0.4968169 , 0.5       ])
    >>> C
    array([0.09999753, 0.7798934 , 0.49989869, 0.4999999 , 0.5       ])

    They are related to the error function `erf`.

    >>> z = np.array([1, 2, 3, 4])
    >>> zeta = 0.5 * np.sqrt(np.pi) * (1 - 1j) * z
    >>> S, C = sc.fresnel(z)
    >>> C + 1j*S
    array([0.7798934 +0.43825915j, 0.48825341+0.34341568j,
           0.60572079+0.496313j  , 0.49842603+0.42051575j])
    >>> 0.5 * (1 + 1j) * sc.erf(zeta)
    array([0.7798934 +0.43825915j, 0.48825341+0.34341568j,
           0.60572079+0.496313j  , 0.49842603+0.42051575j])
    

Vous êtes un professionnel et vous avez besoin d'une formation ? Machine Learning
avec Scikit-Learn Voir le programme détaillé