scipy.special.roots_legendre#

scipy.special.roots_legendre(n, mu=False)[source]#

Gauss-Legendre quadrature.

Compute the sample points and weights for Gauss-Legendre quadrature [GL]. The sample points are the roots of the nth degree Legendre polynomial \(P_n(x)\). These sample points and weights correctly integrate polynomials of degree \(2n - 1\) or less over the interval \([-1, 1]\) with weight function \(w(x) = 1\). See 2.2.10 in [AS] for more details.

Parameters:

nint: quadrature order
mubool, optional: If True, return the sum of the weights, optional.

Returns:

xndarray: Sample points
wndarray: Weights
mufloat: Sum of the weights

See also

scipy.integrate.quadrature
scipy.integrate.fixed_quad
numpy.polynomial.legendre.leggauss

References

[AS]

Milton Abramowitz and Irene A. Stegun, eds. Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables. New York: Dover, 1972.

[GL] (1,2)

Gauss-Legendre quadrature, Wikipedia, https://en.wikipedia.org/wiki/Gauss%E2%80%93Legendre_quadrature

Examples

>>> import numpy as np
>>> from scipy.special import roots_legendre, eval_legendre
>>> roots, weights = roots_legendre(9)

roots holds the roots, and weights holds the weights for Gauss-Legendre quadrature.

>>> roots
array([-0.96816024, -0.83603111, -0.61337143, -0.32425342,  0.        ,
        0.32425342,  0.61337143,  0.83603111,  0.96816024])
>>> weights
array([0.08127439, 0.18064816, 0.2606107 , 0.31234708, 0.33023936,
       0.31234708, 0.2606107 , 0.18064816, 0.08127439])

Verify that we have the roots by evaluating the degree 9 Legendre polynomial at roots. All the values are approximately zero:

>>> eval_legendre(9, roots)
array([-8.88178420e-16, -2.22044605e-16,  1.11022302e-16,  1.11022302e-16,
        0.00000000e+00, -5.55111512e-17, -1.94289029e-16,  1.38777878e-16,
       -8.32667268e-17])

Here we’ll show how the above values can be used to estimate the integral from 1 to 2 of f(t) = t + 1/t with Gauss-Legendre quadrature [GL]. First define the function and the integration limits.

>>> def f(t):
...    return t + 1/t
...
>>> a = 1
>>> b = 2

We’ll use integral(f(t), t=a, t=b) to denote the definite integral of f from t=a to t=b. The sample points in roots are from the interval [-1, 1], so we’ll rewrite the integral with the simple change of variable:

x = 2/(b - a) * t - (a + b)/(b - a)

with inverse:

t = (b - a)/2 * x + (a + 2)/2

Then:

integral(f(t), a, b) =
    (b - a)/2 * integral(f((b-a)/2*x + (a+b)/2), x=-1, x=1)

We can approximate the latter integral with the values returned by roots_legendre.

Map the roots computed above from [-1, 1] to [a, b].

>>> t = (b - a)/2 * roots + (a + b)/2

Approximate the integral as the weighted sum of the function values.

>>> (b - a)/2 * f(t).dot(weights)
2.1931471805599276

Compare that to the exact result, which is 3/2 + log(2):

>>> 1.5 + np.log(2)
2.1931471805599454