# scipy.special.roots_legendre#

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

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

mubool, optional

If True, return the sum of the weights, optional.

Returns
xndarray

Sample points

wndarray

Weights

mufloat

Sum of the weights

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)

Examples

>>> 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