# scipy.special.jacobi#

scipy.special.jacobi(n, alpha, beta, monic=False)[source]#

Jacobi polynomial.

Defined to be the solution of

$(1 - x^2)\frac{d^2}{dx^2}P_n^{(\alpha, \beta)} + (\beta - \alpha - (\alpha + \beta + 2)x) \frac{d}{dx}P_n^{(\alpha, \beta)} + n(n + \alpha + \beta + 1)P_n^{(\alpha, \beta)} = 0$

for $$\alpha, \beta > -1$$; $$P_n^{(\alpha, \beta)}$$ is a polynomial of degree $$n$$.

Parameters:
nint

Degree of the polynomial.

alphafloat

Parameter, must be greater than -1.

betafloat

Parameter, must be greater than -1.

monicbool, optional

If True, scale the leading coefficient to be 1. Default is False.

Returns:
Porthopoly1d

Jacobi polynomial.

Notes

For fixed $$\alpha, \beta$$, the polynomials $$P_n^{(\alpha, \beta)}$$ are orthogonal over $$[-1, 1]$$ with weight function $$(1 - x)^\alpha(1 + x)^\beta$$.

References

[AS]

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

Examples

The Jacobi polynomials satisfy the recurrence relation:

$P_n^{(\alpha, \beta-1)}(x) - P_n^{(\alpha-1, \beta)}(x) = P_{n-1}^{(\alpha, \beta)}(x)$

This can be verified, for example, for $$\alpha = \beta = 2$$ and $$n = 1$$ over the interval $$[-1, 1]$$:

>>> import numpy as np
>>> from scipy.special import jacobi
>>> x = np.arange(-1.0, 1.0, 0.01)
>>> np.allclose(jacobi(0, 2, 2)(x),
...             jacobi(1, 2, 1)(x) - jacobi(1, 1, 2)(x))
True


Plot of the Jacobi polynomial $$P_5^{(\alpha, -0.5)}$$ for different values of $$\alpha$$:

>>> import matplotlib.pyplot as plt
>>> x = np.arange(-1.0, 1.0, 0.01)
>>> fig, ax = plt.subplots()
>>> ax.set_ylim(-2.0, 2.0)
>>> ax.set_title(r'Jacobi polynomials $P_5^{(\alpha, -0.5)}$')
>>> for alpha in np.arange(0, 4, 1):
...     ax.plot(x, jacobi(5, alpha, -0.5)(x), label=rf'$\alpha={alpha}$')
>>> plt.legend(loc='best')
>>> plt.show()