mjo/orthogonal_polynomials.py

   1 from sage.all import *
   2
   3 def legendre_p(n, x, a = -1, b = 1):
   4     """
   5     Returns the ``n``th Legendre polynomial of the first kind over the
   6     interval [a, b] with respect to ``x``.
   7
   8     When [a,b] is not [-1,1], we scale the standard Legendre
   9     polynomial (which is defined over [-1,1]) via an affine map. The
  10     resulting polynomials are still orthogonal and possess the
  11     property that `P(a) = P(b) = 1`.
  12
  13     INPUT:
  14
  15       * ``n`` -- The index of the polynomial.
  16
  17       * ``x`` -- Either the variable to use as the independent
  18         variable in the polynomial, or a point at which to evaluate
  19         the polynomial.
  20
  21       * ``a`` -- The "left" endpoint of the interval. Must be a real number.
  22
  23       * ``b`` -- The "right" endpoint of the interval. Must be a real number.
  24
  25     OUTPUT:
  26
  27     If ``x`` is a variable, a polynomial (symbolic expression) will be
  28     returned. Otherwise, the value of the ``n``th polynomial at ``x``
  29     will be returned.
  30
  31     EXAMPLES:
  32
  33     Create the standard Legendre polynomials in `x`::
  34
  35         sage: legendre_p(0,x)
  36         1
  37         sage: legendre_p(1,x)
  38         x
  39
  40     Reuse the variable from a polynomial ring::
  41         sage: P.<t> = QQ[]
  42         sage: legendre_p(2,t)
  43         3/2*t^2 - 1/2
  44
  45     If ``x`` is a real number, the result should be as well::
  46
  47         sage: legendre_p(3, 1.1)
  48         1.67750000000000
  49
  50     Similarly for complex numbers::
  51
  52         sage: legendre_p(3, 1 + I)
  53         7/2*I - 13/2
  54
  55     Even matrices work::
  56
  57         sage: legendre_p(3, MatrixSpace(ZZ, 2)([1, 2, -4, 7]))
  58         [-179  242]
  59         [-484  547]
  60
  61     And finite field elements::
  62
  63         sage: legendre_P(3, GF(11)(5))
  64         8
  65
  66     TESTS:
  67
  68     We should agree with Maxima for all `n`::
  69
  70         sage: eq = lambda k: bool(legendre_p(k,x) == legendre_P(k,x))
  71         sage: all([eq(k) for k in range(0,20) ]) # long time
  72         True
  73
  74     We can evaluate the result of the zeroth polynomial::
  75
  76         sage: f = legendre_p(0,x)
  77         sage: f(x=10)
  78         1
  79
  80     We should have |P(a)| = |P(b)| = 1 for all a,b::
  81
  82         sage: a = RR.random_element()
  83         sage: b = RR.random_element()
  84         sage: k = ZZ.random_element(20)
  85         sage: P = legendre_p(k, x, a, b)
  86         sage: abs(P(x=a)) # abs tol 1e-12
  87         1
  88         sage: abs(P(x=b)) # abs tol 1e-12
  89         1
  90
  91     Two different polynomials should be orthogonal with respect to the
  92     inner product over `[a,b]`. Note that this test can fail if QQ is
  93     replaced with RR, since integrate() can return NaN::
  94
  95         sage: a = QQ.random_element()
  96         sage: b = QQ.random_element()
  97         sage: j = ZZ.random_element(20)
  98         sage: k = j + 1
  99         sage: Pj = legendre_p(j, x, a, b)
 100         sage: Pk = legendre_p(k, x, a, b)
 101         sage: integrate(Pj*Pk, x, a, b) # abs tol 1e-12
 102         0
 103
 104     The first few polynomials shifted to [0,1] are known to be::
 105
 106         sage: p0 = 1
 107         sage: p1 = 2*x - 1
 108         sage: p2 = 6*x^2 - 6*x + 1
 109         sage: p3 = 20*x^3 - 30*x^2 + 12*x - 1
 110         sage: bool(legendre_p(0, x, 0, 1) == p0)
 111         True
 112         sage: bool(legendre_p(1, x, 0, 1) == p1)
 113         True
 114         sage: bool(legendre_p(2, x, 0, 1) == p2)
 115         True
 116         sage: bool(legendre_p(3, x, 0, 1) == p3)
 117         True
 118
 119     The zeroth polynomial is an element of the ring that we're working
 120     in::
 121
 122         sage: legendre_p(0, MatrixSpace(ZZ, 2)([1, 2, -4, 7]))
 123         [1 0]
 124         [0 1]
 125
 126     """
 127     if not n in ZZ:
 128         raise TypeError('n must be a natural number')
 129
 130     if n < 0:
 131         raise ValueError('n must be nonnegative')
 132
 133     if not (a in RR and b in RR):
 134         raise TypeError('both `a` and `b` must be real numbers')
 135
 136     if n == 0:
 137         # Easy base case, save time. Attempt to return a value in the
 138         # same field/ring as `x`.
 139         return x.parent()(1)
 140
 141     # Even though we know a,b are real we use a different ring. We
 142     # prefer ZZ so that we can support division of finite field
 143     # elements by (a-b). Eventually this should be supported for QQ as
 144     # well, although it does not work at the moment. The preference of
 145     # SR over RR is to return something attractive when e.g. a=pi.
 146     if a in ZZ:
 147         a = ZZ(a)
 148     else:
 149         a = SR(a)
 150
 151     if b in ZZ:
 152         b = ZZ(b)
 153     else:
 154         b = SR(b)
 155
 156     # Ensure that (2**n) is an element of ZZ. This is used later --
 157     # we can divide finite field elements by integers but we can't
 158     # multiply them by rationals at the moment.
 159     n = ZZ(n)
 160
 161     def phi(t):
 162         # This is an affine map from [a,b] into [-1,1] and so
 163         # preserves orthogonality.
 164         return (a + b - 2*t)/(a - b)
 165
 166     def c(m):
 167         return binomial(n,m)*binomial(n, n-m)
 168
 169     def g(m):
 170         # As given in A&S, but with `x` replaced by `phi(x)`.
 171         return ( ((phi(x) - 1)**(n-m)) * (phi(x) + 1)**m )
 172
 173     # From Abramowitz & Stegun, (22.3.2) with alpha = beta = 0.
 174     # Also massaged to support finite field elements.
 175     P = sum([ c(m)*g(m) for m in range(0,n+1) ])/(2**n)
 176
 177     return P