mjo/orthogonal_polynomials.py

   1 from sage.all import *
   2
   3 def legendre_p(n, x, a = -1, b = 1):
   4     r"""
   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     SETUP::
  32
  33         sage: from mjo.orthogonal_polynomials import legendre_p
  34
  35     EXAMPLES:
  36
  37     Create the standard Legendre polynomials in `x`::
  38
  39         sage: legendre_p(0,x)
  40         1
  41         sage: legendre_p(1,x)
  42         x
  43
  44     Reuse the variable from a polynomial ring::
  45         sage: P.<t> = QQ[]
  46         sage: legendre_p(2,t)
  47         3/2*t^2 - 1/2
  48
  49     If ``x`` is a real number, the result should be as well::
  50
  51         sage: legendre_p(3, 1.1)
  52         1.67750000000000
  53
  54     Similarly for complex numbers::
  55
  56         sage: legendre_p(3, 1 + I)
  57         7/2*I - 13/2
  58
  59     Even matrices work::
  60
  61         sage: legendre_p(3, MatrixSpace(ZZ, 2)([1, 2, -4, 7]))
  62         [-179  242]
  63         [-484  547]
  64
  65     And finite field elements::
  66
  67         sage: legendre_p(3, GF(11)(5))
  68         8
  69
  70     Solve a simple least squares problem over `[-\pi, \pi]`::
  71
  72         sage: a = -pi
  73         sage: b = pi
  74         sage: def inner_product(v1, v2):
  75         ....:     return integrate(v1*v2, x, a, b)
  76         sage: def norm(v):
  77         ....:     return sqrt(inner_product(v,v))
  78         sage: def project(basis, v):
  79         ....:     return sum( inner_product(v, b)*b/norm(b)**2
  80         ....:                 for b in basis)
  81         sage: f = sin(x)
  82         sage: legendre_basis = [ legendre_p(k, x, a, b) for k in range(4) ]
  83         sage: proj = project(legendre_basis, f)
  84         sage: proj.simplify_trig()
  85         5/2*(7*(pi^2 - 15)*x^3 - 3*(pi^4 - 21*pi^2)*x)/pi^6
  86
  87     TESTS:
  88
  89     We should agree with Maxima for all `n`::
  90
  91         sage: eq = lambda k: bool(legendre_p(k,x) == legendre_P(k,x))
  92         sage: all( eq(k) for k in range(20) ) # long time
  93         True
  94
  95     We can evaluate the result of the zeroth polynomial::
  96
  97         sage: f = legendre_p(0,x)
  98         sage: f(x=10)
  99         1
 100
 101     We should have |P(a)| = |P(b)| = 1 for all a,b::
 102
 103         sage: a = RR.random_element()
 104         sage: b = RR.random_element()
 105         sage: k = ZZ.random_element(20)
 106         sage: P = legendre_p(k, x, a, b)
 107         sage: abs(P(x=a)) # abs tol 1e-12
 108         1
 109         sage: abs(P(x=b)) # abs tol 1e-12
 110         1
 111
 112     Two different polynomials should be orthogonal with respect to the
 113     inner product over `[a,b]`. Note that this test can fail if QQ is
 114     replaced with RR, since integrate() can return NaN::
 115
 116         sage: a = QQ.random_element()
 117         sage: b = QQ.random_element()
 118         sage: j = ZZ.random_element(20)
 119         sage: k = j + 1
 120         sage: Pj = legendre_p(j, x, a, b)
 121         sage: Pk = legendre_p(k, x, a, b)
 122         sage: integrate(Pj*Pk, x, a, b) # abs tol 1e-12
 123         0
 124
 125     The first few polynomials shifted to [0,1] are known to be::
 126
 127         sage: p0 = 1
 128         sage: p1 = 2*x - 1
 129         sage: p2 = 6*x^2 - 6*x + 1
 130         sage: p3 = 20*x^3 - 30*x^2 + 12*x - 1
 131         sage: bool(legendre_p(0, x, 0, 1) == p0)
 132         True
 133         sage: bool(legendre_p(1, x, 0, 1) == p1)
 134         True
 135         sage: bool(legendre_p(2, x, 0, 1) == p2)
 136         True
 137         sage: bool(legendre_p(3, x, 0, 1) == p3)
 138         True
 139
 140     The zeroth polynomial is an element of the ring that we're working
 141     in::
 142
 143         sage: legendre_p(0, MatrixSpace(ZZ, 2)([1, 2, -4, 7]))
 144         [1 0]
 145         [0 1]
 146
 147     """
 148     if not n in ZZ:
 149         raise TypeError('n must be a natural number')
 150
 151     if n < 0:
 152         raise ValueError('n must be nonnegative')
 153
 154     if not (a in RR and b in RR):
 155         raise TypeError('both `a` and `b` must be real numbers')
 156
 157     if n == 0:
 158         # Easy base case, save time. Attempt to return a value in the
 159         # same field/ring as `x`.
 160         return x.parent()(1)
 161
 162     # Even though we know a,b are real we use a different ring. We
 163     # prefer ZZ so that we can support division of finite field
 164     # elements by (a-b). Eventually this should be supported for QQ as
 165     # well, although it does not work at the moment. The preference of
 166     # SR over RR is to return something attractive when e.g. a=pi.
 167     if a in ZZ:
 168         a = ZZ(a)
 169     else:
 170         a = SR(a)
 171
 172     if b in ZZ:
 173         b = ZZ(b)
 174     else:
 175         b = SR(b)
 176
 177     # Ensure that (2**n) is an element of ZZ. This is used later --
 178     # we can divide finite field elements by integers but we can't
 179     # multiply them by rationals at the moment.
 180     n = ZZ(n)
 181
 182     def phi(t):
 183         # This is an affine map from [a,b] into [-1,1] and so
 184         # preserves orthogonality.
 185         return (a + b - 2*t)/(a - b)
 186
 187     def c(m):
 188         return binomial(n,m)*binomial(n, n-m)
 189
 190     def g(m):
 191         # As given in A&S, but with `x` replaced by `phi(x)`.
 192         return ( ((phi(x) - 1)**(n-m)) * (phi(x) + 1)**m )
 193
 194     # From Abramowitz & Stegun, (22.3.2) with alpha = beta = 0.
 195     # Also massaged to support finite field elements.
 196     P = sum( c(m)*g(m) for m in range(n+1) )/(2**n)
 197
 198     return P