X-Git-Url: http://gitweb.michael.orlitzky.com/?a=blobdiff_plain;f=mjo%2Forthogonal_polynomials.py;h=15c695611fb3e287c357075f45e55eeddf550f7f;hb=HEAD;hp=808a558eb925521148306cb7315fbd9df07bd32a;hpb=9cf845989f17d70c8520f7c4a464f71b10535c0c;p=sage.d.git diff --git a/mjo/orthogonal_polynomials.py b/mjo/orthogonal_polynomials.py index 808a558..15c6956 100644 --- a/mjo/orthogonal_polynomials.py +++ b/mjo/orthogonal_polynomials.py @@ -1,7 +1,7 @@ from sage.all import * def legendre_p(n, x, a = -1, b = 1): - """ + r""" Returns the ``n``th Legendre polynomial of the first kind over the interval [a, b] with respect to ``x``. @@ -28,6 +28,10 @@ def legendre_p(n, x, a = -1, b = 1): returned. Otherwise, the value of the ``n``th polynomial at ``x`` will be returned. + SETUP:: + + sage: from mjo.orthogonal_polynomials import legendre_p + EXAMPLES: Create the standard Legendre polynomials in `x`:: @@ -39,7 +43,7 @@ def legendre_p(n, x, a = -1, b = 1): Reuse the variable from a polynomial ring:: sage: P. = QQ[] - sage: legendre_p(2,t).simplify_rational() + sage: legendre_p(2,t) 3/2*t^2 - 1/2 If ``x`` is a real number, the result should be as well:: @@ -58,12 +62,34 @@ def legendre_p(n, x, a = -1, b = 1): [-179 242] [-484 547] + And finite field elements:: + + sage: legendre_p(3, GF(11)(5)) + 8 + + Solve a simple least squares problem over `[-\pi, \pi]`:: + + sage: a = -pi + sage: b = pi + sage: def inner_product(v1, v2): + ....: return integrate(v1*v2, x, a, b) + sage: def norm(v): + ....: return sqrt(inner_product(v,v)) + sage: def project(basis, v): + ....: return sum( inner_product(v, b)*b/norm(b)**2 + ....: for b in basis) + sage: f = sin(x) + sage: legendre_basis = [ legendre_p(k, x, a, b) for k in range(4) ] + sage: proj = project(legendre_basis, f) + sage: proj.simplify_trig() + 5/2*(7*(pi^2 - 15)*x^3 - 3*(pi^4 - 21*pi^2)*x)/pi^6 + TESTS: We should agree with Maxima for all `n`:: sage: eq = lambda k: bool(legendre_p(k,x) == legendre_P(k,x)) - sage: all([eq(k) for k in range(0,20) ]) # long time + sage: all( eq(k) for k in range(20) ) # long time True We can evaluate the result of the zeroth polynomial:: @@ -133,17 +159,30 @@ def legendre_p(n, x, a = -1, b = 1): # same field/ring as `x`. return x.parent()(1) - # Even though we know a,b are real we use the symbolic ring. This - # lets us return pretty expressions where possible. - a = SR(a) - b = SR(b) - n = ZZ(n) # Ensure that 1/(2**n) is not integer division. - dn = 1/(2**n) + # Even though we know a,b are real we use a different ring. We + # prefer ZZ so that we can support division of finite field + # elements by (a-b). Eventually this should be supported for QQ as + # well, although it does not work at the moment. The preference of + # SR over RR is to return something attractive when e.g. a=pi. + if a in ZZ: + a = ZZ(a) + else: + a = SR(a) + + if b in ZZ: + b = ZZ(b) + else: + b = SR(b) + + # Ensure that (2**n) is an element of ZZ. This is used later -- + # we can divide finite field elements by integers but we can't + # multiply them by rationals at the moment. + n = ZZ(n) def phi(t): - # This is an affine map from [a,b] into [-1,1] and so preserves - # orthogonality. - return (2 / (b-a))*t + 1 - (2*b)/(b-a) + # This is an affine map from [a,b] into [-1,1] and so + # preserves orthogonality. + return (a + b - 2*t)/(a - b) def c(m): return binomial(n,m)*binomial(n, n-m) @@ -153,6 +192,7 @@ def legendre_p(n, x, a = -1, b = 1): return ( ((phi(x) - 1)**(n-m)) * (phi(x) + 1)**m ) # From Abramowitz & Stegun, (22.3.2) with alpha = beta = 0. - P = dn * sum([ c(m)*g(m) for m in range(0,n+1) ]) + # Also massaged to support finite field elements. + P = sum( c(m)*g(m) for m in range(n+1) )/(2**n) return P