/

import math

from ..libmp.backend import xrange

[docs]class QuadratureRule(object): """ Quadrature rules are implemented using this class, in order to simplify the code and provide a common infrastructure for tasks such as error estimation and node caching. You can implement a custom quadrature rule by subclassing :class:QuadratureRule and implementing the appropriate methods. The subclass can then be used by :func:~mpmath.quad by passing it as the *method* argument. :class:QuadratureRule instances are supposed to be singletons. :class:QuadratureRule therefore implements instance caching in :func:~mpmath.__new__. """ def __init__(self, ctx): self.ctx = ctx self.standard_cache = {} self.transformed_cache = {} self.interval_count = {}
[docs] def clear(self): """ Delete cached node data. """ self.standard_cache = {} self.transformed_cache = {} self.interval_count = {}
[docs] def calc_nodes(self, degree, prec, verbose=False): r""" Compute nodes for the standard interval [-1, 1]. Subclasses should probably implement only this method, and use :func:~mpmath.get_nodes method to retrieve the nodes. """ raise NotImplementedError
[docs] def get_nodes(self, a, b, degree, prec, verbose=False): """ Return nodes for given interval, degree and precision. The nodes are retrieved from a cache if already computed; otherwise they are computed by calling :func:~mpmath.calc_nodes and are then cached. Subclasses should probably not implement this method, but just implement :func:~mpmath.calc_nodes for the actual node computation. """ key = (a, b, degree, prec) if key in self.transformed_cache: return self.transformed_cache[key] orig = self.ctx.prec try: self.ctx.prec = prec+20 # Get nodes on standard interval if (degree, prec) in self.standard_cache: nodes = self.standard_cache[degree, prec] else: nodes = self.calc_nodes(degree, prec, verbose) self.standard_cache[degree, prec] = nodes # Transform to general interval nodes = self.transform_nodes(nodes, a, b, verbose) if key in self.interval_count: self.transformed_cache[key] = nodes else: self.interval_count[key] = True finally: self.ctx.prec = orig return nodes
[docs] def transform_nodes(self, nodes, a, b, verbose=False): r""" Rescale standardized nodes (for [-1, 1]) to a general interval [a, b]. For a finite interval, a simple linear change of variables is used. Otherwise, the following transformations are used: .. math :: [a, \infty] : t = \frac{1}{x} + (a-1) [-\infty, b] : t = (b+1) - \frac{1}{x} [-\infty, \infty] : t = \frac{x}{\sqrt{1-x^2}} """ ctx = self.ctx a = ctx.convert(a) b = ctx.convert(b) one = ctx.one if (a, b) == (-one, one): return nodes half = ctx.mpf(0.5) new_nodes = [] if ctx.isinf(a) or ctx.isinf(b): if (a, b) == (ctx.ninf, ctx.inf): p05 = -half for x, w in nodes: x2 = x*x px1 = one-x2 spx1 = px1**p05 x = x*spx1 w *= spx1/px1 new_nodes.append((x, w)) elif a == ctx.ninf: b1 = b+1 for x, w in nodes: u = 2/(x+one) x = b1-u w *= half*u**2 new_nodes.append((x, w)) elif b == ctx.inf: a1 = a-1 for x, w in nodes: u = 2/(x+one) x = a1+u w *= half*u**2 new_nodes.append((x, w)) elif a == ctx.inf or b == ctx.ninf: return [(x,-w) for (x,w) in self.transform_nodes(nodes, b, a, verbose)] else: raise NotImplementedError else: # Simple linear change of variables C = (b-a)/2 D = (b+a)/2 for x, w in nodes: new_nodes.append((D+C*x, C*w)) return new_nodes
[docs] def guess_degree(self, prec): """ Given a desired precision p in bits, estimate the degree m of the quadrature required to accomplish full accuracy for typical integrals. By default, :func:~mpmath.quad will perform up to m iterations. The value of m should be a slight overestimate, so that "slightly bad" integrals can be dealt with automatically using a few extra iterations. On the other hand, it should not be too big, so :func:~mpmath.quad can quit within a reasonable amount of time when it is given an "unsolvable" integral. The default formula used by :func:~mpmath.guess_degree is tuned for both :class:TanhSinh and :class:GaussLegendre. The output is roughly as follows: +---------+---------+ | p | m | +=========+=========+ | 50 | 6 | +---------+---------+ | 100 | 7 | +---------+---------+ | 500 | 10 | +---------+---------+ | 3000 | 12 | +---------+---------+ This formula is based purely on a limited amount of experimentation and will sometimes be wrong. """ # Expected degree # XXX: use mag g = int(4 + max(0, self.ctx.log(prec/30.0, 2))) # Reasonable "worst case" g += 2 return g
[docs] def estimate_error(self, results, prec, epsilon): r""" Given results from integrations [I_1, I_2, \ldots, I_k] done with a quadrature of rule of degree 1, 2, \ldots, k, estimate the error of I_k. For k = 2, we estimate |I_{\infty}-I_2| as |I_2-I_1|. For k > 2, we extrapolate |I_{\infty}-I_k| \approx |I_{k+1}-I_k| from |I_k-I_{k-1}| and |I_k-I_{k-2}| under the assumption that each degree increment roughly doubles the accuracy of the quadrature rule (this is true for both :class:TanhSinh and :class:GaussLegendre). The extrapolation formula is given by Borwein, Bailey & Girgensohn. Although not very conservative, this method seems to be very robust in practice. """ if len(results) == 2: return abs(results[0]-results[1]) try: if results[-1] == results[-2] == results[-3]: return self.ctx.zero D1 = self.ctx.log(abs(results[-1]-results[-2]), 10) D2 = self.ctx.log(abs(results[-1]-results[-3]), 10) except ValueError: return epsilon D3 = -prec D4 = min(0, max(D1**2/D2, 2*D1, D3)) return self.ctx.mpf(10) ** int(D4)
[docs] def summation(self, f, points, prec, epsilon, max_degree, verbose=False): """ Main integration function. Computes the 1D integral over the interval specified by *points*. For each subinterval, performs quadrature of degree from 1 up to *max_degree* until :func:~mpmath.estimate_error signals convergence. :func:~mpmath.summation transforms each subintegration to the standard interval and then calls :func:~mpmath.sum_next. """ ctx = self.ctx I = err = ctx.zero for i in xrange(len(points)-1): a, b = points[i], points[i+1] if a == b: continue # XXX: we could use a single variable transformation, # but this is not good in practice. We get better accuracy # by having 0 as an endpoint. if (a, b) == (ctx.ninf, ctx.inf): _f = f f = lambda x: _f(-x) + _f(x) a, b = (ctx.zero, ctx.inf) results = [] for degree in xrange(1, max_degree+1): nodes = self.get_nodes(a, b, degree, prec, verbose) if verbose: print("Integrating from %s to %s (degree %s of %s)" % \ (ctx.nstr(a), ctx.nstr(b), degree, max_degree)) results.append(self.sum_next(f, nodes, degree, prec, results, verbose)) if degree > 1: err = self.estimate_error(results, prec, epsilon) if err <= epsilon: break if verbose: print("Estimated error:", ctx.nstr(err)) I += results[-1] if err > epsilon: if verbose: print("Failed to reach full accuracy. Estimated error:", ctx.nstr(err)) return I, err
[docs] def sum_next(self, f, nodes, degree, prec, previous, verbose=False): r""" Evaluates the step sum \sum w_k f(x_k) where the *nodes* list contains the (w_k, x_k) pairs. :func:~mpmath.summation will supply the list *results* of values computed by :func:~mpmath.sum_next at previous degrees, in case the quadrature rule is able to reuse them. """ return self.ctx.fdot((w, f(x)) for (x,w) in nodes)
[docs]class TanhSinh(QuadratureRule): r""" This class implements "tanh-sinh" or "doubly exponential" quadrature. This quadrature rule is based on the Euler-Maclaurin integral formula. By performing a change of variables involving nested exponentials / hyperbolic functions (hence the name), the derivatives at the endpoints vanish rapidly. Since the error term in the Euler-Maclaurin formula depends on the derivatives at the endpoints, a simple step sum becomes extremely accurate. In practice, this means that doubling the number of evaluation points roughly doubles the number of accurate digits. Comparison to Gauss-Legendre: * Initial computation of nodes is usually faster * Handles endpoint singularities better * Handles infinite integration intervals better * Is slower for smooth integrands once nodes have been computed The implementation of the tanh-sinh algorithm is based on the description given in Borwein, Bailey & Girgensohn, "Experimentation in Mathematics - Computational Paths to Discovery", A K Peters, 2003, pages 312-313. In the present implementation, a few improvements have been made: * A more efficient scheme is used to compute nodes (exploiting recurrence for the exponential function) * The nodes are computed successively instead of all at once Various documents describing the algorithm are available online, e.g.: * http://crd.lbl.gov/~dhbailey/dhbpapers/dhb-tanh-sinh.pdf * http://users.cs.dal.ca/~jborwein/tanh-sinh.pdf """
[docs] def sum_next(self, f, nodes, degree, prec, previous, verbose=False): """ Step sum for tanh-sinh quadrature of degree m. We exploit the fact that half of the abscissas at degree m are precisely the abscissas from degree m-1. Thus reusing the result from the previous level allows a 2x speedup. """ h = self.ctx.mpf(2)**(-degree) # Abscissas overlap, so reusing saves half of the time if previous: S = previous[-1]/(h*2) else: S = self.ctx.zero S += self.ctx.fdot((w,f(x)) for (x,w) in nodes) return h*S
[docs] def calc_nodes(self, degree, prec, verbose=False): r""" The abscissas and weights for tanh-sinh quadrature of degree m are given by .. math:: x_k = \tanh(\pi/2 \sinh(t_k)) w_k = \pi/2 \cosh(t_k) / \cosh(\pi/2 \sinh(t_k))^2 where t_k = t_0 + hk for a step length h \sim 2^{-m}. The list of nodes is actually infinite, but the weights die off so rapidly that only a few are needed. """ ctx = self.ctx nodes = [] extra = 20 ctx.prec += extra tol = ctx.ldexp(1, -prec-10) pi4 = ctx.pi/4 # For simplicity, we work in steps h = 1/2^n, with the first point # offset so that we can reuse the sum from the previous degree # We define degree 1 to include the "degree 0" steps, including # the point x = 0. (It doesn't work well otherwise; not sure why.) t0 = ctx.ldexp(1, -degree) if degree == 1: #nodes.append((mpf(0), pi4)) #nodes.append((-mpf(0), pi4)) nodes.append((ctx.zero, ctx.pi/2)) h = t0 else: h = t0*2 # Since h is fixed, we can compute the next exponential # by simply multiplying by exp(h) expt0 = ctx.exp(t0) a = pi4 * expt0 b = pi4 / expt0 udelta = ctx.exp(h) urdelta = 1/udelta for k in xrange(0, 20*2**degree+1): # Reference implementation: # t = t0 + k*h # x = tanh(pi/2 * sinh(t)) # w = pi/2 * cosh(t) / cosh(pi/2 * sinh(t))**2 # Fast implementation. Note that c = exp(pi/2 * sinh(t)) c = ctx.exp(a-b) d = 1/c co = (c+d)/2 si = (c-d)/2 x = si / co w = (a+b) / co**2 diff = abs(x-1) if diff <= tol: break nodes.append((x, w)) nodes.append((-x, w)) a *= udelta b *= urdelta if verbose and k % 300 == 150: # Note: the number displayed is rather arbitrary. Should # figure out how to print something that looks more like a # percentage print("Calculating nodes:", ctx.nstr(-ctx.log(diff, 10) / prec)) ctx.prec -= extra return nodes
[docs]class GaussLegendre(QuadratureRule): """ This class implements Gauss-Legendre quadrature, which is exceptionally efficient for polynomials and polynomial-like (i.e. very smooth) integrands. The abscissas and weights are given by roots and values of Legendre polynomials, which are the orthogonal polynomials on [-1, 1] with respect to the unit weight (see :func:~mpmath.legendre). In this implementation, we take the "degree" m of the quadrature to denote a Gauss-Legendre rule of degree 3 \cdot 2^m (following Borwein, Bailey & Girgensohn). This way we get quadratic, rather than linear, convergence as the degree is incremented. Comparison to tanh-sinh quadrature: * Is faster for smooth integrands once nodes have been computed * Initial computation of nodes is usually slower * Handles endpoint singularities worse * Handles infinite integration intervals worse """
[docs] def calc_nodes(self, degree, prec, verbose=False): """ Calculates the abscissas and weights for Gauss-Legendre quadrature of degree of given degree (actually 3 \cdot 2^m). """ ctx = self.ctx # It is important that the epsilon is set lower than the # "real" epsilon epsilon = ctx.ldexp(1, -prec-8) # Fairly high precision might be required for accurate # evaluation of the roots orig = ctx.prec ctx.prec = int(prec*1.5) if degree == 1: x = ctx.mpf(3)/5 w = ctx.mpf(5)/9 nodes = [(-x,w),(ctx.zero,ctx.mpf(8)/9),(x,w)] ctx.prec = orig return nodes nodes = [] n = 3*2**(degree-1) upto = n//2 + 1 for j in xrange(1, upto): # Asymptotic formula for the roots r = ctx.mpf(math.cos(math.pi*(j-0.25)/(n+0.5))) # Newton iteration while 1: t1, t2 = 1, 0 # Evaluates the Legendre polynomial using its defining # recurrence relation for j1 in xrange(1,n+1): t3, t2, t1 = t2, t1, ((2*j1-1)*r*t1 - (j1-1)*t2)/j1 t4 = n*(r*t1- t2)/(r**2-1) t5 = r a = t1/t4 r = r - a if abs(a) < epsilon: break x = r w = 2/((1-r**2)*t4**2) if verbose and j % 30 == 15: print("Computing nodes (%i of %i)" % (j, upto)) nodes.append((x, w)) nodes.append((-x, w)) ctx.prec = orig return nodes
class QuadratureMethods: def __init__(ctx, *args, **kwargs): ctx._gauss_legendre = GaussLegendre(ctx) ctx._tanh_sinh = TanhSinh(ctx) def quad(ctx, f, *points, **kwargs): r""" Computes a single, double or triple integral over a given 1D interval, 2D rectangle, or 3D cuboid. A basic example:: >>> from mpmath import * >>> mp.dps = 15; mp.pretty = True >>> quad(sin, [0, pi]) 2.0 A basic 2D integral:: >>> f = lambda x, y: cos(x+y/2) >>> quad(f, [-pi/2, pi/2], [0, pi]) 4.0 **Interval format** The integration range for each dimension may be specified using a list or tuple. Arguments are interpreted as follows: quad(f, [x1, x2]) -- calculates \int_{x_1}^{x_2} f(x) \, dx quad(f, [x1, x2], [y1, y2]) -- calculates \int_{x_1}^{x_2} \int_{y_1}^{y_2} f(x,y) \, dy \, dx quad(f, [x1, x2], [y1, y2], [z1, z2]) -- calculates \int_{x_1}^{x_2} \int_{y_1}^{y_2} \int_{z_1}^{z_2} f(x,y,z) \, dz \, dy \, dx Endpoints may be finite or infinite. An interval descriptor may also contain more than two points. In this case, the integration is split into subintervals, between each pair of consecutive points. This is useful for dealing with mid-interval discontinuities, or integrating over large intervals where the function is irregular or oscillates. **Options** :func:~mpmath.quad recognizes the following keyword arguments: *method* Chooses integration algorithm (described below). *error* If set to true, :func:~mpmath.quad returns (v, e) where v is the integral and e is the estimated error. *maxdegree* Maximum degree of the quadrature rule to try before quitting. *verbose* Print details about progress. **Algorithms** Mpmath presently implements two integration algorithms: tanh-sinh quadrature and Gauss-Legendre quadrature. These can be selected using *method='tanh-sinh'* or *method='gauss-legendre'* or by passing the classes *method=TanhSinh*, *method=GaussLegendre*. The functions :func:~mpmath.quadts and :func:~mpmath.quadgl are also available as shortcuts. Both algorithms have the property that doubling the number of evaluation points roughly doubles the accuracy, so both are ideal for high precision quadrature (hundreds or thousands of digits). At high precision, computing the nodes and weights for the integration can be expensive (more expensive than computing the function values). To make repeated integrations fast, nodes are automatically cached. The advantages of the tanh-sinh algorithm are that it tends to handle endpoint singularities well, and that the nodes are cheap to compute on the first run. For these reasons, it is used by :func:~mpmath.quad as the default algorithm. Gauss-Legendre quadrature often requires fewer function evaluations, and is therefore often faster for repeated use, but the algorithm does not handle endpoint singularities as well and the nodes are more expensive to compute. Gauss-Legendre quadrature can be a better choice if the integrand is smooth and repeated integrations are required (e.g. for multiple integrals). See the documentation for :class:TanhSinh and :class:GaussLegendre for additional details. **Examples of 1D integrals** Intervals may be infinite or half-infinite. The following two examples evaluate the limits of the inverse tangent function (\int 1/(1+x^2) = \tan^{-1} x), and the Gaussian integral \int_{\infty}^{\infty} \exp(-x^2)\,dx = \sqrt{\pi}:: >>> mp.dps = 15 >>> quad(lambda x: 2/(x**2+1), [0, inf]) 3.14159265358979 >>> quad(lambda x: exp(-x**2), [-inf, inf])**2 3.14159265358979 Integrals can typically be resolved to high precision. The following computes 50 digits of \pi by integrating the area of the half-circle defined by x^2 + y^2 \le 1, -1 \le x \le 1, y \ge 0:: >>> mp.dps = 50 >>> 2*quad(lambda x: sqrt(1-x**2), [-1, 1]) 3.1415926535897932384626433832795028841971693993751 One can just as well compute 1000 digits (output truncated):: >>> mp.dps = 1000 >>> 2*quad(lambda x: sqrt(1-x**2), [-1, 1]) #doctest:+ELLIPSIS 3.141592653589793238462643383279502884...216420198 Complex integrals are supported. The following computes a residue at z = 0 by integrating counterclockwise along the diamond-shaped path from 1 to +i to -1 to -i to 1:: >>> mp.dps = 15 >>> chop(quad(lambda z: 1/z, [1,j,-1,-j,1])) (0.0 + 6.28318530717959j) **Examples of 2D and 3D integrals** Here are several nice examples of analytically solvable 2D integrals (taken from MathWorld [1]) that can be evaluated to high precision fairly rapidly by :func:~mpmath.quad:: >>> mp.dps = 30 >>> f = lambda x, y: (x-1)/((1-x*y)*log(x*y)) >>> quad(f, [0, 1], [0, 1]) 0.577215664901532860606512090082 >>> +euler 0.577215664901532860606512090082 >>> f = lambda x, y: 1/sqrt(1+x**2+y**2) >>> quad(f, [-1, 1], [-1, 1]) 3.17343648530607134219175646705 >>> 4*log(2+sqrt(3))-2*pi/3 3.17343648530607134219175646705 >>> f = lambda x, y: 1/(1-x**2 * y**2) >>> quad(f, [0, 1], [0, 1]) 1.23370055013616982735431137498 >>> pi**2 / 8 1.23370055013616982735431137498 >>> quad(lambda x, y: 1/(1-x*y), [0, 1], [0, 1]) 1.64493406684822643647241516665 >>> pi**2 / 6 1.64493406684822643647241516665 Multiple integrals may be done over infinite ranges:: >>> mp.dps = 15 >>> print(quad(lambda x,y: exp(-x-y), [0, inf], [1, inf])) 0.367879441171442 >>> print(1/e) 0.367879441171442 For nonrectangular areas, one can call :func:~mpmath.quad recursively. For example, we can replicate the earlier example of calculating \pi by integrating over the unit-circle, and actually use double quadrature to actually measure the area circle:: >>> f = lambda x: quad(lambda y: 1, [-sqrt(1-x**2), sqrt(1-x**2)]) >>> quad(f, [-1, 1]) 3.14159265358979 Here is a simple triple integral:: >>> mp.dps = 15 >>> f = lambda x,y,z: x*y/(1+z) >>> quad(f, [0,1], [0,1], [1,2], method='gauss-legendre') 0.101366277027041 >>> (log(3)-log(2))/4 0.101366277027041 **Singularities** Both tanh-sinh and Gauss-Legendre quadrature are designed to integrate smooth (infinitely differentiable) functions. Neither algorithm copes well with mid-interval singularities (such as mid-interval discontinuities in f(x) or f'(x)). The best solution is to split the integral into parts:: >>> mp.dps = 15 >>> quad(lambda x: abs(sin(x)), [0, 2*pi]) # Bad 3.99900894176779 >>> quad(lambda x: abs(sin(x)), [0, pi, 2*pi]) # Good 4.0 The tanh-sinh rule often works well for integrands having a singularity at one or both endpoints:: >>> mp.dps = 15 >>> quad(log, [0, 1], method='tanh-sinh') # Good -1.0 >>> quad(log, [0, 1], method='gauss-legendre') # Bad -0.999932197413801 However, the result may still be inaccurate for some functions:: >>> quad(lambda x: 1/sqrt(x), [0, 1], method='tanh-sinh') 1.99999999946942 This problem is not due to the quadrature rule per se, but to numerical amplification of errors in the nodes. The problem can be circumvented by temporarily increasing the precision:: >>> mp.dps = 30 >>> a = quad(lambda x: 1/sqrt(x), [0, 1], method='tanh-sinh') >>> mp.dps = 15 >>> +a 2.0 **Highly variable functions** For functions that are smooth (in the sense of being infinitely differentiable) but contain sharp mid-interval peaks or many "bumps", :func:~mpmath.quad may fail to provide full accuracy. For example, with default settings, :func:~mpmath.quad is able to integrate \sin(x) accurately over an interval of length 100 but not over length 1000:: >>> quad(sin, [0, 100]); 1-cos(100) # Good 0.137681127712316 0.137681127712316 >>> quad(sin, [0, 1000]); 1-cos(1000) # Bad -37.8587612408485 0.437620923709297 One solution is to break the integration into 10 intervals of length 100:: >>> quad(sin, linspace(0, 1000, 10)) # Good 0.437620923709297 Another is to increase the degree of the quadrature:: >>> quad(sin, [0, 1000], maxdegree=10) # Also good 0.437620923709297 Whether splitting the interval or increasing the degree is more efficient differs from case to case. Another example is the function 1/(1+x^2), which has a sharp peak centered around x = 0:: >>> f = lambda x: 1/(1+x**2) >>> quad(f, [-100, 100]) # Bad 3.64804647105268 >>> quad(f, [-100, 100], maxdegree=10) # Good 3.12159332021646 >>> quad(f, [-100, 0, 100]) # Also good 3.12159332021646 **References** 1. http://mathworld.wolfram.com/DoubleIntegral.html """ rule = kwargs.get('method', 'tanh-sinh') if type(rule) is str: if rule == 'tanh-sinh': rule = ctx._tanh_sinh elif rule == 'gauss-legendre': rule = ctx._gauss_legendre else: raise ValueError("unknown quadrature rule: %s" % rule) else: rule = rule(ctx) verbose = kwargs.get('verbose') dim = len(points) orig = prec = ctx.prec epsilon = ctx.eps/8 m = kwargs.get('maxdegree') or rule.guess_degree(prec) points = [ctx._as_points(p) for p in points] try: ctx.prec += 20 if dim == 1: v, err = rule.summation(f, points[0], prec, epsilon, m, verbose) elif dim == 2: v, err = rule.summation(lambda x: \ rule.summation(lambda y: f(x,y), \ points[1], prec, epsilon, m)[0], points[0], prec, epsilon, m, verbose) elif dim == 3: v, err = rule.summation(lambda x: \ rule.summation(lambda y: \ rule.summation(lambda z: f(x,y,z), \ points[2], prec, epsilon, m)[0], points[1], prec, epsilon, m)[0], points[0], prec, epsilon, m, verbose) else: raise NotImplementedError("quadrature must have dim 1, 2 or 3") finally: ctx.prec = orig if kwargs.get("error"): return +v, err return +v def quadts(ctx, *args, **kwargs): """ Performs tanh-sinh quadrature. The call quadts(func, *points, ...) is simply a shortcut for: quad(func, *points, ..., method=TanhSinh) For example, a single integral and a double integral: quadts(lambda x: exp(cos(x)), [0, 1]) quadts(lambda x, y: exp(cos(x+y)), [0, 1], [0, 1]) See the documentation for quad for information about how points arguments and keyword arguments are parsed. See documentation for TanhSinh for algorithmic information about tanh-sinh quadrature. """ kwargs['method'] = 'tanh-sinh' return ctx.quad(*args, **kwargs) def quadgl(ctx, *args, **kwargs): """ Performs Gauss-Legendre quadrature. The call quadgl(func, *points, ...) is simply a shortcut for: quad(func, *points, ..., method=GaussLegendre) For example, a single integral and a double integral: quadgl(lambda x: exp(cos(x)), [0, 1]) quadgl(lambda x, y: exp(cos(x+y)), [0, 1], [0, 1]) See the documentation for quad for information about how points arguments and keyword arguments are parsed. See documentation for TanhSinh for algorithmic information about tanh-sinh quadrature. """ kwargs['method'] = 'gauss-legendre' return ctx.quad(*args, **kwargs) def quadosc(ctx, f, interval, omega=None, period=None, zeros=None): r""" Calculates .. math :: I = \int_a^b f(x) dx where at least one of a and b is infinite and where f(x) = g(x) \cos(\omega x + \phi) for some slowly decreasing function g(x). With proper input, :func:~mpmath.quadosc can also handle oscillatory integrals where the oscillation rate is different from a pure sine or cosine wave. In the standard case when |a| < \infty, b = \infty, :func:~mpmath.quadosc works by evaluating the infinite series .. math :: I = \int_a^{x_1} f(x) dx + \sum_{k=1}^{\infty} \int_{x_k}^{x_{k+1}} f(x) dx where x_k are consecutive zeros (alternatively some other periodic reference point) of f(x). Accordingly, :func:~mpmath.quadosc requires information about the zeros of f(x). For a periodic function, you can specify the zeros by either providing the angular frequency \omega (*omega*) or the *period* 2 \pi/\omega. In general, you can specify the n-th zero by providing the *zeros* arguments. Below is an example of each:: >>> from mpmath import * >>> mp.dps = 15; mp.pretty = True >>> f = lambda x: sin(3*x)/(x**2+1) >>> quadosc(f, [0,inf], omega=3) 0.37833007080198 >>> quadosc(f, [0,inf], period=2*pi/3) 0.37833007080198 >>> quadosc(f, [0,inf], zeros=lambda n: pi*n/3) 0.37833007080198 >>> (ei(3)*exp(-3)-exp(3)*ei(-3))/2 # Computed by Mathematica 0.37833007080198 Note that *zeros* was specified to multiply n by the *half-period*, not the full period. In theory, it does not matter whether each partial integral is done over a half period or a full period. However, if done over half-periods, the infinite series passed to :func:~mpmath.nsum becomes an *alternating series* and this typically makes the extrapolation much more efficient. Here is an example of an integration over the entire real line, and a half-infinite integration starting at -\infty:: >>> quadosc(lambda x: cos(x)/(1+x**2), [-inf, inf], omega=1) 1.15572734979092 >>> pi/e 1.15572734979092 >>> quadosc(lambda x: cos(x)/x**2, [-inf, -1], period=2*pi) -0.0844109505595739 >>> cos(1)+si(1)-pi/2 -0.0844109505595738 Of course, the integrand may contain a complex exponential just as well as a real sine or cosine:: >>> quadosc(lambda x: exp(3*j*x)/(1+x**2), [-inf,inf], omega=3) (0.156410688228254 + 0.0j) >>> pi/e**3 0.156410688228254 >>> quadosc(lambda x: exp(3*j*x)/(2+x+x**2), [-inf,inf], omega=3) (0.00317486988463794 - 0.0447701735209082j) >>> 2*pi/sqrt(7)/exp(3*(j+sqrt(7))/2) (0.00317486988463794 - 0.0447701735209082j) **Non-periodic functions** If f(x) = g(x) h(x) for some function h(x) that is not strictly periodic, *omega* or *period* might not work, and it might be necessary to use *zeros*. A notable exception can be made for Bessel functions which, though not periodic, are "asymptotically periodic" in a sufficiently strong sense that the sum extrapolation will work out:: >>> quadosc(j0, [0, inf], period=2*pi) 1.0 >>> quadosc(j1, [0, inf], period=2*pi) 1.0 More properly, one should provide the exact Bessel function zeros:: >>> j0zero = lambda n: findroot(j0, pi*(n-0.25)) >>> quadosc(j0, [0, inf], zeros=j0zero) 1.0 For an example where *zeros* becomes necessary, consider the complete Fresnel integrals .. math :: \int_0^{\infty} \cos x^2\,dx = \int_0^{\infty} \sin x^2\,dx = \sqrt{\frac{\pi}{8}}. Although the integrands do not decrease in magnitude as x \to \infty, the integrals are convergent since the oscillation rate increases (causing consecutive periods to asymptotically cancel out). These integrals are virtually impossible to calculate to any kind of accuracy using standard quadrature rules. However, if one provides the correct asymptotic distribution of zeros (x_n \sim \sqrt{n}), :func:~mpmath.quadosc works:: >>> mp.dps = 30 >>> f = lambda x: cos(x**2) >>> quadosc(f, [0,inf], zeros=lambda n:sqrt(pi*n)) 0.626657068657750125603941321203 >>> f = lambda x: sin(x**2) >>> quadosc(f, [0,inf], zeros=lambda n:sqrt(pi*n)) 0.626657068657750125603941321203 >>> sqrt(pi/8) 0.626657068657750125603941321203 (Interestingly, these integrals can still be evaluated if one places some other constant than \pi in the square root sign.) In general, if f(x) \sim g(x) \cos(h(x)), the zeros follow the inverse-function distribution h^{-1}(x):: >>> mp.dps = 15 >>> f = lambda x: sin(exp(x)) >>> quadosc(f, [1,inf], zeros=lambda n: log(n)) -0.25024394235267 >>> pi/2-si(e) -0.250243942352671 **Non-alternating functions** If the integrand oscillates around a positive value, without alternating signs, the extrapolation might fail. A simple trick that sometimes works is to multiply or divide the frequency by 2:: >>> f = lambda x: 1/x**2+sin(x)/x**4 >>> quadosc(f, [1,inf], omega=1) # Bad 1.28642190869861 >>> quadosc(f, [1,inf], omega=0.5) # Perfect 1.28652953559617 >>> 1+(cos(1)+ci(1)+sin(1))/6 1.28652953559617 **Fast decay** :func:~mpmath.quadosc is primarily useful for slowly decaying integrands. If the integrand decreases exponentially or faster, :func:~mpmath.quad will likely handle it without trouble (and generally be much faster than :func:~mpmath.quadosc):: >>> quadosc(lambda x: cos(x)/exp(x), [0, inf], omega=1) 0.5 >>> quad(lambda x: cos(x)/exp(x), [0, inf]) 0.5 """ a, b = ctx._as_points(interval) a = ctx.convert(a) b = ctx.convert(b) if [omega, period, zeros].count(None) != 2: raise ValueError( \ "must specify exactly one of omega, period, zeros") if a == ctx.ninf and b == ctx.inf: s1 = ctx.quadosc(f, [a, 0], omega=omega, zeros=zeros, period=period) s2 = ctx.quadosc(f, [0, b], omega=omega, zeros=zeros, period=period) return s1 + s2 if a == ctx.ninf: if zeros: return ctx.quadosc(lambda x:f(-x), [-b,-a], lambda n: zeros(-n)) else: return ctx.quadosc(lambda x:f(-x), [-b,-a], omega=omega, period=period) if b != ctx.inf: raise ValueError("quadosc requires an infinite integration interval") if not zeros: if omega: period = 2*ctx.pi/omega zeros = lambda n: n*period/2 #for n in range(1,10): # p = zeros(n) # if p > a: # break #if n >= 9: # raise ValueError("zeros do not appear to be correctly indexed") n = 1 s = ctx.quadgl(f, [a, zeros(n)]) def term(k): return ctx.quadgl(f, [zeros(k), zeros(k+1)]) s += ctx.nsum(term, [n, ctx.inf]) return s if __name__ == '__main__': import doctest doctest.testmod()