コード例 #1
0
ファイル: summations.py プロジェクト: DasGespenst/sympy
    def euler_maclaurin(self, m=0, n=0, eps=0, eval_integral=True):
        """
        Return an Euler-Maclaurin approximation of self, where m is the
        number of leading terms to sum directly and n is the number of
        terms in the tail.

        With m = n = 0, this is simply the corresponding integral
        plus a first-order endpoint correction.

        Returns (s, e) where s is the Euler-Maclaurin approximation
        and e is the estimated error (taken to be the magnitude of
        the first omitted term in the tail):

            >>> from sympy.abc import k, a, b
            >>> from sympy import Sum
            >>> Sum(1/k, (k, 2, 5)).doit().evalf()
            1.28333333333333
            >>> s, e = Sum(1/k, (k, 2, 5)).euler_maclaurin()
            >>> s
            -log(2) + 7/20 + log(5)
            >>> from sympy import sstr
            >>> print(sstr((s.evalf(), e.evalf()), full_prec=True))
            (1.26629073187415, 0.0175000000000000)

        The endpoints may be symbolic:

            >>> s, e = Sum(1/k, (k, a, b)).euler_maclaurin()
            >>> s
            -log(a) + log(b) + 1/(2*b) + 1/(2*a)
            >>> e
            Abs(1/(12*b**2) - 1/(12*a**2))

        If the function is a polynomial of degree at most 2n+1, the
        Euler-Maclaurin formula becomes exact (and e = 0 is returned):

            >>> Sum(k, (k, 2, b)).euler_maclaurin()
            (b**2/2 + b/2 - 1, 0)
            >>> Sum(k, (k, 2, b)).doit()
            b**2/2 + b/2 - 1

        With a nonzero eps specified, the summation is ended
        as soon as the remainder term is less than the epsilon.
        """
        from sympy.functions import bernoulli, factorial
        from sympy.integrals import Integral

        m = int(m)
        n = int(n)
        f = self.function
        if len(self.limits) != 1:
            raise ValueError("More than 1 limit")
        i, a, b = self.limits[0]
        if (a > b) == True:
            if a - b == 1:
                return S.Zero,S.Zero
            a, b = b + 1, a - 1
            f = -f
        s = S.Zero
        if m:
            if b.is_Integer and a.is_Integer:
                m = min(m, b - a + 1)
            if not eps or f.is_polynomial(i):
                for k in range(m):
                    s += f.subs(i, a + k)
            else:
                term = f.subs(i, a)
                if term:
                    test = abs(term.evalf(3)) < eps
                    if test == True:
                        return s, abs(term)
                    elif not (test == False):
                        # a symbolic Relational class, can't go further
                        return term, S.Zero
                s += term
                for k in range(1, m):
                    term = f.subs(i, a + k)
                    if abs(term.evalf(3)) < eps and term != 0:
                        return s, abs(term)
                    s += term
            if b - a + 1 == m:
                return s, S.Zero
            a += m
        x = Dummy('x')
        I = Integral(f.subs(i, x), (x, a, b))
        if eval_integral:
            I = I.doit()
        s += I

        def fpoint(expr):
            if b is S.Infinity:
                return expr.subs(i, a), 0
            return expr.subs(i, a), expr.subs(i, b)
        fa, fb = fpoint(f)
        iterm = (fa + fb)/2
        g = f.diff(i)
        for k in range(1, n + 2):
            ga, gb = fpoint(g)
            term = bernoulli(2*k)/factorial(2*k)*(gb - ga)
            if (eps and term and abs(term.evalf(3)) < eps) or (k > n):
                break
            s += term
            g = g.diff(i, 2, simplify=False)
        return s + iterm, abs(term)
コード例 #2
0
ファイル: summations.py プロジェクト: wyom/sympy
    def euler_maclaurin(self, m=0, n=0, eps=0, eval_integral=True):
        """
        Return an Euler-Maclaurin approximation of self, where m is the
        number of leading terms to sum directly and n is the number of
        terms in the tail.

        With m = n = 0, this is simply the corresponding integral
        plus a first-order endpoint correction.

        Returns (s, e) where s is the Euler-Maclaurin approximation
        and e is the estimated error (taken to be the magnitude of
        the first omitted term in the tail):

            >>> from sympy.abc import k, a, b
            >>> from sympy import Sum
            >>> Sum(1/k, (k, 2, 5)).doit().evalf()
            1.28333333333333
            >>> s, e = Sum(1/k, (k, 2, 5)).euler_maclaurin()
            >>> s
            -log(2) + 7/20 + log(5)
            >>> from sympy import sstr
            >>> print(sstr((s.evalf(), e.evalf()), full_prec=True))
            (1.26629073187415, 0.0175000000000000)

        The endpoints may be symbolic:

            >>> s, e = Sum(1/k, (k, a, b)).euler_maclaurin()
            >>> s
            -log(a) + log(b) + 1/(2*b) + 1/(2*a)
            >>> e
            Abs(1/(12*b**2) - 1/(12*a**2))

        If the function is a polynomial of degree at most 2n+1, the
        Euler-Maclaurin formula becomes exact (and e = 0 is returned):

            >>> Sum(k, (k, 2, b)).euler_maclaurin()
            (b**2/2 + b/2 - 1, 0)
            >>> Sum(k, (k, 2, b)).doit()
            b**2/2 + b/2 - 1

        With a nonzero eps specified, the summation is ended
        as soon as the remainder term is less than the epsilon.
        """
        from sympy.functions import bernoulli, factorial
        from sympy.integrals import Integral

        m = int(m)
        n = int(n)
        f = self.function
        if len(self.limits) != 1:
            raise ValueError("More than 1 limit")
        i, a, b = self.limits[0]
        if (a > b) == True:
            if a - b == 1:
                return S.Zero, S.Zero
            a, b = b + 1, a - 1
            f = -f
        s = S.Zero
        if m:
            if b.is_Integer and a.is_Integer:
                m = min(m, b - a + 1)
            if not eps or f.is_polynomial(i):
                for k in range(m):
                    s += f.subs(i, a + k)
            else:
                term = f.subs(i, a)
                if term:
                    test = abs(term.evalf(3)) < eps
                    if test == True:
                        return s, abs(term)
                    elif not (test == False):
                        # a symbolic Relational class, can't go further
                        return term, S.Zero
                s += term
                for k in range(1, m):
                    term = f.subs(i, a + k)
                    if abs(term.evalf(3)) < eps and term != 0:
                        return s, abs(term)
                    s += term
            if b - a + 1 == m:
                return s, S.Zero
            a += m
        x = Dummy('x')
        I = Integral(f.subs(i, x), (x, a, b))
        if eval_integral:
            I = I.doit()
        s += I

        def fpoint(expr):
            if b is S.Infinity:
                return expr.subs(i, a), 0
            return expr.subs(i, a), expr.subs(i, b)

        fa, fb = fpoint(f)
        iterm = (fa + fb) / 2
        g = f.diff(i)
        for k in range(1, n + 2):
            ga, gb = fpoint(g)
            term = bernoulli(2 * k) / factorial(2 * k) * (gb - ga)
            if (eps and term and abs(term.evalf(3)) < eps) or (k > n):
                break
            s += term
            g = g.diff(i, 2, simplify=False)
        return s + iterm, abs(term)
コード例 #3
0
from sympy import *
from sympy import sin, sqrt
from sympy.abc import x, n
from sympy.integrals import Integral

x = Symbol('x')

# example 1: finding the indefinite integral for function: x**2+8 = x**4 + 7*x**(3 + 8)
# reminder: Indefinite integral is an integral with no lower or upper limit specified.
integralex = Integral((x**2) + 8, x)
print(integralex.doit())  # x**3/3 + 8*x

# example 2: integrating the same function above (x**2+8), but we will perform a definite integral with
# respect to a lower limit of 2 and an upper limit of 4.
x = Symbol('x')

integralex = Integral(
    (x**2) + 8, (x, 2, 4)
)  # the second argument reads as: wrt x variable, with lower limit 2, and upper limit 4
print(integralex.doit())