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Python functions.bernoulli函数代码示例

原作者: [db:作者] 来自: [db:来源] 收藏 邀请

本文整理汇总了Python中sympy.functions.bernoulli函数的典型用法代码示例。如果您正苦于以下问题:Python bernoulli函数的具体用法?Python bernoulli怎么用?Python bernoulli使用的例子?那么恭喜您, 这里精选的函数代码示例或许可以为您提供帮助。



在下文中一共展示了bernoulli函数的8个代码示例,这些例子默认根据受欢迎程度排序。您可以为喜欢或者感觉有用的代码点赞,您的评价将有助于我们的系统推荐出更棒的Python代码示例。

示例1: test_Function

def test_Function():
    assert mcode(sin(x) ** cos(x)) == "sin(x).^cos(x)"
    assert mcode(sign(x)) == "sign(x)"
    assert mcode(exp(x)) == "exp(x)"
    assert mcode(log(x)) == "log(x)"
    assert mcode(factorial(x)) == "factorial(x)"
    assert mcode(floor(x)) == "floor(x)"
    assert mcode(atan2(y, x)) == "atan2(y, x)"
    assert mcode(beta(x, y)) == 'beta(x, y)'
    assert mcode(polylog(x, y)) == 'polylog(x, y)'
    assert mcode(harmonic(x)) == 'harmonic(x)'
    assert mcode(bernoulli(x)) == "bernoulli(x)"
    assert mcode(bernoulli(x, y)) == "bernoulli(x, y)"
开发者ID:Lenqth,项目名称:sympy,代码行数:13,代码来源:test_octave.py


示例2: test_genocchi

def test_genocchi():
    genocchis = [1, -1, 0, 1, 0, -3, 0, 17]
    for n, g in enumerate(genocchis):
        assert genocchi(n + 1) == g

    m = Symbol('m', integer=True)
    n = Symbol('n', integer=True, positive=True)
    assert genocchi(m) == genocchi(m)
    assert genocchi(n).rewrite(bernoulli) == 2 * (1 - 2 ** n) * bernoulli(n)
    assert genocchi(2 * n).is_odd
    assert genocchi(4 * n).is_positive
    # This should work for 4 * n - 2, but fails due to some variation of issue
    # 8632 ((4*n-2).is_positive returns None)
    assert genocchi(4 * n + 2).is_negative
开发者ID:nishithshah2211,项目名称:sympy,代码行数:14,代码来源:test_comb_numbers.py


示例3: test_genocchi

def test_genocchi():
    genocchis = [1, -1, 0, 1, 0, -3, 0, 17]
    for n, g in enumerate(genocchis):
        assert genocchi(n + 1) == g

    m = Symbol('m', integer=True)
    n = Symbol('n', integer=True, positive=True)
    assert genocchi(m) == genocchi(m)
    assert genocchi(n).rewrite(bernoulli) == (1 - 2 ** n) * bernoulli(n) * 2
    assert genocchi(2 * n).is_odd
    assert genocchi(4 * n).is_positive
    # these are the only 2 prime Genocchi numbers
    assert genocchi(6, evaluate=False).is_prime == S(-3).is_prime
    assert genocchi(8, evaluate=False).is_prime
    assert genocchi(4 * n + 2).is_negative
    assert genocchi(4 * n - 2).is_negative
开发者ID:A-turing-machine,项目名称:sympy,代码行数:16,代码来源:test_comb_numbers.py


示例4: eval_sum_symbolic

def eval_sum_symbolic(f, limits):
    from sympy.functions import harmonic, bernoulli

    f_orig = f
    (i, a, b) = limits
    if not f.has(i):
        return f*(b - a + 1)

    # Linearity
    if f.is_Mul:
        L, R = f.as_two_terms()

        if not L.has(i):
            sR = eval_sum_symbolic(R, (i, a, b))
            if sR:
                return L*sR

        if not R.has(i):
            sL = eval_sum_symbolic(L, (i, a, b))
            if sL:
                return R*sL

        try:
            f = apart(f, i)  # see if it becomes an Add
        except PolynomialError:
            pass

    if f.is_Add:
        L, R = f.as_two_terms()
        lrsum = telescopic(L, R, (i, a, b))

        if lrsum:
            return lrsum

        lsum = eval_sum_symbolic(L, (i, a, b))
        rsum = eval_sum_symbolic(R, (i, a, b))

        if None not in (lsum, rsum):
            r = lsum + rsum
            if not r is S.NaN:
                return r

    # Polynomial terms with Faulhaber's formula
    n = Wild('n')
    result = f.match(i**n)

    if result is not None:
        n = result[n]

        if n.is_Integer:
            if n >= 0:
                if (b is S.Infinity and not a is S.NegativeInfinity) or \
                   (a is S.NegativeInfinity and not b is S.Infinity):
                    return S.Infinity
                return ((bernoulli(n + 1, b + 1) - bernoulli(n + 1, a))/(n + 1)).expand()
            elif a.is_Integer and a >= 1:
                if n == -1:
                    return harmonic(b) - harmonic(a - 1)
                else:
                    return harmonic(b, abs(n)) - harmonic(a - 1, abs(n))

    if not (a.has(S.Infinity, S.NegativeInfinity) or
            b.has(S.Infinity, S.NegativeInfinity)):
        # Geometric terms
        c1 = Wild('c1', exclude=[i])
        c2 = Wild('c2', exclude=[i])
        c3 = Wild('c3', exclude=[i])
        wexp = Wild('wexp')

        # Here we first attempt powsimp on f for easier matching with the
        # exponential pattern, and attempt expansion on the exponent for easier
        # matching with the linear pattern.
        e = f.powsimp().match(c1 ** wexp)
        if e is not None:
            e_exp = e.pop(wexp).expand().match(c2*i + c3)
            if e_exp is not None:
                e.update(e_exp)

        if e is not None:
            p = (c1**c3).subs(e)
            q = (c1**c2).subs(e)

            r = p*(q**a - q**(b + 1))/(1 - q)
            l = p*(b - a + 1)

            return Piecewise((l, Eq(q, S.One)), (r, True))

        r = gosper_sum(f, (i, a, b))

        if not r in (None, S.NaN):
            return r

    return eval_sum_hyper(f_orig, (i, a, b))
开发者ID:carstimon,项目名称:sympy,代码行数:93,代码来源:summations.py


示例5: euler_maclaurin

    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
#.........这里部分代码省略.........
开发者ID:carstimon,项目名称:sympy,代码行数:101,代码来源:summations.py


示例6: test_bernoulli

def test_bernoulli():
    assert bernoulli(0) == 1
    assert bernoulli(1) == Rational(-1, 2)
    assert bernoulli(2) == Rational(1, 6)
    assert bernoulli(3) == 0
    assert bernoulli(4) == Rational(-1, 30)
    assert bernoulli(5) == 0
    assert bernoulli(6) == Rational(1, 42)
    assert bernoulli(7) == 0
    assert bernoulli(8) == Rational(-1, 30)
    assert bernoulli(10) == Rational(5, 66)
    assert bernoulli(1000001) == 0

    assert bernoulli(0, x) == 1
    assert bernoulli(1, x) == x - Rational(1, 2)
    assert bernoulli(2, x) == x**2 - x + Rational(1, 6)
    assert bernoulli(3, x) == x**3 - (3*x**2)/2 + x/2

    # Should be fast; computed with mpmath
    b = bernoulli(1000)
    assert b.p % 10**10 == 7950421099
    assert b.q == 342999030

    b = bernoulli(10**6, evaluate=False).evalf()
    assert str(b) == '-2.23799235765713e+4767529'
开发者ID:DVNSarma,项目名称:sympy,代码行数:25,代码来源:test_comb_numbers.py


示例7: test_bernoulli

def test_bernoulli():
    assert bernoulli(0) == 1
    assert bernoulli(1) == Rational(-1, 2)
    assert bernoulli(2) == Rational(1, 6)
    assert bernoulli(3) == 0
    assert bernoulli(4) == Rational(-1, 30)
    assert bernoulli(5) == 0
    assert bernoulli(6) == Rational(1, 42)
    assert bernoulli(7) == 0
    assert bernoulli(8) == Rational(-1, 30)
    assert bernoulli(10) == Rational(5, 66)
    assert bernoulli(1000001) == 0

    assert bernoulli(0, x) == 1
    assert bernoulli(1, x) == x - Rational(1, 2)
    assert bernoulli(2, x) == x**2 - x + Rational(1, 6)
    assert bernoulli(3, x) == x**3 - (3*x**2)/2 + x/2

    # Should be fast; computed with mpmath
    b = bernoulli(1000)
    assert b.p % 10**10 == 7950421099
    assert b.q == 342999030

    b = bernoulli(10**6, evaluate=False).evalf()
    assert str(b) == '-2.23799235765713e+4767529'

    # Issue #8527
    l = Symbol('l', integer=True)
    m = Symbol('m', integer=True, nonnegative=True)
    n = Symbol('n', integer=True, positive=True)
    assert isinstance(bernoulli(2 * l + 1), bernoulli)
    assert isinstance(bernoulli(2 * m + 1), bernoulli)
    assert bernoulli(2 * n + 1) == 0
开发者ID:A-turing-machine,项目名称:sympy,代码行数:33,代码来源:test_comb_numbers.py


示例8: eval_sum_symbolic

def eval_sum_symbolic(f, limits):
    from sympy.functions import harmonic, bernoulli

    f_orig = f
    (i, a, b) = limits
    if not f.has(i):
        return f*(b - a + 1)

    # Linearity
    if f.is_Mul:
        L, R = f.as_two_terms()

        if not L.has(i):
            sR = eval_sum_symbolic(R, (i, a, b))
            if sR:
                return L*sR

        if not R.has(i):
            sL = eval_sum_symbolic(L, (i, a, b))
            if sL:
                return R*sL

        try:
            f = apart(f, i)  # see if it becomes an Add
        except PolynomialError:
            pass

    if f.is_Add:
        L, R = f.as_two_terms()
        lrsum = telescopic(L, R, (i, a, b))

        if lrsum:
            return lrsum

        lsum = eval_sum_symbolic(L, (i, a, b))
        rsum = eval_sum_symbolic(R, (i, a, b))

        if None not in (lsum, rsum):
            r = lsum + rsum
            if not r is S.NaN:
                return r

    # Polynomial terms with Faulhaber's formula
    n = Wild('n')
    result = f.match(i**n)

    if result is not None:
        n = result[n]

        if n.is_Integer:
            if n >= 0:
                if (b is S.Infinity and not a is S.NegativeInfinity) or \
                   (a is S.NegativeInfinity and not b is S.Infinity):
                    return S.Infinity
                return ((bernoulli(n + 1, b + 1) - bernoulli(n + 1, a))/(n + 1)).expand()
            elif a.is_Integer and a >= 1:
                if n == -1:
                    return harmonic(b) - harmonic(a - 1)
                else:
                    return harmonic(b, abs(n)) - harmonic(a - 1, abs(n))

    if not (a.has(S.Infinity, S.NegativeInfinity) or
            b.has(S.Infinity, S.NegativeInfinity)):
        # Geometric terms
        c1 = Wild('c1', exclude=[i])
        c2 = Wild('c2', exclude=[i])
        c3 = Wild('c3', exclude=[i])
        wexp = Wild('wexp')

        # Here we first attempt powsimp on f for easier matching with the
        # exponential pattern, and attempt expansion on the exponent for easier
        # matching with the linear pattern.
        e = f.powsimp().match(c1 ** wexp)
        if e is not None:
            e_exp = e.pop(wexp).expand().match(c2*i + c3)
            if e_exp is not None:
                e.update(e_exp)

        if e is not None:
            p = (c1**c3).subs(e)
            q = (c1**c2).subs(e)

            r = p*(q**a - q**(b + 1))/(1 - q)
            l = p*(b - a + 1)

            return Piecewise((l, Eq(q, S.One)), (r, True))

        r = gosper_sum(f, (i, a, b))

        if isinstance(r, (Mul,Add)):
            from sympy import ordered, Tuple
            non_limit = r.free_symbols - Tuple(*limits[1:]).free_symbols
            den = denom(together(r))
            den_sym = non_limit & den.free_symbols
            args = []
            for v in ordered(den_sym):
                try:
                    s = solve(den, v)
                    m = Eq(v, s[0]) if s else S.false
                    if m != False:
#.........这里部分代码省略.........
开发者ID:cklb,项目名称:sympy,代码行数:101,代码来源:summations.py



注:本文中的sympy.functions.bernoulli函数示例由纯净天空整理自Github/MSDocs等源码及文档管理平台,相关代码片段筛选自各路编程大神贡献的开源项目,源码版权归原作者所有,传播和使用请参考对应项目的License;未经允许,请勿转载。


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