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Python polyroots.roots_quadratic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.polys.polyroots.roots_quadratic函数的典型用法代码示例。如果您正苦于以下问题:Python roots_quadratic函数的具体用法?Python roots_quadratic怎么用?Python roots_quadratic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roots_quadratic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_roots_quadratic

def test_roots_quadratic():
    assert roots_quadratic(Poly(2*x**2, x)) == [0, 0]
    assert roots_quadratic(Poly(2*x**2 + 3*x, x)) == [-Rational(3, 2), 0]
    assert roots_quadratic(Poly(2*x**2 + 3, x)) == [-I*sqrt(6)/2, I*sqrt(6)/2]
    assert roots_quadratic(Poly(2*x**2 + 4*x + 3, x)) == [-1 - I*sqrt(2)/2, -1 + I*sqrt(2)/2]

    f = x**2 + (2*a*e + 2*c*e)/(a - c)*x + (d - b + a*e**2 - c*e**2)/(a - c)

    assert roots_quadratic(Poly(f, x)) == \
        [-e*(a + c)/(a - c) - sqrt((a*b + c*d - a*d - b*c + 4*a*c*e**2)/(a - c)**2),
         -e*(a + c)/(a - c) + sqrt((a*b + c*d - a*d - b*c + 4*a*c*e**2)/(a - c)**2)]

    # check for simplification
    f = Poly(y*x**2 - 2*x - 2*y, x)
    assert roots_quadratic(f) == \
        [-sqrt(2*y**2 + 1)/y + 1/y, sqrt(2*y**2 + 1)/y + 1/y]
    f = Poly(x**2 + (-y**2 - 2)*x + y**2 + 1, x)
    assert roots_quadratic(f) == \
        [y**2/2 - sqrt(y**4)/2 + 1, y**2/2 + sqrt(y**4)/2 + 1]

    f = Poly(sqrt(2)*x**2 - 1, x)
    r = roots_quadratic(f)
    assert r == _nsort(r)

    # issue 8255
    f = Poly(-24*x**2 - 180*x + 264)
    assert [w.n(2) for w in f.all_roots(radicals=True)] == \
           [w.n(2) for w in f.all_roots(radicals=False)]
    for _a, _b, _c in cartes((-2, 2), (-2, 2), (0, -1)):
        f = Poly(_a*x**2 + _b*x + _c)
        roots = roots_quadratic(f)
        assert roots == _nsort(roots)
开发者ID:NalinG,项目名称:sympy,代码行数:32,代码来源:test_polyroots.py


示例2: test_roots_quadratic

def test_roots_quadratic():
    assert roots_quadratic(Poly(2*x**2, x)) == [0, 0]
    assert roots_quadratic(Poly(2*x**2 + 3*x, x)) == [-Rational(3, 2), 0]
    assert roots_quadratic(Poly(2*x**2 + 3, x)) == [-I*sqrt(6)/2, I*sqrt(6)/2]
    assert roots_quadratic(Poly(2*x**2 + 4*x+3, x)) == [-1 - I*sqrt(2)/2, -1 + I*sqrt(2)/2]

    f = x**2 + (2*a*e + 2*c*e)/(a - c)*x + (d - b + a*e**2 - c*e**2)/(a - c)

    assert roots_quadratic(Poly(f, x)) == \
        [-e*(a + c)/(a - c) - ((a*b + c*d - a*d - b*c + 4*a*c*e**2)/(a - c)**2)**S.Half,
         -e*(a + c)/(a - c) + ((a*b + c*d - a*d - b*c + 4*a*c*e**2)/(a - c)**2)**S.Half]
开发者ID:qmattpap,项目名称:sympy,代码行数:11,代码来源:test_polyroots.py


示例3: test_roots_binomial

def test_roots_binomial():
    assert roots_binomial(Poly(5*x, x)) == [0]
    assert roots_binomial(Poly(5*x**4, x)) == [0, 0, 0, 0]
    assert roots_binomial(Poly(5*x + 2, x)) == [-Rational(2, 5)]

    A = 10**Rational(3, 4)/10

    assert roots_binomial(Poly(5*x**4 + 2, x)) == \
        [-A - A*I, -A + A*I, A - A*I, A + A*I]

    a1 = Symbol('a1', nonnegative=True)
    b1 = Symbol('b1', nonnegative=True)

    r0 = roots_quadratic(Poly(a1*x**2 + b1, x))
    r1 = roots_binomial(Poly(a1*x**2 + b1, x))

    assert powsimp(r0[0]) == powsimp(r1[0])
    assert powsimp(r0[1]) == powsimp(r1[1])
    for a, b, s, n in cartes((1, 2), (1, 2), (-1, 1), (2, 3, 4, 5)):
        if a == b and a != 1:  # a == b == 1 is sufficient
            continue
        p = Poly(a*x**n + s*b)
        ans = roots_binomial(p)
        assert ans == _nsort(ans)

    # issue 8813
    assert roots(Poly(2*x**3 - 16*y**3, x)) == {
        2*y*(-S(1)/2 - sqrt(3)*I/2): 1,
        2*y: 1,
        2*y*(-S(1)/2 + sqrt(3)*I/2): 1}
开发者ID:Davidjohnwilson,项目名称:sympy,代码行数:30,代码来源:test_polyroots.py


示例4: __new__

    def __new__(cls, expr, func=None, x=None, auto=True, quadratic=False):
        """Construct a new ``RootSum`` instance carrying all roots of a polynomial. """
        coeff, poly = cls._transform(expr, x)

        if not poly.is_univariate:
            raise MultivariatePolynomialError("only univariate polynomials are allowed")

        if func is None:
            func = Lambda(poly.gen, poly.gen)
        else:
            try:
                is_func = func.is_Function
            except AttributeError:
                is_func = False

            if is_func and (func.nargs == 1 or 1 in func.nargs):
                if not isinstance(func, Lambda):
                    func = Lambda(poly.gen, func(poly.gen))
            else:
                raise ValueError("expected a univariate function, got %s" % func)

        var, expr = func.args

        if coeff is not S.One:
            expr = expr.subs(var, coeff*var)

        deg = poly.degree()

        if not expr.has(var):
            return deg*expr

        if expr.is_Add:
            add_const, expr = expr.as_independent(var)
        else:
            add_const = S.Zero

        if expr.is_Mul:
            mul_const, expr = expr.as_independent(var)
        else:
            mul_const = S.One

        func = Lambda(var, expr)

        rational = cls._is_func_rational(poly, func)
        (_, factors), terms = poly.factor_list(), []

        for poly, k in factors:
            if poly.is_linear:
                term = func(roots_linear(poly)[0])
            elif quadratic and poly.is_quadratic:
                term = sum(map(func, roots_quadratic(poly)))
            else:
                if not rational or not auto:
                    term = cls._new(poly, func, auto)
                else:
                    term = cls._rational_case(poly, func)

            terms.append(k*term)

        return mul_const*Add(*terms) + deg*add_const
开发者ID:haz,项目名称:sympy,代码行数:60,代码来源:rootoftools.py


示例5: _roots_trivial

    def _roots_trivial(cls, poly, radicals):
        """Compute roots in linear, quadratic and binomial cases. """
        if poly.degree() == 1:
            return roots_linear(poly)

        if not radicals:
            return None

        if poly.degree() == 2:
            return roots_quadratic(poly)
        elif poly.length() == 2 and poly.TC():
            return roots_binomial(poly)
        else:
            return None
开发者ID:A-turing-machine,项目名称:sympy,代码行数:14,代码来源:rootoftools.py


示例6: roots_trivial

def roots_trivial(poly, radicals=True):
    """Compute roots in linear, quadratic and binomial cases. """
    if poly.degree() == 1:
        return roots_linear(poly)
    else:
        if not radicals:
            return None

        if poly in _rootof_trivial_cache:
            roots = _rootof_trivial_cache[poly]
        else:
            if radicals and poly.degree() == 2:
                roots = roots_quadratic(poly)
            elif radicals and poly.length() == 2 and poly.TC():
                roots = roots_binomial(poly)
            else:
                return None

            _rootof_trivial_cache[poly] = roots

        return roots
开发者ID:haz,项目名称:sympy,代码行数:21,代码来源:rootoftools.py


示例7: _roots_trivial

    def _roots_trivial(cls, poly, radicals):
        """Compute roots in linear, quadratic and binomial cases. """
        if poly.degree() == 1:
            return roots_linear(poly)

        if not radicals:
            return None
        free = len(poly.free_symbols)
        sort = isinstance(poly, PurePoly) and free or free > 1
        if poly.degree() == 2:
            roots = roots_quadratic(poly, sort=sort)
        elif poly.length() == 2 and poly.TC():
            roots = roots_binomial(poly, sort=sort)
        else:
            return None
        # put roots in same order as RootOf instances
        if not sort:
            key = [r.n(2) for r in roots]
            key = [(1 if not r.is_real else 0, C.re(r), C.im(r))
                for r in key]
            _, roots = zip(*sorted(zip(key, roots)))
        return roots
开发者ID:MCGallaspy,项目名称:sympy,代码行数:22,代码来源:rootoftools.py


示例8: test_roots_binomial

def test_roots_binomial():
    assert roots_binomial(Poly(5 * x, x)) == [0]
    assert roots_binomial(Poly(5 * x ** 4, x)) == [0, 0, 0, 0]
    assert roots_binomial(Poly(5 * x + 2, x)) == [-Rational(2, 5)]

    A = 10 ** Rational(3, 4) / 10

    assert roots_binomial(Poly(5 * x ** 4 + 2, x)) == [-A - A * I, -A + A * I, A - A * I, A + A * I]

    a1 = Symbol("a1", nonnegative=True)
    b1 = Symbol("b1", nonnegative=True)

    r0 = roots_quadratic(Poly(a1 * x ** 2 + b1, x))
    r1 = roots_binomial(Poly(a1 * x ** 2 + b1, x))

    assert powsimp(r0[0]) == powsimp(r1[0])
    assert powsimp(r0[1]) == powsimp(r1[1])
    for a, b, s, n in cartes((1, 2), (1, 2), (-1, 1), (2, 3, 4, 5)):
        if a == b and a != 1:  # a == b == 1 is sufficient
            continue
        p = Poly(a * x ** n + s * b)
        roots = roots_binomial(p)
        assert roots == _nsort(roots)
开发者ID:brajeshvit,项目名称:virtual,代码行数:23,代码来源:test_polyroots.py



注:本文中的sympy.polys.polyroots.roots_quadratic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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