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Python miscellaneous.root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.elementary.miscellaneous.root函数的典型用法代码示例。如果您正苦于以下问题:Python root函数的具体用法?Python root怎么用?Python root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root函数的16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: roots_cubic

def roots_cubic(f, trig=False):
    """Returns a list of roots of a cubic polynomial."""
    if trig:
        a, b, c, d = f.all_coeffs()
        p = (3*a*c - b**2)/3/a**2
        q = (2*b**3 - 9*a*b*c + 27*a**2*d)/(27*a**3)
        D = 18*a*b*c*d - 4*b**3*d + b**2*c**2 - 4*a*c**3 - 27*a**2*d**2
        if (D > 0) == True:
            rv = []
            for k in range(3):
                rv.append(2*sqrt(-p/3)*cos(acos(3*q/2/p*sqrt(-3/p))/3 - k*2*pi/3))
            return [i - b/3/a for i in rv]

    _, a, b, c = f.monic().all_coeffs()

    if c is S.Zero:
        x1, x2 = roots([1, a, b], multiple=True)
        return [x1, S.Zero, x2]

    p = b - a**2/3
    q = c - a*b/3 + 2*a**3/27

    pon3 = p/3
    aon3 = a/3

    if p is S.Zero:
        if q is S.Zero:
            return [-aon3]*3
        else:
            if q.is_real:
                if (q > 0) == True:
                    u1 = -root(q, 3)
                else:
                    u1 = root(-q, 3)
            else:
                u1 = root(-q, 3)
    elif q is S.Zero:
        y1, y2 = roots([1, 0, p], multiple=True)
        return [tmp - aon3 for tmp in [y1, S.Zero, y2]]
    elif q.is_real and q < 0:
        u1 = -root(-q/2 + sqrt(q**2/4 + pon3**3), 3)
    else:
        u1 = root(q/2 + sqrt(q**2/4 + pon3**3), 3)

    coeff = I*sqrt(3)/2

    u2 = u1*(-S.Half + coeff)
    u3 = u1*(-S.Half - coeff)

    if p is S.Zero:
        return [u1 - aon3, u2 - aon3, u3 - aon3]

    soln = [
        -u1 + pon3/u1 - aon3,
        -u2 + pon3/u2 - aon3,
        -u3 + pon3/u3 - aon3
    ]

    return soln
开发者ID:NalinG,项目名称:sympy,代码行数:59,代码来源:polyroots.py


示例2: test_nthroot

def test_nthroot():
    assert real_root(-8, 3) == -2
    assert real_root(-16, 4) == root(-16, 4)
    r = root(-7, 4)
    assert real_root(r) == r
    r1 = root(-1, 3)
    r2 = r1**2
    r3 = root(-1, 4)
    assert real_root(r1 + r2 + r3) == -1 + r2 + r3
开发者ID:BDGLunde,项目名称:sympy,代码行数:9,代码来源:test_miscellaneous.py


示例3: test_issue_14000

def test_issue_14000():
    assert isinstance(sqrt(4, evaluate=False), Pow) == True
    assert isinstance(cbrt(3.5, evaluate=False), Pow) == True
    assert isinstance(root(16, 4, evaluate=False), Pow) == True

    assert sqrt(4, evaluate=False) == Pow(4, S.Half, evaluate=False)
    assert cbrt(3.5, evaluate=False) == Pow(3.5, Rational(1, 3), evaluate=False)
    assert root(4, 2, evaluate=False) == Pow(4, Rational(1, 2), evaluate=False)

    assert root(16, 4, 2, evaluate=False).has(Pow) == True
    assert real_root(-8, 3, evaluate=False).has(Pow) == True
开发者ID:aprasanna,项目名称:sympy,代码行数:11,代码来源:test_miscellaneous.py


示例4: taylor_term

 def taylor_term(n, x, *previous_terms):
     if n < 0:
         return S.Zero
     else:
         x = sympify(x)
         if len(previous_terms) > 1:
             p = previous_terms[-1]
             return (3**(S(1)/3)*x * Abs(sin(2*pi*(n + S.One)/S(3))) * C.factorial((n - S.One)/S(3)) /
                     ((n + S.One) * Abs(cos(2*pi*(n + S.Half)/S(3))) * C.factorial((n - 2)/S(3))) * p)
         else:
             return (S.One/(root(3, 6)*pi) * gamma((n + S.One)/S(3)) * Abs(sin(2*pi*(n + S.One)/S(3))) /
                     C.factorial(n) * (root(3, 3)*x)**n)
开发者ID:AdrianPotter,项目名称:sympy,代码行数:12,代码来源:bessel.py


示例5: roots_cyclotomic

def roots_cyclotomic(f, factor=False):
    """Compute roots of cyclotomic polynomials. """
    L, U = _inv_totient_estimate(f.degree())

    for n in range(L, U + 1):
        g = cyclotomic_poly(n, f.gen, polys=True)

        if f == g:
            break
    else:  # pragma: no cover
        raise RuntimeError("failed to find index of a cyclotomic polynomial")

    roots = []

    if not factor:
        # get the indices in the right order so the computed
        # roots will be sorted
        h = n//2
        ks = [i for i in range(1, n + 1) if igcd(i, n) == 1]
        ks.sort(key=lambda x: (x, -1) if x <= h else (abs(x - n), 1))
        d = 2*I*pi/n
        for k in reversed(ks):
            roots.append(exp(k*d).expand(complex=True))
    else:
        g = Poly(f, extension=root(-1, n))

        for h, _ in ordered(g.factor_list()[1]):
            roots.append(-h.TC())

    return roots
开发者ID:bjodah,项目名称:sympy,代码行数:30,代码来源:polyroots.py


示例6: roots_cyclotomic

def roots_cyclotomic(f, factor=False):
    """Compute roots of cyclotomic polynomials. """
    L, U = _inv_totient_estimate(f.degree())

    for n in xrange(L, U + 1):
        g = cyclotomic_poly(n, f.gen, polys=True)

        if f == g:
            break
    else:  # pragma: no cover
        raise RuntimeError("failed to find index of a cyclotomic polynomial")

    roots = []

    if not factor:
        for k in xrange(1, n + 1):
            if igcd(k, n) == 1:
                roots.append(exp(2*k*S.Pi*I/n).expand(complex=True))
    else:
        g = Poly(f, extension=root(-1, n))

        for h, _ in g.factor_list()[1]:
            roots.append(-h.TC())

    return sorted(roots, key=default_sort_key)
开发者ID:cheekujodhpur,项目名称:sympy,代码行数:25,代码来源:polyroots.py


示例7: taylor_term

 def taylor_term(n, x, *previous_terms):
     if n < 0:
         return S.Zero
     else:
         x = sympify(x)
         if len(previous_terms) > 1:
             p = previous_terms[-1]
             return (
                 (3 ** (S(1) / 3) * x) ** (-n)
                 * (3 ** (S(1) / 3) * x) ** (n + 1)
                 * sin(pi * (2 * n / 3 + S(4) / 3))
                 * factorial(n)
                 * gamma(n / 3 + S(2) / 3)
                 / (sin(pi * (2 * n / 3 + S(2) / 3)) * factorial(n + 1) * gamma(n / 3 + S(1) / 3))
                 * p
             )
         else:
             return (
                 S.One
                 / (3 ** (S(2) / 3) * pi)
                 * gamma((n + S.One) / S(3))
                 * sin(2 * pi * (n + S.One) / S(3))
                 / factorial(n)
                 * (root(3, 3) * x) ** n
             )
开发者ID:LuckyStrikes1090,项目名称:sympy,代码行数:25,代码来源:bessel.py


示例8: roots_binomial

def roots_binomial(f):
    """Returns a list of roots of a binomial polynomial. If the domain is ZZ
    then the roots will be sorted with negatives coming before positives.
    The ordering will be the same for any numerical coefficients as long as
    the assumptions tested are correct, otherwise the ordering will not be
    sorted (but will be canonical).
    """
    n = f.degree()

    a, b = f.nth(n), f.nth(0)
    base = -cancel(b/a)
    alpha = root(base, n)

    if alpha.is_number:
        alpha = alpha.expand(complex=True)

    # define some parameters that will allow us to order the roots.
    # If the domain is ZZ this is guaranteed to return roots sorted
    # with reals before non-real roots and non-real sorted according
    # to real part and imaginary part, e.g. -1, 1, -1 + I, 2 - I
    neg = base.is_negative
    even = n % 2 == 0
    if neg:
        if even == True and (base + 1).is_positive:
            big = True
        else:
            big = False

    # get the indices in the right order so the computed
    # roots will be sorted when the domain is ZZ
    ks = []
    imax = n//2
    if even:
        ks.append(imax)
        imax -= 1
    if not neg:
        ks.append(0)
    for i in range(imax, 0, -1):
        if neg:
            ks.extend([i, -i])
        else:
            ks.extend([-i, i])
    if neg:
        ks.append(0)
        if big:
            for i in range(0, len(ks), 2):
                pair = ks[i: i + 2]
                pair = list(reversed(pair))

    # compute the roots
    roots, d = [], 2*I*pi/n
    for k in ks:
        zeta = exp(k*d).expand(complex=True)
        roots.append((alpha*zeta).expand(power_base=False))

    return roots
开发者ID:bjodah,项目名称:sympy,代码行数:56,代码来源:polyroots.py


示例9: test_real_root

def test_real_root():
    assert real_root(-8, 3) == -2
    assert real_root(-16, 4) == root(-16, 4)
    r = root(-7, 4)
    assert real_root(r) == r
    r1 = root(-1, 3)
    r2 = r1**2
    r3 = root(-1, 4)
    assert real_root(r1 + r2 + r3) == -1 + r2 + r3
    assert real_root(root(-2, 3)) == -root(2, 3)
    assert real_root(-8., 3) == -2
开发者ID:sunny94,项目名称:sympy,代码行数:11,代码来源:test_miscellaneous.py


示例10: test_real_root

def test_real_root():
    assert real_root(-8, 3) == -2
    assert real_root(-16, 4) == root(-16, 4)
    r = root(-7, 4)
    assert real_root(r) == r
    r1 = root(-1, 3)
    r2 = r1**2
    r3 = root(-1, 4)
    assert real_root(r1 + r2 + r3) == -1 + r2 + r3
    assert real_root(root(-2, 3)) == -root(2, 3)
    assert real_root(-8., 3) == -2
    x = Symbol('x')
    n = Symbol('n')
    g = real_root(x, n)
    assert g.subs(dict(x=-8, n=3)) == -2
    assert g.subs(dict(x=8, n=3)) == 2
    # give principle root if there is no real root -- if this is not desired
    # then maybe a Root class is needed to raise an error instead
    assert g.subs(dict(x=I, n=3)) == cbrt(I)
    assert g.subs(dict(x=-8, n=2)) == sqrt(-8)
    assert g.subs(dict(x=I, n=2)) == sqrt(I)
开发者ID:Davidjohnwilson,项目名称:sympy,代码行数:21,代码来源:test_miscellaneous.py


示例11: _eval_rewrite_as_hyper

 def _eval_rewrite_as_hyper(self, z):
     pf1 = z ** 2 / (2 * root(3, 6) * gamma(S(2) / 3))
     pf2 = root(3, 6) / gamma(S(1) / 3)
     return pf1 * hyper([], [S(5) / 3], z ** 3 / 9) + pf2 * hyper([], [S(1) / 3], z ** 3 / 9)
开发者ID:LuckyStrikes1090,项目名称:sympy,代码行数:4,代码来源:bessel.py


示例12: roots_cubic

def roots_cubic(f, trig=False):
    """Returns a list of roots of a cubic polynomial.

    References
    ==========
    [1] https://en.wikipedia.org/wiki/Cubic_function, General formula for roots,
    (accessed November 17, 2014).
    """
    if trig:
        a, b, c, d = f.all_coeffs()
        p = (3*a*c - b**2)/3/a**2
        q = (2*b**3 - 9*a*b*c + 27*a**2*d)/(27*a**3)
        D = 18*a*b*c*d - 4*b**3*d + b**2*c**2 - 4*a*c**3 - 27*a**2*d**2
        if (D > 0) == True:
            rv = []
            for k in range(3):
                rv.append(2*sqrt(-p/3)*cos(acos(3*q/2/p*sqrt(-3/p))/3 - k*2*pi/3))
            return [i - b/3/a for i in rv]

    _, a, b, c = f.monic().all_coeffs()

    if c is S.Zero:
        x1, x2 = roots([1, a, b], multiple=True)
        return [x1, S.Zero, x2]

    p = b - a**2/3
    q = c - a*b/3 + 2*a**3/27

    pon3 = p/3
    aon3 = a/3

    u1 = None
    if p is S.Zero:
        if q is S.Zero:
            return [-aon3]*3
        if q.is_real:
            if q.is_positive:
                u1 = -root(q, 3)
            elif q.is_negative:
                u1 = root(-q, 3)
    elif q is S.Zero:
        y1, y2 = roots([1, 0, p], multiple=True)
        return [tmp - aon3 for tmp in [y1, S.Zero, y2]]
    elif q.is_real and q.is_negative:
        u1 = -root(-q/2 + sqrt(q**2/4 + pon3**3), 3)

    coeff = I*sqrt(3)/2
    if u1 is None:
        u1 = S(1)
        u2 = -S.Half + coeff
        u3 = -S.Half - coeff
        a, b, c, d = S(1), a, b, c
        D0 = b**2 - 3*a*c
        D1 = 2*b**3 - 9*a*b*c + 27*a**2*d
        C = root((D1 + sqrt(D1**2 - 4*D0**3))/2, 3)
        return [-(b + uk*C + D0/C/uk)/3/a for uk in [u1, u2, u3]]

    u2 = u1*(-S.Half + coeff)
    u3 = u1*(-S.Half - coeff)

    if p is S.Zero:
        return [u1 - aon3, u2 - aon3, u3 - aon3]

    soln = [
        -u1 + pon3/u1 - aon3,
        -u2 + pon3/u2 - aon3,
        -u3 + pon3/u3 - aon3
    ]

    return soln
开发者ID:bjodah,项目名称:sympy,代码行数:70,代码来源:polyroots.py


示例13: bivariate_type

def bivariate_type(f, x, y, **kwargs):
    """Given an expression, f, 3 tests will be done to see what type
    of composite bivariate it might be, options for u(x, y) are::

        x*y
        x+y
        x*y+x
        x*y+y

    If it matches one of these types, ``u(x, y)``, ``P(u)`` and dummy
    variable ``u`` will be returned. Solving ``P(u)`` for ``u`` and
    equating the solutions to ``u(x, y)`` and then solving for ``x`` or
    ``y`` is equivalent to solving the original expression for ``x`` or
    ``y``. If ``x`` and ``y`` represent two functions in the same
    variable, e.g. ``x = g(t)`` and ``y = h(t)``, then if ``u(x, y) - p``
    can be solved for ``t`` then these represent the solutions to
    ``P(u) = 0`` when ``p`` are the solutions of ``P(u) = 0``.

    Only positive values of ``u`` are considered.

    Examples
    ========

    >>> from sympy.solvers.solvers import solve
    >>> from sympy.solvers.bivariate import bivariate_type
    >>> from sympy.abc import x, y
    >>> eq = (x**2 - 3).subs(x, x + y)
    >>> bivariate_type(eq, x, y)
    (x + y, _u**2 - 3, _u)
    >>> uxy, pu, u = _
    >>> usol = solve(pu, u); usol
    [sqrt(3)]
    >>> [solve(uxy - s) for s in solve(pu, u)]
    [[{x: -y + sqrt(3)}]]
    >>> all(eq.subs(s).equals(0) for sol in _ for s in sol)
    True

    """

    u = Dummy('u', positive=True)

    if kwargs.pop('first', True):
        p = Poly(f, x, y)
        f = p.as_expr()
        _x = Dummy()
        _y = Dummy()
        rv = bivariate_type(Poly(f.subs({x: _x, y: _y}), _x, _y), _x, _y, first=False)
        if rv:
            reps = {_x: x, _y: y}
            return rv[0].xreplace(reps), rv[1].xreplace(reps), rv[2]
        return

    p = f
    f = p.as_expr()

    # f(x*y)
    args = Add.make_args(p.as_expr())
    new = []
    for a in args:
        a = _mexpand(a.subs(x, u/y))
        free = a.free_symbols
        if x in free or y in free:
            break
        new.append(a)
    else:
        return x*y, Add(*new), u

    def ok(f, v, c):
        new = _mexpand(f.subs(v, c))
        free = new.free_symbols
        return None if (x in free or y in free) else new

    # f(a*x + b*y)
    new = []
    d = p.degree(x)
    if p.degree(y) == d:
        a = root(p.coeff_monomial(x**d), d)
        b = root(p.coeff_monomial(y**d), d)
        new = ok(f, x, (u - b*y)/a)
        if new is not None:
            return a*x + b*y, new, u

    # f(a*x*y + b*y)
    new = []
    d = p.degree(x)
    if p.degree(y) == d:
        for itry in range(2):
            a = root(p.coeff_monomial(x**d*y**d), d)
            b = root(p.coeff_monomial(y**d), d)
            new = ok(f, x, (u - b*y)/a/y)
            if new is not None:
                return a*x*y + b*y, new, u
            x, y = y, x
开发者ID:AALEKH,项目名称:sympy,代码行数:93,代码来源:bivariate.py


示例14: test_root

def test_root():
    from sympy.abc import x

    n = Symbol("n", integer=True)
    k = Symbol("k", integer=True)

    assert root(2, 2) == sqrt(2)
    assert root(2, 1) == 2
    assert root(2, 3) == 2 ** Rational(1, 3)
    assert root(2, 3) == cbrt(2)
    assert root(2, -5) == 2 ** Rational(4, 5) / 2

    assert root(-2, 1) == -2

    assert root(-2, 2) == sqrt(2) * I
    assert root(-2, 1) == -2

    assert root(x, 2) == sqrt(x)
    assert root(x, 1) == x
    assert root(x, 3) == x ** Rational(1, 3)
    assert root(x, 3) == cbrt(x)
    assert root(x, -5) == x ** Rational(-1, 5)

    assert root(x, n) == x ** (1 / n)
    assert root(x, -n) == x ** (-1 / n)

    assert root(x, n, k) == x ** (1 / n) * (-1) ** (2 * k / n)
开发者ID:Carreau,项目名称:sympy,代码行数:27,代码来源:test_miscellaneous.py


示例15: test_root

def test_root():
    from sympy.abc import x, y, z
    n = Symbol('n', integer=True)

    assert root(2, 2) == sqrt(2)
    assert root(2, 1) == 2
    assert root(2, 3) == 2**Rational(1,3)
    assert root(2, -5) == 2**Rational(4,5)/2

    assert root(-2, 1) == -2

    assert root(-2, 2) == sqrt(2)*I
    assert root(-2, 1) == -2

    assert root(x, 2) == sqrt(x)
    assert root(x, 1) == x
    assert root(x, 3) == x**Rational(1,3)
    assert root(x, -5) == x**Rational(-1,5)

    assert root(x, n) == x**(1/n)
    assert root(x, -n) == x**(-1/n)
开发者ID:BDGLunde,项目名称:sympy,代码行数:21,代码来源:test_miscellaneous.py


示例16: _eval_rewrite_as_meijerg

 def _eval_rewrite_as_meijerg(self, z):
     return (pi*z**(S(3)/4) / (sqrt(2)*root(z**2, 4)*root(-z, 4))
             * meijerg([], [1], [S(1)/4], [S(3)/4, 0], -pi**2*z**4/16))
开发者ID:Maihj,项目名称:sympy,代码行数:3,代码来源:error_functions.py



注:本文中的sympy.functions.elementary.miscellaneous.root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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